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Description

.

Methods of Analysis.
Outline:
1) Nodal analysis;
2) Supernode;
3) Mesh analysis;
4) Supermesh.
CCNY
ENGR24100
Professor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
1
Nodal Analysis
Ã¢â‚¬Â¢ The objective is to reduce the number of simultaneous equations to solve for.
Ã¢â‚¬Â¢ Given a circuit with Ã°Ââ€™Â nodes, the nodal analysis is accomplished by using the
sequence of steps:
1.
2.
3.
4.
5.
6.
7.
8.
Select one node as the reference node.
Assign potentials V1, V2, Ã¢â‚¬Â¦, Vn to the remaining n-1 nodes, where n-1 voltages of
the non-reference nodes are relative to the reference node.
Assume directions of brunch currents in the circuit.
Apply KCL to each of the n-1 non-reference nodes.
Use OhmÃ¢â‚¬â„¢s law to express the branch currents in terms of voltage/resistance.
Solve n-1 linear equations to obtain n-1 voltages of the non-reference nodes.
Calculate all branch voltages and currents.
Check your solution by using a conservation law.
Ã¢â‚¬Â¢ Usually, the reference is the node that is grounded since its potential is zero
Volts.
2
Example 3.1
Calculate the node voltages in the circuit
3
Steps to determine node voltage:
Node 1 : i1 = i2 + i3 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™eqt1
Node2 : i2 + i4 = i1 + i5 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™eqt2
v1 Ã¢Ë†â€™ v2 v1 Ã¢Ë†â€™ 0
5=
+
20 = v1 Ã¢Ë†â€™ v2 + 2v1
4
2
3v1 Ã¢Ë†â€™ v2 = 20 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™eqt3
v1 Ã¢Ë†â€™ v2
v2 Ã¢Ë†â€™ 0
+ 10 = 5 +
3v1 Ã¢Ë†â€™ 3v2 + 120 = 60 + 2v2
4
6
Ã¢Ë†â€™ 3v1 + 5v2 = 60 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™eqt4
Method 1 (elimination) Add eqt (3) and (4)
4v2 = 80 v2 = 20V
From eqt1 3v1 Ã¢Ë†â€™ 20 = 20 v1 = 13.33V
Ã¢â€¡â€™
Ã¢â€¡Â
Ã¢â€¡â€™
4
Method 2 (CramerÃ¢â‚¬â„¢s rule is described in Appendix A)
3v1 Ã¢Ë†â€™ v2 = 20 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™eqt3
Ã¢Ë†â€™ 3v1 + 5v2 = 60 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eqt4
Eqt (3) and (4) in the matrix form :
Ã¯Æ’Âª Ã¢Ë†â€™ 3 5Ã¯Æ’Âº Ã¯Æ’Âªv Ã¯Æ’Âº = Ã¯Æ’Âª60Ã¯Æ’Âº
Ã¯Æ’Â«
Ã¯Æ’Â»Ã¯Æ’Â« 2 Ã¯Æ’Â» Ã¯Æ’Â« Ã¯Æ’Â»
The determinant of the matrix is
Ã¯Ââ€ž=Ã¯Æ’Âª
= 15 Ã¢Ë†â€™ 3 = 12
Ã¯Æ’Âº
Ã¯Æ’Â« Ã¢Ë†â€™ 3 5Ã¯Æ’Â»
Ã¯Æ’Âª60 5 Ã¯Æ’Âº
Ã¯Æ’Â» = 100 + 60 = 13.33V
v1 and v2 are obtained as v1 = Ã¯Æ’Â«
Ã¯Ââ€ž
12
Ã¯Æ’ÂªÃ¢Ë†â€™ 3 60Ã¯Æ’Âº
Ã¯Æ’Â» = 180 + 60 = 20V
v2 = Ã¯Æ’Â«
Ã¯Ââ€ž
12
v Ã¢Ë†â€™ v2
v
Currents are calculated as i1 = 5 A i2 = 1
= Ã¢Ë†â€™1.67 A, i3 = 1 = 6.67 A,
4
2
v
i4 = 10 A, i5 = 2 = 3.33 A
6
5
7. Calculate all branch voltages and currents.
The voltage drop on the resistor 4Ã°ÂÅ“Â´ is Ã°Ââ€˜Â¼Ã°ÂÅ¸â€™Ã°Ââ€ºâ‚¬ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜ Ã¢â€°Ë† Ã°ÂÅ¸â€. Ã°ÂÅ¸â€¢Ã°Ââ€˜Â½.
The voltage drop on the resistor 6Ã°Ââ€ºâ‚¬ is Ã°Ââ€˜Â¼Ã°ÂÅ¸â€Ã°Ââ€ºâ‚¬ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€˜Â½.
Voltage drop on resistor 2Ã°Ââ€ºâ‚¬ is Ã°Ââ€˜Â¼Ã°ÂÅ¸ÂÃ°Ââ€ºâ‚¬ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜.
8. Check your solution by using a conservation law.
KCL applied to the node #1:
Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸â€˜ Ã¢â€¡â€™ Ã°ÂÅ¸â€œ + Ã°ÂÅ¸Â. Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢ = Ã°ÂÅ¸â€. Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢.
KCL applied to the node #2:
Ã°Ââ€™Å Ã°ÂÅ¸â€™ = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€œ Ã¢â€¡â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°ÂÅ¸â€œ + Ã°ÂÅ¸Â. Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢ + Ã°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜
6
Supernode
(Nodal Analysis with Voltage Sources)
Ã°Ââ€™â€”Ã°ÂÅ¸Â =?,
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ =? .
Ã¢â‚¬Â¢ A supernode is formed by enclosing a (dependent or independent) voltage
source connected between two non-reference nodes and any elements
connected in parallel with it.
Ã¢â‚¬Â¢ Both, KCL and KVL, are applied to determine the node voltage.
7
Applying KVL to a supernode
i1 + i4 = i2 + i3 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq1a
or
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
KVL in the supernode
Ã°Ââ€™â€”Ã°ÂÅ¸Â
v1 Ã¢Ë†â€™ v2 v1 Ã¢Ë†â€™ v3 v2 Ã¢Ë†â€™ 0 v3 Ã¢Ë†â€™ 0
+
=
+
Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq1b
2
4
8
6
Ã¢Ë†â€™ v2 + 5 + v3 = 0 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq2
v2 Ã¢Ë†â€™ v3 = 5 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq2a
v1 = 10 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq3
Ã¢â‚¬Â¢ Properties of a supernode:
Ã¢â‚¬Â¢ The voltage source inside the supernode
provides KVL equation needed to solve
for the node voltages.
Ã¢â‚¬Â¢ A supernode has no voltage of its own.
Ã¢â‚¬Â¢ A supernode requires the application of
both KCL and KVL.
8
Applying KVL to a supernode
i1 + i4 = i2 + i3 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq1a
or
v1 Ã¢Ë†â€™ v2 v1 Ã¢Ë†â€™ v3 v2 Ã¢Ë†â€™ 0 v3 Ã¢Ë†â€™ 0
+
=
+
Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq1b
2
4
8
6
Ã¢Ë†â€™ v2 + 5 + v3 = 0 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq2
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
v2 Ã¢Ë†â€™ v3 = 5 Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq2a
v1 = 10 Ã¢Ë†â€™(From
Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™diagram
Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ of
Ã¢Ë†â€™ Ã¢Ë†â€™the
Ã¢Ë†â€™ circuit).
Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ Ã¢Ë†â€™ eq3
A supernode requires the application of
KVL:
Ã¢Ë†â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œ + Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸Å½ Ã¢â€¡â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ + Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ + Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã¢â€¡â€™
Ã¢â€¡â€™ Ã°ÂÂâ‚¬Ã°ÂÂÂ§Ã°ÂÂÂ¬Ã°ÂÂÂ°Ã°ÂÂÅ¾Ã°ÂÂÂ«: Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸â€™. Ã°ÂÅ¸ÂÃ°Ââ€˜Â½, Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€”. Ã°ÂÅ¸ÂÃ°Ââ€˜Â½
9
Mesh (Loop) Analysis
Ã¢â‚¬Â¢ Mesh analysis only applicable to a circuit that is planar.
Ã¢â‚¬Â¢ A planar circuit is one that can be drawn in a plane with no branches
crossing one another.
Ã¢â‚¬Â¢ A mesh is an elementary loop that does not contain any other loops within
it.
Ã¢â‚¬Â¢ From fundamental theorem of network topology b=l+n-1Ã¢â€¡â€™
n=b-l+1 is number of equations needed for Nodal Analysis;
l=b-n+1 is number of equations needed for Mesh Analysis;
as n>l , one should solve less number of equations by using Mesh
Analysis.
Ã¢â‚¬Â¢ Mesh and branch currents are not always the same.
10
Nonpalanar
vs
Planar
Ã¢Å¸Âº
This circuit is a planar: It can be redrawn to avoid
crossing branches.
11
Mesh Analysis Steps
1) Assign n mesh (loop) currents in to the Ã°Ââ€™Â meshes;
2) Apply KVL to each of the Ã°Ââ€™Â meshes to get Ã°Ââ€™Â linear equations;
3) Use OhmÃ¢â‚¬â„¢s law UR=inR to express the voltages on resistors in terms of the
mesh currents.
4) Solve the resulting Ã°Ââ€™Â linear equations to get the mesh currents in.
5) Use OhmÃ¢â‚¬â„¢s law UR=inR to find the voltage drop on each resistor.
6) Use OhmÃ¢â‚¬â„¢s law IR=UR/R to find the branch currents IR.
12
Example:
Find the branch currents I1, I2 and I3.
Mesh #1 (aÃ¢Å¸Â¶bÃ¢Å¸Â¶eÃ¢Å¸Â¶fÃ¢Å¸Â¶a):
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€˜Â½Ã°ÂÅ¸Â ;
Mesh #2 (bÃ¢Å¸Â¶cÃ¢Å¸Â¶dÃ¢Å¸Â¶eÃ¢Å¸Â¶b):
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€˜Â½Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â + (Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ )Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â
Branch currents:
Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â ;
Fig. 3.17. A circuit with two meshes.
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€˜Â½Ã°ÂÅ¸Â ;
Ã¢Å¸Â¹Ã ÂµÅ¾
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â + (Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ )Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°Ââ€˜Â½Ã°ÂÅ¸Â
=
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸Â , Ã°Ââ€™Å Ã°ÂÅ¸Â , Ã°Ââ€™Å Ã°ÂÅ¸â€˜ are roots of the system with three linear
equations.
Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â ;
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â . Ã¢Å¸Â¹ Mesh and branch currents are not always the same.
13
1Ã°Ââ€˜â€°
2Ã°Ââ€˜â€°
The initial assumption is that
the mech currents Ã°Ââ€™Å Ã°ÂÅ¸Â and Ã°Ââ€™Å Ã°ÂÅ¸Â
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
KVL in the mesh #1:
Ã¢Ë†â€™Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
KVL in the mesh #2:
Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã¢Ë†â€™Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸Â.
14
Ã°ÂÅ¸â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã¢Å¸Â¹Ã¡â€°Â
Ã¢Å¸Â¹
Ã¢Ë†â„¢
=
Ã¢Å¸Â¹
Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€œ
Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸Â.
Ã°ÂÅ¸â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜
= Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â„¢ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Â;
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€œ
Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â =
= Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸Â Ã¢Ë†â„¢ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã¢Ë†â€™Ã°ÂÅ¸Â;
Ã¢Ë†â€™Ã°ÂÅ¸Â Ã°ÂÅ¸â€œ
Ã°ÂÅ¸â€™
Ã°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â =
= Ã¢Ë†â€™Ã°ÂÅ¸â€“ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã¢Ë†â€™Ã°ÂÅ¸â€œ.
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸â€œ
Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨,
Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ“Å¸=
1Ã°Ââ€˜â€°
2Ã°Ââ€˜â€°
The initial assumption is that
the mech currents Ã°Ââ€™Å Ã°ÂÅ¸Â and Ã°Ââ€™Å Ã°ÂÅ¸Â
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨,
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸â€œ
Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨ Ã¢Å¸Â¹
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Directions of both mesh currents should be corrected to show the counter-clockwise direction:
Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨,
Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
Also directions of the branch currents Ã°Ââ€˜Â°Ã°ÂÅ¸Â and Ã°Ââ€˜Â°Ã°ÂÅ¸Â should be changed to opposite ones.
Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â and Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ.
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€ Ã°Ââ€˜Â¨
The original assumption about downward direction of the branch current Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ was correct.
15
1Ã°Ââ€˜â€°
2Ã°Ââ€˜â€°
The initial assumption is that the
mech currents Ã°Ââ€™Å Ã°ÂÅ¸Â and Ã°Ââ€™Å Ã°ÂÅ¸Â are in
opposite directions.
Fig. 3.17c. A circuit with two meshes.
KVL in the mesh #1:
Ã¢Ë†â€™Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
KVL in the mesh #2:
Ã¢Ë†â€™Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
Ã°ÂÅ¸â€™ Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã¢Å¸Â¹Ã¡â€°Â
Ã¢Å¸Â¹
Ã¢Ë†â„¢
=
Ã¢Å¸Â¹
Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€œ Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
Ã¢Ë†â€ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°ÂÅ¸â€” = Ã°ÂÅ¸ÂÃ°ÂÅ¸Â; Ã°ÂÅ¡Â«Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã°ÂÅ¸â€ = Ã¢Ë†â€™Ã°ÂÅ¸Â; Ã°ÂÅ¡Â«Ã°ÂÅ¸Â = Ã°ÂÅ¸â€“ Ã¢Ë†â€™ Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸â€œ
Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
Ã°Ââ€™Å Ã°ÂÅ¸Â =
Ã°ÂÅ¡Â«Ã°ÂÅ¸Â
Ã°ÂÅ¡Â«
=
Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨ and Ã°Ââ€™Å Ã°ÂÅ¸Â =
Ã¢Ë†â€ Ã°ÂÅ¸Â
Ã¢Ë†â€
Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
16
1Ã°Ââ€˜â€°
2Ã°Ââ€˜â€°
The initial assumption is that the
mech currents Ã°Ââ€™Å Ã°ÂÅ¸Â and Ã°Ââ€™Å Ã°ÂÅ¸Â are in
opposite directions.
Fig. 3.17c. A circuit with two meshes.
Ã°Ââ€™Å Ã°ÂÅ¸Â =
Ã°ÂÅ¡Â«Ã°ÂÅ¸Â
Ã°ÂÅ¡Â«
=
Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨ and Ã°Ââ€™Å Ã°ÂÅ¸Â =
Ã¢Ë†â€ Ã°ÂÅ¸Â
Ã¢Ë†â€
Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
Direction of the mesh current Ã°Ââ€™Å Ã°ÂÅ¸Â should be corrected to the counter-clockwise:
Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨ and Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
The branch currents are Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â, Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ,
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€ Ã°Ââ€˜Â¨
17
Check-up by Using Nodal Analysis
Ã°Ââ€˜Â½Ã°Ââ€™â€š = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½
1Ã°Ââ€˜â€°
Ã°Ââ€˜Â½Ã°Ââ€™Æ’
Ã°Ââ€˜Â½Ã°Ââ€™â€ž = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½
2Ã°Ââ€˜â€°
Fig. 3.17d. A circuit with two meshes.
KCL in the node Ã°Ââ€˜Â½Ã°Ââ€™Æ’ :
Ã°Ââ€˜Â°Ã°ÂÅ¸Â
Ã°Ââ€˜Â°Ã°ÂÅ¸Â
Ã°Ââ€˜Â°Ã°ÂÅ¸Â
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜
Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€˜Â½Ã°Ââ€™Æ’ Ã°Ââ€˜Â½Ã°Ââ€™Æ’ Ã¢Ë†â€™ Ã°ÂÅ¸Â Ã°Ââ€˜Â½Ã°Ââ€™Æ’
= Ã°Ââ€˜Â°Ã°ÂÅ¸Â + Ã°Ââ€˜Â° Ã°ÂÅ¸â€˜ Ã¢Å¸Â¹
=
Ã¢Ë†â€™
Ã¢Å¸Â¹ Ã°Ââ€˜Â½Ã°Ââ€™Æ’ = Ã°ÂÅ¸Â. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â½
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸â€˜
Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€˜Â½Ã°Ââ€™Æ’
=
= Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨; Ã¢Å¸Â¹ The initial assumptions for the directions of
Ã°ÂÅ¸Â
branch currents Ã°Ââ€˜Â°Ã°ÂÅ¸Â and Ã°Ââ€˜Â°Ã°ÂÅ¸Â should be corrected
Ã°Ââ€˜Â½Ã°Ââ€™Æ’ Ã¢Ë†â€™ Ã°ÂÅ¸Â
=
= Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨; Ã¢Å¸Â¹ to show both currents from right to left.
Ã°ÂÅ¸Â
Ã°Ââ€˜Â½Ã°Ââ€™Æ’
=
= Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€Ã°ÂÅ¸â€™ Ã°Ââ€˜Â¨.
18
Ã°ÂÅ¸â€˜
Supermesh
Supermesh
Mesh#1
Mesh#2
Ã¢â‚¬Â¢ Current sources (dependent or independent) that are shared by more than
one mesh need a special treatment, because it is not clear what voltage
has to be assigned to the current source;
Ã¢â‚¬Â¢ The two meshes must be joined together, resulting in a supermesh;
Ã¢â‚¬Â¢ The supermesh is constructed by merging the two meshes and excluding
the shared source and any elements in series with it;
Ã¢â‚¬Â¢ KCL should be applied to one node of the supermesh.
19
Supermesh
Example
Ã¢â‚¬Â¢ Apply KVL to the supermesh:
Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ + Ã°ÂÅ¸â€Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ Ã¢Å¸Â¹
or Ã°ÂÅ¸â€Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã¢â‚¬Â¢ Apply KCL to the node Ã¢â‚¬Å“0Ã¢â‚¬Â in the branch where the two meshes
intersect:
Ã°ÂÅ¸â€Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€
Ã¢Å¸Â¹ Ã¡â€°Â
in
out
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€
Ã¢â‚¬Â¢ By solving the system of two equations we get:
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸ÂÃ°ÂÂâ‚¬ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â. Ã°ÂÅ¸â€“Ã°ÂÂâ‚¬ Ã¢Å¸Â¹ Initial assumption for the direction of the
mesh current Ã°Ââ€™Å Ã°ÂÅ¸Â should be corrected.
20
Supermesh
Example
3.2A
6A
2.8A
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸ÂÃ°ÂÂâ‚¬ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â. Ã°ÂÅ¸â€“Ã°ÂÂâ‚¬ Ã¢Å¸Â¹ Initial assumption for the direction of the
mesh current Ã°Ââ€™Å Ã°ÂÅ¸Â should be corrected.
Check up by KCL applied to the node Ã¢â‚¬Å“0Ã¢â‚¬Â: Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸â€ Ã°Ââ€˜Â¨
21
Methods of Analysis.
Outline:
1) Nodal analysis;
2) Supernode;
3) Mesh analysis;
4) Supermesh.
CCNY
ENGR24100
Professor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
1
Nodal Analysis
Ã¢â‚¬Â¢ The objective is to reduce the number of simultaneous equations to solve for.
Ã¢â‚¬Â¢ Given a circuit with Ã°Ââ€™Â nodes, the nodal analysis is accomplished by using the
sequence of steps:
1.
2.
3.
4.
5.
6.
7.
8.
Select one node as the reference node.
Assign potentials V1 , V2 , Ã¢â‚¬Â¦, Vn to the remaining n-1 non-reference nodes.
Assume directions of brunch currents in the circuit.
Apply KCL to each of the n-1 non-reference nodes.
Use OhmÃ¢â‚¬â„¢s law to express the branch currents in terms of voltage/resistance.
Solve n-1 linear equations to obtain n-1 voltages of the non-reference nodes.
Calculate all branch voltages and currents.
Check your solution by using a conservation law.
Ã¢â‚¬Â¢ Usually, the reference is the node that is grounded since its potential is zero
Volts.
2
Example 3.1
Calculate the node voltages in the circuit
3
Steps to determine node voltage:
Ã°Ââ€˜ÂµÃ°Ââ€™ÂÃ°Ââ€™â€¦Ã°Ââ€™â€  Ã°ÂÅ¸Â:
Ã°Ââ€˜ÂµÃ°Ââ€™ÂÃ°Ââ€™â€¦Ã°Ââ€™â€  Ã°ÂÅ¸Â:
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€˜ ,
Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€™ = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€œ ,
Ã°Ââ€™â€” Ã¢Ë†â€™Ã°Ââ€™â€”
Eq.(1)
Eq.(2)
Ã°Ââ€™â€” Ã¢Ë†â€™Ã°ÂÅ¸Å½
Eq.(1)Ã¢â€¡â€™ Ã°ÂÅ¸â€œ = Ã°ÂÅ¸Â Ã°ÂÅ¸â€™ Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°ÂÅ¸Â
Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½.
Eq.(3)
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½
Eq.(2) Ã¢â€¡â€™
+ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°ÂÅ¸â€œ +
Ã°ÂÅ¸â€™
Ã°ÂÅ¸â€
Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°ÂÅ¸â€Ã°ÂÅ¸Å½ + Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€Ã°ÂÅ¸Å½.
Eq.(4)
Add Eqs. (Ã°ÂÅ¸â€˜ and Ã°ÂÅ¸â€™) Ã¢â€¡â€™
Ã°ÂÅ¸â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€“Ã°ÂÅ¸Å½
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€˜Â½
Ã°ÂÂâ€žÃ°ÂÂÂª. Ã°ÂÅ¸â€˜ Ã¢â€¡â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜ Ã°Ââ€˜Â½
4
Method by using matrixes (CramerÃ¢â‚¬â„¢s rule is described in Appendix A)
3Ã°Ââ€˜Â£1 Ã¢Ë†â€™ Ã°Ââ€˜Â£2 = 20
Ã¢Ë†â€™3Ã°Ââ€˜Â£1 + 5Ã°Ââ€˜Â£2 = 60
Eq.(3)
Eq.(4)
Eqs.(3 and 4) in the matrix form:
3 Ã¢Ë†â€™ 1 Ã°Ââ€˜Â£1
20
=
Ã¢Ë†â€™3 5 Ã°Ââ€˜Â£2
60
The determinant of the matrix is
3 Ã¢Ë†â€™1
ÃŽâ€=
= 15 Ã¢Ë†â€™ 3 = 12
Ã¢Ë†â€™3 5
v1 and v2 are obtained as
20 Ã¢Ë†â€™ 1
100 + 60
Ã°Ââ€˜Â£1 = 60 5
=
= 13.33 Ã°Ââ€˜â€°
ÃŽâ€
12
3 20
180 + 60
Ã°Ââ€˜Â£2 = Ã¢Ë†â€™3 60 =
= 20 Ã°Ââ€˜â€°
ÃŽâ€
12
5
7. Calculate all branch voltages and currents.
The voltage drop on the resistor 4Ã°Ââ€ºâ‚¬ is Ã°Ââ€˜Â¼Ã°ÂÅ¸â€™Ã°Ââ€ºâ‚¬ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜ Ã¢â€°Ë† Ã°ÂÅ¸â€. Ã°ÂÅ¸â€¢Ã°Ââ€˜Â½.
The voltage drop on the resistor 6Ã°Ââ€ºâ‚¬ is Ã°Ââ€˜Â¼Ã°ÂÅ¸â€Ã°Ââ€ºâ‚¬ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€˜Â½.
Voltage drop on resistor 2Ã°Ââ€ºâ‚¬ is Ã°Ââ€˜Â¼Ã°ÂÅ¸ÂÃ°Ââ€ºâ‚¬ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜.
8. Check your solution by using a conservation law.
KCL applied to the node #1:
Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸â€˜ Ã¢â€¡â€™ Ã°ÂÅ¸â€œ + Ã°ÂÅ¸Â. Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢ = Ã°ÂÅ¸â€. Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢.
KCL applied to the node #2:
Ã°Ââ€™Å Ã°ÂÅ¸â€™ = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€œ Ã¢â€¡â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°ÂÅ¸â€œ + Ã°ÂÅ¸Â. Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢ + Ã°ÂÅ¸â€˜. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜
6
Supernode
(Nodal Analysis with Voltage Sources)
Ã°Ââ€™â€”Ã°ÂÅ¸Â =?,
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ =? .
Ã¢â‚¬Â¢ A supernode is formed by enclosing a (dependent or independent) voltage
source connected between two non-reference nodes and any elements
connected in parallel with it.
Ã¢â‚¬Â¢ Both, KCL and KVL, are applied to determine the node voltage.
7
Applying KVL to the Supernode
Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€™ = Ã°Ââ€™Å  Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Eq.(1a)
or
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½ Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ Ã¢Ë†â€™ Ã°ÂÅ¸Å½
+
=
+
Eq.(1b)
Ã°ÂÅ¸Â
Ã°ÂÅ¸â€™
Ã°ÂÅ¸â€“
Ã°ÂÅ¸â€
The voltage source inside the supernode
provides KVL equation needed to solve for
the node voltages. KVL applied to the
supernode:
Ã¢Ë†â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œ + Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸Å½ Ã°ÂÂâ€žÃ°ÂÂÂª. (Ã°ÂÅ¸Â)
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ + Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã°ÂÂâ€žÃ°ÂÂÂª. (Ã°ÂÅ¸ÂÃ°ÂÂÅ¡)
From the diagram one might see that
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°ÂÂâ€žÃ°ÂÂÂª. (Ã°ÂÅ¸â€˜)
8
Applying KVL to the Supernode
From Eqs.(1b, 2a and 3) one might see that
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã¡â€°Â
Ã¢Å¸Â¹
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ + Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã°ÂÂâ‚¬Ã°ÂÂÂ§Ã°ÂÂÂ¬Ã°ÂÂÂ°Ã°ÂÂÅ¾Ã°ÂÂÂ«: Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸â€™. Ã°ÂÅ¸Â Ã°Ââ€˜Â½, Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸â€”. Ã°ÂÅ¸Â Ã°Ââ€˜Â½
9
Mesh (Loop) Analysis
Ã¢â‚¬Â¢ Mesh analysis only applicable to a circuit that is planar.
Ã¢â‚¬Â¢ A planar circuit is one that can be drawn in a plane with no branches
crossing one another.
Ã¢â‚¬Â¢ A mesh is an elementary loop that does not contain any other loops within
it.
Ã¢â‚¬Â¢ From fundamental theorem of network topology b=l+n-1Ã¢â€¡â€™
n=b-l+1 is number of equations needed for Nodal Analysis;
l=b-n+1 is number of equations needed for Mesh Analysis;
as n>l , one should solve less number of equations by using Mesh
Analysis.
Ã¢â‚¬Â¢ Mesh and branch currents are not always the same.
10
Nonpalanar
vs
Planar
Ã¢Å¸Âº
This circuit is a planar: It can be redrawn to avoid
crossing branches.
11
Mesh Analysis Steps
1) Assign n mesh (loop) currents in to the Ã°Ââ€™Â meshes;
2) Apply KVL to each of the Ã°Ââ€™Â meshes to get Ã°Ââ€™Â linear equations;
3) Use OhmÃ¢â‚¬â„¢s law UR=inR to express the voltages on resistors in terms of the
mesh currents.
4) Solve the resulting Ã°Ââ€™Â linear equations to get the mesh currents in.
5) Use OhmÃ¢â‚¬â„¢s law UR=inR to find the voltage drop on each resistor.
6) Use OhmÃ¢â‚¬â„¢s law IR=UR/R to find the branch currents IR.
12
Example:
Find the branch currents I1, I2 and I3.
Mesh #1 (aÃ¢Å¸Â¶bÃ¢Å¸Â¶eÃ¢Å¸Â¶fÃ¢Å¸Â¶a):
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€˜Â½Ã°ÂÅ¸Â ;
Mesh #2 (bÃ¢Å¸Â¶cÃ¢Å¸Â¶dÃ¢Å¸Â¶eÃ¢Å¸Â¶b):
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€˜Â½Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â + (Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ )Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â
Branch currents:
Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â ;
Fig. 3.17. A circuit with two meshes.
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€˜Â½Ã°ÂÅ¸Â ;
Ã¢Å¸Â¹Ã ÂµÅ¾
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â + (Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ )Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°Ââ€˜Â½Ã°ÂÅ¸Â
=
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸Â , Ã°Ââ€™Å Ã°ÂÅ¸Â , Ã°Ââ€™Å Ã°ÂÅ¸â€˜ are roots of the system with three linear
equations.
Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â ;
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â . Ã¢Å¸Â¹ Mesh and branch currents are not always the same.
13
1Ã°Ââ€˜â€°
2Ã°Ââ€˜â€°
The initial assumption is that
the mech currents Ã°Ââ€™Å Ã°ÂÅ¸Â and Ã°Ââ€™Å Ã°ÂÅ¸Â
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
KVL in the mesh #1:
Ã¢Ë†â€™Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â.
KVL in the mesh #2:
Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½;
Ã¢Ë†â€™Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸Â.
14
Ã°ÂÅ¸â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã¢Å¸Â¹Ã¡â€°Â
Ã¢Å¸Â¹
Ã¢Ë†â„¢
=
Ã¢Å¸Â¹
Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€œ
Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã¢Ë†â€™Ã°ÂÅ¸â€˜Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€œÃ°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸Â.
Ã°ÂÅ¸â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜
= Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â„¢ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Â;
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€œ
Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â =
= Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸Â Ã¢Ë†â„¢ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã¢Ë†â€™Ã°ÂÅ¸Â;
Ã¢Ë†â€™Ã°ÂÅ¸Â Ã°ÂÅ¸â€œ
Ã°ÂÅ¸â€™
Ã°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â =
= Ã¢Ë†â€™Ã°ÂÅ¸â€“ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã¢Ë†â€™Ã°ÂÅ¸â€œ.
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸â€œ
Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨,
Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ“Å¸=
1Ã°Ââ€˜â€°
2Ã°Ââ€˜â€°
The initial assumption is that
the mech currents Ã°Ââ€™Å Ã°ÂÅ¸Â and Ã°Ââ€™Å Ã°ÂÅ¸Â
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨,
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã°ÂÅ“Å¸Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°ÂÅ¸â€œ
Ã°Ââ€™Å Ã°ÂÅ¸Â =
=
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨ Ã¢Å¸Â¹
Ã°ÂÅ“Å¸
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â
Directions of both mesh currents should be corrected to show the counter-clockwise direction:
Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã°Ââ€˜Â¨,
Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â¨
Also directions of the branch currents Ã°Ââ€˜Â°Ã°ÂÅ¸Â and Ã°Ââ€˜Â°Ã°ÂÅ¸Â should be changed to opposite ones.
Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â and Ã°Ââ€˜Â°Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ.
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€ Ã°Ââ€˜Â¨
The original assumption about downward direction of the branch current Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜ was correct.
15
Supermesh
Supermesh
Mesh#1
Mesh#2
Ã¢â‚¬Â¢ Current sources (dependent or independent) that are shared by more than
one mesh need a special treatment, because it is not clear what voltage
has to be assigned to the current source;
Ã¢â‚¬Â¢ The two meshes must be joined together, resulting in a supermesh;
Ã¢â‚¬Â¢ The supermesh is constructed by merging the two meshes and excluding
the shared source and any elements in series with it;
Ã¢â‚¬Â¢ KCL should be applied to one node of the supermesh.
16
Supermesh
Example
Ã¢â‚¬Â¢ Apply KVL to the supermesh:
Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ + Ã°ÂÅ¸â€Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ Ã¢Å¸Â¹
or Ã°ÂÅ¸â€Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã¢â‚¬Â¢ Apply KCL to the node Ã¢â‚¬Å“0Ã¢â‚¬Â in the branch where the two meshes
intersect:
Ã°ÂÅ¸â€Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€
Ã¢Å¸Â¹ Ã¡â€°Â
in
out
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€
Ã¢â‚¬Â¢ By solving the system of two equations we get:
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸ÂÃ°ÂÂâ‚¬ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â. Ã°ÂÅ¸â€“Ã°ÂÂâ‚¬ Ã¢Å¸Â¹ Initial assumption for the direction of the
mesh current Ã°Ââ€™Å Ã°ÂÅ¸Â should be corrected.
17
Supermesh
Example
3.2A
6A
2.8A
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸ÂÃ°ÂÂâ‚¬ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Â. Ã°ÂÅ¸â€“Ã°ÂÂâ‚¬ Ã¢Å¸Â¹ Initial assumption for the direction of the
mesh current Ã°Ââ€™Å Ã°ÂÅ¸Â should be corrected.
18
Circuit Theorems.
Outline:
1) Linearity Property;
2) Superposition;
3) Source Transformation;
4) TheveninÃ¢â‚¬â„¢s Theorem;
5) NortonÃ¢â‚¬â„¢s Theorem;
6) Value of Maximum Power.
CCNY
ENGR20400
Professor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
1
Linearity Property for Network of Resistors
Ã¢â€“Âª
Ã¢â€“Âª
Ã¢â€“Âª
The homogeneity property requires that if the input is
multiplied by a constant Ã¢â‚¬Å“Ã°Ââ€˜Ëœ”, then the output is
multiplied by the same constant:
Ã°Ââ€˜Ëœ Ã¢Ë†â„¢ Ã°Ââ€˜â€“Ã°Ââ€˜â€¦ = Ã°Ââ€˜Ëœ Ã¢Ë†â„¢ Ã°Ââ€˜â€°.
The additivity property requires that the response to a
sum of inputs is the sum of the response to each input
applied separately:
Ã°Ââ€˜â€° = Ã°Ââ€˜â€“1 + Ã°Ââ€˜â€“2 Ã°Ââ€˜â€¦ = Ã°Ââ€˜â€“1 Ã°Ââ€˜â€¦ + Ã°Ââ€˜â€“2 Ã°Ââ€˜â€¦ = Ã°Ââ€˜â€°1 + Ã°Ââ€˜â€°2 .
In general a linear circuit is one whose output is
linearly related (or directly proportional) to its input.
2
Superposition
Ã¢â‚¬Â¢ The superposition principle states that the voltage across (or current
through) an element in a linear circuit is the algebraic sum of the voltage
across (or currents through) that element due to each independent source
acting alone.
Ã¢â‚¬Â¢ Steps:
1) Turn off all independent sources except one source. Find the output
(voltage or current) due to that active source;
Ã¢â‚¬Â¢ Replace a voltage source by short circuit.
Ã¢â‚¬Â¢ Replace a current sources by open circuit.
2) Dependent sources should be left intact;
3) Repeat this for each of the other independent sources;
4) Find the total contribution by adding algebraically all the
contributions due to the independent sources.
Ã¢â‚¬Â¢ Advantage: reduce a complex circuit to simpler circuits.
3
2) Switch Ã¢â‚¬â€œOFF 25V voltage source.
Ã°Ââ€˜Â£Ã°Ââ€˜Â¥2
Ã°Ââ€™Å Ã°ÂÅ¸Â
0.1Ã°Ââ€˜Â£Ã°Ââ€˜Â¥2
Ã°Ââ€™Å Ã°ÂÅ¸Â
KCL: Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ + Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â ;
Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â
+
= Ã°ÂÅ¸â€œ + Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â ;
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°ÂÅ¸â€™
Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€˜Â½.
1) Switch Ã¢â‚¬â€œOFF 5A current source.
Ã°Ââ€˜Â£Ã°Ââ€˜Â¥1
Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸Â
0.1Ã°Ââ€˜Â£Ã°Ââ€˜Â¥1
KCL: Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â ;
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â
+ Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â =
;
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°ÂÅ¸â€™
Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â = Ã°ÂÅ¸â€. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€˜Â½.
3) Superposition:
Ã°Ââ€™â€”Ã°Ââ€™â„¢ = Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â + Ã°Ââ€™â€”Ã°Ââ€™â„¢Ã°ÂÅ¸Â = Ã°ÂÅ¸â€. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ + Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ = Ã°ÂÅ¸â€˜Ã°ÂÅ¸Â. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€˜Â½
4
Source Transformation
Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ is a voltage between
terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â
provided by the voltage
source Ã°Ââ€™â€”Ã°Ââ€™â€ and resistor R.
We should be able to find
a current source Ã°Ââ€™Å Ã°Ââ€™â€ , which is going
to provide the same voltage drop
Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ on the resistor R.
Similarly, to the Ã¢Ë†â€ -Y transformation for the resistorsÃ¢â‚¬â„¢ networks, it is
possible to transform a circuit with a voltage source to a circuit with a
current source and vice versa. This is a useful technique to simplify
analysis of the electrical circuits.
5
Circuit analysis with a load Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ .
Ã°Ââ€™Å Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â£Ã°Ââ€˜Å½Ã°Ââ€˜Â
(a)
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â£Ã°Ââ€˜Å½Ã°Ââ€˜Â
(b)
(a)Ã¢Å¸Â¹ Voltage division:
Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™â€”Ã°Ââ€™â€
;
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
(b) LetÃ¢â‚¬â„¢s use a current source with
Ã°Ââ€™â€”
Ã°Ââ€™Å Ã°Ââ€™â€ = Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã¢â€¡â€™ Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹;
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â¹
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹
= Ã°Ââ€™Å Ã°Ââ€™â€
;
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€™Å Ã°Ââ€™â€ =
Ã°Ââ€™â€”
(a)Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€˜Â¹+Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™â€šÃ°Ââ€™Æ’
=
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹
= Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹+Ã°Ââ€˜Â¹ .
Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â¹
Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
For the circuit with the voltage source and resistor R:
Ã°Ââ€™Å Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
6
Circuit analysis with a load Ã°Ââ€˜â€¦Ã°Ââ€˜Å½Ã°Ââ€˜Â .
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â£Ã°Ââ€˜Å½Ã°Ââ€˜Â
(a)
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â£Ã°Ââ€˜Å½Ã°Ââ€˜Â
(b)
Ã°Ââ€™â€”
(b) LetÃ¢â‚¬â„¢s use a current source Ã°Ââ€™Å Ã°Ââ€™â€ that Ã°Ââ€™Å Ã°Ââ€™â€ = Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã¢Å¸Â¹
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹;
Ã°Ââ€˜Â¹
(b) Ã¢Å¸Â¹Current division: Ã°Ââ€™Å Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹+Ã°Ââ€˜Â¹ ;
Ã°Ââ€™â€šÃ°Ââ€™Æ’
For the circuit with the current source and resistor R:
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹
Ã°Ââ€™â€šÃ°Ââ€™Æ’
OhmÃ¢â‚¬â„¢s law: Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€™Å Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹+Ã°Ââ€˜Â¹
Ã°Ââ€™â€šÃ°Ââ€™Æ’
Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’ Ã°Ââ€˜Â¹
Ã°Ââ€™â€”Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
For the circuit with the voltage source and resistor R:
Ã°Ââ€™Å Ã°Ââ€™â€šÃ°Ââ€™Æ’ = Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
7
Ã¢â‚¬Â¢ Source transformation also applies to dependent sources. The
transformation between the dependent sources:
Ã¢â‚¬Â¢ Note that the arrow of the current source is directed towards the
positive terminal of the voltage source.
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹
Ã¢â‚¬Â¢ Statement that Ã¢â‚¬Å“LetÃ¢â‚¬â„¢s use a current source Ã°Ââ€™Å Ã°Ââ€™â€ =
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€ =
Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹ Ã¢â‚¬Å“ determines main two formulas for the source
transformation.
Ã¢â‚¬Â¢ The transformation is not possible for Ã°Ââ€˜Â¹ = Ã°ÂÅ¸Å½ because Ã°Ââ€™Å Ã°Ââ€™â€ =
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°ÂÅ¸Å½
Ã¢Å¸Â¶ Ã¢Ë†Å¾;
Ã¢â‚¬Â¢ Ã°Ââ€˜Â¹ = Ã¢Ë†Å¾ is also prohibited because Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°Ââ€™â€ Ã¢Ë†â„¢ Ã¢Ë†Å¾ Ã¢Å¸Â¶ Ã¢Ë†Å¾.
8
1)
Ã°ÂÅ¸Â
Ã°Ââ€˜Â°
Ã°ÂÅ¸â€œ Ã°Ââ€™â„¢
2)
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™ Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°ÂÅ¸Â
Ã¢Ë†â€™ Ã°Ââ€˜Â°Ã°Ââ€™â„¢
Ã°ÂÅ¸â€œ
Ã°ÂÅ¸â€œÃ°ÂÅ“Â´
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ“Â´
Ã°ÂÅ¸â€œÃ°ÂÅ“Â´
3)
+
Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°Ââ€˜Â°Ã°Ââ€™â„¢ Ã¢Ë†â€™
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ“Â´
Ã°Ââ€˜Â°Ã°Ââ€™â„¢
KVL:
Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°Ââ€˜Â°Ã°Ââ€™â„¢ + Ã°ÂÅ¸â€œÃ°Ââ€˜Â°Ã°Ââ€™â„¢ + Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€˜Â°Ã°Ââ€™â„¢ = Ã°ÂÅ¸Å½
Ã°Ââ€˜Â°Ã°Ââ€™â„¢
Ã°Ââ€˜Â°Ã°Ââ€™â„¢ = Ã°ÂÅ¸â€¢. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€œÃ°ÂÅ¸â€“Ã°ÂÅ¸â€“ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢â€°Ë† Ã°ÂÅ¸â€¢. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™Å½Ã°Ââ€˜Â¨
9
TheveninÃ¢â‚¬â„¢s Theorem
Ã¢â‚¬Â¢ TheveninÃ¢â‚¬â„¢s theorem states that a linear two-terminal circuit can be
replaced by an equivalent TheveninÃ¢â‚¬â„¢s circuit consisting of a voltage
source Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° in series with a resistor Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€°, where Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° is the opencircuit voltage at the terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â and Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° is the input (or
equivalent) resistance at the terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â when the
independent source Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° is turned -off.
TheveninÃ¢â‚¬â„¢s circuit
10
Ã¢â‚¬Â¢ Source transformation, application of probe voltage (or current)
source and other techniques help to simplify a circuit to the
TheveninÃ¢â‚¬â„¢s equivalent circuit contained of voltage source Ã°Ââ€˜â€°Ã°Ââ€˜â€¡Ã¢â€žÅ½ and
resistor Ã°Ââ€˜â€¦Ã°Ââ€˜â€¡Ã¢â€žÅ½ connected in the series network.
Ã¢â‚¬Â¢ Consider that a linear circuit is loaded by the variable resistor Ã°Ââ€˜â€¦Ã°ÂÂÂ¿.
Ã¢â‚¬Â¢ An advantage to use TheveninÃ¢â‚¬â„¢s circuit: The currents Ã°ÂÂÂ¼Ã°ÂÂÂ¿ and voltages
Ã°Ââ€˜â€°Ã°ÂÂÂ¿ across the different loads Ã°Ââ€˜â€¦Ã°ÂÂÂ¿ can be easily determined by using
TheveninÃ¢â‚¬â„¢s equivalent circuit with the effective voltage source Ã°Ââ€˜â€°Ã°Ââ€˜â€¡Ã¢â€žÅ½ and
resistor Ã°Ââ€˜â€¦Ã°Ââ€˜â€¡Ã¢â€žÅ½ .
Ã¢Å¸Â¹
IL =
OhmÃ¢â‚¬â„¢s law:
VTh
RTh + RL
RL
VL = RL I L =
VTh
RTh + RL
11
To find RThevenin consider two possible case scenarios:
Ã¢â‚¬Â¢ Case1: If the circuit has no dependent source, one should turn -off all
independent sources, then Ã°Ââ€˜â€¦Ã°Ââ€˜â€¡Ã¢â€žÅ½ is the effective resistance of the
remaining resistor network measured between terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â.
Ã¢â‚¬Â¢ Case2: If the network has dependent sources, one should turn -off
all independent sources. As with superposition, dependent sources
are not to be turned -off because they are controlled by circuit
variables. One should apply a probe voltage v0 =1V (or probe current
Ã°Ââ€˜â€“0=1A) to the terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â and determine the current Ã°Ââ€˜â€“0
(voltage Ã°Ââ€˜Â£0). TheveninÃ¢â‚¬â„¢s resistance can be found by using OhmÃ¢â‚¬â„¢s law:
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€°
If Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½ then Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° =
Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â
=
Ã°Ââ€™Å Ã°Ââ€™Â
If Ã°Ââ€™Å Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°Ââ€˜Â¨ then Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° =
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸Â
12
LetÃ¢â‚¬â„¢s apply the source transformation to
TheveninÃ¢â‚¬â„¢s equivalent circuit:
TheveninÃ¢â‚¬â„¢s
Equivalent
circuit
Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€°
NortonÃ¢â‚¬â„¢s
Equivalent
circuit
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ
Ã¢Å¸Âº
Ã°Ââ€™â€”Ã°Ââ€™â€š = Ã°Ââ€˜Â°Ã°Ââ€˜Âµ Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ = Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€°
Ã¢Å¸Â¸ Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°Ââ€˜Â°Ã°Ââ€˜Âµ Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ
Ã°Ââ€˜Â°Ã°Ââ€˜Âµ =
Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€°
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€°
Potential difference between terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â is Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° .
13
Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°ÂÅ¸Å½Ã°Ââ€˜Â½
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ = Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€°
2)
Ã°Ââ€˜Â°Ã°ÂÅ¸Â
Ã°Ââ€˜Â°Ã°ÂÅ¸Â
Ã°Ââ€˜Â°Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â²Ã°Ââ€˜ÂªÃ°Ââ€˜Â³
Ã°Ââ€˜Â²Ã°Ââ€˜ÂªÃ°Ââ€˜Â³
1)
3)
= Ã°Ââ€˜Â¹Ã°Ââ€™â€šÃ°Ââ€™Æ’
14
NortonÃ¢â‚¬â„¢s Theorem
Ã¢â‚¬Â¢ NortonÃ¢â‚¬â„¢s theorem states that a linear two-terminal circuit can
be replaced by an equivalent circuit consisting of a current
source Ã°Ââ€˜Â°Ã°Ââ€˜Âµ in parallel with a resistor Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ , where Ã°Ââ€˜Â°Ã°Ââ€˜Âµ is the shortcircuit current through the terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â and Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ is
the input or equivalent resistance at the terminals when the
independent sources are turned -off.
15
Ã¢â‚¬Â¢ The NortonÃ¢â‚¬â„¢s resistance Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ is equivalent (or input) resistance
at terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â when all independent sources are
turned Ã¢â‚¬â€œoff, thus Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ = Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° .
Ã¢â‚¬Â¢ The NortonÃ¢â‚¬â„¢s current source Ã°Ââ€˜Â°Ã°Ââ€˜Âµ can be found by using the shortcircuit current flowing from terminal Ã¢â‚¬Å“aÃ¢â‚¬Â to Ã¢â‚¬Å“bÃ¢â‚¬Â, where Ã°Ââ€˜Â°Ã°Ââ€˜Âµ = Ã°Ââ€™Å Ã°Ââ€™â€Ã°Ââ€™â€ž .
Ã°Ââ€™Å Ã°Ââ€™â€Ã°Ââ€™â€ž = Ã°Ââ€˜Â°Ã°Ââ€˜Âµ
In this particular case the resistor Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ is shorten, thus
Ã°Ââ€™Å Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ = Ã°ÂÅ¸Å½.
16
Terminology for TheveninÃ¢â‚¬â„¢s and NortonÃ¢â‚¬â„¢
Transformations
Ã¢â‚¬Â¢ The open-circuit voltage Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â€ž across terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â:
Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â€ž = Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° ;
Ã¢â‚¬Â¢ The short-circuit current Ã°Ââ€™Å Ã°Ââ€™â€Ã°Ââ€™â€ž at terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â:
Ã°Ââ€™Å Ã°Ââ€™â€Ã°Ââ€™â€ž = Ã°Ââ€˜Â°Ã°Ââ€˜Âµ ;
Ã¢â‚¬Â¢ The equivalent or input resistance Ã°Ââ€˜Â¹Ã°Ââ€™Å Ã°Ââ€™Â at terminals Ã¢â‚¬Å“aÃ¢â‚¬Â and Ã¢â‚¬Å“bÃ¢â‚¬Â
when all independent sources are turned off:
Ã°Ââ€˜Â¹Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ .
Ã¢â‚¬Â¢ It often occurs that an effective resistance Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° ( or Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ ) takes a
negative value. In this case, the negative effective resistance
implies that the circuit is supplying power. This is a highly
possible situation in a circuit with dependent sources.
17
Example 4.11.
Find the Norton equivalent circuit:
1) Ã°Ââ€˜Â¹Ã°Ââ€˜Âµ = Ã°ÂÅ¸â€œ||(Ã°ÂÅ¸â€“ + Ã°ÂÅ¸â€™ + Ã°ÂÅ¸â€“) = Ã°ÂÅ¸â€œ||Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ =
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸â€œ
= Ã°ÂÅ¸â€™Ã°ÂÅ“Â´
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ
2) To find Ã°Ââ€˜Â° Ã°Ââ€˜Âµ, one should short Ã°ÂÂÂ­Ã°ÂÂÂ¡Ã°ÂÂÅ¾ circuit
terminals a and b.
Ignore the 5Ã°ÂÅ“Â´ resistor because it was
shortÃ¢Ë†â€™circuited.
Ã°ÂÂÂÃ°ÂÂÂ² applying mesh analysis, one can get
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°Ââ€˜Â¨,
Ã°ÂÅ¸â€“ + Ã°ÂÅ¸â€“ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸Â + Ã°ÂÅ¸â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½
Ã°Ââ€˜ÂºÃ°Ââ€™Â Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°Ââ€˜Â¨ = Ã°Ââ€™Å Ã°Ââ€™â€Ã°Ââ€™â€ž = Ã°Ââ€˜Â°Ã°Ââ€˜Âµ
18
Maximum Power Transfer
Ã¢â‚¬Â¢ In many applications, a circuit is designed to power a load.
Among those applications there are many cases where one
would wish to maximize the power transferred to the load.
Ã¢â‚¬Â¢ TheveninÃ¢â‚¬â„¢s equivalent circuit can be used to find the
condition for maximum power transferred to the load.
Ã¢â‚¬Â¢ One can see that if Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ approaches 0 or Ã¯â€šÂ¥ the power
transferred goes to zero.
Ã°ÂÅ¸Â
Ã°Ââ€˜Â·Ã°Ââ€˜Â³ =
Ã°Ââ€™Å Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€°
Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€°
= Ã°Ââ€™Å =
=
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ ;
Ã°Ââ€™ÂÃ°Ââ€™Å Ã°Ââ€™Å½
Ã°Ââ€˜Â· = Ã°ÂÅ¸Å½;
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã¢Å¸Â¶ Ã°ÂÅ¸Å½ Ã°Ââ€˜Â³
Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€˜Â»Ã°Ââ€™â€°
Ã°Ââ€™ÂÃ°Ââ€™Å Ã°Ââ€™Å½
Ã°Ââ€˜Â· Ã¢â€°Ë†
= Ã°ÂÅ¸Å½;
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã¢Å¸Â¶ Ã¢Ë†Å¾ Ã°Ââ€˜Â³
Ã¢Ë†Å¾
19
Maximum power transferred if Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ = Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€˜Â¯
FYI:
Ã°Ââ€™â€¦ Ã°Ââ€™â€¡Ã°ÂÅ¸Â Ã°Ââ€™â„¢
Ã°Ââ€™â€¦Ã°Ââ€™â„¢ Ã°Ââ€™â€¡Ã°ÂÅ¸Â Ã°Ââ€™â„¢
=
Ã°Ââ€™â€¡Ã¢â‚¬Â²Ã°ÂÅ¸Â Ã°Ââ€™â„¢ Ã°Ââ€™â€¡Ã°ÂÅ¸Â Ã°Ââ€™â„¢ Ã¢Ë†â€™Ã°Ââ€™â€¡Ã°ÂÅ¸Â Ã°Ââ€™â„¢ Ã°Ââ€™â€¡Ã¢â‚¬Â²Ã°ÂÅ¸Â Ã°Ââ€™â„¢
Ã°Ââ€™â€¡Ã°ÂÅ¸ÂÃ°ÂÅ¸Â Ã°Ââ€™â„¢
LetÃ¢â‚¬â„¢s find when the power transferred to Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ has an extreme value:
Ã°Ââ€™â€¦Ã°Ââ€˜Â·
Ã°Ââ€™â€¦
= Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€˜Â»Ã°Ââ€™â€°
Ã°Ââ€™â€¦Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€™â€¦Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°ÂÅ¸Â
=
= Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€˜Â»Ã°Ââ€™â€°
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
=
Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°ÂÅ¸â€™
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LetÃ¢â‚¬â„¢s check if this extremum is a maximum:
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< Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ this extremum is maximum. 20 Value of Maximum Power Ã°Ââ€˜Â·Ã°Ââ€˜Â³Ã°Ââ€™Å½Ã°Ââ€™â€šÃ°Ââ€™â„¢ = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°Ââ€™Å = = Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ = Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€˜Â»Ã°Ââ€™â€° = Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ = Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° = , Ã°ÂÅ¸â€™Ã°Ââ€˜Â¹ Ã°Ââ€˜Â»Ã°Ââ€™â€° Thus, Maximum power is transferred to the load Ã°Ââ€˜â€¦Ã°ÂÂÂ¿ when that load resistance is equal to the Thevenin resistance Ã°Ââ€˜â€¦Ã°Ââ€˜â€¡Ã¢â€žÅ½ . 21 1) Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â„¢ Ã°Ââ€™Å Ã°ÂÅ¸Å½ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â„¢ KVL in the mesh #1: Ã°ÂÅ¸â€Ã°ÂÅ¸Å½Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â + Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â + Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â„¢ = Ã°ÂÅ¸Å½ Ã ÂµÅ¾ Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â = Ã°ÂÅ¸â€”Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â OhmÃ¢â‚¬â„¢s law: Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â„¢ = Ã°ÂÅ¸â€Ã°ÂÅ¸Å½Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â KVL in the mesh #2: Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â„¢ + Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â = Ã°ÂÅ¸Å½ ; Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â = Ã°ÂÅ¸Â Ã ÂµÅ¾ Mesh #1: Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â = Ã°ÂÅ¸â€”Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°ÂÅ¸Â Ã°ÂÅ¸â€” Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â = Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ã°Ââ€˜Â¨ , Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â = Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ã°Ââ€˜Â¨. Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°Ââ€˜Â½ = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½ = Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸â€. Ã°ÂÅ¸â€ Ã°ÂÅ¸â€ Ã°ÂÅ“Â´ Ã°Ââ€™Å Ã°Ââ€™Â = Ã¢Ë†â€™Ã°Ââ€™Å Ã°Ââ€™ÂÃ°ÂÅ¸Â Ã°Ââ€™Å Ã°Ââ€™Â 22 Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° 2) Ã°Ââ€™Å Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°ÂÅ¸Â KVL in the mesh #1: Ã¢Ë†â€™Ã°ÂÅ¸â€” + Ã°ÂÅ¸â€Ã°ÂÅ¸Å½Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™â„¢ = Ã°ÂÅ¸Å½ Ã°ÂÅ¸Â Ã ÂµÅ¾ Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€˜Â¨. Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½ Ã°Ââ€™â€”Ã°Ââ€™â„¢ = Ã°ÂÅ¸â€Ã°ÂÅ¸Å½Ã°Ââ€™Å Ã°ÂÅ¸Â Ã°ÂÂÅ¡Ã°ÂÂÂ§Ã°ÂÂÂ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ KVL in the mesh #2: Ã¢Ë†â€™Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™â„¢ + Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½ Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°ÂÅ¸Å½ Ã ÂµÅ¾ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ Ã°Ââ€™â€šÃ°Ââ€™ÂÃ°Ââ€™â€¦ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€˜Â·Ã°Ââ€˜Â³Ã°Ââ€™Å½Ã°Ââ€™â€šÃ°Ââ€™â„¢ Ã°ÂÅ¸Â Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°ÂÅ¸â€¢ Ã°Ââ€˜Â½. Ã°Ââ€˜Â¨; Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€˜Â»Ã°Ââ€™â€° Ã°Ââ€˜Â½Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°ÂÅ¸â€¢Ã°Ââ€˜Â½ = = = Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€”Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢Ã°ÂÅ¸Â = Ã°ÂÅ¸â€”Ã°ÂÅ¸â€. Ã°ÂÅ¸â€¢Ã°ÂÅ¸ÂÃ°ÂÂÂ¦Ã°ÂÂâ€“. Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° = Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸â€. Ã°ÂÅ¸â€ Ã°ÂÅ¸â€ Ã°ÂÅ“Â´ Ã°ÂÅ¸â€™Ã°Ââ€˜Â¹Ã°Ââ€˜Â»Ã°Ââ€™â€° 23 Alternating Current and Voltage Outline: 1) 2) 3) 4) DC and AC signals; RMS amplitude; LCR circuit; Resonance in a LCR circuit. 1 Direct and Alternating Currents Current from a battery flows steadily in one direction (direct current, DC). Current from a power plant varies sinusoidally (alternating current, AC). 2 Alternating Current, Voltage and Power The voltage varies sinusoidally with time: as does the current: By multiplying the current square and resistance one finds the power: 3 Average Power of the AC Current Usually we are interested in the average power: Ã°Ââ€˜Â°Ã°ÂÅ¸ÂÃ°Ââ€™Â Ã°Ââ€˜Â°Ã°ÂÅ¸ÂÃ°Ââ€™Â Ã Â´Â¥Ã°ÂÅ¸Â = Ã°Ââ€˜Â°Ã Â´Â¥Ã°ÂÅ¸Â Ã°Ââ€˜Â¹ = Ã¢Ë†â„¢Ã°Ââ€˜Â¹= = Ã°Ââ€˜Â° Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€™Â Ã°ÂÅ¸Â Ã°Ââ€˜Â½Ã°ÂÅ¸Â Ã°Ââ€˜Â½Ã°ÂÅ¸ÂÃ°Ââ€™Â Ã°ÂÅ¸Â = = Ã¢Ë†â„¢ = = Ã°Ââ€˜Â½ Ã°ÂÅ¸Â Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹ Ã°ÂÅ¸Â RMS Current and Voltage The current and voltage both have average values of zero, so we square them, take the average, then take the square root, yielding the root mean square (rms) value. 4 RMS amplitudes https://en.wikipedia.org/wiki/Root_mean_square 5 Digital Signal In order to convert an analog signal to digital, the signal must be sampled. A higher sampling rate reproduces the signal more precisely. Before it is sent to a loudspeaker or headset, a digital audio signal must be converted back to analog. Signals with a noise Noise can easily corrupt an analog signal; a digital signal is much less sensitive to noise. AC Circuit with a Resistance only The current through a resistor is in phase with the voltage. Ã°Ââ€˜â€šÃ¢â€žÅ½Ã°Ââ€˜Å¡Ã¢â‚¬Â² Ã°Ââ€˜Â  Ã°Ââ€˜â„¢Ã°Ââ€˜Å½Ã°Ââ€˜Â¤: Ã°Ââ€˜Ë†(Ã°Ââ€˜Â¡) = Ã°ÂÂÂ¼(Ã°Ââ€˜Â¡) Ã¢Ë†â„¢ Ã°Ââ€˜â€¦ 8 AC Circuit with a Capacitor only In a capacitor, the voltage VC lags the current by 90Ã‚Â°. charging discharging Ã°Ââ€˜â€˜Ã°Ââ€˜â€° Ã°ÂÂÂ¼=Ã°ÂÂÂ¶ Ã°Ââ€˜â€˜Ã°Ââ€˜Â¡ U charging 90o U discharging 9 AC Circuit with Inductance only The voltage on the inductor VL leads current through an inductor by 90Ã‚Â°. 90o charging discharging U U charging U discharging 10 LRC Series AC Circuit Only one current Ã°Ââ€˜Â° exists in this serial circuit, however there are three voltages Ã°Ââ€˜Â½Ã°Ââ€˜Â¹ , Ã°Ââ€˜Â½Ã°Ââ€˜Â³ , Ã°Ââ€˜Â½Ã°Ââ€˜Âª and three effective resistances (ratio of voltage to current), called the resistance R and reactances XL and XC , where Ã°Ââ€˜â€°Ã°Ââ€˜â€¦ = Ã°ÂÂÂ¼ Ã¢Ë†â„¢ Ã°Ââ€˜â€¦, Ã°Ââ€˜â€°Ã°ÂÂÂ¿ = Ã°ÂÂÂ¼ Ã¢Ë†â„¢ Ã°Ââ€˜â€¹Ã°ÂÂÂ¿ and Ã°Ââ€˜â€°Ã°ÂÂÂ¶ = Ã°ÂÂÂ¼ Ã¢Ë†â„¢ Ã°Ââ€˜â€¹Ã°ÂÂÂ¶ . Note that both reactances are frequency depend: 11 LRC Series AC Circuit (Phasor diagram) The voltage on the inductor VL leads current by 90Ã‚Â°. The voltage of resistor VR and current are in phase. We calculate all voltages in respect to the AC current by using what are called phasors, these are vectors representing the individual voltages, where Ã°ÂÂÂ¼Ã°Ââ€˜Å“ is an amplitude of that AC current. The voltage VC lags the current by 90Ã‚Â°. t=0 As time goes on, the phasors will rotate counterclockwise. Thus, some time t later on, the phasors have rotated (b) to new positions. 12 t>0
LRC Series AC Circuit
The voltages Ã°Ââ€˜â€°Ã°Ââ€˜â€¦,Ã°ÂÂÂ¿,Ã°ÂÂÂ¶ Ã°Ââ€˜Â¡ across each device are given by the xcomponent of each, and the current Ã°ÂÂÂ¼ Ã°Ââ€˜Â¡ by its x-component.
The current Ã°ÂÂÂ¼ Ã°Ââ€˜Â¡ is the same throughout the circuit.
Ã°ÂÂÂ¼ = Ã°ÂÂÂ¼0 cos 2Ã°ÂÅ“â€¹Ã°Ââ€˜â€œÃ°Ââ€˜Â¡
Ã°Ââ€˜â€° = Ã°Ââ€˜â€°Ã°Ââ€˜â€¦ + Ã°Ââ€˜â€°Ã°ÂÂÂ¿ + Ã°Ââ€˜â€°Ã°ÂÂÂ¶
(KVL)
13
Impedance Ã°Ââ€˜Â Ã°ÂÂÅ½ of LRC Circuit
y
V0
VR0
VL0-VC0
x
By using the Pythagorean theorem one can find that
Ã°Ââ€˜â€°0 =
2
Ã°Ââ€˜â€°Ã°Ââ€˜â€¦0
+ Ã°Ââ€˜â€°Ã°ÂÂÂ¿0 Ã¢Ë†â€™ Ã°Ââ€˜â€°Ã°ÂÂÂ¶0
2
= Ã°ÂÂÂ¼0
Ã°Ââ€˜â€¦2
+ Ã°Ââ€˜â€¹Ã°ÂÂÂ¿ Ã°ÂÅ“â€ Ã¢Ë†â€™ Ã°Ââ€˜â€¹Ã°ÂÂÂ¶ Ã°ÂÅ“â€
Ã°Ââ€˜â€°0 = Ã°Ââ€˜Â Ã°ÂÅ“â€ Ã¢Ë†â„¢ Ã°ÂÂÂ¼0
14
2
LRC Series AC Circuit
We find from the ratio of voltage to current that the
effective (equivalent) resistance, called the impedance,
of the circuit is given by:
where Ã°Ââ€˜â€¦ =
Ã°Ââ€˜Ë†
=Constant
Ã°ÂÂÂ¼
15
Current in AC Circuits
The rms current in an ac circuit is:
Clearly, Irms is a frequency depend current: Ã°ÂÂÂ¼Ã°Ââ€˜Å¸Ã°Ââ€˜Å¡Ã°Ââ€˜Â  = Ã°ÂÂÂ¼Ã°Ââ€˜Å¸Ã°Ââ€˜Å¡Ã°Ââ€˜Â  Ã°Ââ€˜â€œ
16
Current Resonance in AC Circuits
We see that Irms demonstrates a maximum
Ã°Ââ€˜â€¹Ã°ÂÂÂ¶ = Ã°Ââ€˜â€¹Ã°ÂÂÂ¿
when XC = XL;
the frequency at which this occurs is
this is called the resonant frequency.
Ã°Ââ€˜â€¹Ã°ÂÂÂ¿ Ã¢â€ â€™ Ã¢Ë†Å¾
Ã°Ââ€˜â€¹Ã°ÂÂÂ¶ Ã¢â€ â€™ 0
Ã°Ââ€˜â€¹Ã°ÂÂÂ¿ Ã¢â€ â€™ 0
Ã°Ââ€˜â€¹Ã°ÂÂÂ¶ Ã¢â€ â€™ Ã¢Ë†Å¾
17
Resonance in LC Circuits
In a capacitor, the
voltage VC lags the
current by 90Ã‚Â°.
The voltage on the
current through an
inductor by 90Ã‚Â°.
https://en.wikipedia.org/wiki/LC_circuit
18
Frequency Spectrum of Signals
Ã°Ââ€™â€”Ã°Ââ€™â€š Ã°Ââ€™â€¢ = Ã°Ââ€˜Â½Ã°Ââ€™â€š Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÂÅ½Ã°Ââ€™Â Ã°Ââ€™â€¢ is the time dependent voltage with peak amplitude Ã°Ââ€˜Â½Ã°Ââ€™â€š and
Ã°ÂÅ¸Â
angular frequency Ã°ÂÂÅ½Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡Ã°Ââ€™Â , Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€ , where Ã°Ââ€™â€¡Ã°Ââ€™Â = Ã°Ââ€˜Â» , Ã°Ââ€˜Â¯Ã°Ââ€™â€º is a temporal frequency
Ã°Ââ€™Â
and Ã°Ââ€˜Â»Ã°Ââ€™Â , Ã°Ââ€™â€ is period of the sinusoid. This signal might be analyzed by using a
frequency spectrum calculated by using Fourier transformation:
Ã°Ââ€˜Âº Ã°ÂÂÅ½ = Ã°Ââ€˜Â­ Ã°Ââ€™â€¡ Ã°Ââ€™â€¢ is Fourier spectrum (or spectral density) of the signal Ã°Ââ€™â€¡ Ã°Ââ€™â€¢ =
Ã¢Ë†Å¾
Ã°ÂÅ¸Â Ã¢Ë†Å¾
Ã°Ââ€˜Â­Ã¢Ë†â€™Ã°ÂÅ¸Â Ã°Ââ€˜Âº Ã°ÂÂÅ½ , where Ã°Ââ€˜Â­ = Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã¢Ë†Å¾ Ã¢â‚¬Â¦ Ã°Ââ€™â€ Ã¢Ë†â€™Ã°Ââ€™Å Ã°ÂÂÅ½Ã°Ââ€™â€¢ Ã°Ââ€™â€¦Ã°Ââ€™â€¢ and Ã°Ââ€˜Â­Ã¢Ë†â€™Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°ÂÂâ€¦ Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã¢Ë†Å¾ Ã¢â‚¬Â¦ Ã°Ââ€™â€ Ã°Ââ€™Å Ã°ÂÂÅ½Ã°Ââ€™â€¢ Ã°Ââ€™â€¦Ã°Ââ€™â€¢ are forward
and inverse Fourier transformations, respectively.
Thus, Ã°Ââ€˜ÂºÃ°Ââ€™â€š Ã°ÂÂÅ½ =
Ã¢Ë†Å¾ Ã°Ââ€™â€ Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™Å Ã°Ââ€™â€¡Ã°Ââ€™Â Ã°Ââ€™â€¢ Ã¢Ë†â€™Ã°Ââ€™â€ Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™Å Ã°Ââ€™â€¡Ã°Ââ€™Â Ã°Ââ€™â€¢
Ã°Ââ€˜Â½Ã°Ââ€™â€š Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã¢Ë†Å¾
Ã°ÂÅ¸ÂÃ°Ââ€™Å
Ã°Ââ€™Å
Ã¢Ë†Å¾
Ã°Ââ€™â€ Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡Ã°Ââ€™â€¢ Ã°Ââ€™â€¦Ã°Ââ€™â€¢ = Ã°ÂÅ¸Â Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã¢Ë†Å¾ Ã¢Ë†â€™Ã°Ââ€™â€ Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™Å
Ã°Ââ€™â€¡Ã¢Ë†â€™Ã°Ââ€™â€¡Ã°Ââ€™Â Ã°Ââ€™â€¢
+ Ã°Ââ€™â€ Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™Å
Ã°Ââ€™â€¡+Ã°Ââ€™â€¡Ã°Ââ€™Â Ã°Ââ€™â€¢
Ã°Ââ€™Å
= Ã°ÂÅ¸Â Ã°ÂÅ“Â¹ Ã°ÂÂÅ½ + Ã°ÂÂÅ½Ã°Ââ€™Â Ã¢Ë†â€™ Ã°ÂÅ“Â¹ Ã°ÂÂÅ½ Ã¢Ë†â€™ Ã°ÂÂÅ½Ã°Ââ€™Â .
19
=
Fourier series
A symmetrical square-wave signal of
amplitude V.
The frequency spectrum (also known as
the line spectrum).
+Ã¢Ë†Å¾
Ã°Ââ€™â€” Ã°Ââ€™â€¢ =
where
Ã°Ââ€™â€”Ã°Ââ€™Â
+ Ã Â·Â (Ã°Ââ€™â€šÃ°Ââ€™Å’ Ã°Ââ€™â€žÃ°Ââ€™ÂÃ°Ââ€™â€ Ã°Ââ€™Å’Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡Ã°Ââ€™â€¢ + Ã°Ââ€™Æ’Ã°Ââ€™Å’ Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°Ââ€™Å’Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡Ã°Ââ€™â€¢ ) ,
Ã°ÂÅ¸Â
Ã°Ââ€™Å’=Ã°ÂÅ¸Â
Ã°ÂÅ¸Â Ã°ÂÂâ€¦
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÂâ€¦ Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã°ÂÂâ€¦ Ã°Ââ€™â€¡
Ã°ÂÅ¸Â
Ã°ÂÂâ€¦
Ã°ÂÅ¸Â
Ã°ÂÂâ€¦
Ã°Ââ€™â€¢ Ã°Ââ€™â€¦Ã°Ââ€™â€¢, Ã°Ââ€™â€šÃ°Ââ€™Å’ = Ã°ÂÂâ€¦ Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã°ÂÂâ€¦ Ã°Ââ€™â€¡ Ã°Ââ€™â€¢ Ã°Ââ€™â€žÃ°Ââ€™ÂÃ°Ââ€™â€ Ã°Ââ€™ÂÃ°Ââ€™â€¢ Ã°Ââ€™â€¦Ã°Ââ€™â€¢ , Ã°Ââ€™Æ’Ã°Ââ€™Å’ = Ã°ÂÂâ€¦ Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬Ã¢Ë†â€™Ã°ÂÂâ€¦ Ã°Ââ€™â€¡ Ã°Ââ€™â€¢ Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°Ââ€™ÂÃ°Ââ€™â€¢ Ã°Ââ€™â€¦Ã°Ââ€™â€¢
and Ã°Ââ€™Å’Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡ = Ã°Ââ€™Å’Ã°ÂÂÅ½Ã°Ââ€™Â is an angular frequency of Ã°Ââ€™Å’ harmonic oscillation.
Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
Ã°ÂÅ¸Â Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
Ã°ÂÅ¸Â Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
Ã°ÂÅ¸Â Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
Ã°Ââ€™â€” Ã°Ââ€™â€¢ =
Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÂÅ½Ã°Ââ€™Â Ã°Ââ€™â€¢ + Ã¢Ë†â„¢
Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸â€˜Ã°ÂÂÅ½Ã°Ââ€™Â Ã°Ââ€™â€¢ + Ã¢Ë†â„¢
Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸â€œÃ°ÂÂÅ½Ã°Ââ€™Â Ã°Ââ€™â€¢ + Ã¢Ë†â„¢
Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸â€¢Ã°ÂÂÅ½Ã°Ââ€™Â Ã°Ââ€™â€¢ + Ã¢â€¹Â¯ ,
Ã°ÂÂâ€¦
Ã°ÂÅ¸â€˜ Ã°ÂÂâ€¦
Ã°ÂÅ¸â€œ Ã°ÂÂâ€¦
Ã°ÂÅ¸â€¢ Ã°ÂÂâ€¦
where Ã°ÂÂÅ½Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡ is called the Fundamental frequency
20
Examples
E1.5. Find the frequencies f and Ãâ€° of a sine-wave signal with a period of 1 ms.
Ã°ÂÅ¸Â
Ã°Ââ€˜Â»
=
Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°Ââ€™Å’Ã°Ââ€˜Â¯Ã°Ââ€™â€º and Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÂâ€¦Ã°Ââ€™â€¡ = Ã°ÂÅ¸ÂÃ°ÂÂâ€¦ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€.
E1.7. The UHF (ultra high frequency) television broadcast band begins with
channel 14 and extends from 470 MHz to 608 MHz. If 6 MHz is allocated for
each channel, how many channels can this band accommodate?
Answer: Ã¢Ë†â€ Ã°Ââ€˜Â­ = Ã°ÂÅ¸â€Ã°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã¢Ë†â€™ Ã°ÂÅ¸â€™Ã°ÂÅ¸â€¢Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜Ã°ÂÅ¸â€“ Ã°Ââ€˜Â´Ã°Ââ€˜Â¯Ã°Ââ€™â€º Ã¢Å¸Â¹ Ã°Ââ€˜Âµ =
Ã¢Ë†â€ Ã°Ââ€˜Â­
Ã¢Ë†â€ Ã°Ââ€™â€¡
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜Ã°ÂÅ¸â€“
Ã°ÂÅ¸â€
= Ã°ÂÅ¸ÂÃ°ÂÅ¸â€˜
Channels from 14 to 36:
14, 15, 16,Ã¢â‚¬Â¦, 36.
21
E1.8. When the square-wave signal of Fig. 1.5, whose Fourier series is given
in Eq. (1.2), is applied to a resistor, the total power dissipated may be
Ã°ÂÅ¸Â Ã°Ââ€˜Â» Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€¦Ã°Ââ€™â€¢ or
Ã¢â‚¬Â«Ã—Â¬Ã¢â‚¬Â¬
Ã°Ââ€˜Â» Ã°ÂÅ¸Å½ Ã°Ââ€˜Â¹
calculated directly using the relationship Ã°Ââ€˜Â· =
indirectly by
summing the contribution of each of the harmonic components, that is,
Ã°Ââ€˜Â· = Ã°Ââ€˜Â·Ã°ÂÅ¸Â + Ã°Ââ€˜Â·Ã°ÂÅ¸â€˜ + Ã°Ââ€˜Â·Ã°ÂÅ¸â€œ + Ã¢â‚¬Â¦ , which may be found directly from rms values.
Verify that the two approaches are equivalent.
a) What fraction of the energy of a square wave is in its fundamental?
b) In its first five harmonics? c)In its first seven? d) First nine? e) In what
number of harmonics is 90% of the energy?
22
E1.8.
a) What fraction of the energy of a square wave is in its fundamental?
b) In its first five harmonics? c)In its first seven? d) First nine? e) In what
number of harmonics is 90% of the energy?
a)
b)
c)
d)
e)
23
Operational Amplifiers.
Outline:
1) Introduction;
2) Ideal Operational Amplifier;
3) Inverting and Noninverting circuits of amplifiers;
4) Examples.
CCNY
ENGR24100
Professor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
1
Introduction
Ã¢â‚¬Â¢ An Operational Amplifier (OPA) is a high-gain electronic
voltage amplifier. It is a) DC-coupled, b) with a differential input c)
and a single-ended output.
Ã¢â‚¬Â¢ OPA produces an output potential (in reference to the circuit ground)
that is typically hundreds of thousands of times larger than the
potential difference between its input terminals.
Ã¢â‚¬Â¢ OPA had their origins in analog computers, where they were used to
perform mathematical operations in many linear, non-linear, and
frequency-dependent circuits.
Ã¢â‚¬Â¢ OPA are building blocks in analog circuits. By using negative
feedback, the characteristics of an OA circuit, its gain, input
and output impedance, bandwidth etc. are determined by external
components;
Ã¢â‚¬Â¢ OPA have a relatively little dependence on temperature.
Ã¢â‚¬Â¢ OPA are among the most widely used electronic devices today.
2
OPAs are devices consisting of amplifiers
designed by using resistors, transistors,
capacitors, and diodes:
OPA 627 and 637 from Burr-Brown:
3
+Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
Type equation here.
+Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
(a) typical pin configuration, (b) circuit symbol.
4
+Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
+Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
Ã¢Ë†â€™Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž
5
OPA output voltage vo as a function of
the differential input voltage vd :
The output voltage is limited by source Ã°Ââ€˜Â½Ã°Ââ€™â€žÃ°Ââ€™â€ž .
6
Equivalent circuit
OPA is a high-gain electronic
voltage amplifier:
a) DC-coupled;
b) differential input;
c) single-ended output.
d) produces an output potential in
reference to the circuit ground.
Input: Ã°Ââ€™â€”Ã°Ââ€™â€¦ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â
Output: Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™â€¦ = Ã°Ââ€˜Â¨ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â
Note:
AÃ°Ââ€™â€”Ã°Ââ€™â€¦ is a voltage controlled
source.
Voltage gain can be
expressed in decibels
A dB = 10 log10 A
Typical Ranges of OPAs
Ideal Values
Open-loop voltage gain
A
105 — 108
Ã¢Ë†Å¾
Input resistance
Ã°Ââ€˜Â¹Ã°Ââ€™Å
103 — 1013 Ã°ÂÅ“Â´
Ã¢Ë†Å¾
Output resistance
Ã°Ââ€˜Â¹Ã°Ââ€™Â
10 — 100 Ã°ÂÅ“Â´
0
Supply Voltage
Ã°Ââ€˜Â¼Ã°Ââ€™â€žÃ°Ââ€™â€ž
Ã‚Â±5, Ã‚Â±Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ, Ã‚Â±Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™ Ã°Ââ€˜Â½
7
Ideal OPA
Ã°ÂÂÂ´ Ã¢Å¸Â¶ Ã¢Ë†Å¾, Ã°Ââ€˜â€¦Ã°Ââ€˜â€“ Ã¢Å¸Â¶ Ã¢Ë†Å¾, Ã°Ââ€˜â€¦Ã°Ââ€˜Å“ Ã¢â€°Ë†0
Fig. 5.8
Fig. 5.8 and 5.4
Ã°Ââ€™â€”Ã°Ââ€™â€¦
Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã¢Å¸Â¶ Ã¢Ë†Å¾ Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â =
Ã¢â€°Ë†Ã°ÂÅ¸Å½
Ã°Ââ€˜Â¹Ã°Ââ€™Å
OPA is designed to amplify a small signal,
thus Ã°Ââ€™â€”Ã°Ââ€™â€¦ Ã¢Å¸Â¶ Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€¦ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½ Ã°ÂÂÅ¡Ã°ÂÂÂ§Ã°ÂÂÂ Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½
8
Inverting Amplifier
An inverting amplifier reverses the polarity of the input signal
while amplifying it.
Ã°Ââ€™â€”Ã°Ââ€™Å  Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Å  Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Å¸Â¹
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½ Ã¢Å¸Â¹
=
Ã¢Å¸Â¹
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°ÂÅ¸Å½ = Ã¢Ë†â€™
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Å  and Ã°Ââ€˜Â¨Ã°Ââ€™â€” =
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Å
=Ã¢Ë†â€™
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
, thus, the gain is the feedback resistance
divided by the input resistance which means that the gain depends only on the
external elements connected to OPA.
9
Ã°Ââ€™Å Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸Â
0A
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€˜Â½
Ã°Ââ€™â€”Ã°ÂÅ¸Å½ = Ã¢Ë†â€™
Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã¢Ë†â€™
Ã¢Ë†â„¢
Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œ
Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â , Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Å½ Ã¢Å¸Â¹
Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã¢Ë†â€™ Ã°ÂÅ¸Å½
Ã°Ââ€™Å Ã°ÂÅ¸Â =
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Â. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°ÂÂÂÃ°Ââ€˜Â¨
Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ
Ã°Ââ€™Å Ã°ÂÅ¸Â =
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Â. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°ÂÂÂÃ°Ââ€˜Â¨
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Gain: Ã°Ââ€˜Â¨Ã°Ââ€™â€” =
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Å
=Ã¢Ë†â€™
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
=
Ã¢Ë†â€™Ã°ÂÅ¸â€˜.Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ
Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“Ã°ÂÅ¸Å½Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã¢Ë†â€™
Ã°ÂÅ¸â€™Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
= Ã¢Ë†â€™Ã°ÂÅ¸â€¢Ã°ÂÅ¸Å½
10
Noninverting Amplifier
A noninverting amplifier is an op amp circuit designed to
provide a positive voltage gain.
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™Å  Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢Å¸Â¹
=
Ã¢Å¸Â¹Ã¢Ë†â€™
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÅ¸Â +
Ã°Ââ€™â€”
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€™â€”Ã°Ââ€™Â
Thus, the gain depends only on the external resistors Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÂÅ¡Ã°ÂÂÂ§Ã°ÂÂÂ Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ .
Ã°Ââ€˜Â¨Ã°Ââ€™â€” =
=Ã°ÂÅ¸Â+
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
If Ã°Ââ€˜â€¦Ã°Ââ€˜â€œ = 0 and Ã°Ââ€˜â€¦1 Ã¢Å¸Â¶ Ã¢Ë†Å¾, then Ã°ÂÂÂ´Ã°Ââ€˜Â£ = 1 (such a
circuit is called as a voltage follower that
minimizes interaction between the two stages.
11
Solution by using superposition:
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€™â€”Ã°Ââ€™Â,Ã°ÂÅ¸â€Ã°Ââ€˜Â½ + Ã°Ââ€™â€”Ã°Ââ€™Â,Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
If 4V is Ã¢â‚¬â€œOFF:
then Ã°Ââ€™â€”Ã°ÂÅ¸Å½,Ã°ÂÅ¸â€Ã°Ââ€˜Â½ = Ã¢Ë†â€™
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Å  =
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã¢Ë†â€™
Ã°ÂÅ¸â€
Ã°ÂÅ¸â€™Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
= Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°Ââ€˜Â½
If 6V is Ã¢â‚¬â€œOFF:
then Ã°Ââ€™â€”Ã°Ââ€™Â,Ã°ÂÅ¸â€™Ã°Ââ€˜Â½ = Ã°ÂÅ¸Â +
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Â +
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€™Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€™ = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°Ââ€˜Â½
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€™â€”Ã°Ââ€™Â,Ã°ÂÅ¸â€Ã°Ââ€˜Â½ + Ã°Ââ€™â€”Ã°Ââ€™Â,Ã°ÂÅ¸â€™Ã°Ââ€˜Â½ = Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ + Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™ = Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÂâ€¢
(thus, to get positive Ã°Ââ€™â€”Ã°Ââ€™Â , one should increase 4V
source or decrease 6V source).
Check-up: applying KCL at node a, one can get
Ã°ÂÅ¸â€Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™â€š
Ã°ÂÅ¸â€™Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
=
Ã°Ââ€™â€”Ã°Ââ€™â€š Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â„¢Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°Ââ€™â€¦ Ã¢â€°Ë† Ã°ÂÅ¸Å½
Ã¢Å¸Â¹
Ã¢Å¸Â¹
Ã°Ââ€™â€”Ã°Ââ€™â€š Ã¢â€°Ë† Ã°Ââ€™â€”Ã°Ââ€™Æ’ = Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
Ã°ÂÅ¸â€ Ã¢Ë†â€™ Ã°ÂÅ¸â€™ Ã°ÂÅ¸â€™ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°Ââ€˜Â½
Ã°ÂÅ¸â€™
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
12
Example 5.1.
A 741 OPA has an open-loop voltage gain A=2 x 105, input resistance Ã°Ââ€˜Â¹Ã°Ââ€™Å  =2 MÃ¯Ââ€” and
output resistance Ã°Ââ€˜Â¹Ã°Ââ€™Â =50 Ã¯Ââ€”. OPA is used in the circuit below. Find closed-loop gain
Ã°Ââ€™â€”Ã°Ââ€™Â/Ã°Ââ€™â€”Ã°Ââ€˜Âº . Determine current Ã°Ââ€™Å  when Ã°Ââ€™â€”Ã°Ââ€˜Âº =2V.
Ideal OPA: Inverting Amplifier
Ã°Ââ€™â€”Ã°Ââ€™Å
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= Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°ÂÂÂ¦Ã°ÂÂâ‚¬ Ã¢Å¸Â¹ Ã°Ââ€™Å  Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™Å½Ã°Ââ€˜Â¨
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Ã°Ââ€˜Â¨Ã°Ââ€™â€” Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€” Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã¢Ë†â€™Ã°ÂÅ¸Â Ã¢Ë†â„¢ Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸â€™Ã°Ââ€˜Â½ Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸â€™Ã°Ââ€˜Â½
13
Example 5.1.
A 741 OPA has an open-loop voltage gain A=2 x 105, input resistance Ã°Ââ€˜Â¹Ã°Ââ€™Å  =2 MÃ¯Ââ€” and
output resistance Ã°Ââ€˜Â¹Ã°Ââ€™Â =50 Ã¯Ââ€”. OPA is used in the circuit below. Find closed-loop gain
Ã°Ââ€™â€”Ã°Ââ€™Â/Ã°Ââ€™â€”Ã°Ââ€˜Âº . Determine current Ã°Ââ€™Å  when Ã°Ââ€™â€”Ã°Ââ€˜Âº =2V.
14
Node 1: Ã°Ââ€™Å Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å
Ã°Ââ€™â€”Ã°Ââ€™â€ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
=
+
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ¸Â Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½
Ã°Ââ€˜Â¨ = Ã°ÂÅ¸Â Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°Ââ€™â€ + Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°Ââ€™â€ Ã¢â€°â€¦ Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â => Ã°Ââ€™â€”Ã°ÂÅ¸Â =
(Ã°ÂÂâ€žÃ°ÂÂÂª. Ã°ÂÅ¸Â)
Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Ë†â€™ Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™â€¦
Node 0: Ã°Ââ€™Å  = Ã°Ââ€™Å Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½ Ã¢Å¸Â¹
=
,
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½
Ã°Ââ€˜â€“
Ã°Ââ€˜â€“10
Ã°Ââ€™â€”Ã°Ââ€˜Âº
Ã°Ââ€˜Â£1
Ã°Ââ€˜â€“2
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™â€¦
Ã°Ââ€˜â€“50
where vÃ°Ââ€™â€¦ = Ã¢Ë†â€™vÃ°ÂÅ¸Â and A = 200 Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â£2
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã°Ââ€™â€”Ã°Ââ€™Â + Ã°ÂÅ¸Â Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ãƒâ€” Ã°Ââ€™â€”Ã°ÂÅ¸Â (Ã°ÂÂâ€žÃ°ÂÂÂª. Ã°ÂÅ¸Â)
From Eqs.1 and 2: Ã°ÂÅ¸ÂÃ°ÂÅ¸â€, Ã°ÂÅ¸â€Ã°ÂÅ¸â€Ã°ÂÅ¸â€, Ã°ÂÅ¸Å½Ã°ÂÅ¸â€Ã°ÂÅ¸â€¢Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸â€˜, Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜, Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™â€ Ã¢Å¸Â¹
Ã°Ââ€™â€”
The closedÃ¢Ë†â€™loop gain Ã°ÂÂÂ¢Ã°ÂÂÂ¬ Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™Ã°ÂÅ¸Â. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€”Ã°ÂÅ¸â€” and Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€” Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã¢Ë†â€™Ã°ÂÅ¸Â. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€”Ã°ÂÅ¸â€” Ã¢Ë†â„¢ Ã°ÂÅ¸Â = Ã¢Ë†â€™Ã°ÂÅ¸â€˜. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€”Ã°ÂÅ¸â€“V
Ã°Ââ€™â€
From (Eq.1) Ã°Ââ€™â€”Ã°ÂÅ¸Â = 20.0667Ã°ÂÂÂV.
The input current for OPA: Ã°Ââ€™Å Ã°ÂÅ¸Â =
Thus, Ã°Ââ€™Å  =
Ã°Ââ€™â€”Ã°Ââ€™â€¦
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ãƒâ€”Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
= Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€”Ã°ÂÅ¸â€”Ã°ÂÅ¸â€”Ã°Ââ€™Å½Ã°Ââ€˜Â¨
Ã°Ââ€™â€”Ã°Ââ€™â€¦ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â ,
Ã¢Ë†â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ãƒâ€”Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€
= Ã°ÂÂÂ°Ã°ÂÂÂ¡Ã°ÂÂÅ¾Ã°ÂÂÂ«Ã°ÂÂÅ¾ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ = Ã°Ââ€˜Â¹ = Ã°ÂÅ¸ÂÃƒâ€”Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã¢â€°Ë† Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸Â Ã°Ââ€˜Â¨ Ã¢Å¸Â¶ Ã°ÂÅ¸Å½
Ã°Ââ€™Å
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€¦ = Ã¢Ë†â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â
15
Ã¢â‚¬Å“YÃ¢â‚¬Â Ã¢Å¸Â¹ “Ã°ÂÅ¡Â«Ã¢â‚¬Å“ transformation:
Ã°Ââ€˜Â¹Ã°Ââ€™â€š = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™Å’Ã°ÂÅ“Â´,
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸â€˜. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€“Ã°Ââ€™Å’Ã°ÂÅ“Â´,
Ã ÂµÅ¾ Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ = Ã°ÂÅ¸â€œÃ°Ââ€™Å’Ã°ÂÅ“Â´, Ã¢Å¸Â¹ Ã¡â€°ÂÃ°Ââ€˜Â¹Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸ÂÃ°ÂÅ¸Â. Ã°ÂÅ¸â€˜Ã°ÂÅ¸ÂÃ°Ââ€™Å’Ã°ÂÅ“Â´,
Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã¢â€°Ë† Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°Ââ€™Å’Ã°ÂÅ“Â´.
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž = Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½Ã°Ââ€™Å’Ã°ÂÅ“Â´,
Ã°ÂÂÂ´Ã°Ââ€˜Â£
Ideal OPA: Noninverting Amplifier
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Â. Ã°ÂÅ¸â€˜Ã°ÂÅ¸ÂÃ°Ââ€™Å’Ã°ÂÅ“Â´
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â = Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™Ã°Ââ€™Å’Ã°ÂÅ“Â´
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Â. Ã°ÂÅ¸â€˜Ã°ÂÅ¸Â Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â¨Ã°Ââ€™â€” =
=Ã°ÂÅ¸Â+
=Ã°ÂÅ¸Â+
Ã¢â€°Ë† Ã°ÂÅ¸â€“. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€”
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œÃ°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½ Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€” Ã°Ââ€™â€”Ã°Ââ€™Å  Ã¢â€°Ë† Ã°ÂÅ¸â€“. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€”Ã°Ââ€˜Â½
16
Given: Ã°Ââ€™â€”Ã°Ââ€˜Âº = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½; Ã°Ââ€˜Â¹Ã°Ââ€™Å  = Ã°ÂÅ¸ÂÃ°Ââ€˜Â´Ã°ÂÅ“Â´,
Ã°Ââ€˜Â¨ = Ã°ÂÅ¸Â Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ , Ã°Ââ€˜Â¹Ã°Ââ€™Â = Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°ÂÅ“Â´ .
Find: Ã°Ââ€˜Â¨Ã°Ââ€™â€” , Ã°Ââ€™Å Ã°Ââ€˜Â¶ .
Ã°ÂÂÂ´Ã°Ââ€˜Â£
17
KCL in the node Ã°Ââ€™â€”Ã°ÂÅ¸Â :
Ã°Ââ€™Å Ã°Ââ€˜Â¹Ã°Ââ€™Å  = Ã°Ââ€™Å Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ + Ã°Ââ€™Å Ã°ÂÅ¸â€œ Ã¢Å¸Â¹
Ã°Ââ€™â€”Ã°Ââ€™â€ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€˜Âº + Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°Ââ€™â€”Ã°Ââ€™Â
=
+
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°ÂÅ¸Â =
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€œ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸Â
Given: Ã°Ââ€™â€”Ã°Ââ€˜Âº = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½; Ã°Ââ€˜Â¹Ã°Ââ€™Å  = Ã°ÂÅ¸ÂÃ°Ââ€˜Â´Ã°ÂÅ“Â´,
Ã°Ââ€˜Â¨ = Ã°ÂÅ¸Â Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ , Ã°Ââ€˜Â¹Ã°Ââ€™Â = Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°ÂÅ“Â´ .
Find: Ã°Ââ€˜Â¨Ã°Ââ€™â€” , Ã°Ââ€™Å Ã°Ââ€˜Â¶ .
Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™â€¦
Ã°Ââ€˜â€“Ã°Ââ€˜â€¦Ã°Ââ€˜â€“
KCL in the node Ã°Ââ€™â€”Ã°Ââ€™Â :
Ã°Ââ€™Å Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ + Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€™Å Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Å¸Â¹
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™â€¦ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â
+
=
,
Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
where Ã°Ââ€™â€”Ã°Ââ€™â€¦ = Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°Ââ€™â€”Ã°Ââ€˜Âº Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â .
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½
Ã°Ââ€™Å Ã°ÂÅ¸â€œ
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™Å Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã¢Å¸Â¹ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€¢ Ã°Ââ€™â€”Ã°Ââ€˜Âº Ã¢Ë†â€™ Ã°ÂÅ¸â€“Ã°ÂÅ¸Å½Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½.
18
Ã°Ââ€™â€”Ã°Ââ€˜Âº + Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°ÂÅ¸Â =
;
Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸Â
Ã¢Å¸Â¹
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€¢ Ã°Ââ€™â€”Ã°Ââ€˜Âº Ã¢Ë†â€™ Ã°ÂÅ¸â€“Ã°ÂÅ¸Å½Ã°ÂÅ¸â€˜Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Ë†â€™ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Â Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ ;
Given: Ã°Ââ€™â€”Ã°Ââ€˜Âº = Ã°ÂÅ¸ÂÃ°Ââ€˜Â½; Ã°Ââ€˜Â¹Ã°Ââ€™Å  = Ã°ÂÅ¸ÂÃ°Ââ€˜Â´Ã°ÂÅ“Â´,
Ã°Ââ€˜Â¨ = Ã°ÂÅ¸Â Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ , Ã°Ââ€˜Â¹Ã°Ââ€™Â = Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°ÂÅ“Â´ .
Find: Ã°Ââ€˜Â¨Ã°Ââ€™â€” , Ã°Ââ€™Å Ã°Ââ€˜Â¶ .
Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã¢Å¸Â¹ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°Ââ€™â€”Ã°Ââ€˜Âº Ã¢Ë†â€™ Ã°ÂÅ¸Â. Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÅ¸Å½ ;
Ã°Ââ€™â€”Ã°Ââ€™Â
= Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã°ÂÅ¸â€”. Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€¢(Ã°ÂÂÅ“Ã°ÂÂÂ¥Ã°ÂÂÂ¨Ã°ÂÂÂ¬Ã°ÂÂÅ¾Ã°ÂÂÂ Ã°ÂÂÂ¥Ã°ÂÂÂ¨Ã°ÂÂÂ¨Ã°ÂÂÂ© Ã°ÂÂÂ Ã°ÂÂÅ¡Ã°ÂÂÂ¢Ã°ÂÂÂ§).
Ã°Ââ€™â€”Ã°Ââ€˜Âº
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™â€¦
If Ã°Ââ€™â€”Ã°Ââ€˜Âº = Ã°ÂÅ¸Â, Ã°ÂÂÂ­Ã°ÂÂÂ¡Ã°ÂÂÅ¾Ã°ÂÂÂ§ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€™â€”Ã°Ââ€™â€ Ã°Ââ€˜Â¨Ã°Ââ€™â€” Ã¢â€°Ë† Ã°ÂÅ¸â€”Ã°Ââ€˜Â½.
Ã°Ââ€™â€”Ã°ÂÅ¸Â =
Ã°Ââ€™â€”Ã°Ââ€˜Âº +Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°ÂÅ¸Â
Ã°Ââ€˜â€“Ã°Ââ€˜â€¦Ã°Ââ€˜â€“
Ã¢â€°Ë† Ã°ÂÅ¸ÂÃ°Ââ€˜Â½.
Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€™Å Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½Ã°ÂÂÂ§Ã°ÂÂÅ¾Ã°ÂÂÂ° + Ã°Ââ€™Å Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ =
Ã°Ââ€™â€”Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Â
=
+
Ã¢â€°Ë†
Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€“
Ã°ÂÅ¸â€”
Ã¢â€°Ë†
+
= Ã°ÂÅ¸â€Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½Ã°ÂÂÂÃ°Ââ€˜Â¨
Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™Å Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½
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Ã°Ââ€™Å Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
19
Frequency Response of Amplifiers
Outline:
1. Linear Voltage Amplifier;
2. Amplifier Bandwidth;
3. Amplifier Transfer Function;
4. Ex. 1.5.
ENGR24100, Electronics I
Instructor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
CCNY
1
Linear Voltage Amplifier
A linear voltage amplifier fed at its input with a sine-wave signal of amplitude Ã°Ââ€˜Â½Ã°Ââ€™Å  and
frequency Ã°ÂÂÅ½. Whenever a sine-wave signal is applied to a linear circuit, the resulting output is
sinusoidal with the same frequency as the input. In fact, the sine wave is the only signal that
does not change shape as it passes through a linear circuit. However, that the output sinusoid
will in general have a different amplitude and will be shifted in phase relative to the input.
If we denote the amplifier transmission, or transfer function as it is more commonly known, by
Ã°Ââ€˜Â»(Ã°ÂÂÅ½), then
Ã°Ââ€˜Â» Ã°ÂÂÅ½
=
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
and Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ .
Plots of Ã¢â‚¬Å“ 20log Ã°Ââ€˜Â» Ã°ÂÂÅ½ vs. Ã°ÂÂÅ½ Ã¢â‚¬Å“ and Ã¢â‚¬Å“ Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ vs. Ã°ÂÂÅ½ Ã¢â‚¬Å“ constitute the magnitude (or amplitude)
and phase frequency response, respectively.
2
Amplifier Bandwidth
Figure shows the magnitude response of an amplifier. It indicates that the gain is almost
constant over a wide frequency range, roughly between Ã°ÂÂÅ½Ã°ÂÅ¸Â and Ã°ÂÂÅ½Ã°ÂÅ¸Â . Signals whose frequencies
are below Ã°ÂÂÅ½Ã°ÂÅ¸Â or above Ã°ÂÂÅ½Ã°ÂÅ¸Â will experience lower gain causing some distortions to the input
signal. The band of frequencies over which the gain of the amplifier is almost constant, to
within a certain number of decibels (usually 3 dB), is called the amplifier bandwidth.
Ã°Ââ€˜Â» Ã°ÂÂÅ½ =
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
is a complex function called as Ã¢â‚¬Å“amplifier transfer functionÃ¢â‚¬Â, where Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½ and
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½ denote the input and output signals, respectively. A similar description might be
provided by using by using the complex frequency variable Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°ÂÂÅ½ = Ã°Ââ€™â€¹Ã°ÂÂÅ½. Thus, one might see
that Ã°Ââ€˜Â» Ã°Ââ€™â€ =
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°Ââ€™â€
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°Ââ€™â€
.
3
Single-Time-Constant (STC) Networks
(a) a low-pass filter, it passes lowfrequency, sine-wave inputs Ã°Ââ€˜Â½Ã°Ââ€™Å
with little or no attenuation (at
Ã°ÂÂÅ½ = Ã°ÂÅ¸Å½, the transmission is unity)
and attenuates high-frequency
input sinusoids.
Ã°ÂÅ¸Â
Ã°Ââ€™Â=
Ã Â¸Â­
Ã°Ââ€™Å Ã°ÂÂÅ½Ã°Ââ€˜Âª
Ã°ÂÅ¸Â
Ã°Ââ€™Â=
Ã Â¸Â­
Ã°Ââ€™Å Ã°ÂÂÅ½Ã°Ââ€˜Âª
(b) a high-pass filter, its
transmission for Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½ is
unity at Ã°ÂÂÅ½ = Ã¢Ë†Å¾ and
decreases as Ã°ÂÂÅ½ is reduced,
reaching 0 for Ã°ÂÂÅ½ = Ã°ÂÅ¸Å½.
Ã¢â€°Ë†Ã°ÂÅ¸Å½
Ã°ÂÂÅ½Ã¢Å¸Â¶Ã¢Ë†Å¾
Ã¢â€°Ë†Ã¢Ë†Å¾
Ã°ÂÂÅ½Ã¢Å¸Â¶Ã°ÂÅ¸Å½
4
Cutoff frequency
Ã¢â‚¬Â¢ Definition: a cutoff frequency (is also known as 3-dB frequency, corner frequency, break
frequency or pole frequency) is a boundary in a system’s frequency response at which
energy flowing through the system begins to be reduced (attenuated or reflected) rather
than passing through.
Ã¢â‚¬Â¢ In electronic filters a cutoff frequency is a frequency characterizing a boundary between a
passband and a stopband, it applies to an edge characteristic plots of a lowpass, highpass,
bandpass, or band-stop filters.
Ã¢â‚¬Â¢ It is the point in the filter response where a transition band and passband meet, for
example, as defined by a half-power point (a frequency for which the output of the circuit is
Ã¢Ë†â€™3 dB of the nominal passband value).
Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ ,
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
=Ã°Ââ€™â€
Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
= Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ , = Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°Ââ€™â€¦Ã°Ââ€˜Â© = Ã°Ââ€™â€
Ã¢â€°Ë†0.74
https://en.wikipedia.org/wiki/Cutoff_frequency
5
Ã¢â‚¬Â¢ A single-time-constant (STC) network formed of an inductance Ã°Ââ€˜Â³ and a resistance
Ã°Ââ€˜Â¹ has a time constant Ã°ÂÂâ€° = Ã°Ââ€˜Â³/Ã°Ââ€˜Â¹. The time constant Ã°ÂÂâ€° of an STC network composed
of a capacitance Ã°Ââ€˜Âª and a resistance Ã°Ââ€˜Â¹ is given by Ã°ÂÂâ€° = Ã°Ââ€˜Â¹Ã°Ââ€˜Âª.
Ã¢â‚¬Â¢ Definition: an octave is the interval between one musical pitch and another with
Ã°Ââ€™â€¡
double its frequency. Thus, if 1 Ã°Ââ€™ÂÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™â€šÃ°Ââ€™â€”Ã°Ââ€™â€  = Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸Â Ã°ÂÅ¸Â , then Ã°Ââ€™â€¡Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°Ââ€™â€¡Ã°ÂÅ¸Â
Ã°Ââ€™â€¡Ã°ÂÅ¸Â
Ã¢â‚¬Â¢ The number of octaves between two frequencies is given by the formula:
Ã°Ââ€™â€¡Ã°ÂÅ¸Â
Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™Å½Ã°Ââ€™Æ’Ã°Ââ€™â€ Ã°Ââ€™â€œ Ã°Ââ€™ÂÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™â€šÃ°Ââ€™â€”Ã°Ââ€™â€  = Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸Â
Ã°Ââ€™â€¡Ã°ÂÅ¸Â
https://en.wikipedia.org/wiki/Phasor
6
Bode plots
(a) a low-pass filter.
(b) a high-pass filter.
The two asymptotes meet at the frequency Ã°ÂÂÅ½Ã°Ââ€™Â , which is called the corner frequency. 7
Precision High-speed OPA 627,637
(from Burr-Brown)
8
Ex. 1.5a Figure shows a voltage amplifier having an input resistance Ã°Ââ€˜Â¹Ã°Ââ€™Å  , an input
capacitance Ã°Ââ€˜ÂªÃ°Ââ€™Å  , a gain factor Ã°ÂÂÂ, and an output resistance Ro. The amplifier is fed
with a voltage source VÃ°Ââ€™â€ having a source resistance Ã°Ââ€˜Â¹Ã°Ââ€™â€ , and a load of resistance RL
is connected to the output. Derive an expression for the amplifier voltage gain
Vo /Vs as a function of frequency. From this find expressions for the DC gain and
the 3-dB frequency.
The voltage division applied to input circuit gives
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°Ââ€˜Â½Ã°Ââ€™Å
Ã°Ââ€™ÂÃ°Ââ€™Å
Ã°Ââ€™ÂÃ°Ââ€™Å
Ã°ÂÅ¸Â
Ã°Ââ€™ÂÃ°Ââ€™Å
=
=
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=
= Ã°Ââ€™ÂÃ°Ââ€™Å  = Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°ÂÅ¸Â = Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€™â€Ã°Ââ€˜ÂªÃ°Ââ€™Å  =
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â½Ã°Ââ€™â€ Ã°Ââ€™ÂÃ°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™ÂÃ°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+ Ã°Ââ€™â€
Ã°Ââ€™â€Ã°Ââ€˜ÂªÃ°Ââ€™Å
Ã°Ââ€™ÂÃ°Ââ€™Å
Ã°Ââ€™ÂÃ°Ââ€™Å
=
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°Ââ€™â€
9
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
=
Ã¢Å¸Â¹
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+
+ Ã°Ââ€™â€Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€
+ Ã°Ââ€™â€Ã°Ââ€˜ÂªÃ°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ex. 1.5a Derive an expression for the amplifier voltage gain Vo /Vs as a function of
frequency. From this find expressions for the DC gain and the 3-dB frequency.
The voltage division applied to input circuit gives
Ã°Ââ€˜Â½Ã°Ââ€™Å
Ã°ÂÅ¸Â
Ã¢Å¸Â¹
=
=
Ã°Ââ€˜Â¹Ã°Ââ€™â€
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Ã°ÂÅ¸Â
+
Ã°Ââ€™Å  Ã°Ââ€™â€
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹
Ã°Ââ€™Å
Ã°Ââ€™Å
Ã°ÂÅ¸Â
Ã°Ââ€™â€Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°ÂÅ¸Â+
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+ Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Å
=
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™Å
Ã°Ââ€™â€Ã°Ââ€˜Âª Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹ Ã°Ââ€™Å + Ã°Ââ€™â€Ã°Ââ€˜Â¹ Ã°Ââ€™Å
Ã°Ââ€™Å
Ã°Ââ€™â€
Also, by applying the voltage division to output circuit one get that
Thus,
Ã°Ââ€˜Â½Ã°Ââ€™Å
Ã°Ââ€˜Â½Ã°Ââ€™â€
Ã¢Ë†â„¢
Ã°Ââ€˜Â½Ã°Ââ€™Â
Ã°ÂÂÂÃ°Ââ€˜Â½Ã°Ââ€™Å
=
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=Ã°ÂÂÂ
Ã°Ââ€˜Â¹
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Ã°Ââ€™Å
=
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Ã°Ââ€™Å
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Ã°ÂÅ¸Â+ Ã°Ââ€˜Â¹Ã°Ââ€™Å +Ã°Ââ€˜Â¹
Ã°Ââ€™Å  Ã°Ââ€™â€
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Ã°Ââ€™â€Ã°Ââ€˜Âª Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹
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Ã°Ââ€™Å
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Ã°ÂÂÂÃ°Ââ€˜Â½Ã°Ââ€™Å
=
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
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.
Ã¢Å¸Â¹
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹
Ã¢Ë†â„¢
Ã¢Ë†â„¢ Ã°Ââ€˜Â³=
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
+
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Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°ÂÂÂ
Ã°Ââ€™â€Ã°Ââ€˜Âª Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹ Ã°Ââ€™Å + Ã°Ââ€™â€Ã°Ââ€˜Â¹ Ã°Ââ€™Å
Ã°Ââ€™Å
Ã°Ââ€™â€
Ã¢Å¸Â¹
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â³
10
Ex. 1.5a Derive an expression for the amplifier voltage gain Vo /Vs as a function of
frequency. From this find expressions for the DC gain and the 3-dB frequency.
Ã°Ââ€˜Â½Ã°Ââ€™Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã¢Å¸Â¹
=Ã°ÂÂÂÃ¢Ë†â„¢
Ã¢Ë†â„¢
Ã¢Ë†â„¢
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€™â€Ã°Ââ€˜Âª Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹
Ã°Ââ€˜Â½Ã°Ââ€™â€
Ã°ÂÅ¸Â+Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹ Ã°Ââ€™Å + Ã°Ââ€™â€Ã°Ââ€˜Â¹ Ã°Ââ€™Å
=Ã°ÂÂÂÃ¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™Å
Ã°Ââ€™Å
Ã°Ââ€˜Â³
Ã¢Ë†â„¢
Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°ÂÂÅ½
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÂÅ¡Ã°ÂÂÂ§Ã°ÂÂÂ
Ã¢Ë†â„¢
=
=
Ã°ÂÂÂ
Ã¢Ë†â„¢
Ã¢Ë†â„¢
Ã¢Ë†â„¢
.
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°ÂÂÅ½
Ã°ÂÅ¸Â + Ã°Ââ€™â€Ã°ÂÂâ€°
Ã°ÂÂâ€°Ã¢â€°â€¦
Ã°ÂÅ¸Â+Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÂÅ½
Ã°ÂÂÅ½Ã°ÂÅ¸Å½
Ã°Ââ€™Å
Ã°Ââ€˜Â³
Ã°Ââ€™Â
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â + Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â³
Ã°Ââ€™Å
Ã°ÂÅ¸Â
Ã°ÂÂâ€°
Thus, the 3-dB frequency is Ã°ÂÂÅ½Ã°Ââ€™Â Ã¢â€°â€¦ =
Ã°Ââ€™â€
Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™Å
= Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã¢Ë†Â¥ = Ã°ÂÂâ€°,
Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
=
=
Ã°ÂÂÂ°Ã°ÂÂÂ¡Ã°ÂÂÅ¾Ã°ÂÂÂ«Ã°ÂÂÅ¾
Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã¢Ë†Â¥ =
Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°ÂÅ¸Â
Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã¢Ë†Â¥
=
Ã°Ââ€˜Â¹Ã°Ââ€™Å  +Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™Å
and DC gain Ã°ÂÂÅ½ = Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€™â€ = Ã°ÂÅ¸Å½ is
Ã°Ââ€˜Â²=
11
Ã°Ââ€˜Â½Ã°Ââ€™Â
Ã¡â€°Â¤
Ã°Ââ€˜Â½Ã°Ââ€™â€ Ã°ÂÂÅ½
=Ã°ÂÅ¸Å½
=Ã°ÂÂÂÃ¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+ Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™Å
Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â³
.
Ex. 1.5b Calculate , the 3-dB frequency, and the frequency at which the gain becomes
0 dB (i.e., unity) for the case Rs =20Ã°Ââ€™Å’Ã°Ââ€ºâ‚¬, Ri =100Ã°Ââ€™Å’Ã°Ââ€ºâ‚¬, Ci =60Ã°Ââ€™â€˜Ã°Ââ€˜Â­, Ã°ÂÂÂ = 144Ã°Ââ€˜Â½/Ã°Ââ€˜Â½, Ro = 200Ã°Ââ€ºâ‚¬,
and RL = 1 Ã°Ââ€™Å’Ã°Ââ€ºâ‚¬.
Ã°Ââ€˜Â½Ã°Ââ€™Â
Ã°Ââ€˜Â½Ã°Ââ€™â€
=Ã°ÂÂÂÃ¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+ Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+ Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÂÅ½
Ã°ÂÅ¸Â+Ã°Ââ€™Å Ã°ÂÂÅ½
, Ã°ÂÂÅ½Ã°Ââ€™Â Ã¢â€°â€¦
Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Å  +Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜ÂªÃ°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™Å
and Ã°Ââ€˜Â² =
Ã°Ââ€˜Â½Ã°Ââ€™Â
Ã¡â€°Â¤
Ã°Ââ€˜Â½Ã°Ââ€™â€ Ã°ÂÂÅ½
=Ã°ÂÅ¸Å½
=Ã°ÂÂÂÃ¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+ Ã°Ââ€˜Â¹
Ã°ÂÅ¸Â+Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã¢Ë†â„¢
Ã°Ââ€™Å
Ã°Ââ€˜Â¹
.
Ã°Ââ€˜Â³
The values of the DC gain is
Ã°Ââ€˜Â²=
Ã°Ââ€˜Â½Ã°Ââ€™Â
Ã¡â€°Â¤
Ã°Ââ€˜Â½Ã°Ââ€™â€ Ã°ÂÂÅ½
=Ã°ÂÅ¸Å½
= Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™Ã°ÂÅ¸â€™ Ã¢Ë†â„¢
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Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸Â+
Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸Â+
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ V/V.
The 3-dB frequency is
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ + Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÅ¸â€
Ã°ÂÂÅ½Ã°Ââ€™Â Ã¢â€°â€¦
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã¢Å¸Â¹
Ã°Ââ€™â€¡
=
Ã¢â€°â€¦ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°ÂÅ¸â€”. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°ÂÅ¸â€œ Ã°Ââ€™Å’Ã°Ââ€˜Â¯Ã°Ââ€™â€º
Ã°Ââ€™Â
Ã°ÂÅ¸â€Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸Â Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™â€
Ã°ÂÅ¸ÂÃ°ÂÂâ€¦
If the gain falls off at the rate of -20 dB/decade, the frequency at which the gain becomes 0 dB
(i.e., unity) should be reached with the frequency 100 times higher than Ã°ÂÂÅ½Ã°Ââ€™Â (see diagrams
Bode plots), thus
Ã°ÂÂÅ½Ã°Ââ€™â€“Ã°Ââ€™ÂÃ°Ââ€™Å Ã°Ââ€™â€¢Ã°Ââ€™Å¡
Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦
Ã°ÂÅ¸â€“
Ã°ÂÂÅ½Ã°Ââ€™â€“Ã°Ââ€™ÂÃ°Ââ€™Å Ã°Ââ€™â€¢Ã°Ââ€™Å¡ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÂÅ½Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã¢Å¸Â¹ Ã°Ââ€™â€¡Ã°Ââ€™â€“Ã°Ââ€™ÂÃ°Ââ€™Å Ã°Ââ€™â€¢Ã°Ââ€™Å¡ =
Ã¢â€°â€¦ Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ. Ã°ÂÅ¸â€”Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œÃ°ÂÅ¸â€œ Ã°Ââ€˜Â´Ã°Ââ€˜Â¯Ã°Ââ€™â€º
Ã°Ââ€™â€
Ã°ÂÅ¸ÂÃ°ÂÂâ€¦
12
13
Ex. 1.5c Find Ã°Ââ€™â€”Ã°Ââ€™Â (Ã°Ââ€™â€¢) for each of the following inputs:
(i) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(ii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iv) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
By the definition of transfer function is
Ã°Ââ€˜Â» Ã°ÂÂÅ½
=
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
and Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
Ã°Ââ€˜Â» Ã°ÂÂÅ½ =
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
=Ã°Ââ€˜Â²Ã¢Ë†â„¢
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
Ã°ÂÅ¸Â
Ã°ÂÂÅ½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÅ¸Â
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
and the phase response is Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€™â€¢ Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
Ã¢Å¸Â¹
.
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
(i) Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã¢â€°Ë† Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â Ã°ÂÅ¸Å½ Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½ Ã¢Å¸Â¹
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Å½ Ã¢Ë†â„¢
Ã¢â€°Ë† Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€¢ .
Ã°ÂÅ¸Â+
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
14
Ex. 1.5c Find Ã°Ââ€™â€”Ã°Ââ€™Â (Ã°Ââ€™â€¢) for each of the following inputs:
(i) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(ii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iv) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
By the definition of transfer function is
Ã°Ââ€˜Â» Ã°ÂÂÅ½
=
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
and Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
Ã°Ââ€˜Â» Ã°ÂÂÅ½ =
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
=Ã°Ââ€˜Â²Ã¢Ë†â„¢
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
Ã°ÂÅ¸Â
Ã°ÂÂÅ½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÅ¸Â
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
and the phase response is Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€™â€¢ Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
Ã¢Å¸Â¹
.
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸Â Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦ Ã¢Å¸Â¹ Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°ÂÅ¸â€œ. Ã°ÂÅ¸â€¢Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸â€œ
Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸Â Ã¢Ë†â„¢
Ã¢â€°Ë† Ã°ÂÅ¸â€”. Ã°ÂÅ¸â€”Ã°ÂÅ¸â€œÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸Â .
(ii) Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ
Ã°ÂÅ¸Â+
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
15
Ex. 1.5c Find Ã°Ââ€™â€”Ã°Ââ€™Â (Ã°Ââ€™â€¢) for each of the following inputs:
(i) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(ii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iv) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
By the definition of transfer function is
Ã°Ââ€˜Â» Ã°ÂÂÅ½
=
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
and Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
Ã°Ââ€˜Â» Ã°ÂÂÅ½ =
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
=Ã°Ââ€˜Â²Ã¢Ë†â„¢
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
Ã°ÂÅ¸Â
Ã°ÂÂÅ½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÅ¸Â
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
and the phase response is Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€™â€¢ Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
Ã¢Å¸Â¹
.
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€” Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦ Ã¢Å¸Â¹ Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°ÂÅ¸â€™Ã°ÂÅ¸â€œÃ°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸â€
Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€” Ã¢Ë†â„¢
Ã¢â€°Ë† Ã°ÂÅ¸â€¢. Ã°ÂÅ¸Å½Ã°ÂÅ¸â€¢Ã°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€” .
(iii) Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â+
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
16
Ex. 1.5c Find Ã°Ââ€™â€”Ã°Ââ€™Â (Ã°Ââ€™â€¢) for each of the following inputs:
(i) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(ii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iii) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
(iv) Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€¢ , Ã°Ââ€˜Â½
Ã¢Å¸Â¹ Ã°ÂÂÅ½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“ Ã°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦/Ã°Ââ€™â€
By the definition of transfer function is
Ã°Ââ€˜Â» Ã°ÂÂÅ½
=
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
and Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
Ã°Ââ€˜Â» Ã°ÂÂÅ½ =
Ã°Ââ€˜Â½Ã°Ââ€™Â Ã°ÂÂÅ½
=Ã°Ââ€˜Â²Ã¢Ë†â„¢
Ã°Ââ€˜Â½Ã°Ââ€™Å  Ã°ÂÂÅ½
Ã°ÂÅ¸Â
Ã°ÂÂÅ½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÅ¸Â
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
and the phase response is Ã¢Ë†Â Ã°Ââ€˜Â» Ã°ÂÂÅ½ = Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€™â€¢ Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸Â+
Ã°ÂÂÅ½
Ã°ÂÂÅ½Ã°Ââ€™Â
Ã°ÂÂÅ½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
Ã¢Å¸Â¹
.
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã¢â€°Ë† Ã¢Ë†â€™Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œÃ°Ââ€™â€œÃ°Ââ€™â€šÃ°Ââ€™â€¦ Ã¢Å¸Â¹ Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°ÂÅ¸â€”Ã°ÂÅ¸Å½Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸â€
Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€¢ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œ Ã¢Ë†â„¢
Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°Ââ€™â€Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ Ã°Ââ€™â€¢ Ã¢Ë†â€™ Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œ .
(iv) Ã°ÂÂâ€œ Ã°ÂÂÅ½ = Ã¢Ë†â€™Ã°Ââ€™â€šÃ°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€¢Ã°Ââ€™Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€“
Ã°ÂÅ¸Â+
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
Ã°ÂÅ¸Â
17
Ex.2.5 The circuit shown in Fig.(a) can be used to implement a transresistance amplifier (see
Table 1.1 in Section 1.5). Such circuits are often used to amplify weak current signals, such as
those from photodiodes. Find the value of the input resistance Ri , the transresistance Rm , and
the output resistance Ro of the transresistance amplifier. If the signal source shown in Fig.(b) is
connected to the input of the transresistance amplifier, find the amplifier output voltage.
Ã°Ââ€™Å Ã°Ââ€™â€¡
Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€™Å Ã°ÂÅ¸Â
(Ã°Ââ€™â€š)
(Ã°Ââ€™Æ’)
Ã°Ââ€™Å Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°Ââ€™â€¡ Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™â€ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€œ Ã¢Ë†â„¢
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜ Ã°Ââ€˜Â¨
Ã¢Å¸Â¹ Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€œ Ã¢Ë†â„¢

Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
=
+
Ã¢Å¸Â¹ Ã°ÂÅ¸â€œ = Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
If Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°ÂÅ¸Å½ then Ã°Ââ€™â€”Ã°ÂÅ¸Å½ also should be 0. Thus, Ã°Ââ€™â€”Ã°ÂÅ¸Â = Ã°ÂÅ¸Å½ and Ã°ÂÅ¸â€œ = Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™Ã°ÂÅ¸â€œ Ã°Ââ€˜Â½.
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã¢Ë†â€™Ã°ÂÅ¸â€œ
Ã°Ââ€˜Â¹Ã°Ââ€™Å½ =
=
= Ã¢Ë†â€™Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€™Å’Ã°ÂÅ“Â´
Ã°Ââ€™Å Ã°Ââ€™Å
Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€œ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜
18
Voltage Follower ( Buffer Amplifier )
The buffer amplifier is not required to provide any voltage gain; rather, it is used
mainly as an impedance transformer or a power amplifier to obtain the unity gain
Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã°ÂÅ¸Â, Fig. 2.14. This circuit is referred to as a voltage follower, since the output
voltage Ã¢â‚¬Å“followsÃ¢â‚¬Â the input voltage, where Ã°Ââ€˜Â¹Ã°ÂÅ¸Â = Ã°Ââ€˜Â¹Ã°Ââ€™Å  = Ã¢Ë†Å¾ , Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ = Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€™â€”Ã°Ââ€™Å  .
Ã°ÂÂâ€¦Ã°ÂÂÂ¢Ã°ÂÂÂ . Ã°ÂÅ¸Â. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€™(Ã°ÂÂÅ¡)
Noninverting OPA Amplifier with Ã°Ââ€˜Â¹Ã°Ââ€™Å  = Ã¢Ë†Å¾ , Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ = Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã°ÂÅ¸Â
Ã¢Å¸Â¹
Ã°Ââ€™Å Ã°ÂÅ¸Â = Ã°Ââ€™Å Ã°ÂÅ¸Â Ã¢â€¡â€
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã°Ââ€™â€” Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÅ¸Â +
Ã°Ââ€™â€”Ã°Ââ€™Å  Ã¢Å¸Â¹ Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ = Ã°ÂÅ¸Å½, Ã°Ââ€˜Â¹Ã°ÂÅ¸Â = Ã¢Ë†Å¾ Ã¢Å¸Â¹
= Ã°ÂÅ¸Â Ã¢Å¸Â¹ Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã°ÂÅ¸Â.
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Å
19
The Weighted Summer
Fig. 2.10
By using KCL one may get that Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã°Ââ€™Å Ã°ÂÅ¸Â + Ã¢â€¹Â¯ Ã°Ââ€™Å Ã°Ââ€™Â = ÃÆ’Ã°Ââ€™ÂÃ°Ââ€™Å =Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°Ââ€™Å  = Ã°Ââ€™Å  , where Ã°Ââ€™Å Ã°Ââ€™Å  =
Ã°Ââ€™Å =
Ã°ÂÅ¸Å½Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
. Thus, ÃÆ’Ã°Ââ€™ÂÃ°Ââ€™Å =Ã°ÂÅ¸Â Ã°Ââ€™Å Ã°Ââ€™Å  = Ã°Ââ€™Å  Ã¢Å¸Âº ÃÆ’Ã°Ââ€™ÂÃ°Ââ€™Å =Ã°ÂÅ¸Â
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
=
Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™ ÃÆ’Ã°Ââ€™ÂÃ°Ââ€™Å =Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
and
Ã°Ââ€™â€”Ã°Ââ€™Å  .
In the weighted summer of Fig. 2.10 all the summing coefficients must be of the same sign.
20
D2.7 Design the weighted sum circuit Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â . Choose values for Ã°Ââ€˜Â¹Ã°ÂÅ¸Â , Ã°Ââ€˜Â¹Ã°ÂÅ¸Â ,
and Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ so that for a maximum output voltage of -4 V the current in the feedback
resistor will not exceed 1 mA.
Ã°Ââ€™Å Ã°Ââ€™â€¡Ã°Ââ€™Å½Ã°Ââ€™â€šÃ°Ââ€™â„¢ =
Ã°ÂÅ¸Å½ Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™Å½Ã°Ââ€™â€šÃ°Ââ€™â„¢
Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™Å½Ã°Ââ€™â€šÃ°Ââ€™â„¢
Ã¢Ë†â€™Ã°ÂÅ¸â€™
Ã¢Å¸Â¹ Ã°Ââ€˜Â¹Ã°Ââ€™â€¡Ã°Ââ€™Å½Ã°Ââ€™Å Ã°Ââ€™Â = Ã¢Ë†â€™
=Ã¢Ë†â€™
Ã¢Å¸Â¹ Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ Ã¢â€°Â¥ Ã°ÂÅ¸â€™Ã°Ââ€™Å’Ã°ÂÅ“Â´
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡Ã°Ââ€™Å½Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€™Å Ã°Ââ€™â€¡Ã°Ââ€™Å½Ã°Ââ€™â€šÃ°Ââ€™â„¢
Ã°ÂÅ¸Â Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™ Ã°Ââ€™â€”Ã°ÂÅ¸Â + Ã°ÂÅ¸â€™Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ = Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ“Â´
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€™ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã¢Ë†â€™ Ã Â·Â Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã¢Ë†â€™
Ã°Ââ€™â€”Ã°ÂÅ¸Â +
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€™Å =Ã°ÂÅ¸Â
Ã¢Å¸Â¹ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â = Ã°ÂÅ¸â€™Ã°Ââ€™Å’Ã°ÂÅ“Â´, Ã°Ââ€˜Â¹Ã°ÂÅ¸Â = Ã°ÂÅ¸ÂÃ°Ââ€™Å’Ã°ÂÅ“Â´
21
The Weighted Summer with Opposite Signs
0
Ã°Ââ€˜â€“Ã°Ââ€˜Â
0
Ã°Ââ€™Å
Ã°Ââ€˜â€“3
Ã°Ââ€˜â€“4
Fig. 2.11
Ã°Ââ€˜Â¹
Ã°Ââ€™Å  = Ã°Ââ€™Å Ã°Ââ€™Æ’ + Ã°Ââ€™Å Ã°ÂÅ¸â€˜ + Ã°Ââ€™Å Ã°ÂÅ¸â€™ , where Ã°Ââ€™Å Ã°Ââ€™Æ’ =
Ã°Ââ€™â€”
Ã°Ââ€™â€”
Ã°Ââ€™Å Ã°ÂÅ¸â€˜ = Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ , Ã°Ââ€™Å Ã°ÂÅ¸â€™ = Ã°Ââ€˜Â¹Ã°ÂÅ¸â€™ and Ã°Ââ€™Å  =
Ã°ÂÅ¸â€˜
Ã°ÂÅ¸â€™
Ã°Ââ€™â€”Ã°Ââ€™Â =
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹
Ã°ÂÅ¸Å½Ã¢Ë†â€™Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž
Ã°Ââ€˜Â¹
Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°Ââ€™â€š Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™Ã°Ââ€˜Â¹Ã°Ââ€™â€š Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™Æ’
Ã°Ââ€™â€”
=Ã¢Ë†â€™
Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€™â€”
Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â
Ã°ÂÅ¸Â
. Thus, Ã¢Ë†â€™ Ã°Ââ€˜Â¹Ã°Ââ€™Â = Ã¢Ë†â€™ Ã°Ââ€˜Â¹
Ã°Ââ€™â€ž
Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€™â€”Ã°ÂÅ¸Â
Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
+
Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€™â€”
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹
+ Ã°Ââ€˜Â¹Ã°Ââ€™â€š Ã°Ââ€™â€”Ã°ÂÅ¸Â +
Ã°ÂÅ¸Â
Ã°ÂÅ¸â€™
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã°Ââ€™â€”
+ Ã°Ââ€˜Â¹Ã°ÂÅ¸â€™ Ã¢Å¸Â¹
Ã°ÂÅ¸â€™
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž
Ã°Ââ€™â€”Ã°ÂÅ¸Â +
Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™
Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ Ã¢Ë†â€™
Ã°Ââ€™â€”
Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â¹Ã°ÂÅ¸â€™ Ã°ÂÅ¸â€™
Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹
Voltages Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ and Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€™â€”Ã°ÂÅ¸â€™ would be subtracted from Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€˜Â¹Ã°Ââ€™â€š Ã°Ââ€™â€”Ã°ÂÅ¸Â and
Ã°ÂÅ¸â€˜
,
Ã°Ââ€™Æ’
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€™â€” .
Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â
22
D2.8 Use the idea presented in Fig. 2.11 to design a weighted summer that provides
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°ÂÅ¸ÂÃ°Ââ€™â€”Ã°ÂÅ¸Â + Ã°Ââ€™â€”Ã°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸â€™Ã°Ââ€™â€”Ã°ÂÅ¸â€˜ .
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž
Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€˜Â¹Ã°Ââ€™â€š
Ã°Ââ€˜Â¹Ã°Ââ€™â€ž
Ã°Ââ€™â€”Ã°Ââ€™Â =
Ã°Ââ€™â€” +
Ã°Ââ€™â€” Ã¢Ë†â€™
Ã°Ââ€™â€” = Ã°Ââ€˜Â¹Ã°Ââ€™â€ž = Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ =
Ã°Ââ€™â€” +
Ã°Ââ€™â€” Ã¢Ë†â€™
Ã°Ââ€™â€” =
Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°Ââ€™Æ’ Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€˜
= Ã°Ââ€˜Â¹Ã°Ââ€™â€š = Ã°Ââ€˜Â¹Ã°Ââ€™â€ž = Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã¢â‚¬Â²
Ã°ÂÂÂ¥Ã°ÂÂÅ¾Ã°ÂÂÂ­
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡
Ã°ÂÂÂ¨Ã°ÂÂÅ¸ Ã°ÂÂÆ’Ã°ÂÅ¸Â. Ã°ÂÅ¸â€¢
=
Ã°Ââ€™â€” +
Ã°Ââ€™â€” Ã¢Ë†â€™
Ã°Ââ€™â€” =
=
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°ÂÅ¸Â Ã°ÂÅ¸Â Ã°Ââ€˜Â¹Ã°ÂÅ¸â€˜ Ã°ÂÅ¸â€˜
Ã°Ââ€˜Â¹Ã°Ââ€™â€¡ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€™Å’Ã°Ââ€ºâ‚¬
Ã°Ââ€˜Â¹Ã°ÂÅ¸Â = Ã°ÂÅ¸â€œ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°Ââ€ºâ‚¬
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
=
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23
Amplifiers
Outline:
1. Signal Amplification;
2. Amplifier Power Efficiency;
3. Amplifier Saturation;
4. Voltage Amplifier.
ENGR24100, Electronics I
Instructor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
CCNY
1
Signal Amplification
Ã¢â‚¬Â¢ An electronic amplifier (or amp) is an electronic device that can increase the power
of a signal (a time-varying voltage or current). It is a two-port electronic circuit that
uses electric power from a power supply to increase the amplitude of a signal
applied to its input terminals, producing a proportionally greater amplitude signal
at its output.
Ã¢â‚¬Â¢ The amount of amplification provided by an amplifier is measured by its gain: the
ratio of output voltage, current, or power to input.
Ã¢â‚¬Â¢ An amplifier is a circuit that has a power gain greater than one. An attenuator is a
circuit that has a power gain less than one.
Ã¢â‚¬Â¢ The Ã¢â‚¬Å“wigglesÃ¢â‚¬Â in the output waveform must be identical to those in the input
waveform. Any change in waveform added from operation of an amplifier is
considered to be distortion and is obviously undesirable.
Ã¢â‚¬Â¢ Mathematically a linear amplifier that preserves the details of the signal waveform
is characterized by the linear relationship
Ã°Ââ€˜Â¨=
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Ã°Ââ€™â€”Ã°Ââ€™Å Ã°Ââ€™Â
where Ã°Ââ€™â€”Ã°Ââ€™Å Ã°Ââ€™Â and Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ are the input and output signals,
respectively, and A is a constant representing the
magnitude of amplification, known as amplifier gain.
2
Ã¢â‚¬Â¢ Voltage amplifiers is an electrical circuit to increase a power of signal by
increasing an amplitude of voltage without a significant change of current.
Ã¢â‚¬Â¢ A direct-coupled amplifier (or Direct Current amplifier) is a type of
amplifier in which the output of one stage of the amplifier is coupled to the
input of the next stage in such a way as to permit signals with zero
frequency, also referred to as direct current, to pass from input to output.
Ã¢â‚¬Â¢ An audio power amplifier (or power amp) is an amplifier that amplifies
or electric guitar pickup to a level that is high enough for driving
loudspeakers or headphones. It is the final electronic stage in a typical
audio playback chain before the signal is sent to the loudspeakers.
Ã¢â‚¬Â¢ A preamplifier, also known as a preamp, is an electronic amplifier that
converts a weak electrical signal into an output signal strong enough to be
noise-tolerant and strong enough for further processing, or for sending to
a power amplifier and a loudspeaker. Without this, the final signal would
be noisy or distorted.
3
Ã¢â‚¬Â¢ Power gain of an amplifier is
Gain
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where Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™ÂÃ°Ââ€™Â
of logarithms.
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, Ã°Ââ€˜ÂµÃ°Ââ€™â€˜ = [Ã°Ââ€˜ÂµÃ°Ââ€™â€ Ã°Ââ€™â€˜Ã°Ââ€™â€ Ã°Ââ€™â€œ] after John Napier, the inventor
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Ã¢â‚¬Â¢ Voltage gain is defined from Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ as the next
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4
Ã¢â‚¬Â¢ Current gain is defined in similar way
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Def.: Ã°Ââ€˜Â¨Ã°Ââ€™Å  =
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Assignment: Find values of Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ .
Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ [ ]
0
?
1
?
10
?
100
?
5
Assignment: Find values of Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ .
= Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°Ââ€˜Â¨Ã°Ââ€™â€˜ ,
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1
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10
Ã°Ââ€™â€ Ã°ÂÅ¸Â Ã¢â€°Ë† Ã°ÂÅ¸Â. Ã°ÂÅ¸â€¢Ã°ÂÅ¸Â
100
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6
Ã¢â‚¬Â¢ Ex.1.10. An amplifier has a voltage gain of 100 V/V and a current gain of 1000 A/A.
Express the voltage and current gains in decibels and find the power gain.
Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
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= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ = Ã°ÂÅ¸â€™Ã°ÂÅ¸Å½ Ã°Ââ€™â€¦Ã°Ââ€˜Â©;
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= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½ = Ã°ÂÅ¸â€œÃ°ÂÅ¸Å½ Ã°Ââ€™â€¦Ã°Ââ€˜Â©.
Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â
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7
Amplifier Power Efficiency
The power delivered to the amplifier from the DC power supply is
Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª = Ã°Ââ€˜Â½Ã°Ââ€˜ÂªÃ°Ââ€˜Âª Ã°Ââ€˜Â°Ã°Ââ€˜ÂªÃ°Ââ€˜Âª + Ã°Ââ€˜Â½Ã°Ââ€˜Â¬Ã°Ââ€˜Â¬ Ã°Ââ€˜Â°Ã°Ââ€˜Â¬Ã°Ââ€˜Â¬ .
The power-balance equation for the amplifier can be written as
Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª + Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ + Ã°Ââ€˜Â·Ã°Ââ€™â€¦Ã°Ââ€™Å Ã°Ââ€™â€ ,
where Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â is the power drawn from the signal source, Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ is the power
delivered to the load and Ã°Ââ€˜Â·Ã°Ââ€™â€¦Ã°Ââ€™Å Ã°Ââ€™â€ is the power dissipated in the amplifier circuit
itself. By assuming Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â Ã¢Å¸Â¶ Ã°ÂÅ¸Å½ one may define the amplifier power efficiency as
Ã°Ââ€˜Â·
Ã°ÂÅ“Â¼ = Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½.
Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª
8
Ã¢â‚¬Â¢ Ex.1.11. An amplifier operating from a single 15-V supply provides a 12-V peak-topeak sine-wave signal to a 1- kÃŽÂ© load and draws negligible input current from the
signal source. The dc current drawn from the 15-V supply is 8 mA. What is the
power dissipated in the amplifier, and what is the amplifier efficiency?
Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª + Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ + Ã°Ââ€˜Â·Ã°Ââ€™â€¦Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª = Ã°Ââ€˜Â½Ã°Ââ€˜ÂªÃ°Ââ€˜Âª Ã°Ââ€˜Â°Ã°Ââ€˜ÂªÃ°Ââ€˜Âª = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ Ã¢Ë†â„¢ Ã°ÂÅ¸â€“ Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€˜ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°ÂÅ¸Â Ã°Ââ€˜Â¾;
Ã°Ââ€™Å Ã°Ââ€™Å Ã°Ââ€™Â Ã¢â€°Ë† Ã°ÂÅ¸Å½ Ã¢Å¸Â¹ Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â Ã¢â€°Ë† Ã°ÂÅ¸Å½ Ã°Ââ€˜Â¾ Ã¢Å¸Â¹ Ã°Ââ€˜Â·Ã°Ââ€™â€¦Ã°Ââ€™Å Ã°Ââ€™â€ = Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª + Ã°Ââ€˜Â·Ã°Ââ€™Å Ã°Ââ€™Â Ã¢Ë†â€™ Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ Ã¢â€°Ë† Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª Ã¢Ë†â€™ Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ ;
Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢
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Ã°Ââ€˜Â·Ã°Ââ€™â€¦Ã°Ââ€™Å Ã°Ââ€™â€ Ã¢â€°Ë† Ã°Ââ€˜Â·Ã°Ââ€˜Â«Ã°Ââ€˜Âª Ã¢Ë†â€™ Ã°Ââ€˜Â·Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°ÂÅ¸Â Ã¢Ë†â€™ Ã°ÂÅ¸Å½. Ã°ÂÅ¸Å½Ã°ÂÅ¸ÂÃ°ÂÅ¸â€“ = Ã°ÂÅ¸Å½. Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°ÂÂâ€“ = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Â Ã°Ââ€™Å½Ã°Ââ€˜Â¾;
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Ã¢Ë†â„¢ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½ = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ %.
9
Amplifier Saturation
Practically speaking, the amplifier transfer characteristic remains linear over only a
limited range of input and output voltages.
In order to avoid distorting
the output signal waveform,
the input signal swing must
be kept within the linear
range of operation:
Ã°Ââ€˜Â³Ã¢Ë†â€™ Ã¢â€°Â¤ Ã°Ââ€˜Â½Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢ Ã¢â€°Â¤ Ã°Ââ€˜Â³+
or
Ã°Ââ€˜Â³Ã¢Ë†â€™
Ã°Ââ€˜Â³+
Ã¢â€°Â¤ Ã°Ââ€˜Â½Ã°Ââ€™Å Ã°Ââ€™Â Ã¢â€°Â¤
Ã°Ââ€˜Â¨Ã°Ââ€™â€”
Ã°Ââ€˜Â¨Ã°Ââ€™â€”
10
Symbols and Terminology
Ã°ÂÂÂ¼Ã°ÂÂÂ¶ is DC component;
Ã°ÂÂÂ¼Ã°Ââ€˜Â is the amplitude of AC component;
Ã°Ââ€˜â€“Ã°Ââ€˜Â Ã°Ââ€˜Â¡ = Ã°ÂÂÂ¼Ã°Ââ€˜Â Ã°Ââ€˜Â Ã°Ââ€˜â€“Ã°Ââ€˜â€º Ã°ÂÅ“â€ºÃ°Ââ€˜Â¡ is the sinusoidal AC signal;
Ã°Ââ€˜â€“Ã°ÂÂÂ¶ Ã°Ââ€˜Â¡ = Ã°ÂÂÂ¼Ã°ÂÂÂ¶ + Ã°Ââ€˜â€“Ã°Ââ€˜Â Ã°Ââ€˜Â¡ is the total instantaneous signal.
11
Circuit Models: Voltage Amplifier
Ã°Ââ€˜Â£Ã°Ââ€˜Â
The model of the voltage amplifier consists of a voltage-controlled voltage source
having a gain factor Avo (which is an open-circuit voltage gain), an input resistance Ri
that accounts for the fact that the amplifier draws an input current from the signal
source, and an output resistance Ro that accounts for the change in output voltage
as the amplifier is called to supply output current Ã°Ââ€™Å Ã°Ââ€™Â and voltage Ã°Ââ€™â€”Ã°Ââ€™Â to a load Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ .
The nonzero output resistance Ro causes only a fraction of the voltage Ã°Ââ€˜Â¨vo vi to
appear across the output load Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ .
By using the voltage-divider one might see that
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å
.
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ + Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ + Ã°Ââ€˜Â¹Ã°Ââ€™Â
Thus, the voltage gain of the voltage amplifier is given by Ã°Ââ€˜Â¨Ã°Ââ€™â€” =
For the ideal OPA Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã¢Å¸Â¶ Ã°ÂÅ¸Å½. Also, if Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã¢Å¸Â¶ Ã¢Ë†Å¾, then Ã°Ââ€˜Â¨Ã°Ââ€™â€” Ã¢â€°Ë† Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â .
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Å
= Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
.
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ +Ã°Ââ€˜Â¹Ã°Ââ€™Â
12
Circuit Models: Voltage Amplifier
Ã°Ââ€˜Â£Ã°Ââ€˜Â
One may apply the voltage-divider to the input of amplifier getting that only a
fraction of the source signal vs actually reaches the input terminals of the amplifier
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™â€”Ã°Ââ€™Å
.
Ã°Ââ€™â€”Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Thus, in order not to lose a significant portion of the input signal, the amplifier
must be designed to have an input resistance Ri Ã¢â€°Â« Ã°Ââ€˜Â¹Ã°Ââ€™â€ and Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã¢Å¸Â¶ Ã°ÂÅ¸Å½, Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã¢Å¸Â¶ Ã¢Ë†Å¾.
The overall voltage gain of the voltage amplifier is
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¨
Ã°Ââ€™â€”
Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™Å  Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â³
Ã°Ââ€™Â
=
= Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Ë†â„¢
Ã¢Ë†â„¢
.
Ã°Ââ€˜Â¹
+
Ã°Ââ€˜Â¹
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ + Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
13
Circuit Models: Voltage Amplifier
Ã°Ââ€˜Â£Ã°Ââ€˜Â
The overall voltage gain of the voltage amplifier is
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¨
Ã°Ââ€™â€”
Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™Å  Ã°Ââ€˜Â¹ + Ã°Ââ€˜Â¹
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â³
Ã°Ââ€™Â
=
= Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Ë†â„¢
Ã¢Ë†â„¢
.
Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹
+
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹
+
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â³
Ã°Ââ€™Â
Ã°Ââ€™Å
Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å
There are situations in which one is interested not in voltage gain but only in a
Ã°Ââ€™â€”
significant power gain. For instance, if Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã¢â€°Â« Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ , Ã°Ââ€˜Â¹Ã°Ââ€™Å  then Ã°Ââ€™Â Ã¢Å¸Â¶ Ã°ÂÅ¸Å½. In such a case,
Ã°Ââ€™â€”Ã°Ââ€™â€
one requires an additional amplifier (in between the source and voltage amplifier)
Ã°Ââ€™â€”
with modest voltage gain ( Ã°Ââ€™Â Ã¢â€°Â¤ Ã°ÂÅ¸Â) because of Ã°Ââ€˜Â¹Ã°Ââ€™â€ Ã¢â€°Â« Ã°Ââ€˜Â¹Ã°Ââ€™Å  and Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã¢â€°Â« Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ . Such
Ã°Ââ€™â€”Ã°Ââ€™â€
additional amplifier is referred to as a buffer amplifier.
14
Ã°ÂÂÂ´Ã°Ââ€˜Â£Ã°Ââ€˜Å“ = 1;
Ã°ÂÂÂ´Ã°Ââ€˜Â£Ã°Ââ€˜Å“ = 100;
Ã°ÂÂÂ´Ã°Ââ€˜Â£Ã°Ââ€˜Å“ = 10;
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Å
= Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã¢Ë†â„¢
Ã¢Ë†â„¢
=
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ + Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€™Å  + Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™â€
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã¢â€°Ë† Ã°ÂÅ¸â€”Ã°ÂÅ¸Å½.
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™â€
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Â
Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€“.
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸â€œ
Ã¢Ë†â„¢
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ + Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€˜ Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€ + Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸â€œ
=
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½
Ã¢â€°Ë† Ã°ÂÅ¸â€”.
Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸Â
15
Ã°ÂÂÂ´Ã°Ââ€˜Â£Ã°Ââ€˜Å“ = 100;
Ã°ÂÂÂ´Ã°Ââ€˜Â£Ã°Ââ€˜Å“ = 10;
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã¢â€°Ë† Ã°ÂÅ¸â€”.
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã¢â€°Ë† Ã°ÂÅ¸â€”Ã°ÂÅ¸Å½.
Ã°ÂÂÂ´Ã°Ââ€˜Â£Ã°Ââ€˜Å“ = 1;
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€“.
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã¡â€°Â®
Ã¢â€°Ë† Ã°ÂÅ¸â€” Ã¢Ë†â„¢ Ã°ÂÅ¸â€”Ã°ÂÅ¸Å½ Ã¢Ë†â„¢ Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€“ = Ã°ÂÅ¸â€Ã°ÂÅ¸â€™Ã°ÂÅ¸â€“.
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€™ÂÃ°Ââ€™â€ Ã°Ââ€™â€¢
Ã°Ââ€˜Â¨Ã°Ââ€™â€” = Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½
Ã°Ââ€™â€”Ã°Ââ€™ÂÃ°Ââ€™â€“Ã°Ââ€™â€¢
= Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°Ââ€™ÂÃ°Ââ€™ÂÃ°Ââ€™Ë†Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ Ã°ÂÅ¸â€Ã°ÂÅ¸â€™Ã°ÂÅ¸â€“ Ã¢â€°Ë† Ã°ÂÅ¸â€œÃ°ÂÅ¸â€ Ã°Ââ€™â€¦Ã°Ââ€˜Â©
Ã°Ââ€™â€”Ã°Ââ€™Å Ã°Ââ€™Â
16
Other Amplifier Types
From the TheveninÃ¢â‚¬â„¢s
theorem one might see
that
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã¢Å¸Â¶ Ã¢Ë†Å¾ Ã¢Å¸Â¹
Ã°Ââ€™Å Ã°Ââ€™Â = Ã°ÂÅ¸Å½ and Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€™Å Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹
Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€™Â
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Â
= Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Å
17
A transconductance amplifier (Ã°Ââ€˜Â®Ã°Ââ€™Å½ =
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Å
slop amplifier )
puts out a current Ã°Ââ€™Å Ã°Ââ€™Â proportional to its input voltage Ã°Ââ€™â€”Ã°Ââ€™Å  .
Ã°Ââ€˜Â®Ã°Ââ€™Å½ =
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°Ââ€˜Â®Ã°Ââ€™Å½ Ã°Ââ€™â€”Ã°Ââ€™Å  .
18
Other Amplifier Types
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã¢Å¸Â¶ Ã¢Ë†Å¾ Ã¢Å¸Â¹
Ã°Ââ€™Å Ã°Ââ€™Â = Ã°ÂÅ¸Å½ and Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€™Å Ã°Ââ€™Å  Ã¢â€°Ë† Ã°ÂÅ¸Å½
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â®Ã°Ââ€™Å½ Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°Ââ€˜Â®Ã°Ââ€™Å½ Ã°Ââ€™â€”Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â®Ã°Ââ€™Å½ Ã°Ââ€˜Â¹Ã°Ââ€™Â
19
A transresistance amplifier outputs a voltage Ã°Ââ€™â€”Ã°Ââ€™Â proportional
to its input current Ã°Ââ€™Å Ã°Ââ€™Å  .
Ã°Ââ€˜Â¹Ã°Ââ€™Å½ =
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™Å Ã°Ââ€™Å
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™Å Ã°Ââ€™Å  = Ã°Ââ€˜Â¹Ã°Ââ€™Å½ Ã°Ââ€™Å Ã°Ââ€™Å  .
20
Other Amplifier Types
From the TheveninÃ¢â‚¬â„¢s
theorem one might see
that
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã¢Å¸Â¶ Ã¢Ë†Å¾ Ã¢Å¸Â¹
Ã°Ââ€™Å Ã°Ââ€™Â = Ã°ÂÅ¸Å½ and Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å
From the TheveninÃ¢â‚¬â„¢s
theorem one might see
that
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¹Ã°Ââ€™Å½ Ã°Ââ€™Å Ã°Ââ€™Å
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°Ââ€˜Â¹Ã°Ââ€™Å½ Ã°Ââ€™Å Ã°Ââ€™Å
Ã°Ââ€˜Â¨Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Å
21
Determining Ri and Ro
From the Table 1.1:
One can find Ã°Ââ€˜Â¹Ã°Ââ€™Å  and Ã°Ââ€˜Â¹Ã°Ââ€™Â by using OhmÃ¢â‚¬â„¢s law:
Ã°Ââ€˜Â¹Ã°Ââ€™Å  =
Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€™Å Ã°Ââ€™Å
, where Ã°Ââ€™â€”Ã°Ââ€™Å  and Ã°Ââ€™Å Ã°Ââ€™Å  are parameters measured by using meters;
and
Ã°Ââ€˜Â¹Ã°Ââ€™Â =
Ã°Ââ€˜Â¨Ã¢Ë†Å¾ Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€™Å Ã°Ââ€™Â,Ã°Ââ€™â€Ã°Ââ€™â€°Ã°Ââ€™ÂÃ°Ââ€™â€œÃ°Ââ€™â€¢_Ã°Ââ€™â€žÃ°Ââ€™Å Ã°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€“Ã°Ââ€™Å Ã°Ââ€™â€¢
, where Ã°Ââ€˜Â¨Ã¢Ë†Å¾ Ã°Ââ€™â€”Ã°Ââ€™Å  and Ã°Ââ€™Å Ã°Ââ€™Â,Ã°Ââ€™â€Ã°Ââ€™â€°Ã°Ââ€™ÂÃ°Ââ€™â€œÃ°Ââ€™â€¢_Ã°Ââ€™â€žÃ°Ââ€™Å Ã°Ââ€™â€œÃ°Ââ€™â€žÃ°Ââ€™â€“Ã°Ââ€™Å Ã°Ââ€™â€¢ are open-circuit output voltage to
the short-circuit output current measured by using meters.
22
Determining Ro
From the Table 1.1:
(alternative technique)
Alternatively, the output resistance can be found by
eliminating the input signal source (then Ã°Ââ€™Å Ã°Ââ€™Å  = Ã°ÂÅ¸Å½ and Ã°Ââ€™â€”Ã°Ââ€™Å  = Ã°ÂÅ¸Å½)
and applying a voltage signal Ã°Ââ€™â€”Ã°Ââ€™â„¢ to the output of the
amplifier. If one denotes the current drawn from Ã°Ââ€™â€”Ã°Ââ€™â„¢ as Ã°Ââ€™Å Ã°Ââ€™â„¢
Ã°Ââ€™â€”
(note that it is opposite in direction of Ã°Ââ€™Å Ã°Ââ€™Â ), then Ã°Ââ€˜Â¹Ã°Ââ€™Â = Ã°Ââ€™Å  Ã°Ââ€™â„¢.
Ã°Ââ€™â„¢
These techniques are conceptually correct for a theoretical
calculations, but it is not employed in actual practice to
characterize Ã°Ââ€˜Â¹Ã°Ââ€™Â .
23
Unilateral Models
Those amplifier models are unilateral; that
is, signal flow is unidirectional, from input to
output. Whereas the unilateral model
suggests that an amplifierÃ¢â‚¬â„¢s input current
and voltage are completely independent of
what is connected to the output, this may
not be the case. For example, portions of
signals at the amplifier output to appear at
its input because of feedback or unintended
coupling .
24
EXERCISES
E1.18. Consider a current amplifier having the model shown in the second row of Table 1.1.
Let the amplifier be fed with a signal current-source is having a resistance Rs, and let the
output be connected to a load resistance RL. Show that the overall current gain is given by
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â
= Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã¢Ë†â„¢
Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹ Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¹ Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€™Å Ã°Ââ€™Å  = Ã°Ââ€™Å Ã°Ââ€™â€ =
Ã°Ââ€˜Â¹ Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€™Å Ã°Ââ€™Å
Ã°Ââ€˜â€¦Ã°Ââ€˜Â
Ã°Ââ€˜â€¦Ã°ÂÂÂ¿
Ã°Ââ€˜Â£Ã°Ââ€˜Â
Ã°Ââ€™Å Ã°Ââ€™â€
Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
=
= Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€™Å Ã°Ââ€™Å
Ã¢Ë†â„¢
=
= Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã¢Ë†â„¢
Ã°Ââ€™â€”Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™â€
25
E1.19. Consider the transconductance amplifier whose model is shown in the third row of
Table 1.1. Let a voltage signal source vs with a source resistance Ã°Ââ€˜Â¹Ã°Ââ€™â€ be connected to the input
and a load resistance Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ be connected to the output. Show that the overall voltage gain is
given by
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹ Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
=
Ã°Ââ€˜Â®
Ã¢Ë†â„¢
Ã°Ââ€™Å½ Ã°Ââ€˜Â¹ +Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹ +Ã°Ââ€˜Â¹ .
Ã°Ââ€™â€”
Ã°Ââ€™â€
Ã°Ââ€™Å
Ã°Ââ€™â€
Ã°Ââ€™Â
Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹ Ã°Ââ€˜Â¹
Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€™Å Ã°Ââ€™Å  Ã°Ââ€˜Â¹ Ã°Ââ€™Â+Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ and Ã°Ââ€™â€”Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€™Â
Ã°Ââ€˜Â³
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°ÂÅ¸Â
= Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€™Å Ã°Ââ€™Å
Ã¢Ë†â„¢
=
Ã°Ââ€™â€”Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°Ââ€™Å Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
= Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜â€¦Ã°Ââ€˜Â
Ã°Ââ€˜â€¦Ã°ÂÂÂ¿
Ã°Ââ€˜Â£Ã°Ââ€˜Â
Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€™Å Ã°Ââ€™Â Ã°ÂÅ¸Â
Ã°Ââ€™Å Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°ÂÅ¸Â
Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€˜Â¨Ã°Ââ€™Å Ã°Ââ€™â€ Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Å
= Ã¢Ë†â„¢
= = Ã°Ââ€˜Â®Ã°Ââ€™Å½ =
Ã¢Ë†â„¢
Ã¢Ë†â„¢
=
Ã¢Ë†â„¢
Ã¢Ë†â„¢
=
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€™Å  Ã°Ââ€™â€ Ã°Ââ€˜Â¹ Ã°Ââ€™Å  Ã°Ââ€™â€”Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
= Ã°Ââ€˜Â®Ã°Ââ€™Å½ Ã¢Ë†â„¢
Ã°Ââ€˜Â¹Ã°Ââ€™Â Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã¢Ë†â„¢
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹ Ã°Ââ€˜Â³ Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
26
E1.20. Consider a transresistance amplifier having the model shown in the fourth row of
Table 1.1. Let the amplifier be fed with a signal current source Ã°Ââ€™Å Ã°Ââ€™â€ having a resistance Ã°Ââ€˜Â¹Ã°Ââ€™â€ , and
let the output be connected to a load resistance Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ . Show that the overall gain is given by
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
= Ã°Ââ€˜Â¹Ã°Ââ€™Å½
Ã¢Ë†â„¢
.
Ã°Ââ€™Å Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ + Ã°Ââ€˜Â¹Ã°Ââ€™Â
Ã°Ââ€™Å Ã°Ââ€™â€
The voltage division gives
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³ , Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Å½ Ã°Ââ€˜Â¹ Ã°Ââ€˜Â³
=
Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™Â = Ã°Ââ€™Å Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å½ Ã°Ââ€™Å Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹ Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Â + Ã°Ââ€˜Â¹ Ã°Ââ€˜Â³
The current division gives
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€™Å Ã°Ââ€™Å  = Ã°Ââ€™Å Ã°Ââ€™â€
Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™â€ = Ã°Ââ€™Å Ã°Ââ€™Å
.
Ã°Ââ€˜Â¹Ã°Ââ€™â€ + Ã°Ââ€˜Â¹Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Then
Ã°Ââ€™â€”Ã°Ââ€™Â
Ã°Ââ€™Å Ã°Ââ€™â€
= Ã°Ââ€™Å Ã°Ââ€™Å
Ã°Ââ€˜Â¹Ã°Ââ€™Å½ Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Â +Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã¢Ë†â„¢
Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™Å Ã°Ââ€™Å  Ã°Ââ€˜Â¹Ã°Ââ€™â€ +Ã°Ââ€˜Â¹Ã°Ââ€™Å
= Ã°Ââ€˜Â¹Ã°Ââ€™Å½
Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã°Ââ€˜Â¹Ã°Ââ€™Â +Ã°Ââ€˜Â¹Ã°Ââ€˜Â³
Ã¢Ë†â„¢
Ã°Ââ€˜Â¹Ã°Ââ€™â€
.
Ã°Ââ€˜Â¹Ã°Ââ€™â€ +Ã°Ââ€˜Â¹Ã°Ââ€™Å
27
E1.21. Find the input resistance between terminals B (base) and G (ground) in the circuit
shown in Fig. The voltage Ã°Ââ€™â€”Ã°Ââ€™â„¢ is a test voltage (base – ground) with the input resistance
Ã°Ââ€™â€”
Ã°Ââ€˜Â¹Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€˜Â¹Ã°Ââ€˜Â©Ã°Ââ€˜Â® defined as Ã°Ââ€˜Â¹Ã°Ââ€™Å Ã°Ââ€™Â = Ã°Ââ€™Å  Ã°Ââ€™â„¢ .
Ã°Ââ€™â„¢
Ã°Ââ€™Å Ã°Ââ€™â€
From KCL one may get that
Ã°Ââ€™Å Ã°Ââ€™â€  = Ã°Ââ€™Å Ã°Ââ€™Æ’ + Ã°ÂÅ“Â·Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™â€  = Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã°ÂÅ¸Â + Ã°ÂÅ“Â·
From KVL one may get that
Ã°Ââ€™â€”Ã°Ââ€™â„¢ = Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã°Ââ€™â€œÃ°ÂÂâ€¦ + Ã°Ââ€™Å Ã°Ââ€™â€  Ã°Ââ€˜Â¹Ã°Ââ€™â€  Ã¢Å¸Â¹ Ã°Ââ€™Å Ã°Ââ€™â„¢ = Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã¢Å¸Â¹ Ã°Ââ€™â€”Ã°Ââ€™â„¢ = Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã°Ââ€™â€œÃ°ÂÂâ€¦ + Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã°ÂÅ¸Â + Ã°ÂÅ“Â· Ã°Ââ€˜Â¹Ã°Ââ€™â€  = Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã°Ââ€™â€œÃ°ÂÂâ€¦ + Ã°ÂÅ¸Â + Ã°ÂÅ“Â· Ã°Ââ€˜Â¹Ã°Ââ€™â€
Thus, Ã°Ââ€˜Â¹Ã°Ââ€™Å Ã°Ââ€™Â =
Ã°Ââ€™â€”Ã°Ââ€™â„¢
Ã°Ââ€™Å Ã°Ââ€™â„¢
=
Ã°Ââ€™â€”Ã°Ââ€™â„¢
Ã°Ââ€™Å Ã°Ââ€™Æ’
=
Ã°Ââ€™Å Ã°Ââ€™Æ’ Ã°Ââ€™â€œÃ°ÂÂâ€¦ + Ã°ÂÅ¸Â+Ã°ÂÅ“Â· Ã°Ââ€˜Â¹Ã°Ââ€™â€
Ã°Ââ€™Å Ã°Ââ€™Æ’
= Ã°Ââ€™â€œÃ°ÂÂâ€¦ + Ã°ÂÅ¸Â + Ã°ÂÅ“Â· Ã°Ââ€˜Â¹Ã°Ââ€™â€  .
28
Semiconductors
Outline:
1. Intrinsic semiconductor;
2. Doped (Extrinsic) semiconductors;
3. Diodes;
4. BJT.
ENGR24100, Electronics I
Instructor: Andrii Golovin
E-mail: agolovin@ccny.cuny.edu
CCNY
1
Ã¢â‚¬Â¢ Semiconductor is a material that has an electrical conductivity falling between that of a
conductor, such as metallic copper, and an insulator, such as glass.
Ã¢â‚¬Â¢ Its resistivity falls as its temperature rises; metals behave in the opposite way.
Ã¢â‚¬Â¢ Its conducting properties may be altered in useful ways by introducing impurities
(“doping”) into the crystal structure. When two differently doped regions exist in the
same sample, a semiconductor junction is created.
Ã¢â‚¬Â¢ The behavior of charge carriers, which include electrons, ions and electron holes, at
these junctions is the basis of diodes, transistors and most modern electronics.
Ã¢â‚¬Â¢ Some examples of semiconductors are single element crystals such as silicon,
germanium and compound crystals such as gallium arsenide. After silicon, gallium
arsenide is the second-most common semiconductor and is used in laser diodes, solar
cells, microwave-frequency integrated circuits, and others. Silicon is a critical element
for fabricating most electronic circuits.
https://en.wikipedia.org/wiki/Semiconductor
Ã¢â‚¬Â¢ A semiconductor device is an electronic component that relies on the electronic
properties of a semiconductor material for its function.
Ã¢â‚¬Â¢ Semiconductor devices have replaced vacuum tubes in most applications. They use
electrical conduction in the solid state rather than the gaseous state or thermionic
emission in a vacuum.
Ã¢â‚¬Â¢ Semiconductor devices are manufactured both as single discrete devices and
as integrated circuit (IC) chips, which consist of two or more devices manufactured and
interconnected on a single semiconductor wafer (also called a substrate).
https://en.wikipedia.org/wiki/Semiconductor_device
2
Ã¢â‚¬Â¢ An intrinsic (pure) semiconductor, also called an undoped semiconductor or
semiconductor without any significant dopant species present. The number of
charge carriers is therefore determined by the properties of the material itself
instead of the amount of impurities. In intrinsic semiconductors the number of
excited electrons and the number of holes are equal: n = p.
Ã¢â‚¬Â¢ A silicon atom has four valence electrons, and thus it requires another four to
complete its outermost shell. This is achieved by sharing one of its valence
electrons with each of its four neighboring atoms. Each pair of shared electrons
forms a covalent bond.
Figure 3.1 Two-dimensional representation of the silicon crystal. The circles represent the silicon
atoms, with +4 indicating its positive charge of +4q, which is neutralized by the charge of the four
valence electrons. Observe how the covalent bonds are formed by sharing of the valence
3
electrons. At 0 K, all bonds are intact and no free electrons are available for current conduction.
At room temperature, sufficient thermal energy exists to break some of the covalent bonds, a
process known as thermal generation. As shown in Fig. 3.2, when a covalent bond is broken,
an electron is freed. The free electron can wander away from its parent atom, and it becomes
available to conduct electric current if an electric field is applied to the crystal. As the electron
leaves its parent atom, it leaves behind a net positive charge Ã¢â‚¬Å“holeÃ¢â‚¬Â, equal to the magnitude of
the electron charge. An electrons from a neighboring atom may be attracted to this positive
charge, and leaves its parent atom to fill up the Ã¢â‚¬Å“holeÃ¢â‚¬Â. This process may repeat itself, with the
result in an electric current of holes. As temperature increases, more covalent bonds are
broken and electronÃ¢â‚¬â€œhole pairs are generated.
Figure 3.2 At room temperature, some
of the covalent bonds are broken by
thermal generation. Each broken bond
gives rise to a free electron and a hole,
both are available for current
conductions. Some electrons may fill
some of the holes, this is called as
Ã¢â‚¬Å“recombinationÃ¢â‚¬Â. The recombination
rate is proportional to the number
of free electrons and holes, which in
turn is determined by the thermal
Ã¢â‚¬Å“generationÃ¢â‚¬Â rate.
4
Ã¢â‚¬Â¢ A thermal equilibrium is a state of intrinsic s/c crystal that is featuring by the
recombination rate equal to the generation rate
Ã¢â‚¬Â¢ In thermal equilibrium, the concentration of free electrons Ã°Ââ€™Â is equal to the
concentration of holes Ã°Ââ€™â€˜,
Ã°Ââ€™Â = Ã°Ââ€™â€˜ = Ã°Ââ€™ÂÃ°Ââ€™Å
where Ã°Ââ€™ÂÃ°Ââ€™Å  = Ã°Ââ€˜Â©Ã°Ââ€˜Â»Ã°ÂÅ¸â€˜/Ã°ÂÅ¸Â Ã°Ââ€™â€ Ã¢Ë†â€™Ã°Ââ€˜Â¬Ã°Ââ€™Ë† /Ã°ÂÅ¸ÂÃ°Ââ€™Å’Ã°Ââ€˜Â» is the number of free electrons and holes in a unit
volume (Ã°Ââ€™â€žÃ°Ââ€™Å½Ã°ÂÅ¸â€˜ ) of intrinsic silicon at a given temperature Ã°Ââ€˜Â» Ã°ÂÂÅ  ; Ã°Ââ€˜Â© = Ã°ÂÅ¸â€¢. Ã°ÂÅ¸â€˜ Ãƒâ€”
Ã°ÂÅ¸â€˜
Ã¢Ë†â€™Ã°ÂÅ¸Â
Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ
Ã¢Ë†â€™Ã°ÂÅ¸â€˜
Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ [Ã°Ââ€™â€žÃ°Ââ€™Å½ Ã°Ââ€˜Â² ] is a material-dependent parameter that is given for silicon; Ã°Ââ€˜Â¬Ã°Ââ€™Ë† is
a material parameter known as the bandgap energy, it is the minimum energy
required to break a covalent bond and thus generate an electron-hole pair, for
silicon Ã°Ââ€˜Â¬Ã°Ââ€™Ë† =1.12 [eV] ; and Ã°Ââ€™Å’ = Ã°ÂÅ¸â€“. Ã°ÂÅ¸â€Ã°ÂÅ¸Â Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€œ Ã°Ââ€™â€ Ã°Ââ€˜Â½/Ã°Ââ€˜Â² is BoltzmannÃ¢â‚¬â„¢s constant.
Ã¢â‚¬Â¢ It is useful to express the product of the hole and free-electron concentration as
Ã°Ââ€™â€˜Ã°Ââ€™Â = Ã°Ââ€™ÂÃ°ÂÅ¸ÂÃ°Ââ€™Å  ,
where for the intrinsic silicon at the room temperature, Ã°Ââ€™ÂÃ°Ââ€™Å  Ã¢â€°Ë† Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ /Ã°Ââ€™â€žÃ°Ââ€™Å½Ã°ÂÅ¸â€˜ .
5
Energy Bands, Band Gap
(K. Kano, Ã¢â‚¬Å“Semiconductor DevicesÃ¢â‚¬Â)
Conductance
band
Valence
band
The band gap is temperature dependent Ã°Ââ€™Å½Ã°ÂÂÅ¡Ã°ÂÂÂ­Ã°ÂÂÅ¾Ã°ÂÂÂ«Ã°ÂÂÂ¢Ã°ÂÂÅ¡Ã°ÂÂÂ¥ Ã°ÂÂÂ©Ã°ÂÂÅ¡Ã°ÂÂÂ«Ã°ÂÂÅ¡Ã°ÂÂÂ¦Ã°ÂÂÅ¾Ã°ÂÂÂ­Ã°ÂÂÅ¾Ã°ÂÂÂ« Ã°Ââ€˜Â¬Ã°Ââ€™Ë† Ã°Ââ€˜Â» = Ã°Ââ€˜Â¬Ã°Ââ€™Ë† Ã°ÂÅ¸Å½
Ã°ÂÅ“Â¶Ã°Ââ€˜Â»Ã°ÂÅ¸Â
Ã¢Ë†â€™ Ã°Ââ€˜Â»+Ã°ÂÅ“Â·
Silicon: Ã°Ââ€˜Â¬Ã°Ââ€™Ë† Ã°ÂÅ¸Å½ = Ã°ÂÅ¸Â. Ã°ÂÅ¸ÂÃ°ÂÅ¸â€¢ Ã°Ââ€™â€ Ã°Ââ€˜Â½, Ã°ÂÅ“Â¶ = Ã°ÂÅ¸â€™. Ã°ÂÅ¸â€¢Ã°ÂÅ¸â€˜ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã¢Ë†â€™Ã°ÂÅ¸â€™ , Ã°ÂÅ“Â· = Ã°ÂÅ¸â€Ã°ÂÅ¸â€˜Ã°ÂÅ¸â€, Ã°Ââ€˜Â¬Ã°Ââ€™Ë† Ã°ÂÅ¸â€˜Ã°ÂÅ¸Å½Ã°ÂÅ¸Å½Ã°Ââ€˜Â² = Ã°ÂÅ¸Â. Ã°ÂÅ¸ÂÃ°ÂÅ¸ÂÃ°ÂÅ¸â€œ Ã°Ââ€™â€ Ã°Ââ€˜Â½.
6
[Ã°Ââ€™â€ Ã°Ââ€˜Â½],
Doped (Extrinsic) Semiconductors
Ã¢â‚¬Â¢ Doping involves introducing impurity atoms into the semiconductor crystal in
sufficient numbers to substantially increase the concentration of either free
electrons or holes but with little or no change in the crystal properties of silicon.
Ã¢â‚¬Â¢ To increase the concentration of free electrons, Ã°Ââ€™Â, silicon is doped with an element
with a valence of 5, such as phosphorus. The resulting doped silicon is then said to
be of Ã°Ââ€™Â type.
Ã¢â‚¬Â¢ To increase the concentration of holes, Ã°Ââ€™â€˜, silicon is doped with an element having
a valence of 3, such as boron, and the resulting doped silicon is said to be of p
type.
N-type semiconductor:
silicon doped by donors (phosphorus atoms)
P-type semiconductor:
silicon doped by acceptors (boron atoms)
7
N-type Semiconductor:
e.g. Silicon Doped by Donors (Phosphorus Atoms)
1s2 2s2 2p6 3s2 3p2
1s2 2s2 2p6 3s2 3p3
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p3
P-type Semiconductor:
e.g. Silicon Doped by Acceptors (Boron Atoms)
Si
1s2 2s2 2p6 3s2 3p2
1s2 2s2 2p1
1s2 2s2 2p6 3s2 3p1
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p1
Electron transitions in doped semiconductors
n-type
p-type
The impurity provides additional energy states to the semiconductor,
which makes their conductance higher and controllable.
10
pÃ¢â‚¬â€œn junction
Two electrically neutral
crystals separated by a
parallel air gap.
Two crystals are connected
by pÃ¢â‚¬â€œn junction:
Ã¢â‚¬Â¢ Electron
and
hole
concentration
are
plotted with blue and
red lines, respectively.
Ã¢â‚¬Â¢ The gray regions are
charge-neutral.
The
light-red
zone
is
positively charged.
Ã¢â‚¬Â¢ The light-blue zone is
negatively charged.
Ã¢â‚¬Â¢ Ã°Ââ€˜Â¬ is inner electric field
of the depleted zone.
11
The inner field Ã°Ââ€˜Â¬- field stops both diffusion currents of
electrons and holes throughout p-n junction
https://en.wikipedia.org/
space charge region
or
depleted zone
12
P-N junction
13
P-N junction
The barrier voltage Ã°Ââ€˜Â½Ã°Ââ€™Æ’Ã°Ââ€™Å  across the p-n
junction is given by
Ã°Ââ€˜Â½Ã°Ââ€™Æ’Ã°Ââ€™Å  = Ã°Ââ€˜Â½Ã°Ââ€˜Â» Ã°Ââ€™ÂÃ°Ââ€™Â
Ã°Ââ€˜ÂµÃ°Ââ€˜Â¨ Ã°Ââ€˜ÂµÃ°Ââ€˜Â«
Ã°Ââ€™ÂÃ°ÂÅ¸ÂÃ°Ââ€™Å
Ã¢â€°Ë† Ã°ÂÅ¸Å½. Ã°ÂÅ¸â€¢ [Ã°Ââ€˜Â½],
where Ã°Ââ€˜ÂµÃ°Ââ€˜Â¨ ~Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸ÂÃ°ÂÅ¸â€¢ /Ã°Ââ€™â€žÃ°Ââ€™Å½Ã°ÂÅ¸â€˜ and Ã°Ââ€˜ÂµÃ°Ââ€˜Â« ~Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ /Ã°Ââ€™â€žÃ°Ââ€™Å½Ã°ÂÅ¸â€˜
are the doping concentrations of the Ã°Ââ€™â€˜ side
and Ã°Ââ€™Â side of the junction, respectively;
Ã°Ââ€™ÂÃ°Ââ€™Å  Ã¢â€°Ë† Ã°ÂÅ¸Â. Ã°ÂÅ¸â€œ Ãƒâ€” Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½Ã°ÂÅ¸ÂÃ°ÂÅ¸Å½ /Ã°Ââ€™â€žÃ°Ââ€™Å½Ã°ÂÅ¸â€˜ is concentration of
free electrons Ã°Ââ€™Â and holes in intrinsic s/c;
and Ã°Ââ€˜Â½Ã°Ââ€˜Â» = Ã°Ââ€™Å’Ã°Ââ€˜Â»ÃŽÂ¤Ã°Ââ€™â€™ = Ã°ÂÅ¸ÂÃ°ÂÅ¸â€œ. Ã°ÂÅ¸â€” Ã°Ââ€™Å½Ã°Ââ€˜Â½ (when
T=300K) is thermal voltage, it provides a
measure of how much the spatial
distribution of electrons or holes is affected
by a boundary held at a fixed voltage.
14
Ã¢â‚¬Â¢ The Diffusion Current Ã°Ââ€˜Â°Ã°Ââ€˜Â« There are two diffusing components of that current,
namely, holes diffuse across the p-n junction from the p side to the n side and,
similarly, electrons diffuse across the same from the n side to the p side. Direction
of the Ã°Ââ€˜Â°Ã°Ââ€˜Â« current is from the p side to the n side.
Ã¢â‚¬Â¢ The Drift Current Ã°Ââ€˜Â°Ã°Ââ€˜Âº . In addition to the majority-carrier diffusion, components due
to minority-carrier drifts exists across the p-n junction. Specifically, some of the
thermally generated holes in the n side move toward the junction and reach the
edge of the depletion region, then the electric field of the depletion region sweeps
them across that region into the p side. Similarly, some of the minority thermally
generated electrons in the p side move to the edge of the depletion region and get
swept by the electric field of the depletion region across that region into the n
side. These two current components form the drift current Ã°Ââ€˜Â°Ã°Ââ€˜Âº , whose direction is
from the n side to the p side of the junction.
Ã¢â‚¬Â¢ Under open-circuit conditions with no external electrical field applied across p-n
junction, the two opposite currents across the junction are equal in magnitude:
Ã°Ââ€˜Â°Ã°Ââ€˜Â« = Ã°Ââ€˜Â°Ã°Ââ€˜Âº .
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Ã°Ââ€˜Â°Ã°Ââ€˜Â«
Ã°Ââ€˜Â°Ã°Ââ€˜Âº
metal
metal
Ã¢â‚¬Â¢ When the p-n junction terminals are left open-circuited, the voltage measured
between them will be zero. That is, the voltage Ã°Ââ€˜Â½Ã°Ââ€™Æ’Ã°Ââ€™Å  across the depletion region
does not appear between the junction terminals. This is because similar built-in
voltages arise at the metalÃ¢â‚¬â€œsemiconductor junctions at the terminals, which
counter and exactly balance the barrier voltage. If this were not the case, we
would be able to draw energy from the isolated p-n junction, which would clearly
violate the principle of conservation of energy.
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Ã¢â‚¬Â¢ Ã°Ââ€˜ÂµÃ°Ââ€˜Â¨ > Ã°Ââ€˜ÂµÃ°Ââ€˜Â« is a common situation in practice;
Ã¢â‚¬Â¢ The depletion region extends in both the p and n sides and equal amounts of
charge exist on both sides Ã°Ââ€˜Â¸+ = Ã°Ââ€˜Â¸Ã¢Ë†â€™ ;
Ã¢â‚¬Â¢ Since Ã°Ââ€˜ÂµÃ°Ââ€˜Â¨ > Ã°Ââ€˜ÂµÃ°Ââ€˜Â« , the width of the depletion layer will not be the same on the two
sides. Rather, to uncover the same amount of charge, …