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.

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
• The objective is to reduce the number of simultaneous equations to solve for.
• Given a circuit with 𝒏 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, …, 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 Ohm’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.
• 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 − − − −eqt1
Node2 : i2 + i4 = i1 + i5 − − − − − −eqt2
v1 − v2 v1 − 0
5=
+
20 = v1 − v2 + 2v1
4
2
3v1 − v2 = 20 − − − −eqt3
v1 − v2
v2 − 0
+ 10 = 5 +
3v1 − 3v2 + 120 = 60 + 2v2
4
6
− 3v1 + 5v2 = 60 − − − − − −eqt4
Method 1 (elimination) Add eqt (3) and (4)
4v2 = 80 v2 = 20V
From eqt1 3v1 − 20 = 20 v1 = 13.33V
⇒
⇐
⇒
4
Method 2 (Cramer’s rule is described in Appendix A)
3v1 − v2 = 20 − − − −eqt3
− 3v1 + 5v2 = 60 − − − eqt4
Eqt (3) and (4) in the matrix form :
 3 − 1  v1  20
 − 3 5 v  = 60

 2   
The determinant of the matrix is
 3 −1
=
= 15 − 3 = 12

 − 3 5
 20 − 1
60 5 
 = 100 + 60 = 13.33V
v1 and v2 are obtained as v1 = 

12
 3 20 
− 3 60
 = 180 + 60 = 20V
v2 = 

12
v − v2
v
Currents are calculated as i1 = 5 A i2 = 1
= −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 4𝜴 is 𝑼𝟒𝛀 = 𝒗𝟏 − 𝒗𝟐 = 𝟐𝟎 − 𝟏𝟑. 𝟑𝟑 ≈ 𝟔. 𝟕𝑽.
The voltage drop on the resistor 6𝛀 is 𝑼𝟔𝛀 = 𝒗𝟐 − 𝟎 = 𝟐𝟎𝑽.
Voltage drop on resistor 2𝛀 is 𝑼𝟐𝛀 = 𝒗𝟏 − 𝟎 = 𝟏𝟑. 𝟑𝟑.
8. Check your solution by using a conservation law.
KCL applied to the node #1:
𝒊𝟏 + 𝒊𝟐 = 𝒊𝟑 ⇒ 𝟓 + 𝟏. 𝟔𝟕 = 𝟔. 𝟔𝟕.
KCL applied to the node #2:
𝒊𝟒 = 𝒊𝟏 + 𝒊𝟐 + 𝒊𝟓 ⇒ 𝟏𝟎 = 𝟓 + 𝟏. 𝟔𝟕 + 𝟑. 𝟑𝟑
6
Supernode
(Nodal Analysis with Voltage Sources)
𝒗𝟐 =?,
𝒗𝟑 =? .
• 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.
• Both, KCL and KVL, are applied to determine the node voltage.
7
Applying KVL to a supernode
i1 + i4 = i2 + i3 − − − − − − − − − − − − − eq1a
or
𝒗𝟑
KVL in the supernode
𝒗𝟐
v1 − v2 v1 − v3 v2 − 0 v3 − 0
+
=
+
− − − eq1b
2
4
8
6
− v2 + 5 + v3 = 0 − − − − − − − − − − − eq2
v2 − v3 = 5 − − − − − − − − − − − − − eq2a
v1 = 10 − − − − − − − − − − − − − − − eq3
• Properties of a supernode:
• The voltage source inside the supernode
provides KVL equation needed to solve
for the node voltages.
• A supernode has no voltage of its own.
• A supernode requires the application of
both KCL and KVL.
8
Applying KVL to a supernode
i1 + i4 = i2 + i3 − − − − − − − − − − − − − eq1a
or
v1 − v2 v1 − v3 v2 − 0 v3 − 0
+
=
+
− − − eq1b
2
4
8
6
− v2 + 5 + v3 = 0 − − − − − − − − − − − eq2
𝟏𝟖𝒗𝟏 − 𝟏𝟓𝒗𝟐 = 𝟏𝟎𝒗𝟑
v2 − v3 = 5 − − − − − − − − − − − − − eq2a
v1 = 10 −(From
− − −diagram
− − − − of
− −the
− circuit).
− − − − eq3
A supernode requires the application of
KVL:
−𝒗𝟐 + 𝟓 + 𝒗𝟑 = 𝟎 ⇒ 𝒗𝟐 = 𝟓 + 𝒗𝟑
𝟏𝟖𝟎 − 𝟏𝟓𝒗𝟐 = 𝟏𝟎𝒗𝟑
𝒗𝟐 = 𝟓 + 𝒗𝟑
⇒
⇒ 𝐀𝐧𝐬𝐰𝐞𝐫: 𝒗𝟑 = 𝟒. 𝟐𝑽, 𝒗𝟐 = 𝟗. 𝟐𝑽
9
Mesh (Loop) Analysis
• Mesh analysis only applicable to a circuit that is planar.
• A planar circuit is one that can be drawn in a plane with no branches
crossing one another.
• A mesh is an elementary loop that does not contain any other loops within
it.
• From fundamental theorem of network topology b=l+n-1⇒
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.
• Mesh and branch currents are not always the same.
10
Nonpalanar
vs
Planar
⟺
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 𝒏 meshes;
2) Apply KVL to each of the 𝒏 meshes to get 𝒏 linear equations;
3) Use Ohm’s law UR=inR to express the voltages on resistors in terms of the
mesh currents.
4) Solve the resulting 𝒏 linear equations to get the mesh currents in.
5) Use Ohm’s law UR=inR to find the voltage drop on each resistor.
6) Use Ohm’s law IR=UR/R to find the branch currents IR.
12
Example:
Find the branch currents I1, I2 and I3.
Mesh #1 (a⟶b⟶e⟶f⟶a):
−𝑽𝟏 + 𝑹𝟏 𝒊𝟏 + 𝑹𝟑 𝒊𝟏 − 𝒊𝟐 = 𝟎;
𝑹𝟏 + 𝑹𝟑 𝒊𝟏 − 𝑹𝟑 𝒊𝟐 = 𝑽𝟏 ;
Mesh #2 (b⟶c⟶d⟶e⟶b):
𝑹𝟐 𝒊𝟐 + 𝑽𝟐 + 𝑹𝟑 𝒊𝟐 − 𝒊𝟏 = 𝟎;
−𝑹𝟑 𝒊𝟏 + (𝑹𝟐 + 𝑹𝟑 )𝒊𝟐 = −𝑽𝟐
Branch currents:
𝑰𝟏 = 𝒊𝟏 ;
Fig. 3.17. A circuit with two meshes.
𝑹𝟏 + 𝑹𝟑 𝒊𝟏 − 𝑹𝟑 𝒊𝟐 = 𝑽𝟏 ;
⟹൞
−𝑹𝟑 𝒊𝟏 + (𝑹𝟐 + 𝑹𝟑 )𝒊𝟐 = −𝑽𝟐
𝑹𝟏 + 𝑹𝟑
−𝑹𝟑
−𝑹𝟑
𝒊𝟏
𝑽𝟏
=
𝑹𝟐 + 𝑹𝟑 𝒊𝟐
−𝑽𝟐
𝒊𝟏 , 𝒊𝟐 , 𝒊𝟑 are roots of the system with three linear
equations.
𝑰𝟐 = 𝒊𝟐 ;
𝑰𝟑 = 𝒊𝟏 − 𝒊𝟐 . ⟹ Mesh and branch currents are not always the same.
13
1Ω
1𝑉
2Ω
2𝑉
3Ω
The initial assumption is that
the mech currents 𝒊𝟏 and 𝒊𝟐
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
KVL in the mesh #1:
−𝟏 + 𝟏𝒊𝟏 + 𝟑 𝒊𝟏 − 𝒊𝟐 = 𝟎;
𝟒𝒊𝟏 − 𝟑𝒊𝟐 = 𝟏.
𝟒𝒊𝟏 − 𝟑𝒊𝟐 = 𝟏.
KVL in the mesh #2:
𝟐 + 𝟑 𝒊𝟐 − 𝒊𝟏 + 𝟐𝒊𝟐 = 𝟎;
−𝟑𝒊𝟏 + 𝟓𝒊𝟐 = −𝟐.
14
𝟒 −𝟑 𝒊𝟏
𝟏
⟹ቐ
⟹
∙
=
⟹
𝒊𝟐
−𝟑 𝟓
−𝟐
−𝟑𝒊𝟏 + 𝟓𝒊𝟐 = −𝟐.
𝟒 −𝟑
= 𝟒 ∙ 𝟓 − −𝟑 ∙ −𝟑 = 𝟏𝟏;
−𝟑 𝟓
𝟏 −𝟑
𝜟𝟏 =
= 𝟓 − −𝟐 ∙ −𝟑 = −𝟏;
−𝟐 𝟓
𝟒
𝟏
𝜟𝟐 =
= −𝟖 − −𝟑 = −𝟓.
−𝟑 −𝟐
𝜟𝟏 −𝟏
𝜟𝟐 −𝟓
⟹ 𝒊𝟏 =
=
≈ −𝟎. 𝟎𝟗𝟏 𝑨,
𝒊𝟐 =
=
≈ −𝟎. 𝟒𝟓𝟒𝟓 𝑨
𝜟
𝟏𝟏
𝜟
𝟏𝟏
𝜟=
1Ω
1𝑉
2Ω
2𝑉
3Ω
The initial assumption is that
the mech currents 𝒊𝟏 and 𝒊𝟐
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
𝜟𝟏 −𝟏
𝒊𝟏 =
=
≈ −𝟎. 𝟎𝟗𝟏 𝑨,
𝜟
𝟏𝟏
𝜟𝟐 −𝟓
𝒊𝟐 =
=
≈ −𝟎. 𝟒𝟓𝟒𝟓 𝑨 ⟹
𝜟
𝟏𝟏
Directions of both mesh currents should be corrected to show the counter-clockwise direction:
𝒊𝟏 ≈ 𝟎. 𝟎𝟗𝟏 𝑨,
𝒊𝟐 ≈ 𝟎. 𝟒𝟓𝟒𝟓 𝑨
Also directions of the branch currents 𝑰𝟏 and 𝑰𝟐 should be changed to opposite ones.
𝑰𝟏 = 𝒊𝟏 ≈ 𝟎. 𝟎𝟗𝟏 and 𝑰𝟐 = 𝒊𝟐 ≈ 𝟎. 𝟒𝟓𝟒𝟓.
𝑰𝟑 = 𝒊𝟐 − 𝒊𝟏 ≈ 𝟎. 𝟒𝟓𝟒𝟓 − 𝟎. 𝟎𝟗𝟏 ≈ 𝟎. 𝟑𝟔 𝑨
The original assumption about downward direction of the branch current 𝑰𝟑 was correct.
15
1Ω
1𝑉
2Ω
2𝑉
3Ω
The initial assumption is that the
mech currents 𝒊𝟏 and 𝒊𝟐 are in
opposite directions.
Fig. 3.17c. A circuit with two meshes.
KVL in the mesh #1:
−𝟏 + 𝟏𝒊𝟏 + 𝟑 𝒊𝟏 + 𝒊𝟐 = 𝟎;
𝟒𝒊𝟏 + 𝟑𝒊𝟐 = 𝟏.
KVL in the mesh #2:
−𝟐 + 𝟐𝒊𝟐 + 𝟑 𝒊𝟐 + 𝒊𝟏 = 𝟎;
𝟒𝒊𝟏 + 𝟑𝒊𝟐 = 𝟏.
𝟒 𝟑 𝒊𝟏
𝟏
⟹ቐ
⟹
∙
=
⟹
𝟑 𝟓 𝒊𝟐
𝟐
𝟑𝒊𝟏 + 𝟓𝒊𝟐 = 𝟐.
∆= 𝟐𝟎 − 𝟗 = 𝟏𝟏; 𝚫𝟏 = 𝟓 − 𝟔 = −𝟏; 𝚫𝟐 = 𝟖 − 𝟑 = 𝟓
𝟑𝒊𝟏 + 𝟓𝒊𝟐 = 𝟐.
𝒊𝟏 =
𝚫𝟏
𝚫
=
−𝟏
𝟏𝟏
≈ −𝟎. 𝟎𝟗𝟏 𝑨 and 𝒊𝟐 =
∆𝟐
∆
≈ 𝟎. 𝟒𝟓𝟒𝟓 𝑨
16
1Ω
2Ω
1𝑉
2𝑉
3Ω
The initial assumption is that the
mech currents 𝒊𝟏 and 𝒊𝟐 are in
opposite directions.
Fig. 3.17c. A circuit with two meshes.
𝒊𝟏 =
𝚫𝟏
𝚫
=
−𝟏
𝟏𝟏
≈ −𝟎. 𝟎𝟗𝟏 𝑨 and 𝒊𝟐 =
∆𝟐
∆
≈ 𝟎. 𝟒𝟓𝟒𝟓 𝑨
Direction of the mesh current 𝒊𝟏 should be corrected to the counter-clockwise:
𝒊𝟏 ≈ 𝟎. 𝟎𝟗𝟏 𝑨 and 𝒊𝟐 ≈ 𝟎. 𝟒𝟓𝟒𝟓 𝑨
The branch currents are 𝑰𝟏 = 𝒊𝟏 ≈ 𝟎. 𝟎𝟗𝟏, 𝑰𝟐 = 𝒊𝟐 ≈ 𝟎. 𝟒𝟓𝟒𝟓,
𝑰𝟑 = 𝒊𝟐 − 𝒊𝟏 ≈ 𝟎. 𝟒𝟓𝟒𝟓 − 𝟎. 𝟎𝟗𝟏 ≈ 𝟎. 𝟑𝟔 𝑨
17
Check-up by Using Nodal Analysis
𝑽𝒂 = 𝟏𝑽
1𝑉
1Ω
2Ω
𝑽𝒃
3Ω
𝑽𝒄 = 𝟐𝑽
2𝑉
Fig. 3.17d. A circuit with two meshes.
KCL in the node 𝑽𝒃 :
𝑰𝟏
𝑰𝟏
𝑰𝟐
𝑰𝟑
𝟏 − 𝑽𝒃 𝑽𝒃 − 𝟐 𝑽𝒃
= 𝑰𝟐 + 𝑰 𝟑 ⟹
=
−
⟹ 𝑽𝒃 = 𝟏. 𝟎𝟗𝟏 𝑽
𝟏
𝟐
𝟑
𝟏 − 𝑽𝒃
=
= −𝟎. 𝟎𝟗𝟏 𝑨; ⟹ The initial assumptions for the directions of
𝟏
branch currents 𝑰𝟏 and 𝑰𝟐 should be corrected
𝑽𝒃 − 𝟐
=
= −𝟎. 𝟒𝟓𝟒𝟓 𝑨; ⟹ to show both currents from right to left.
𝟐
𝑽𝒃
=
= 𝟎. 𝟑𝟔𝟒 𝑨.
18
𝟑
Supermesh
Supermesh
Mesh#1
Mesh#2
• 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;
• The two meshes must be joined together, resulting in a supermesh;
• The supermesh is constructed by merging the two meshes and excluding
the shared source and any elements in series with it;
• KCL should be applied to one node of the supermesh.
19
Supermesh
Example
• Apply KVL to the supermesh:
−𝟐𝟎 + 𝟔𝒊𝟏 + 𝟏𝟎𝒊𝟐 + 𝟒𝒊𝟐 = 𝟎 ⟹
or 𝟔𝒊𝟏 + 𝟏𝟒𝒊𝟐 = 𝟐𝟎
• Apply KCL to the node “0” in the branch where the two meshes
intersect:
𝟔𝒊𝟏 + 𝟏𝟒𝒊𝟐 = 𝟐𝟎
𝒊𝟐 = 𝒊𝟏 + 𝟔
⟹ ቐ
in
out
𝒊𝟐 = 𝒊𝟏 + 𝟔
• By solving the system of two equations we get:
𝒊𝟏 = −𝟑. 𝟐𝐀 𝒊𝟐 = 𝟐. 𝟖𝐀 ⟹ Initial assumption for the direction of the
mesh current 𝒊𝟏 should be corrected.
20
Supermesh
Example
3.2A
6A
2.8A
𝒊𝟏 = −𝟑. 𝟐𝐀 𝒊𝟐 = 𝟐. 𝟖𝐀 ⟹ Initial assumption for the direction of the
mesh current 𝒊𝟏 should be corrected.
Check up by KCL applied to the node “0”: 𝒊𝟏 + 𝒊𝟐 = 𝟔 𝑨
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
• The objective is to reduce the number of simultaneous equations to solve for.
• Given a circuit with 𝒏 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 , …, 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 Ohm’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.
• 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:
𝑵𝒐𝒅𝒆 𝟐:
𝒊𝟏 = 𝒊𝟐 + 𝒊𝟑 ,
𝒊𝟐 + 𝒊𝟒 = 𝒊𝟏 + 𝒊𝟓 ,
𝒗 −𝒗
Eq.(1)
Eq.(2)
𝒗 −𝟎
Eq.(1)⇒ 𝟓 = 𝟏 𝟒 𝟐 + 𝟏𝟐
𝟐𝟎 = 𝒗𝟏 − 𝒗𝟐 + 𝟐𝒗𝟏
𝟑𝒗𝟏 − 𝒗𝟐 = 𝟐𝟎.
Eq.(3)
𝒗𝟏 − 𝒗𝟐
𝒗𝟐 − 𝟎
Eq.(2) ⇒
+ 𝟏𝟎 = 𝟓 +
𝟒
𝟔
𝟑𝒗𝟏 − 𝟑𝒗𝟐 + 𝟏𝟐𝟎 = 𝟔𝟎 + 𝟐𝒗𝟐
−𝟑𝒗𝟏 + 𝟓𝒗𝟐 = 𝟔𝟎.
Eq.(4)
Add Eqs. (𝟑 and 𝟒) ⇒
𝟒𝒗𝟐 = 𝟖𝟎
𝒗𝟐 = 𝟐𝟎 𝑽
𝐄𝐪. 𝟑 ⇒ 𝟑𝒗𝟏 − 𝟐𝟎 = 𝟐𝟎
𝒗𝟏 = 𝟏𝟑. 𝟑𝟑 𝑽
4
Method by using matrixes (Cramer’s rule is described in Appendix A)
3𝑣1 − 𝑣2 = 20
−3𝑣1 + 5𝑣2 = 60
Eq.(3)
Eq.(4)
Eqs.(3 and 4) in the matrix form:
3 − 1 𝑣1
20
=
−3 5 𝑣2
60
The determinant of the matrix is
3 −1
Δ=
= 15 − 3 = 12
−3 5
v1 and v2 are obtained as
20 − 1
100 + 60
𝑣1 = 60 5
=
= 13.33 𝑉
Δ
12
3 20
180 + 60
𝑣2 = −3 60 =
= 20 𝑉
Δ
12
5
7. Calculate all branch voltages and currents.
The voltage drop on the resistor 4𝛀 is 𝑼𝟒𝛀 = 𝒗𝟏 − 𝒗𝟐 = 𝟐𝟎 − 𝟏𝟑. 𝟑𝟑 ≈ 𝟔. 𝟕𝑽.
The voltage drop on the resistor 6𝛀 is 𝑼𝟔𝛀 = 𝒗𝟐 − 𝟎 = 𝟐𝟎𝑽.
Voltage drop on resistor 2𝛀 is 𝑼𝟐𝛀 = 𝒗𝟏 − 𝟎 = 𝟏𝟑. 𝟑𝟑.
8. Check your solution by using a conservation law.
KCL applied to the node #1:
𝒊𝟏 + 𝒊𝟐 = 𝒊𝟑 ⇒ 𝟓 + 𝟏. 𝟔𝟕 = 𝟔. 𝟔𝟕.
KCL applied to the node #2:
𝒊𝟒 = 𝒊𝟏 + 𝒊𝟐 + 𝒊𝟓 ⇒ 𝟏𝟎 = 𝟓 + 𝟏. 𝟔𝟕 + 𝟑. 𝟑𝟑
6
Supernode
(Nodal Analysis with Voltage Sources)
𝒗𝟐 =?,
𝒗𝟑 =? .
• 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.
• Both, KCL and KVL, are applied to determine the node voltage.
7
Applying KVL to the Supernode
𝒊𝟏 + 𝒊𝟒 = 𝒊 𝟐 + 𝒊𝟑
𝒗𝟐
𝒗𝟑
Eq.(1a)
or
𝒗𝟏 − 𝒗𝟐 𝒗𝟏 − 𝒗𝟑 𝒗𝟐 − 𝟎 𝒗𝟑 − 𝟎
+
=
+
Eq.(1b)
𝟐
𝟒
𝟖
𝟔
The voltage source inside the supernode
provides KVL equation needed to solve for
the node voltages. KVL applied to the
supernode:
−𝒗𝟐 + 𝟓 + 𝒗𝟑 = 𝟎 𝐄𝐪. (𝟐)
𝒗𝟐 = 𝟓 + 𝒗𝟑
𝐄𝐪. (𝟐𝐚)
From the diagram one might see that
𝒗𝟏 = 𝟏𝟎
𝐄𝐪. (𝟑)
8
Applying KVL to the Supernode
From Eqs.(1b, 2a and 3) one might see that
𝟏𝟖𝟎 − 𝟏𝟓𝒗𝟐 = 𝟏𝟎𝒗𝟑
ቐ
⟹
𝒗𝟐 = 𝟓 + 𝒗𝟑
𝐀𝐧𝐬𝐰𝐞𝐫: 𝒗𝟑 = 𝟒. 𝟐 𝑽, 𝒗𝟐 = 𝟗. 𝟐 𝑽
9
Mesh (Loop) Analysis
• Mesh analysis only applicable to a circuit that is planar.
• A planar circuit is one that can be drawn in a plane with no branches
crossing one another.
• A mesh is an elementary loop that does not contain any other loops within
it.
• From fundamental theorem of network topology b=l+n-1⇒
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.
• Mesh and branch currents are not always the same.
10
Nonpalanar
vs
Planar
⟺
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 𝒏 meshes;
2) Apply KVL to each of the 𝒏 meshes to get 𝒏 linear equations;
3) Use Ohm’s law UR=inR to express the voltages on resistors in terms of the
mesh currents.
4) Solve the resulting 𝒏 linear equations to get the mesh currents in.
5) Use Ohm’s law UR=inR to find the voltage drop on each resistor.
6) Use Ohm’s law IR=UR/R to find the branch currents IR.
12
Example:
Find the branch currents I1, I2 and I3.
Mesh #1 (a⟶b⟶e⟶f⟶a):
−𝑽𝟏 + 𝑹𝟏 𝒊𝟏 + 𝑹𝟑 𝒊𝟏 − 𝒊𝟐 = 𝟎;
𝑹𝟏 + 𝑹𝟑 𝒊𝟏 − 𝑹𝟑 𝒊𝟐 = 𝑽𝟏 ;
Mesh #2 (b⟶c⟶d⟶e⟶b):
𝑹𝟐 𝒊𝟐 + 𝑽𝟐 + 𝑹𝟑 𝒊𝟐 − 𝒊𝟏 = 𝟎;
−𝑹𝟑 𝒊𝟏 + (𝑹𝟐 + 𝑹𝟑 )𝒊𝟐 = −𝑽𝟐
Branch currents:
𝑰𝟏 = 𝒊𝟏 ;
Fig. 3.17. A circuit with two meshes.
𝑹𝟏 + 𝑹𝟑 𝒊𝟏 − 𝑹𝟑 𝒊𝟐 = 𝑽𝟏 ;
⟹൞
−𝑹𝟑 𝒊𝟏 + (𝑹𝟐 + 𝑹𝟑 )𝒊𝟐 = −𝑽𝟐
𝑹𝟏 + 𝑹𝟑
−𝑹𝟑
−𝑹𝟑
𝒊𝟏
𝑽𝟏
=
𝑹𝟐 + 𝑹𝟑 𝒊𝟐
−𝑽𝟐
𝒊𝟏 , 𝒊𝟐 , 𝒊𝟑 are roots of the system with three linear
equations.
𝑰𝟐 = 𝒊𝟐 ;
𝑰𝟑 = 𝒊𝟏 − 𝒊𝟐 . ⟹ Mesh and branch currents are not always the same.
13
1Ω
1𝑉
2Ω
2𝑉
3Ω
The initial assumption is that
the mech currents 𝒊𝟏 and 𝒊𝟐
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
KVL in the mesh #1:
−𝟏 + 𝟏𝒊𝟏 + 𝟑 𝒊𝟏 − 𝒊𝟐 = 𝟎;
𝟒𝒊𝟏 − 𝟑𝒊𝟐 = 𝟏.
𝟒𝒊𝟏 − 𝟑𝒊𝟐 = 𝟏.
KVL in the mesh #2:
𝟐 + 𝟑 𝒊𝟐 − 𝒊𝟏 + 𝟐𝒊𝟐 = 𝟎;
−𝟑𝒊𝟏 + 𝟓𝒊𝟐 = −𝟐.
14
𝟒 −𝟑 𝒊𝟏
𝟏
⟹ቐ
⟹
∙
=
⟹
𝒊𝟐
−𝟑 𝟓
−𝟐
−𝟑𝒊𝟏 + 𝟓𝒊𝟐 = −𝟐.
𝟒 −𝟑
= 𝟒 ∙ 𝟓 − −𝟑 ∙ −𝟑 = 𝟏𝟏;
−𝟑 𝟓
𝟏 −𝟑
𝜟𝟏 =
= 𝟓 − −𝟐 ∙ −𝟑 = −𝟏;
−𝟐 𝟓
𝟒
𝟏
𝜟𝟐 =
= −𝟖 − −𝟑 = −𝟓.
−𝟑 −𝟐
𝜟𝟏 −𝟏
𝜟𝟐 −𝟓
⟹ 𝒊𝟏 =
=
≈ −𝟎. 𝟎𝟗𝟏 𝑨,
𝒊𝟐 =
=
≈ −𝟎. 𝟒𝟓𝟒𝟓 𝑨
𝜟
𝟏𝟏
𝜟
𝟏𝟏
𝜟=
1Ω
1𝑉
2Ω
2𝑉
3Ω
The initial assumption is that
the mech currents 𝒊𝟏 and 𝒊𝟐
are both clockwise directed.
Fig. 3.17b. A circuit with two meshes.
𝜟𝟏 −𝟏
𝒊𝟏 =
=
≈ −𝟎. 𝟎𝟗𝟏 𝑨,
𝜟
𝟏𝟏
𝜟𝟐 −𝟓
𝒊𝟐 =
=
≈ −𝟎. 𝟒𝟓𝟒𝟓 𝑨 ⟹
𝜟
𝟏𝟏
Directions of both mesh currents should be corrected to show the counter-clockwise direction:
𝒊𝟏 ≈ 𝟎. 𝟎𝟗𝟏 𝑨,
𝒊𝟐 ≈ 𝟎. 𝟒𝟓𝟒𝟓 𝑨
Also directions of the branch currents 𝑰𝟏 and 𝑰𝟐 should be changed to opposite ones.
𝑰𝟏 = 𝒊𝟏 ≈ 𝟎. 𝟎𝟗𝟏 and 𝑰𝟐 = 𝒊𝟐 ≈ 𝟎. 𝟒𝟓𝟒𝟓.
𝑰𝟑 = 𝒊𝟐 − 𝒊𝟏 ≈ 𝟎. 𝟒𝟓𝟒𝟓 − 𝟎. 𝟎𝟗𝟏 ≈ 𝟎. 𝟑𝟔 𝑨
The original assumption about downward direction of the branch current 𝑰𝟑 was correct.
15
Supermesh
Supermesh
Mesh#1
Mesh#2
• 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;
• The two meshes must be joined together, resulting in a supermesh;
• The supermesh is constructed by merging the two meshes and excluding
the shared source and any elements in series with it;
• KCL should be applied to one node of the supermesh.
16
Supermesh
Example
• Apply KVL to the supermesh:
−𝟐𝟎 + 𝟔𝒊𝟏 + 𝟏𝟎𝒊𝟐 + 𝟒𝒊𝟐 = 𝟎 ⟹
or 𝟔𝒊𝟏 + 𝟏𝟒𝒊𝟐 = 𝟐𝟎
• Apply KCL to the node “0” in the branch where the two meshes
intersect:
𝟔𝒊𝟏 + 𝟏𝟒𝒊𝟐 = 𝟐𝟎
𝒊𝟐 = 𝒊𝟏 + 𝟔
⟹ ቐ
in
out
𝒊𝟐 = 𝒊𝟏 + 𝟔
• By solving the system of two equations we get:
𝒊𝟏 = −𝟑. 𝟐𝐀 𝒊𝟐 = 𝟐. 𝟖𝐀 ⟹ Initial assumption for the direction of the
mesh current 𝒊𝟏 should be corrected.
17
Supermesh
Example
3.2A
6A
2.8A
𝒊𝟏 = −𝟑. 𝟐𝐀 𝒊𝟐 = 𝟐. 𝟖𝐀 ⟹ Initial assumption for the direction of the
mesh current 𝒊𝟏 should be corrected.
18
Circuit Theorems.
Outline:
1) Linearity Property;
2) Superposition;
3) Source Transformation;
4) Thevenin’s Theorem;
5) Norton’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
â–ª
â–ª
â–ª
The homogeneity property requires that if the input is
multiplied by a constant “𝑘”, then the output is
multiplied by the same constant:
𝑘 ∙ 𝑖𝑅 = 𝑘 ∙ 𝑉.
The additivity property requires that the response to a
sum of inputs is the sum of the response to each input
applied separately:
𝑉 = 𝑖1 + 𝑖2 𝑅 = 𝑖1 𝑅 + 𝑖2 𝑅 = 𝑉1 + 𝑉2 .
In general a linear circuit is one whose output is
linearly related (or directly proportional) to its input.
2
Superposition
• 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.
• Steps:
1) Turn off all independent sources except one source. Find the output
(voltage or current) due to that active source;
• Replace a voltage source by short circuit.
• 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.
• Disadvantage: involve more work.
• Advantage: reduce a complex circuit to simpler circuits.
3
2) Switch –OFF 25V voltage source.
𝑣𝑥2
𝒊𝟏
0.1𝑣𝑥2
𝒊𝟐
KCL: 𝒊𝟏 + 𝒊𝟐 = 𝟓 + 𝟎. 𝟏𝒗𝒙𝟐 ;
𝒗𝒙𝟐 𝒗𝒙𝟐
+
= 𝟓 + 𝟎. 𝟏𝒗𝒙𝟐 ;
𝟐𝟎
𝟒
𝒗𝒙𝟐 = 𝟐𝟓𝑽.
1) Switch –OFF 5A current source.
𝑣𝑥1
𝒊𝟏
𝒊𝟐
0.1𝑣𝑥1
KCL: 𝒊𝟏 + 𝟎. 𝟏𝒗𝒙𝟏 = 𝒊𝟐 ;
𝟐𝟓 − 𝒗𝒙𝟏
𝒗𝒙𝟏
+ 𝟎. 𝟏𝒗𝒙𝟏 =
;
𝟐𝟎
𝟒
𝒗𝒙𝟏 = 𝟔. 𝟐𝟓𝑽.
3) Superposition:
𝒗𝒙 = 𝒗𝒙𝟏 + 𝒗𝒙𝟐 = 𝟔. 𝟐𝟓 + 𝟐𝟓 = 𝟑𝟏. 𝟐𝟓𝑽
4
Source Transformation
𝒊𝒔
𝒗𝒂𝒃 is a voltage between
terminals “a” and “b”
provided by the voltage
source 𝒗𝒔 and resistor R.
We should be able to find
a current source 𝒊𝒔 , which is going
to provide the same voltage drop
𝒗𝒂𝒃 on the resistor R.
Similarly, to the ∆-Y transformation for the resistors’ 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 𝑹𝒂𝒃 .
𝒊𝒂𝒃
𝑹𝒂𝒃 𝑣𝑎𝑏
(a)
𝑹𝒂𝒃 𝑣𝑎𝑏
(b)
(a)⟹ Voltage division:
𝒗𝒂𝒃
𝑹𝒂𝒃
𝑹𝒂𝒃
=
⟹ 𝒗𝒂𝒃 = 𝒗𝒔
;
𝒗𝒔
𝑹 + 𝑹𝒂𝒃
𝑹 + 𝑹𝒂𝒃
(b) Let’s use a current source with
𝒗
𝒊𝒔 = 𝑹𝒔 ⇒ 𝒗𝒔 = 𝒊𝒔 𝑹;
𝑹𝒂𝒃
𝑹𝒂𝒃 𝑹
⟹ 𝒗𝒂𝒃 = 𝒊𝒔 𝑹
= 𝒊𝒔
;
𝑹 + 𝑹𝒂𝒃
𝑹 + 𝑹𝒂𝒃
𝒊𝒔 =
𝒗
(a)⟹ 𝒊𝒂𝒃 = 𝑹+𝑹𝒔
𝒂𝒃
=
𝒗𝒔
𝑹
𝑹
= 𝒊𝒔 𝑹+𝑹 .
𝒂𝒃
𝒗𝒔 = 𝒊𝒔 𝑹
𝑹𝒂𝒃 𝑹
𝒗𝒂𝒃 = 𝒊𝒔
𝑹 + 𝑹𝒂𝒃
For the circuit with the voltage source and resistor R:
𝒊𝒂𝒃 = 𝒊𝒔
𝑹
𝑹 + 𝑹𝒂𝒃
6
Circuit analysis with a load 𝑅𝑎𝑏 .
𝑹𝒂𝒃 𝑣𝑎𝑏
(a)
𝑹𝒂𝒃 𝑣𝑎𝑏
(b)
𝒗
(b) Let’s use a current source 𝒊𝒔 that 𝒊𝒔 = 𝑹𝒔 ⟹
⟹ 𝒗𝒔 = 𝒊𝒔 𝑹;
𝑹
(b) ⟹Current division: 𝒊𝒂𝒃 = 𝒊𝒔 𝑹+𝑹 ;
𝒂𝒃
For the circuit with the current source and resistor R:
𝑹
𝑹
𝒂𝒃
Ohm’s law: 𝒗𝒂𝒃 = 𝑹𝒂𝒃 𝒊𝒂𝒃 = 𝒊𝒔 𝑹+𝑹
𝒂𝒃
𝑹𝒂𝒃 𝑹
𝒗𝒂𝒃 = 𝒊𝒔
𝑹 + 𝑹𝒂𝒃
For the circuit with the voltage source and resistor R:
𝒊𝒂𝒃 = 𝒊𝒔
𝑹
𝑹 + 𝑹𝒂𝒃
7
• Source transformation also applies to dependent sources. The
transformation between the dependent sources:
• Note that the arrow of the current source is directed towards the
positive terminal of the voltage source.
𝒗𝒔
𝑹
• Statement that “Let’s use a current source 𝒊𝒔 =
⟹ 𝒗𝒔 =
𝒊𝒔 𝑹 “ determines main two formulas for the source
transformation.
• The transformation is not possible for 𝑹 = 𝟎 because 𝒊𝒔 =
𝒗𝒔
𝟎
⟶ ∞;
• 𝑹 = ∞ is also prohibited because 𝒗𝒔 = 𝒊𝒔 ∙ ∞ ⟶ ∞.
8
1)
𝟐
𝑰
𝟓 𝒙
2)
𝟐𝟒 ∙
𝟏𝟎−𝟑
𝟐
− 𝑰𝒙
𝟓
𝟓𝜴
𝟏𝟎𝜴
𝟓𝜴
3)
+
𝟏𝟐𝟎 ∙ 𝟏𝟎−𝟑 − 𝟐𝑰𝒙 −
𝟏𝟎𝜴
𝑰𝒙
KVL:
− 𝟏𝟐𝟎 ∙ 𝟏𝟎−𝟑 − 𝟐𝑰𝒙 + 𝟓𝑰𝒙 + 𝟏𝟎𝑰𝒙 = 𝟎
𝑰𝒙
𝑰𝒙 = 𝟕. 𝟎𝟓𝟖𝟖 ∙ 𝟏𝟎−𝟑 ≈ 𝟕. 𝟎𝟔 𝒎𝑨
9
Thevenin’s Theorem
• Thevenin’s theorem states that a linear two-terminal circuit can be
replaced by an equivalent Thevenin’s circuit consisting of a voltage
source 𝑽𝑻𝒉 in series with a resistor 𝑹𝑻𝒉, where 𝑽𝑻𝒉 is the opencircuit voltage at the terminals “a” and “b” and 𝑹𝑻𝒉 is the input (or
equivalent) resistance at the terminals “a” and “b” when the
independent source 𝑽𝑻𝒉 is turned -off.
Variable load
Thevenin’s circuit
10
• Source transformation, application of probe voltage (or current)
source and other techniques help to simplify a circuit to the
Thevenin’s equivalent circuit contained of voltage source 𝑉𝑇ℎ and
resistor 𝑅𝑇ℎ connected in the series network.
• Consider that a linear circuit is loaded by the variable resistor 𝑅𝐿.
• An advantage to use Thevenin’s circuit: The currents 𝐼𝐿 and voltages
𝑉𝐿 across the different loads 𝑅𝐿 can be easily determined by using
Thevenin’s equivalent circuit with the effective voltage source 𝑉𝑇ℎ and
resistor 𝑅𝑇ℎ .
⟹
IL =
Ohm’s law:
VTh
RTh + RL
RL
VL = RL I L =
VTh
RTh + RL
11
To find RThevenin consider two possible case scenarios:
• Case1: If the circuit has no dependent source, one should turn -off all
independent sources, then 𝑅𝑇ℎ is the effective resistance of the
remaining resistor network measured between terminals “a” and “b”.
• 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
𝑖0=1A) to the terminals “a” and “b” and determine the current 𝑖0
(voltage 𝑣0). Thevenin’s resistance can be found by using Ohm’s law:
𝑹𝑻𝒉
If 𝒗𝒐 = 𝟏𝑽 then 𝑹𝑻𝒉 =
𝟏
𝒊𝒐
𝒗𝒐
=
𝒊𝒐
If 𝒊𝒐 = 𝟏𝑨 then 𝑹𝑻𝒉 =
𝒗𝒐
𝟏
12
Let’s apply the source transformation to
Thevenin’s equivalent circuit:
Thevenin’s
Equivalent
circuit
𝑽𝑻𝒉
Norton’s
Equivalent
circuit
𝑹𝑻𝒉 = 𝑹𝑵
⟺
𝒗𝒂 = 𝑰𝑵 𝑹𝑵 = 𝑽𝑻𝒉
⟸ 𝑽𝑻𝒉 = 𝑰𝑵 𝑹𝑵
𝑰𝑵 =
𝑽𝑻𝒉
𝑹𝑻𝒉
Potential difference between terminals “a” and “b” is 𝑽𝑻𝒉 .
13
𝒗𝟏 = 𝟏𝟖𝟎𝑽
𝒗𝟐
𝒗𝟑 = 𝑽𝑻𝒉
2)
𝑰𝟏
𝑰𝟐
𝑰𝟑
𝑲𝑪𝑳
𝑲𝑪𝑳
1)
3)
= 𝑹𝒂𝒃
14
Norton’s Theorem
• Norton’s theorem states that a linear two-terminal circuit can
be replaced by an equivalent circuit consisting of a current
source 𝑰𝑵 in parallel with a resistor 𝑹𝑵 , where 𝑰𝑵 is the shortcircuit current through the terminals “a” and “b” and 𝑹𝑵 is
the input or equivalent resistance at the terminals when the
independent sources are turned -off.
15
• The Norton’s resistance 𝑹𝑵 is equivalent (or input) resistance
at terminals “a” and “b” when all independent sources are
turned –off, thus 𝑹𝑵 = 𝑹𝑻𝒉 .
• The Norton’s current source 𝑰𝑵 can be found by using the shortcircuit current flowing from terminal “a” to “b”, where 𝑰𝑵 = 𝒊𝒔𝒄 .
𝒊𝒔𝒄 = 𝑰𝑵
In this particular case the resistor 𝑹𝑵 is shorten, thus
𝒊𝑹𝑵 = 𝟎.
16
Terminology for Thevenin’s and Norton’
Transformations
• The open-circuit voltage 𝒗𝒐𝒄 across terminals “a” and “b”:
𝒗𝒐𝒄 = 𝑽𝑻𝒉 ;
• The short-circuit current 𝒊𝒔𝒄 at terminals “a” and “b”:
𝒊𝒔𝒄 = 𝑰𝑵 ;
• The equivalent or input resistance 𝑹𝒊𝒏 at terminals “a” and “b”
when all independent sources are turned off:
𝑹𝒊𝒏 = 𝑹𝑻𝒉 = 𝑹𝑵 .
• It often occurs that an effective resistance 𝑹𝑻𝒉 ( or 𝑹𝑵 ) 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) 𝑹𝑵 = 𝟓||(𝟖 + 𝟒 + 𝟖) = 𝟓||𝟐𝟎 =
𝟐𝟎 ∙ 𝟓
= 𝟒𝜴
𝟐𝟓
2) To find 𝑰 𝑵, one should short 𝐭𝐡𝐞 circuit
terminals a and b.
Ignore the 5𝜴 resistor because it was
short−circuited.
𝐁𝐲 applying mesh analysis, one can get
𝒊𝟏 = 𝟐𝑨,
𝟖 + 𝟖 𝒊𝟐 − 𝟏𝟐 + 𝟒 𝒊𝟐 − 𝒊𝟏 = 𝟎
𝑺𝒐 𝒊𝟐 = 𝟏𝑨 = 𝒊𝒔𝒄 = 𝑰𝑵
18
Maximum Power Transfer
• 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.
• Thevenin’s equivalent circuit can be used to find the
condition for maximum power transferred to the load.
• One can see that if 𝑹𝑳 approaches 0 or  the power
transferred goes to zero.
𝟐
𝑷𝑳 =
𝒊𝟐 𝑹𝑳
𝑽𝑻𝒉
𝑽𝑻𝒉
= 𝒊=
=
𝑹𝑻𝒉 + 𝑹𝑳
𝑹𝑻𝒉 + 𝑹𝑳
𝑹𝑳 ;
𝒍𝒊𝒎
𝑷 = 𝟎;
𝑹𝑳 ⟶ 𝟎 𝑳
𝑽𝟐𝑻𝒉
𝒍𝒊𝒎
𝑷 ≈
= 𝟎;
𝑹𝑳 ⟶ ∞ 𝑳
∞
19
Maximum power transferred if 𝑹𝑳 = 𝑹𝑻𝑯
FYI:
𝒅 𝒇𝟏 𝒙
𝒅𝒙 𝒇𝟐 𝒙
=
𝒇′𝟏 𝒙 𝒇𝟐 𝒙 −𝒇𝟏 𝒙 𝒇′𝟐 𝒙
𝒇𝟐𝟐 𝒙
Let’s find when the power transferred to 𝑹𝑳 has an extreme value:
𝒅𝑷
𝒅
= 𝑽𝟐𝑻𝒉
𝒅𝑹𝑳
𝒅𝑹𝑳
𝑹𝑳
𝑹𝑻𝒉 + 𝑹𝑳
𝟐
=
= 𝑽𝟐𝑻𝒉
𝑹𝑻𝒉 + 𝑹𝑳 𝟐 − 𝟐𝑹𝑳 𝑹𝑻𝒉 + 𝑹𝑳
=
𝑹𝑻𝒉 + 𝑹𝑳 𝟒
𝑹𝟐𝑻𝒉 −𝑹𝟐𝑳
𝟐
𝑽𝑻𝒉
𝑹𝑻𝒉 +𝑹𝑳 𝟒
= 𝑽𝟐𝑻𝒉
𝑹𝑻𝒉 −𝑹𝑳
𝑹𝑻𝒉 +𝑹𝑳 𝟑
⟹
𝒅𝑷
𝒅𝑹𝑳
= 𝟎 𝐢𝐟 𝑹𝑻𝒉 = 𝑹𝑳 .
Let’s check if this extremum is a maximum:
𝒅𝟐 𝑷
𝒅
𝟐
𝟐 = 𝑽𝑻𝒉 𝒅𝑹
𝒅𝑹𝑳
𝑳
𝑹𝑻𝒉 − 𝑹𝑳
𝑹𝑻𝒉 + 𝑹𝑳
=
=
𝟑
𝟐 𝟐𝑹𝑻𝒉
−𝑽𝑻𝒉
𝟐𝑹𝑻𝒉 𝟔
=−
𝟐 −
𝑽𝑻𝒉
𝑽𝟐𝑻𝒉
𝟖𝑹𝟑𝑻𝒉
𝟑
=
𝑹𝑻𝒉 + 𝑹𝑳
𝟑
− 𝟑 𝑹𝑻𝒉 − 𝑹𝑳 𝑹𝑻𝒉 + 𝑹𝑳
𝑹𝑻𝒉 + 𝑹𝑳 𝟔
𝟐
= 𝑹𝑳 = 𝑹𝑻𝒉 =
< 𝟎 ⟹ this extremum is maximum. 20 Value of Maximum Power 𝑷𝑳𝒎𝒂𝒙 = 𝒊𝟐 𝑹𝑳 𝑽𝑻𝒉 𝑽𝑻𝒉 = 𝒊= = 𝑹𝑻𝒉 + 𝑹𝑳 𝑹𝑻𝒉 + 𝑹𝑳 𝟐 𝑹𝑳 = 𝑽𝟐𝑻𝒉 = 𝑹𝑳 = 𝑹𝑻𝒉 = , 𝟒𝑹 𝑻𝒉 Thus, Maximum power is transferred to the load 𝑅𝐿 when that load resistance is equal to the Thevenin resistance 𝑅𝑇ℎ . 21 1) 𝒗𝒐𝒙 𝒊𝟎 𝒊𝒐𝟏 𝒊𝒐𝟐 𝒗𝒐𝒙 KVL in the mesh #1: 𝟔𝟎𝒊𝒐𝟏 + 𝟑𝟎 𝒊𝒐𝟏 − 𝒊𝒐𝟐 + 𝟑𝒗𝒐𝒙 = 𝟎 ൞ ⟹ 𝒊𝒐𝟐 = 𝟗𝒊𝒐𝟏 Ohm’s law: 𝒗𝒐𝒙 = 𝟔𝟎𝒊𝒐𝟏 KVL in the mesh #2: 𝟏 − 𝟑𝒗𝒐𝒙 + 𝟑𝟎 𝒊𝒐𝟐 − 𝒊𝒐𝟏 + 𝟏𝟐𝟎𝒊𝒐𝟐 = 𝟎 ; 𝟐𝟏𝟎𝒊𝒐𝟏 − 𝟏𝟓𝟎𝒊𝒐𝟐 = 𝟏 ൞ Mesh #1: 𝒊𝒐𝟐 = 𝟗𝒊𝒐𝟏 𝑹𝑻𝒉 = 𝟏 𝟗 ⟹ 𝒊𝒐𝟏 = − 𝟏𝟏𝟒𝟎 𝑨 , 𝒊𝒐𝟐 = − 𝟏𝟏𝟒𝟎 𝑨. 𝑽𝒐 𝑽 = 𝟏𝑽 = 𝒐 = 𝟏𝟐𝟔. 𝟔 𝟔 𝜴 𝒊𝒐 = −𝒊𝒐𝟐 𝒊𝒐 22 𝑽𝑻𝒉 2) 𝒊𝟏 𝒊𝟐 KVL in the mesh #1: −𝟗 + 𝟔𝟎𝒊𝟏 + 𝟑𝟎 𝒊𝟏 − 𝒊𝟐 + 𝟑𝒗𝒙 = 𝟎 𝟏 ൞ ⟹ 𝒊𝟏 = 𝑨. 𝟑𝟎 𝒗𝒙 = 𝟔𝟎𝒊𝟏 𝐚𝐧𝐝 𝒊𝟐 = 𝟎 KVL in the mesh #2: −𝟑𝒗𝒙 + 𝟑𝟎 𝒊𝟐 − 𝒊𝟏 + 𝟏𝟐𝟎𝒊𝟐 + 𝑽𝑻𝒉 = 𝟎 ൞ 𝒊𝟐 = 𝟎 𝒂𝒏𝒅 𝒊𝟏 = 𝑷𝑳𝒎𝒂𝒙 𝟏 𝟑𝟎 ⟹ 𝑽𝑻𝒉 = 𝟕 𝑽. 𝑨; 𝑽𝟐𝑻𝒉 𝑽𝑻𝒉 = 𝟕𝑽 = = = 𝟎. 𝟎𝟗𝟔𝟕𝟏 = 𝟗𝟔. 𝟕𝟏𝐦𝐖. 𝑹𝑻𝒉 = 𝟏𝟐𝟔. 𝟔 𝟔 𝜴 𝟒𝑹𝑻𝒉 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: 𝑰𝟐𝒐 𝑰𝟐𝒐 ഥ𝟐 = 𝑰ഥ𝟐 𝑹 = ∙𝑹= = 𝑰 𝟐 𝟐 𝑽𝟐𝒐 𝟏 𝑽𝟐 𝑽𝟐𝒐 𝟐 = = ∙ = = 𝑽 𝟐 𝑹 𝑹 𝟐 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. 𝑂ℎ𝑚′ 𝑠 𝑙𝑎𝑤: 𝑈(𝑡) = 𝐼(𝑡) ∙ 𝑅 8 AC Circuit with a Capacitor only In a capacitor, the voltage VC lags the current by 90°. charging discharging 𝑑𝑉 𝐼=𝐶 𝑑𝑡 U charging 90o U discharging 9 AC Circuit with Inductance only The voltage on the inductor VL leads current through an inductor by 90°. 90o charging discharging U U charging U discharging 10 LRC Series AC Circuit Only one current 𝑰 exists in this serial circuit, however there are three voltages 𝑽𝑹 , 𝑽𝑳 , 𝑽𝑪 and three effective resistances (ratio of voltage to current), called the resistance R and reactances XL and XC , where 𝑉𝑅 = 𝐼 ∙ 𝑅, 𝑉𝐿 = 𝐼 ∙ 𝑋𝐿 and 𝑉𝐶 = 𝐼 ∙ 𝑋𝐶 . Note that both reactances are frequency depend: 11 LRC Series AC Circuit (Phasor diagram) The voltage on the inductor VL leads current by 90°. 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 𝐼𝑜 is an amplitude of that AC current. The voltage VC lags the current by 90°. 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 𝑉𝑅,𝐿,𝐶 𝑡 across each device are given by the xcomponent of each, and the current 𝐼 𝑡 by its x-component.
The current 𝐼 𝑡 is the same throughout the circuit.
𝐼 = 𝐼0 cos 2𝜋𝑓𝑡
𝑉 = 𝑉𝑅 + 𝑉𝐿 + 𝑉𝐶
(KVL)
13
Impedance 𝑍 𝝎 of LRC Circuit
y
V0
VR0
VL0-VC0
x
By using the Pythagorean theorem one can find that
𝑉0 =
2
𝑉𝑅0
+ 𝑉𝐿0 − 𝑉𝐶0
2
= 𝐼0
𝑅2
+ 𝑋𝐿 𝜔 − 𝑋𝐶 𝜔
𝑉0 = 𝑍 𝜔 ∙ 𝐼0
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 𝑅 =
𝑈
=Constant
𝐼
15
Current in AC Circuits
The rms current in an ac circuit is:
Clearly, Irms is a frequency depend current: 𝐼𝑟𝑚𝑠 = 𝐼𝑟𝑚𝑠 𝑓
16
Current Resonance in AC Circuits
We see that Irms demonstrates a maximum
𝑋𝐶 = 𝑋𝐿
when XC = XL;
the frequency at which this occurs is
this is called the resonant frequency.
𝑋𝐿 → ∞
𝑋𝐶 → 0
𝑋𝐿 → 0
𝑋𝐶 → ∞
17
Resonance in LC Circuits
In a capacitor, the
voltage VC lags the
current by 90°.
The voltage on the
inductor VL leads
current through an
inductor by 90°.
https://en.wikipedia.org/wiki/LC_circuit
18
Frequency Spectrum of Signals
𝒗𝒂 𝒕 = 𝑽𝒂 𝒔𝒊𝒏 𝝎𝒐 𝒕 is the time dependent voltage with peak amplitude 𝑽𝒂 and
𝟏
angular frequency 𝝎𝒐 = 𝟐𝝅𝒇𝒐 , 𝒓𝒂𝒅/𝒔 , where 𝒇𝒐 = 𝑻 , 𝑯𝒛 is a temporal frequency
𝒐
and 𝑻𝒐 , 𝒔 is period of the sinusoid. This signal might be analyzed by using a
frequency spectrum calculated by using Fourier transformation:
𝑺 𝝎 = 𝑭 𝒇 𝒕 is Fourier spectrum (or spectral density) of the signal 𝒇 𝒕 =
∞
𝟏 ∞
𝑭−𝟏 𝑺 𝝎 , where 𝑭 = ‫׬‬−∞ … 𝒆−𝒊𝝎𝒕 𝒅𝒕 and 𝑭−𝟏 = 𝟐𝝅 ‫׬‬−∞ … 𝒆𝒊𝝎𝒕 𝒅𝒕 are forward
and inverse Fourier transformations, respectively.
Thus, 𝑺𝒂 𝝎 =
∞ 𝒆𝟐𝝅𝒊𝒇𝒐 𝒕 −𝒆−𝟐𝝅𝒊𝒇𝒐 𝒕
𝑽𝒂 ‫׬‬−∞
𝟐𝒊
𝒊
∞
𝒆−𝟐𝝅𝒇𝒕 𝒅𝒕 = 𝟐 ‫׬‬−∞ −𝒆−𝟐𝝅𝒊
𝒇−𝒇𝒐 𝒕
+ 𝒆−𝟐𝝅𝒊
𝒇+𝒇𝒐 𝒕
𝒊
= 𝟐 𝜹 𝝎 + 𝝎𝒐 − 𝜹 𝝎 − 𝝎𝒐 .
19
=
Fourier series
A symmetrical square-wave signal of
amplitude V.
The frequency spectrum (also known as
the line spectrum).
+∞
𝒗 𝒕 =
where
𝒗𝒐
+ ෍ (𝒂𝒌 𝒄𝒐𝒔 𝒌𝟐𝝅𝒇𝒕 + 𝒃𝒌 𝒔𝒊𝒏 𝒌𝟐𝝅𝒇𝒕 ) ,
𝟐
𝒌=𝟏
𝟏 𝝅
𝒗𝒐 = 𝝅 ‫׬‬−𝝅 𝒇
𝟏
𝝅
𝟏
𝝅
𝒕 𝒅𝒕, 𝒂𝒌 = 𝝅 ‫׬‬−𝝅 𝒇 𝒕 𝒄𝒐𝒔 𝒏𝒕 𝒅𝒕 , 𝒃𝒌 = 𝝅 ‫׬‬−𝝅 𝒇 𝒕 𝒔𝒊𝒏 𝒏𝒕 𝒅𝒕
and 𝒌𝟐𝝅𝒇 = 𝒌𝝎𝒐 is an angular frequency of 𝒌 harmonic oscillation.
𝟒𝑽
𝟏 𝟒𝑽
𝟏 𝟒𝑽
𝟏 𝟒𝑽
𝒗 𝒕 =
𝒔𝒊𝒏 𝝎𝒐 𝒕 + ∙
𝒔𝒊𝒏 𝟑𝝎𝒐 𝒕 + ∙
𝒔𝒊𝒏 𝟓𝝎𝒐 𝒕 + ∙
𝒔𝒊𝒏 𝟕𝝎𝒐 𝒕 + ⋯ ,
𝝅
𝟑 𝝅
𝟓 𝝅
𝟕 𝝅
where 𝝎𝒐 = 𝟐𝝅𝒇 is called the Fundamental frequency
20
Examples
E1.5. Find the frequencies f and ω of a sine-wave signal with a period of 1 ms.
Answer: 𝒇 =
𝟏
𝑻
=
𝟏
𝟏∙𝟏𝟎−𝟑
= 𝟏𝟎𝟎𝟎 = 𝟏𝒌𝑯𝒛 and 𝝎 = 𝟐𝝅𝒇 = 𝟐𝝅 ∙ 𝟏𝟎𝟑 𝒓𝒂𝒅/𝒔.
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: ∆𝑭 = 𝟔𝟎𝟖 − 𝟒𝟕𝟎 = 𝟏𝟑𝟖 𝑴𝑯𝒛 ⟹ 𝑵 =
∆𝑭
∆𝒇
=
𝟏𝟑𝟖
𝟔
= 𝟐𝟑
Channels from 14 to 36:
14, 15, 16,…, 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
𝟏 𝑻 𝒗𝟐
𝒅𝒕 or
‫׬‬
𝑻 𝟎 𝑹
calculated directly using the relationship 𝑷 =
indirectly by
summing the contribution of each of the harmonic components, that is,
𝑷 = 𝑷𝟏 + 𝑷𝟑 + 𝑷𝟓 + … , 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
• 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.
• 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.
• OPA had their origins in analog computers, where they were used to
perform mathematical operations in many linear, non-linear, and
frequency-dependent circuits.
• 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;
• OPA have a relatively little dependence on temperature.
• 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
+𝑽𝒄𝒄
−𝑽𝒄𝒄
Type equation here.
+𝑽𝒄𝒄
−𝑽𝒄𝒄
(a) typical pin configuration, (b) circuit symbol.
4
+𝑽𝒄𝒄
+𝑽𝒄𝒄
−𝑽𝒄𝒄
−𝑽𝒄𝒄
5
OPA output voltage vo as a function of
the differential input voltage vd :
The output voltage is limited by source 𝑽𝒄𝒄 .
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: 𝒗𝒅 = 𝒗𝟐 − 𝒗𝟏
Output: 𝒗𝒐 = 𝑨𝒗𝒅 = 𝑨 𝒗𝟐 − 𝒗𝟏
Note:
A𝒗𝒅 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
∞
Input resistance
𝑹𝒊
103 — 1013 𝜴
∞
Output resistance
𝑹𝒐
10 — 100 𝜴
0
Supply Voltage
𝑼𝒄𝒄
±5, ±𝟏𝟓, ±𝟐𝟒 𝑽
7
Ideal OPA
𝐴 ⟶ ∞, 𝑅𝑖 ⟶ ∞, 𝑅𝑜 ≈0
Fig. 5.8
Fig. 5.8 and 5.4
𝒗𝒅
𝑹𝒊 ⟶ ∞ ⟹ 𝒊𝟏 = 𝒊𝟐 =
≈𝟎
𝑹𝒊
OPA is designed to amplify a small signal,
thus 𝒗𝒅 ⟶ 𝟎 ⟹ 𝒗𝒅 = 𝒗𝟏 − 𝒗𝟐 ≈ 𝟎 𝐚𝐧𝐝 𝒊𝟏 = 𝒊𝟐 ≈ 𝟎 ⟹ 𝒗𝟏 = 𝒗𝟐 ≈ 𝟎
8
Inverting Amplifier
An inverting amplifier reverses the polarity of the input signal
while amplifying it.
𝒗𝒊 − 𝒗𝟏 𝒗𝟏 − 𝒗𝒐
𝒗𝒊 −𝒗𝒐
𝒊𝟏 = 𝒊𝟐 ⟹
=
⟹ 𝒗𝟏 = 𝒗𝟐 ≈ 𝟎 ⟹
=
⟹
𝑹𝟏
𝑹𝒇
𝑹𝟏
𝑹𝒇
⟹ 𝒗𝟎 = −
𝑹𝒇
𝑹𝟏
𝒗𝒊 and 𝑨𝒗 =
𝒗𝒐
𝒗𝒊
=−
𝑹𝒇
𝑹𝟏
, 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
𝒊𝟐
𝒊𝟏
0A
𝑹𝒇
𝟐𝟖𝟎 ∙ 𝟏𝟎𝟑
−𝟑 = −𝟑. 𝟏𝟓𝑽
𝒗𝟎 = −
𝒗𝒊 = −
∙
𝟒𝟓
∙
𝟏𝟎
𝑹𝟏
𝟒 ∙ 𝟏𝟎𝟑
𝒊𝟏 = 𝒊𝟐 , 𝒗𝟏 ≈ 𝒗𝟐 ≈ 𝟎 ⟹
𝟒𝟓 ∙ 𝟏𝟎−𝟑 − 𝟎
𝒊𝟏 =
= 𝟏𝟏. 𝟐𝟓𝝁𝑨
𝟒 ∙ 𝟏𝟎𝟑
𝟎 − −𝟑. 𝟏𝟓
𝒊𝟐 =
= 𝟏𝟏. 𝟐𝟓𝝁𝑨
𝟐𝟖𝟎 ∙ 𝟏𝟎𝟑
Gain: 𝑨𝒗 =
𝒗𝒐
𝒗𝒊
=−
𝑹𝒇
𝑹𝟏
=
−𝟑.𝟏𝟓
𝟒𝟓∙𝟏𝟎−𝟑
=
𝟐𝟖𝟎∙𝟏𝟎𝟑
−
𝟒∙𝟏𝟎𝟑
= −𝟕𝟎
10
Noninverting Amplifier
A noninverting amplifier is an op amp circuit designed to
provide a positive voltage gain.
𝑹𝒇
𝟎 − 𝒗𝟏 𝒗𝟏 − 𝒗𝒐
𝒗𝟏 𝒗𝟏 − 𝒗𝒐
𝒊 𝟏 = 𝒊𝟐 ⟹
=
⟹−
=
⟹ 𝒗𝒐 = 𝟏 +
𝒗
𝑹𝟏
𝑹𝒇
𝑹𝟏
𝑹𝒇
𝑹𝟏 𝒊
𝑹𝒇
𝒗𝒐
Thus, the gain depends only on the external resistors 𝑹𝟏 𝐚𝐧𝐝 𝑹𝒇 .
𝑨𝒗 =
=𝟏+
𝒗𝒊
𝑹𝟏
If 𝑅𝑓 = 0 and 𝑅1 ⟶ ∞, then 𝐴𝑣 = 1 (such a
circuit is called as a voltage follower that
minimizes interaction between the two stages.
11
Solution by using superposition:
𝒗𝒐 = 𝒗𝒐,𝟔𝑽 + 𝒗𝒐,𝟒𝑽
If 4V is –OFF:
then 𝒗𝟎,𝟔𝑽 = −
𝑹𝒇
𝑹𝟏
𝒗𝒊 =
𝟏𝟎∙𝟏𝟎𝟑
−
𝟔
𝟒∙𝟏𝟎𝟑
= −𝟏𝟓𝑽
If 6V is –OFF:
then 𝒗𝒐,𝟒𝑽 = 𝟏 +
𝑹𝒇
𝑹𝟏
𝒗𝒊 = 𝟏 +
𝟏𝟎∙𝟏𝟎𝟑
𝟒∙𝟏𝟎𝟑
𝟒 = 𝟏𝟒𝑽
𝒗𝒐 = 𝒗𝒐,𝟔𝑽 + 𝒗𝒐,𝟒𝑽 = −𝟏𝟓 + 𝟏𝟒 = −𝟏𝐕
(thus, to get positive 𝒗𝒐 , one should increase 4V
source or decrease 6V source).
Check-up: applying KCL at node a, one can get
𝟔−𝒗𝒂
𝟒∙𝟏𝟎𝟑
=
𝒗𝒂 −𝒗𝒐
𝟏𝟎∙𝟏𝟎𝟑
𝒗𝒅 ≈ 𝟎
⟹
⟹
𝒗𝒂 ≈ 𝒗𝒃 = 𝟒𝑽
𝟔 − 𝟒 𝟒 − 𝒗𝒐
=
⟹ 𝒗𝒐 = −𝟏𝑽
𝟒
𝟏𝟎
12
Example 5.1.
A 741 OPA has an open-loop voltage gain A=2 x 105, input resistance 𝑹𝒊 =2 M and
output resistance 𝑹𝒐 =50 . OPA is used in the circuit below. Find closed-loop gain
𝒗𝒐/𝒗𝑺 . Determine current 𝒊 when 𝒗𝑺 =2V.
Ideal OPA: Inverting Amplifier
𝒗𝒊
𝒗𝒐
𝟐
𝒊𝟏 = 𝒊𝟐 =
=−
=
= 𝟎. 𝟐𝐦𝐀 ⟹ 𝒊 ≈ 𝟎. 𝟐𝒎𝑨
𝑹𝟏
𝑹𝒇 𝟏𝟎 ∙ 𝟏𝟎𝟑
𝑹𝒇
𝒗𝒐
𝟐𝟎 ∙ 𝟏𝟎𝟑
𝑨𝒗 =
=−
=−
= −𝟐 ⟹
𝒗𝒊
𝑹𝟏
𝟏𝟎 ∙ 𝟏𝟎𝟑
𝑨𝒗 ≈ −𝟐
𝒗𝒐 = 𝑨𝒗 𝒗𝒊 = −𝟐 ∙ 𝟐 = −𝟒𝑽 ⟹ 𝒗𝒐 ≈ −𝟒𝑽
13
Example 5.1.
A 741 OPA has an open-loop voltage gain A=2 x 105, input resistance 𝑹𝒊 =2 M and
output resistance 𝑹𝒐 =50 . OPA is used in the circuit below. Find closed-loop gain
𝒗𝒐/𝒗𝑺 . Determine current 𝒊 when 𝒗𝑺 =2V.
14
Node 1: 𝒊𝟏𝟎 = 𝒊𝟐 + 𝒊
𝒗𝒔 − 𝒗𝟏
𝒗𝟏
𝒗𝟏 − 𝒗𝒐
=
+
𝟏𝟎 × 𝟏𝟎𝟑 𝟐 × 𝟏𝟎𝟔 𝟐𝟎 × 𝟏𝟎𝟑
𝒗𝒔 = 𝟐𝑽
𝑨 = 𝟐 ∙ 𝟏𝟎𝟓
𝟐𝟎𝟎𝒗𝒔 = 𝟑𝟎𝟏𝒗𝟏 − 𝟏𝟎𝟎𝒗𝒐
𝟐𝒗𝒔 + 𝒗𝒐
𝟐𝒗𝒔 ≅ 𝟑𝒗𝟏 − 𝒗𝒐 => 𝒗𝟏 =
(𝐄𝐪. 𝟏)
𝟑
𝒗𝟏 − 𝒗𝒐
𝒗𝒐 − 𝑨𝒗𝒅
Node 0: 𝒊 = 𝒊𝟓𝟎 ⟹
=
,
𝟐𝟎 × 𝟏𝟎𝟑
𝟓𝟎
𝑖
𝑖10
𝒗𝑺
𝑣1
𝑖2
𝑨𝒗𝒅
𝑖50
where v𝒅 = −v𝟏 and A = 200 × 𝟏𝟎𝟑
𝑣2
𝒗𝟏 − 𝒗𝒐 = 𝟒𝟎𝟎 𝒗𝒐 + 𝟐 × 𝟏𝟎𝟓 × 𝒗𝟏 (𝐄𝐪. 𝟐)
From Eqs.1 and 2: 𝟐𝟔, 𝟔𝟔𝟔, 𝟎𝟔𝟕𝒗𝒐 ≈ −𝟓𝟑, 𝟑𝟑𝟑, 𝟑𝟑𝟑𝒗𝒔 ⟹
𝒗
The closed−loop gain 𝐢𝐬 𝑨𝒗 = − 𝒗𝒐 = −𝟏. 𝟗𝟗𝟗 and 𝒗𝒐 = 𝑨𝒗 𝒗𝒊 = −𝟏. 𝟗𝟗𝟗 ∙ 𝟐 = −𝟑. 𝟗𝟗𝟖V
𝒔
From (Eq.1) 𝒗𝟏 = 20.0667𝝁V.
The input current for OPA: 𝒊𝟐 =
Thus, 𝒊 =
𝒗𝒅
𝑹𝒊
𝒗𝟏 −𝒗𝒐
𝟐𝟎×𝟏𝟎𝟑
= 𝟎. 𝟏𝟗𝟗𝟗𝒎𝑨
𝒗𝒅 = 𝒗𝟐 − 𝒗𝟏 ,
−𝒗𝟏
𝟐𝟎×𝟏𝟎−𝟔
= 𝐰𝐡𝐞𝐫𝐞 𝒗𝟐 = 𝟎 = 𝑹 = 𝟐×𝟏𝟎𝟔 ≈ 𝟏𝟎−𝟏𝟏 𝑨 ⟶ 𝟎
𝒊
⟹ 𝒗𝒅 = −𝒗𝟏
15
“Y” ⟹ “𝚫“ transformation:
𝑹𝒂 = 𝟐𝟎𝒌𝜴,
𝑹𝟏 ≈ 𝟑. 𝟎𝟖𝒌𝜴,
൞ 𝑹𝒃 = 𝟓𝒌𝜴, ⟹ ቐ𝑹𝟐 ≈ 𝟏𝟐. 𝟑𝟏𝒌𝜴,
𝑹𝟑 ≈ 𝟏. 𝟓𝟒𝒌𝜴.
𝑹𝒄 = 𝟒𝟎𝒌𝜴,
𝐴𝑣
Ideal OPA: Noninverting Amplifier
𝑹𝒇 = 𝟏𝟐. 𝟑𝟏𝒌𝜴
𝑹𝟏 = 𝟏. 𝟓𝟒𝒌𝜴
𝑹𝒇
𝒗𝒐
𝟏𝟐. 𝟑𝟏 ∙ 𝟏𝟎𝟑
𝑨𝒗 =
=𝟏+
=𝟏+
≈ 𝟖. 𝟗𝟗
𝒗𝒊
𝑹𝟏
𝟏. 𝟓𝟒 ∙ 𝟏𝟎𝟑
𝒗𝒔 = 𝒗𝒊 = 𝟏𝑽 ⟹ 𝒗𝒐 = 𝑨𝒗 𝒗𝒊 ≈ 𝟖. 𝟗𝟗𝑽
16
Given: 𝒗𝑺 = 𝟏𝑽; 𝑹𝒊 = 𝟐𝑴𝜴,
𝑨 = 𝟐 × 𝟏𝟎𝟓 , 𝑹𝒐 = 𝟓𝟎𝜴 .
Find: 𝑨𝒗 , 𝒊𝑶 .
𝐴𝑣
17
KCL in the node 𝒗𝟏 :
𝒊𝑹𝒊 = 𝒊𝟒𝟎 + 𝒊𝟓 ⟹
𝒗𝒔 − 𝒗𝟏
𝒗𝟏 − 𝒗𝒐
𝒗𝟏
𝒗𝑺 + 𝟓𝟎𝒗𝒐
=
+
⟹ 𝒗𝟏 =
𝑹𝒊
𝟒𝟎 × 𝟏𝟎𝟑 𝟓 × 𝟏𝟎𝟑
𝟒𝟓𝟏
Given: 𝒗𝑺 = 𝟏𝑽; 𝑹𝒊 = 𝟐𝑴𝜴,
𝑨 = 𝟐 × 𝟏𝟎𝟓 , 𝑹𝒐 = 𝟓𝟎𝜴 .
Find: 𝑨𝒗 , 𝒊𝑶 .
𝒗𝒔 = 𝒗𝟐
𝑨𝒗𝒅
𝑖𝑅𝑖
KCL in the node 𝒗𝒐 :
𝒊𝟒𝟎 + 𝒊𝒐 = 𝒊𝟐𝟎 ⟹
𝒗𝟏 − 𝒗𝒐
𝑨𝒗𝒅 − 𝒗𝒐
𝒗𝒐
+
=
,
𝟒𝟎 × 𝟏𝟎𝟑
𝟓𝟎
𝟐𝟎 × 𝟏𝟎𝟑
where 𝒗𝒅 = 𝒗𝟐 − 𝒗𝟏 = 𝒗𝑺 − 𝒗𝟏 .
𝒗𝟏
𝒊𝟒𝟎
𝒊𝟓
𝒊𝒐
𝒗𝒐
𝒊𝟐𝟎
⟹ 𝟏𝟔 × 𝟏𝟎𝟕 𝒗𝑺 − 𝟖𝟎𝟑𝒗𝒐 − 𝟏𝟔 × 𝟏𝟎𝟕 − 𝟏 𝒗𝟏 = 𝟎.
18
𝒗𝑺 + 𝟓𝟎𝒗𝒐
𝒗𝟏 =
;
𝟒𝟓𝟏
⟹
𝟏𝟔 × 𝟏𝟎𝟕 𝒗𝑺 − 𝟖𝟎𝟑𝒗𝒐 − 𝟏𝟔 × 𝟏𝟎𝟕 − 𝟏 𝒗𝟏 = 𝟎 ;
Given: 𝒗𝑺 = 𝟏𝑽; 𝑹𝒊 = 𝟐𝑴𝜴,
𝑨 = 𝟐 × 𝟏𝟎𝟓 , 𝑹𝒐 = 𝟓𝟎𝜴 .
Find: 𝑨𝒗 , 𝒊𝑶 .
𝒗𝒔 = 𝒗𝟐
⟹ 𝟏𝟓. 𝟗𝟔𝟒𝟓𝒗𝑺 − 𝟏. 𝟕𝟕𝟑𝟕𝟔𝒗𝒐 = 𝟎 ;
𝒗𝒐
= 𝑨𝒗 = 𝟗. 𝟎𝟎𝟎𝟑𝟕(𝐜𝐥𝐨𝐬𝐞𝐝 𝐥𝐨𝐨𝐩 𝐠𝐚𝐢𝐧).
𝒗𝑺
𝑨𝒗𝒅
If 𝒗𝑺 = 𝟏, 𝐭𝐡𝐞𝐧 𝒗𝒐 = 𝒗𝒔 𝑨𝒗 ≈ 𝟗𝑽.
𝒗𝟏 =
𝒗𝑺 +𝟓𝟎𝒗𝒐
𝟒𝟓𝟏
𝑖𝑅𝑖
≈ 𝟏𝑽.
𝒊𝒐 = 𝒊𝟒𝟎𝐧𝐞𝐰 + 𝒊𝟐𝟎 =
𝒗𝟎 − 𝒗𝟏
𝒗𝒐
=
+
≈
𝟒𝟎 ∙ 𝟏𝟎𝟑 𝟐𝟎 ∙ 𝟏𝟎𝟑
𝟖
𝟗
≈
+
= 𝟔𝟓𝟎𝝁𝑨
𝟒𝟎 ∙ 𝟏𝟎𝟑 𝟐𝟎 ∙ 𝟏𝟎𝟑
𝒊𝟒𝟎
𝒗𝟏 = 𝟏𝑽
𝒊𝟓
𝒊𝒐
𝒗𝒐 = 𝟗𝐕
𝒊𝟐𝟎
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 𝑽𝒊 and
frequency 𝝎. 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
𝑻(𝝎), then
𝑻 𝝎
=
𝑽𝒐 𝝎
𝑽𝒊 𝝎
and ∠𝑻 𝝎 = 𝝓 𝝎 .
Plots of “ 20log 𝑻 𝝎 vs. 𝝎 “ and “ ∠𝑻 𝝎 vs. 𝝎 “ 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 𝝎𝟏 and 𝝎𝟐 . Signals whose frequencies
are below 𝝎𝟏 or above 𝝎𝟐 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.
𝑻 𝝎 =
𝑽𝒐 𝝎
𝑽𝒊 𝝎
is a complex function called as “amplifier transfer function”, where 𝑽𝒊 𝝎 and
𝑽𝒐 𝝎 denote the input and output signals, respectively. A similar description might be
provided by using by using the complex frequency variable 𝒔 = 𝒊𝝎 = 𝒋𝝎. Thus, one might see
that 𝑻 𝒔 =
𝑽𝒐 𝒔
𝑽𝒊 𝒔
.
3
Single-Time-Constant (STC) Networks
(a) a low-pass filter, it passes lowfrequency, sine-wave inputs 𝑽𝒊
with little or no attenuation (at
𝝎 = 𝟎, the transmission is unity)
and attenuates high-frequency
input sinusoids.
𝟏
𝒁=
อ
𝒊𝝎𝑪
𝟏
𝒁=
อ
𝒊𝝎𝑪
(b) a high-pass filter, its
transmission for 𝑽𝒊 𝝎 is
unity at 𝝎 = ∞ and
decreases as 𝝎 is reduced,
reaching 0 for 𝝎 = 𝟎.
≈𝟎
𝝎⟶∞
≈∞
𝝎⟶𝟎
4
Cutoff frequency
• 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.
• 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.
• 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
−3 dB of the nominal passband value).
𝑨𝒑 ,
𝑨𝒑 , 𝒅𝑩
𝟏𝟎
=𝒆
−𝟑
𝟏𝟎
= 𝑨𝒑 , = −𝟑 𝒅𝑩 = 𝒆
≈0.74
https://en.wikipedia.org/wiki/Cutoff_frequency
5
• A single-time-constant (STC) network formed of an inductance 𝑳 and a resistance
𝑹 has a time constant 𝝉 = 𝑳/𝑹. The time constant 𝝉 of an STC network composed
of a capacitance 𝑪 and a resistance 𝑹 is given by 𝝉 = 𝑹𝑪.
• Definition: an octave is the interval between one musical pitch and another with
𝒇
double its frequency. Thus, if 1 𝒐𝒄𝒕𝒂𝒗𝒆 = 𝒍𝒐𝒈𝟐 𝟐 , then 𝒇𝟐 = 𝟐𝒇𝟏
𝒇𝟏
• The number of octaves between two frequencies is given by the formula:
𝒇𝟐
𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒄𝒕𝒂𝒗𝒆 = 𝒍𝒐𝒈𝟐
𝒇𝟏
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 𝝎𝒐 , 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 𝑹𝒊 , an input
capacitance 𝑪𝒊 , a gain factor 𝝁, and an output resistance Ro. The amplifier is fed
with a voltage source V𝒔 having a source resistance 𝑹𝒔 , 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
𝟏
𝟏
𝟏
𝟏
𝟏
𝑽𝒊
𝒁𝒊
𝒁𝒊
𝟏
𝒁𝒊
=
=
∙
=
= 𝒁𝒊 = 𝑹𝒊 + 𝟏 = 𝑹𝒊 + 𝒔𝑪𝒊 =
𝑹
𝑽𝒔 𝒁𝒊 + 𝑹𝒔
𝒁𝒊 + 𝑹𝒔 𝟏
𝟏+ 𝒔
𝒔𝑪𝒊
𝒁𝒊
𝒁𝒊
=
𝟏 + 𝑹𝒔
9
𝟏
𝟏
=
⟹
𝑹𝒔
𝟏
𝟏+
+ 𝒔𝑪𝒊 𝑹𝒔
+ 𝒔𝑪𝒊
𝑹𝒊
𝑹𝒊
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
𝑽𝒊
𝟏
⟹
=
=
𝑹𝒔
𝑽𝒔 𝟏 + 𝑹𝒔 + 𝒔𝑪 𝑹
𝟏
+
𝒊 𝒔
𝑹
𝑹
𝒊
𝒊
𝟏
𝒔𝑪𝒊 𝑹𝒔
𝟏+
𝑹
𝟏+ 𝒔
𝑹𝒊
=
𝟏
𝑹
𝟏 + 𝑹𝒔
𝒊
𝒔𝑪 𝑹 𝑹
𝟏 + 𝑹 𝒊+ 𝒔𝑹 𝒊
𝒊
𝒔
Also, by applying the voltage division to output circuit one get that
Thus,
𝑽𝒊
𝑽𝒔
∙
𝑽𝒐
𝝁𝑽𝒊
=
𝑽𝒐
𝝁𝑽𝒔
𝑽𝒐
=𝝁
𝑹
𝑽𝒔
𝟏 + 𝑹𝒔
𝒊
=
𝟏
𝑹
𝟏+ 𝑹𝒔
𝒊
𝒔𝑪 𝑹𝒔 𝑹𝒊
𝟏+ 𝑹𝒊+𝑹
𝒊 𝒔
𝟏
𝒔𝑪 𝑹 𝑹
𝟏 + 𝑹 𝒊+ 𝒔𝑹 𝒊
𝒊
𝒔
∙
𝑹𝑳
𝑹𝒐 +𝑹𝑳
𝑽𝒐
𝝁𝑽𝒊
=
𝑹𝑳
𝑹𝒐 +𝑹𝑳
.
⟹
𝟏
𝑹𝑳
𝑹
∙
∙ 𝑳=
𝑹𝒔
𝑹𝒐 + 𝑹𝑳 𝟏
𝟏
+
𝑹𝑳
𝑹𝒊
𝝁
𝒔𝑪 𝑹 𝑹
𝟏 + 𝑹 𝒊+ 𝒔𝑹 𝒊
𝒊
𝒔
⟹
𝑹
𝟏 + 𝑹𝒐
𝑳
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.
𝑽𝒐
𝟏
𝟏
𝟏
⟹
=𝝁∙
∙
∙
𝑹𝒔
𝑹𝒐
𝒔𝑪 𝑹 𝑹
𝑽𝒔
𝟏+𝑹
𝟏+𝑹
𝟏 + 𝑹 𝒊+ 𝒔𝑹 𝒊
=𝝁∙
𝟏
𝑹
𝟏 + 𝑹𝒔
𝒊
𝒊
𝑳
∙
𝒔 = 𝒊𝝎
𝟏
𝟏
𝟏
𝟏
𝐚𝐧𝐝
∙
=
=
𝝁
∙
∙
∙
.
𝟏
𝑹𝒔
𝑹𝒐
𝝎
𝟏 + 𝒔𝝉
𝝉≅
𝟏+𝑹
𝟏+𝑹
𝟏 + 𝒊𝝎
𝝎𝟎
𝒊
𝑳
𝒐
𝟏
𝑹
𝟏 + 𝑹𝒐
𝑳
𝒊
𝟏
𝝉
Thus, the 3-dB frequency is 𝝎𝒐 ≅ =
𝒔
𝑪𝒊 𝑹𝒔 𝑹𝒊
= 𝑪𝒊 𝑹∥ = 𝝉,
𝑹𝒊 + 𝑹𝒔
=
=
𝐰𝐡𝐞𝐫𝐞
𝑹𝒔 𝑹𝒊
𝑹∥ =
𝑹𝒊 + 𝑹𝒔
𝟏
𝑪𝒊 𝑹∥
=
𝑹𝒊 +𝑹𝒔
𝑪𝒊 𝑹𝒔 𝑹𝒊
and DC gain 𝝎 = 𝟎 ⟹ 𝒔 = 𝟎 is
𝑲=
11
𝑽𝒐
ቤ
𝑽𝒔 𝝎
=𝟎
=𝝁∙
𝟏
𝑹
𝟏+ 𝑹𝒔
𝒊
∙
𝟏
𝑹
𝟏+𝑹𝒐
𝑳
.
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 =20𝒌𝛀, Ri =100𝒌𝛀, Ci =60𝒑𝑭, 𝝁 = 144𝑽/𝑽, Ro = 200𝛀,
and RL = 1 𝒌𝛀.
𝑽𝒐
𝑽𝒔
=𝝁∙
𝟏
𝑹
𝟏+ 𝒔
𝑹𝒊
∙
𝟏
𝑹
𝟏+ 𝒐
𝑹𝑳
∙
𝟏
𝝎
𝟏+𝒊𝝎
, 𝝎𝒐 ≅
𝒐
𝑹𝒊 +𝑹𝒔
𝑪𝒊 𝑹𝒔 𝑹𝒊
and 𝑲 =
𝑽𝒐
ቤ
𝑽𝒔 𝝎
=𝟎
=𝝁∙
𝟏
𝟏
𝟏+ 𝑹
𝟏+𝑹𝒐
𝑹𝒔 ∙
𝒊
𝑹
.
𝑳
The values of the DC gain is
𝑲=
𝑽𝒐
ቤ
𝑽𝒔 𝝎
=𝟎
= 𝟏𝟒𝟒 ∙
𝟏
𝟐𝟎
𝟏𝟎𝟎
𝟏+
∙
𝟏
𝟐𝟎𝟎
𝟏𝟎𝟎𝟎
𝟏+
= 𝟏𝟎𝟎 V/V.
The 3-dB frequency is
𝟏𝟎𝟎 ∙ 𝟏𝟎𝟑 + 𝟐𝟎 ∙ 𝟏𝟎𝟑
𝒓𝒂𝒅
𝝎𝒐
𝟔
𝝎𝒐 ≅
=
𝟏𝟎
⟹
𝒇
=
≅ 𝟏𝟓𝟗. 𝟏𝟓𝟓 𝒌𝑯𝒛
𝒐
𝟔𝟎 ∙ 𝟏𝟎−𝟏𝟐 ∙ 𝟐𝟎 ∙ 𝟏𝟎𝟑 ∙ 𝟏𝟎𝟎 ∙ 𝟏𝟎𝟑
𝒔
𝟐𝝅
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 𝝎𝒐 (see diagrams
Bode plots), thus
𝝎𝒖𝒏𝒊𝒕𝒚
𝒓𝒂𝒅
𝟖
𝝎𝒖𝒏𝒊𝒕𝒚 = 𝟏𝟎𝟎𝝎𝒐 = 𝟏𝟎
⟹ 𝒇𝒖𝒏𝒊𝒕𝒚 =
≅ 𝟏𝟓. 𝟗𝟏𝟓𝟓 𝑴𝑯𝒛
𝒔
𝟐𝝅
12
13
Ex. 1.5c Find 𝒗𝒐 (𝒕) for each of the following inputs:
(i) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟐 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟐 𝒓𝒂𝒅/𝒔
(ii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟓 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟓 𝒓𝒂𝒅/𝒔
(iii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟔 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟔 𝒓𝒂𝒅/𝒔
(iv) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟖 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟖 𝒓𝒂𝒅/𝒔
By the definition of transfer function is
𝑻 𝝎
=
𝑽𝒐 𝝎
𝑽𝒊 𝝎
and ∠𝑻 𝝎 = 𝝓 𝝎 .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
𝑻 𝝎 =
𝑽𝒐 𝝎
=𝑲∙
𝑽𝒊 𝝎
𝟏
𝝎
𝟏+
𝝎𝒐
𝟏
= 𝟏𝟎𝟎 ∙
𝟐
𝟏+
𝝎
𝟏𝟎𝟔
𝟐
and the phase response is ∠𝑻 𝝎 = 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
⟹ 𝒗𝒐 𝒕 = 𝒗𝒊 𝒕 ∙
𝟏𝟎𝟎
𝟏+
𝝎
𝝎𝒐
𝝎
𝟏𝟎𝟔
𝟐
⟹
.
𝟏𝟎𝟐
𝟏𝟎𝟔
𝟐
(i) 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
≈ −𝒂𝒓𝒄𝒕𝒏 𝟎 ≈ −𝟎 ⟹
𝟏𝟎𝟎
𝒗𝒐 𝒕 ≈ 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎 𝒕 − 𝟎 ∙
≈ 𝟏𝟎𝒔𝒊𝒏 𝟏𝟎𝟐 𝒕 .
𝟏+
𝟏𝟎𝟐
𝟏𝟎𝟔
𝟐
14
Ex. 1.5c Find 𝒗𝒐 (𝒕) for each of the following inputs:
(i) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟐 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟐 𝒓𝒂𝒅/𝒔
(ii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟓 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟓 𝒓𝒂𝒅/𝒔
(iii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟔 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟔 𝒓𝒂𝒅/𝒔
(iv) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟖 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟖 𝒓𝒂𝒅/𝒔
By the definition of transfer function is
𝑻 𝝎
=
𝑽𝒐 𝝎
𝑽𝒊 𝝎
and ∠𝑻 𝝎 = 𝝓 𝝎 .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
𝑻 𝝎 =
𝑽𝒐 𝝎
=𝑲∙
𝑽𝒊 𝝎
𝟏
𝝎
𝟏+
𝝎𝒐
𝟏
= 𝟏𝟎𝟎 ∙
𝟐
𝟏+
𝝎
𝟏𝟎𝟔
𝟐
and the phase response is ∠𝑻 𝝎 = 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
⟹ 𝒗𝒐 𝒕 = 𝒗𝒊 𝒕 ∙
𝟏𝟎𝟎
𝟏+
𝝎
𝝎𝒐
𝝎
𝟏𝟎𝟔
𝟐
⟹
.
𝟏𝟎𝟓
𝟏𝟎𝟔
≈ −𝟎. 𝟏 𝒓𝒂𝒅 ⟹ 𝝓 𝝎 = −𝟓. 𝟕𝒐
𝟏𝟎𝟎
𝟓
𝒗𝒐 𝒕 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎 𝒕 − 𝟎. 𝟏 ∙
≈ 𝟗. 𝟗𝟓𝒔𝒊𝒏 𝟏𝟎𝟓 𝒕 − 𝟎. 𝟏 .
(ii) 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
𝟏𝟎𝟓
𝟏+
𝟏𝟎𝟔
𝟐
15
Ex. 1.5c Find 𝒗𝒐 (𝒕) for each of the following inputs:
(i) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟐 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟐 𝒓𝒂𝒅/𝒔
(ii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟓 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟓 𝒓𝒂𝒅/𝒔
(iii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟔 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟔 𝒓𝒂𝒅/𝒔
(iv) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟖 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟖 𝒓𝒂𝒅/𝒔
By the definition of transfer function is
𝑻 𝝎
=
𝑽𝒐 𝝎
𝑽𝒊 𝝎
and ∠𝑻 𝝎 = 𝝓 𝝎 .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
𝑻 𝝎 =
𝑽𝒐 𝝎
=𝑲∙
𝑽𝒊 𝝎
𝟏
𝝎
𝟏+
𝝎𝒐
𝟏
= 𝟏𝟎𝟎 ∙
𝟐
𝟏+
𝝎
𝟏𝟎𝟔
𝟐
and the phase response is ∠𝑻 𝝎 = 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
⟹ 𝒗𝒐 𝒕 = 𝒗𝒊 𝒕 ∙
𝟏𝟎𝟎
𝟏+
𝝎
𝝎𝒐
𝝎
𝟏𝟎𝟔
𝟐
⟹
.
𝟏𝟎𝟔
𝟏𝟎𝟔
≈ −𝟎. 𝟕𝟗 𝒓𝒂𝒅 ⟹ 𝝓 𝝎 = −𝟒𝟓𝒐
𝟏𝟎𝟎
𝟔
𝒗𝒐 𝒕 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎 𝒕 − 𝟎. 𝟕𝟗 ∙
≈ 𝟕. 𝟎𝟕𝒔𝒊𝒏 𝟏𝟎𝟔 𝒕 − 𝟎. 𝟕𝟗 .
(iii) 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
𝟏𝟎𝟔
𝟏+
𝟏𝟎𝟔
𝟐
16
Ex. 1.5c Find 𝒗𝒐 (𝒕) for each of the following inputs:
(i) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟐 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟐 𝒓𝒂𝒅/𝒔
(ii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟓 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟓 𝒓𝒂𝒅/𝒔
(iii) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟔 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟔 𝒓𝒂𝒅/𝒔
(iv) 𝒗𝒊 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟖 𝒕 , 𝑽
⟹ 𝝎 = 𝟏𝟎𝟖 𝒓𝒂𝒅/𝒔
By the definition of transfer function is
𝑻 𝝎
=
𝑽𝒐 𝝎
𝑽𝒊 𝝎
and ∠𝑻 𝝎 = 𝝓 𝝎 .
Thus, from 1.5a, 1.5b and Table 1.2 one may get that the magnitude response is
𝑻 𝝎 =
𝑽𝒐 𝝎
=𝑲∙
𝑽𝒊 𝝎
𝟏
𝝎
𝟏+
𝝎𝒐
𝟏
= 𝟏𝟎𝟎 ∙
𝟐
𝟏+
𝝎
𝟏𝟎𝟔
𝟐
and the phase response is ∠𝑻 𝝎 = 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
⟹ 𝒗𝒐 𝒕 = 𝒗𝒊 𝒕 ∙
𝟏𝟎𝟎
𝟏+
𝝎
𝝎𝒐
𝝎
𝟏𝟎𝟔
𝟐
⟹
.
𝟏𝟎𝟖
𝟏𝟎𝟔
≈ −𝟏. 𝟓𝒓𝒂𝒅 ⟹ 𝝓 𝝎 = −𝟗𝟎𝒐
𝟏𝟎𝟎
𝟔
𝒗𝒐 𝒕 = 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎 𝒕 − 𝟏. 𝟓 ∙
≈ 𝟎. 𝟏𝒔𝒊𝒏 𝟏𝟎𝟔 𝒕 − 𝟏. 𝟓 .
(iv) 𝝓 𝝎 = −𝒂𝒓𝒄𝒕𝒏
𝟏𝟎𝟖
𝟏+
𝟏𝟎𝟔
𝟐
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.
𝒊𝒇
𝒊𝒔
𝒊𝟏
(𝒂)
(𝒃)
𝒊𝒔 = 𝒊𝟏 + 𝒊𝒇 ⟹ 𝒊𝒔 = 𝟎. 𝟓 ∙
𝟏𝟎−𝟑 𝑨
⟹ 𝟎. 𝟓 ∙

𝟏𝟎−𝟑
𝒗𝟏 − 𝟎
𝒗𝟏 − 𝒗𝒐
=
+
⟹ 𝟓 = 𝟐𝒗𝟏 − 𝒗𝟎
𝟏𝟎 ∙ 𝟏𝟎𝟑 𝟏𝟎 ∙ 𝟏𝟎𝟑
If 𝒗𝒔 = 𝟎 then 𝒗𝟎 also should be 0. Thus, 𝒗𝟏 = 𝟎 and 𝟓 = −𝒗𝒐 ⟹ 𝒗𝒐 = −𝟓 𝑽.
𝒗𝒐
−𝟓
𝑹𝒎 =
=
= −𝟏𝟎 𝒌𝜴
𝒊𝒊
𝟎. 𝟓 ∙ 𝟏𝟎−𝟑
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
𝑨𝒗 = 𝟏, Fig. 2.14. This circuit is referred to as a voltage follower, since the output
voltage “follows” the input voltage, where 𝑹𝟏 = 𝑹𝒊 = ∞ , 𝑹𝒇 = 𝟎 ⟹ 𝒗𝒐 = 𝒗𝒊 .
𝐅𝐢𝐠. 𝟐. 𝟏𝟒(𝐚)
Noninverting OPA Amplifier with 𝑹𝒊 = ∞ , 𝑹𝒇 = 𝟎 ⟹ 𝑨𝒗 = 𝟏
⟹
𝒊𝟏 = 𝒊𝟐 ⇔
𝑹𝒇
𝟎 − 𝒗𝟏 𝒗 𝟏 − 𝒗𝒐
𝒗𝒐
=
⟹ 𝒗𝒐 = 𝟏 +
𝒗𝒊 ⟹ 𝑹𝒇 = 𝟎, 𝑹𝟏 = ∞ ⟹
= 𝟏 ⟹ 𝑨𝒗 = 𝟏.
𝑹𝟏
𝑹𝒇
𝑹𝟏
𝒗𝒊
19
The Weighted Summer
Fig. 2.10
By using KCL one may get that 𝒊𝟏 + 𝒊𝟐 + ⋯ 𝒊𝒏 = σ𝒏𝒊=𝟏 𝒊𝒊 = 𝒊 , where 𝒊𝒊 =
𝒊=
𝟎−𝒗𝒐
𝑹𝒇
. Thus, σ𝒏𝒊=𝟏 𝒊𝒊 = 𝒊 ⟺ σ𝒏𝒊=𝟏
𝒗𝒊
𝑹𝒊
=
−𝒗𝒐
𝑹𝒇
⟹ 𝒗𝒐 = − σ𝒏𝒊=𝟏
𝑹𝒇
𝑹𝒊
𝒗𝒊
𝑹𝒊
and
𝒗𝒊 .
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 𝒗𝒐 = − 𝒗𝟏 + 𝟒𝒗𝟐 . Choose values for 𝑹𝟏 , 𝑹𝟐 ,
and 𝑹𝒇 so that for a maximum output voltage of -4 V the current in the feedback
resistor will not exceed 1 mA.
𝒊𝒇𝒎𝒂𝒙 =
𝟎 − 𝒗𝒐𝒎𝒂𝒙
𝒗𝒐𝒎𝒂𝒙
−𝟒
⟹ 𝑹𝒇𝒎𝒊𝒏 = −
=−
⟹ 𝑹𝒇 ≥ 𝟒𝒌𝜴
𝑹𝒇𝒎𝒊𝒏
𝒊𝒇𝒎𝒂𝒙
𝟏 ∙ 𝟏𝟎−𝟑
𝒗𝒐 = − 𝒗𝟏 + 𝟒𝒗𝟐
𝑹𝒇 = 𝟒 ∙ 𝟏𝟎𝟑 𝜴
𝟐
𝑹𝒇
𝟒 ∙ 𝟏𝟎𝟑
𝟒 ∙ 𝟏𝟎𝟑
𝒗𝒐 = − ෍ 𝒗𝒊 = −
𝒗𝟏 +
𝒗𝟐
𝑹𝒊
𝑹𝟏
𝑹𝟐
𝒊=𝟏
⟹ 𝑹𝟏 = 𝟒𝒌𝜴, 𝑹𝟐 = 𝟏𝒌𝜴
21
The Weighted Summer with Opposite Signs
0
𝑖𝑏
0
𝒊
𝑖3
𝑖4
Fig. 2.11
𝑹
𝒊 = 𝒊𝒃 + 𝒊𝟑 + 𝒊𝟒 , where 𝒊𝒃 =
𝒗
𝒗
𝒊𝟑 = 𝑹𝟑 , 𝒊𝟒 = 𝑹𝟒 and 𝒊 =
𝟑
𝟒
𝒗𝒐 =
𝑹
𝑹
𝟎−𝒗𝒐
𝑹𝒄
𝑹
−𝑹𝒂 𝒗𝟏 −𝑹𝒂 𝒗𝟐
𝟏
𝟐
𝑹𝒃
𝒗
=−
𝟏 𝑹𝒂
𝒗
𝑹𝒃 𝑹𝟏 𝟏
𝟏
. Thus, − 𝑹𝒐 = − 𝑹
𝒄
𝑹𝒂
𝒗𝟏
𝒃 𝑹𝟏
+
𝑹𝒂
𝒗
𝑹𝟐 𝟐
𝑹
+ 𝑹𝒂 𝒗𝟐 +
𝟐
𝟒
𝒗𝟑
𝑹𝟑
𝒗
+ 𝑹𝟒 ⟹
𝟒
𝑹𝒄 𝑹𝒂
𝑹𝒄 𝑹𝒂
𝑹𝒄
𝑹𝒄
𝒗𝟏 +
𝒗𝟐 −
𝒗𝟑 −
𝒗
𝑹𝒃 𝑹𝟏
𝑹𝒃 𝑹𝟐
𝑹𝟑
𝑹𝟒 𝟒
𝑹 𝑹
Voltages 𝑹𝒄 𝒗𝟑 and 𝑹𝒄 𝒗𝟒 would be subtracted from 𝑹𝒄 𝑹𝒂 𝒗𝟏 and
𝟑
,
𝒃
𝟏
𝑹𝒄 𝑹𝒂
𝒗 .
𝑹𝒃 𝑹𝟐 𝟐
22
D2.8 Use the idea presented in Fig. 2.11 to design a weighted summer that provides
𝒗𝒐 = 𝟐𝒗𝟏 + 𝒗𝟐 − 𝟒𝒗𝟑 .
𝑹𝒄 𝑹𝒂
𝑹𝒄 𝑹𝒂
𝑹𝒄
𝑹𝒂
𝑹𝒂
𝑹𝒄
𝒗𝒐 =
𝒗 +
𝒗 −
𝒗 = 𝑹𝒄 = 𝑹𝒃 =
𝒗 +
𝒗 −
𝒗 =
𝑹𝒃 𝑹𝟏 𝟏 𝑹𝒃 𝑹𝟐 𝟐 𝑹𝟑 𝟑
𝑹𝟏 𝟏 𝑹𝟐 𝟐 𝑹𝟑 𝟑
= 𝑹𝒂 = 𝑹𝒄 = 𝑹𝒇
′
𝐥𝐞𝐭
𝐬 𝐮𝐬𝐞 𝐭𝐡𝐞 𝐚𝐩𝐩𝐫𝐨𝐚𝐜𝐡
𝑹𝒇
𝑹𝒇
𝑹𝒇
𝐨𝐟 𝐃𝟐. 𝟕
=
𝒗 +
𝒗 −
𝒗 =
=
𝑹𝟏 𝟏 𝑹𝟐 𝟐 𝑹𝟑 𝟑
𝑹𝒇 = 𝟏𝟎 𝒌𝛀
𝑹𝟏 = 𝟓 ∙ 𝟏𝟎𝟑 𝛀
𝟏𝟎 ∙ 𝟏𝟎
𝟏𝟎 ∙ 𝟏𝟎
𝟏𝟎 ∙ 𝟏𝟎
=
𝒗𝟏 +
𝒗𝟐 −
𝒗𝟑 = 𝑹𝟐 = 𝟏𝟎 ∙ 𝟏𝟎𝟑 𝛀 = 𝟐𝒗𝟏 + 𝒗𝟐 − 𝟒𝒗𝟑
𝑹𝟏
𝑹𝟐
𝑹𝟑
𝑹𝟑 = 𝟐. 𝟓 ∙ 𝟏𝟎𝟑 𝛀
𝟑
𝟑
𝟑
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
• 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.
• The amount of amplification provided by an amplifier is measured by its gain: the
ratio of output voltage, current, or power to input.
• 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.
• The “wiggles” 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.
• Mathematically a linear amplifier that preserves the details of the signal waveform
is characterized by the linear relationship
𝑨=
𝒗𝒐𝒖𝒕
𝒗𝒊𝒏
where 𝒗𝒊𝒏 and 𝒗𝒐𝒖𝒕 are the input and output signals,
respectively, and A is a constant representing the
magnitude of amplification, known as amplifier gain.
2
• Voltage amplifiers is an electrical circuit to increase a power of signal by
increasing an amplitude of voltage without a significant change of current.
• 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.
• An audio power amplifier (or power amp) is an amplifier that amplifies
low-power electronic audio signals such as the signal from radio receiver
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.
• 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
• Power gain of an amplifier is
Gain
𝑷𝒐𝒖𝒕 𝒗𝒐𝒖𝒕 𝒊𝒐𝒖𝒕
𝑨𝒑 =
=
,[ ]
𝑷𝒊𝒏
𝒗𝒊𝒏 𝒊𝒊𝒏
Definition: 𝑨𝒑 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
where 𝑨𝒑 = 𝟏𝟎𝒍𝒏
of logarithms.
𝑷𝒐𝒖𝒕
𝑷𝒊𝒏
𝑷𝒐𝒖𝒕
𝑷𝒊𝒏
, 𝒅𝑩
, 𝑵𝒑 = [𝑵𝒆𝒑𝒆𝒓] after John Napier, the inventor
𝟏 𝒅𝑩 ≈ 𝟎. 𝟏𝟏𝟓𝟏𝟑 𝑵𝒑 or 𝟏 𝑵𝒑 ≈ 𝟖. 𝟔𝟖𝟓𝟖𝟗 𝒅𝑩.
• Voltage gain is defined from 𝑨𝒑 as the next
𝒊𝒇
𝒗𝟐𝒐𝒖𝒕
𝑷𝒐𝒖𝒕
𝑹𝒊𝒏 ≈ 𝑹𝒐𝒖𝒕
𝑹𝒐𝒖𝒕
𝑨𝒑 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
= 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
⟹
𝒗𝒐𝒖𝒕
𝟐
𝑷𝒊𝒏
𝒗𝒊𝒏
𝑨𝒗 =
,
𝒗
𝑹𝒊𝒏
𝒊𝒏
Def.: 𝑨𝒗 =
𝒗𝒐𝒖𝒕 𝟐
𝟏𝟎𝒍𝒐𝒈𝟏𝟎
𝒗𝒊𝒏
= 𝟐𝟎𝒍𝒐𝒈𝟏𝟎
𝒗𝒐𝒖𝒕
𝒗𝒊𝒏
⟹
, 𝒅𝑩
4
• Current gain is defined in similar way
𝑨𝒑 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
Def.: 𝑨𝒊 =
𝑷𝒐𝒖𝒕
𝒊𝟐𝒐𝒖𝒕 𝑹𝒐𝒖𝒕
= 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
𝑷𝒊𝒏
𝒊𝟐𝒊𝒏 𝑹𝒊𝒏
𝒊𝒐𝒖𝒕 𝟐
𝟏𝟎𝒍𝒐𝒈𝟏𝟎
𝒊𝒊𝒏
= 𝟐𝟎𝒍𝒐𝒈𝟏𝟎
𝒊𝒐𝒖𝒕
𝒊𝒊𝒏
𝒊𝒇
𝑹𝒊𝒏 ≈ 𝑹𝒐𝒖𝒕
⟹
𝒊𝒐𝒖𝒕
𝑨𝒊 =
,
𝒊𝒊𝒏
⟹
, 𝒅𝑩 .
Assignment: Find values of 𝑨𝒑 .
𝑨𝒑 𝒅𝑩
𝑨𝒑 [ ]
0
?
1
?
10
?
100
?
5
Assignment: Find values of 𝑨𝒑 .
𝑨𝒑 , 𝒅𝑩 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
𝑨𝒑 , 𝒅𝑩
= 𝒍𝒐𝒈𝟏𝟎 𝑨𝒑 ,
𝟏𝟎
𝑷𝒐𝒖𝒕
𝑷𝒊𝒏
⟹ 𝑨𝒑 , 𝒅𝑩 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎 𝑨𝒑 ,
⟹ 𝑨𝒑 ,
=
𝑨𝒑 , 𝒅𝑩
𝒆 𝟏𝟎
𝑨𝒑 𝒅𝑩
𝑨𝒑 [ ]
0
𝒆𝟎 =1
1
⟹
𝟏
𝒆𝟏𝟎
≈ 𝟏. 𝟏𝟏
10
𝒆𝟏 ≈ 𝟐. 𝟕𝟐
100
𝒆𝟏𝟎 ≈ 𝟐𝟐 ∙ 𝟏𝟎𝟑
6
• 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.
𝑨𝒗 = 𝟐𝟎𝒍𝒐𝒈𝟏𝟎
𝒗𝒐𝒖𝒕
= 𝟐𝟎𝒍𝒐𝒈𝟏𝟎 𝟏𝟎𝟎 = 𝟒𝟎 𝒅𝑩;
𝒗𝒊𝒏
𝑨𝒊 = 𝟐𝟎𝒍𝒐𝒈𝟏𝟎
𝒊𝒐𝒖𝒕
= 𝟐𝟎𝒍𝒐𝒈𝟏𝟎 𝟏𝟎𝟎𝟎 = 𝟔𝟎 𝒅𝑩;
𝒊𝒊𝒏
𝑨𝒑 = 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
𝑷𝒐𝒖𝒕
𝒗𝒐𝒖𝒕 𝒊𝒐𝒖𝒕
𝒗𝒐𝒖𝒕 𝒊𝒐𝒖𝒕
= 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
= 𝟏𝟎𝒍𝒐𝒈𝟏𝟎
∙
= 𝟏𝟎𝒍𝒐𝒈𝟏𝟎 𝟏𝟎𝟎 ∙ 𝟏𝟎𝟎𝟎 = 𝟓𝟎 𝒅𝑩.
𝑷𝒊𝒏
𝒗𝒊𝒏 𝒊𝒊𝒏
𝒗𝒊𝒏 𝒊𝒊𝒏
7
Amplifier Power Efficiency
The power delivered to the amplifier from the DC power supply is
𝑷𝑫𝑪 = 𝑽𝑪𝑪 𝑰𝑪𝑪 + 𝑽𝑬𝑬 𝑰𝑬𝑬 .
The power-balance equation for the amplifier can be written as
𝑷𝑫𝑪 + 𝑷𝒊𝒏 = 𝑷𝒐𝒖𝒕 + 𝑷𝒅𝒊𝒔 ,
where 𝑷𝒊𝒏 is the power drawn from the signal source, 𝑷𝒐𝒖𝒕 is the power
delivered to the load and 𝑷𝒅𝒊𝒔 is the power dissipated in the amplifier circuit
itself. By assuming 𝑷𝒊𝒏 ⟶ 𝟎 one may define the amplifier power efficiency as
𝑷
𝜼 = 𝒐𝒖𝒕 ∙ 𝟏𝟎𝟎.
𝑷𝑫𝑪
8
• Ex.1.11. An amplifier operating from a single 15-V supply provides a 12-V peak-topeak sine-wave signal to a 1- kΩ 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?
𝑷𝑫𝑪 + 𝑷𝒊𝒏 = 𝑷𝒐𝒖𝒕 + 𝑷𝒅𝒊𝒔
𝑷𝑫𝑪 = 𝑽𝑪𝑪 𝑰𝑪𝑪 = 𝟏𝟓 ∙ 𝟖 ∙ 𝟏𝟎−𝟑 = 𝟎. 𝟏𝟐 𝑾;
𝒊𝒊𝒏 ≈ 𝟎 ⟹ 𝑷𝒊𝒏 ≈ 𝟎 𝑾 ⟹ 𝑷𝒅𝒊𝒔 = 𝑷𝑫𝑪 + 𝑷𝒊𝒏 − 𝑷𝒐𝒖𝒕 ≈ 𝑷𝑫𝑪 − 𝑷𝒐𝒖𝒕 ;
𝑷𝒐𝒖𝒕
𝒗𝒓𝒎𝒔
=
𝑹𝑳
𝟐
= 𝒗
𝒓𝒎𝒔
𝒗𝒑𝒌−𝒑𝒌
𝟐
=
=
𝟐
𝟔
𝟐
𝟐
∙ 𝟏 ∙ 𝟏𝟎−𝟑 = 𝟎. 𝟎𝟏𝟖 𝑾;
𝑷𝒅𝒊𝒔 ≈ 𝑷𝑫𝑪 − 𝑷𝒐𝒖𝒕 = 𝟎. 𝟏𝟐 − 𝟎. 𝟎𝟏𝟖 = 𝟎. 𝟏𝟎𝟐 𝐖 = 𝟏𝟎𝟐 𝒎𝑾;
𝜼=
𝑷𝒐𝒖𝒕
𝑷𝑫𝑪
∙ 𝟏𝟎𝟎 =
𝟎.𝟎𝟏𝟖
𝟎.𝟏𝟐
∙ 𝟏𝟎𝟎 = 𝟏𝟓 %.
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:
𝑳− ≤ 𝑽𝒐𝒖𝒕 ≤ 𝑳+
or
𝑳−
𝑳+
≤ 𝑽𝒊𝒏 ≤
𝑨𝒗
𝑨𝒗
10
Symbols and Terminology
𝐼𝐶 is DC component;
𝐼𝑐 is the amplitude of AC component;
𝑖𝑐 𝑡 = 𝐼𝑐 𝑠𝑖𝑛 𝜛𝑡 is the sinusoidal AC signal;
𝑖𝐶 𝑡 = 𝐼𝐶 + 𝑖𝑐 𝑡 is the total instantaneous signal.
11
Circuit Models: Voltage Amplifier
𝑣𝑠
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 𝒊𝒐 and voltage 𝒗𝒐 to a load 𝑹𝑳 .
The nonzero output resistance Ro causes only a fraction of the voltage 𝑨vo vi to
appear across the output load 𝑹𝑳 .
By using the voltage-divider one might see that
𝒗𝒐
𝑹𝑳
𝑹𝑳
=
⟹ 𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊
.
𝑨𝒗𝒐 𝒗𝒊 𝑹𝑳 + 𝑹𝒐
𝑹𝑳 + 𝑹𝒐
Thus, the voltage gain of the voltage amplifier is given by 𝑨𝒗 =
For the ideal OPA 𝑹𝒐 ⟶ 𝟎. Also, if 𝑹𝑳 ⟶ ∞, then 𝑨𝒗 ≈ 𝑨𝒗𝒐 .
𝒗𝒐
𝒗𝒊
= 𝑨𝒗𝒐
𝑹𝑳
.
𝑹𝑳 +𝑹𝒐
12
Circuit Models: Voltage Amplifier
𝑣𝑠
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
𝒗𝒊
𝑹𝒊
𝑹𝒊 + 𝑹𝒔
=
⟹ 𝒗𝒔 = 𝒗𝒊
.
𝒗𝒔 𝑹𝒊 + 𝑹𝒔
𝑹𝒊
Thus, in order not to lose a significant portion of the input signal, the amplifier
must be designed to have an input resistance Ri ≫ 𝑹𝒔 and 𝑹𝒔 ⟶ 𝟎, 𝑹𝒊 ⟶ ∞.
The overall voltage gain of the voltage amplifier is
𝑹𝑳
𝑨
𝒗
𝒗𝒐 𝒊 𝑹 + 𝑹
𝒗𝒐
𝑹𝑳
𝑹𝒊
𝑳
𝒐
=
= 𝑨𝒗𝒐 ∙
∙
.
𝑹
+
𝑹
𝒗𝒔
𝑹𝑳 + 𝑹𝒐 𝑹𝒊 + 𝑹𝒔
𝒔
𝒗𝒊 𝒊
𝑹𝒊
13
Circuit Models: Voltage Amplifier
𝑣𝑠
The overall voltage gain of the voltage amplifier is
𝑹𝑳
𝑨
𝒗
𝒗𝒐 𝒊 𝑹 + 𝑹
𝒗𝒐
𝑹𝑳
𝑹𝒊
𝑳
𝒐
=
= 𝑨𝒗𝒐 ∙
∙
.
𝑹𝒊 + 𝑹𝒔
𝒗𝒔
𝑹
+
𝑹
𝑹
+
𝑹
𝑳
𝒐
𝒊
𝒔
𝒗𝒊
𝑹𝒊
There are situations in which one is interested not in voltage gain but only in a
𝒗
significant power gain. For instance, if 𝑹𝒔 ≫ 𝑹𝑳 , 𝑹𝒊 then 𝒐 ⟶ 𝟎. In such a case,
𝒗𝒔
one requires an additional amplifier (in between the source and voltage amplifier)
𝒗
with modest voltage gain ( 𝒐 ≤ 𝟏) because of 𝑹𝒔 ≫ 𝑹𝒊 and 𝑹𝒐 ≫ 𝑹𝑳 . Such
𝒗𝒔
additional amplifier is referred to as a buffer amplifier.
14
Cascaded Amplifiers
𝐴𝑣𝑜 = 1;
𝑅𝐿 = 102 Ω;
𝑅𝑜 = 10 Ω;
𝑅𝑖 = 104 Ω;
𝑅𝑠 = 103 Ω.
𝐴𝑣𝑜 = 100;
𝑅𝐿 = 104 Ω;
𝑅𝑜 = 103 Ω;
𝑅𝑖 = 105 Ω;
𝑅𝑠 = 103 Ω.
𝐴𝑣𝑜 = 10;
𝑅𝐿 = 105 Ω;
𝑅𝑜 = 103 Ω;
𝑅𝑖 = 106 Ω;
𝑅𝑠 = 105 Ω.
𝒗𝒐
𝑹𝑳
𝑹𝒊
= 𝑨𝒗𝒐 ∙
∙
=
𝒗𝒔
𝑹𝑳 + 𝑹𝒐 𝑹𝒊 + 𝑹𝒔
𝒗𝒐
𝒗𝒔
=
𝟏𝟎𝟎𝟎𝟎𝟎
𝟏𝟏𝟏𝟏
≈ 𝟗𝟎.
𝒗𝒐
𝒗𝒔
=
𝟏𝟎𝟎
𝟏𝟐𝟏
≈ 𝟎. 𝟖.
𝟏𝟎𝟓
𝟏𝟎𝟔
= 𝟏𝟎 ∙ 𝟓
∙
=
𝟏𝟎 + 𝟏𝟎𝟑 𝟏𝟎𝟔 + 𝟏𝟎𝟓
=
𝟏𝟎𝟎𝟎𝟎
≈ 𝟗.
𝟏𝟏𝟏𝟏
15
Cascaded Amplifiers
𝐴𝑣𝑜 = 100;
𝑅𝐿 = 104 Ω;
𝑅𝑜 = 103 Ω;
𝑅𝑖 = 105 Ω;
𝑅𝑠 = 103 Ω.
𝐴𝑣𝑜 = 10;
𝑅𝐿 = 105 Ω;
𝑅𝑜 = 103 Ω;
𝑅𝑖 = 106 Ω;
𝑅𝑠 = 105 Ω.
𝒗𝒐
𝒗𝒔
𝒗𝒐
≈ 𝟗.
𝒗𝒔
≈ 𝟗𝟎.
𝐴𝑣𝑜 = 1;
𝑅𝐿 = 102 Ω;
𝑅𝑜 = 10 Ω;
𝑅𝑖 = 104 Ω;
𝑅𝑠 = 103 Ω.
𝒗𝒐
𝒗𝒔
≈ 𝟎. 𝟖.
𝒗𝒐
ቮ
≈ 𝟗 ∙ 𝟗𝟎 ∙ 𝟎. 𝟖 = 𝟔𝟒𝟖.
𝒗𝒔
𝒏𝒆𝒕
𝑨𝒗 = 𝟐𝟎𝒍𝒐𝒈𝟏𝟎
𝒗𝒐𝒖𝒕
= 𝟐𝟎𝒍𝒐𝒈𝟏𝟎 𝟔𝟒𝟖 ≈ 𝟓𝟔 𝒅𝑩
𝒗𝒊𝒏
16
Other Amplifier Types
From the Thevenin’s
theorem one might see
that
𝑹𝑳 ⟶ ∞ ⟹
𝒊𝒐 = 𝟎 and 𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊
𝒗𝒐 = 𝑨𝒊𝒔 𝒊𝒊 𝑹𝒐 = 𝑨𝒊𝒔
𝒗𝒊
𝑹
𝑹𝒊 𝒐
𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊 = 𝑨𝒊𝒔
𝒗𝒊
𝑹
𝑹𝒊 𝒐
𝑨𝒗𝒐
𝑹𝒐
= 𝑨𝒊𝒔
𝑹𝒊
17
A transconductance amplifier (𝑮𝒎 =
𝒊𝒐
𝒗𝒊
slop amplifier )
puts out a current 𝒊𝒐 proportional to its input voltage 𝒗𝒊 .
𝑮𝒎 =
𝒊𝒐
𝒗𝒊
⟹ 𝒊𝒐 𝒗𝒊 = 𝑮𝒎 𝒗𝒊 .
18
Other Amplifier Types
𝑹𝑳 ⟶ ∞ ⟹
𝒊𝒐 = 𝟎 and 𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊
𝒊𝒊 ≈ 𝟎
𝒗𝒐 = 𝑮𝒎 𝒗𝒊 𝑹𝒐
𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊 = 𝑮𝒎 𝒗𝒊 𝑹𝒐
𝑨𝒗𝒐 = 𝑮𝒎 𝑹𝒐
19
A transresistance amplifier outputs a voltage 𝒗𝒐 proportional
to its input current 𝒊𝒊 .
𝑹𝒎 =
𝒗𝒐
𝒊𝒊
⟹ 𝒗𝒐 𝒊𝒊 = 𝑹𝒎 𝒊𝒊 .
20
Other Amplifier Types
From the Thevenin’s
theorem one might see
that
𝑹𝑳 ⟶ ∞ ⟹
𝒊𝒐 = 𝟎 and 𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊
From the Thevenin’s
theorem one might see
that
𝒗𝒐 = 𝑹𝒎 𝒊𝒊
𝒗𝒐 = 𝑨𝒗𝒐 𝒗𝒊 = 𝑹𝒎 𝒊𝒊
𝑨𝒗𝒐 = 𝑨𝒊𝒔
𝑹𝒐
𝑹𝒊
21
Determining Ri and Ro
From the Table 1.1:
One can find 𝑹𝒊 and 𝑹𝒐 by using Ohm’s law:
𝑹𝒊 =
𝒗𝒊
𝒊𝒊
, where 𝒗𝒊 and 𝒊𝒊 are parameters measured by using meters;
and
𝑹𝒐 =
𝑨∞ 𝒗𝒊
𝒊𝒐,𝒔𝒉𝒐𝒓𝒕_𝒄𝒊𝒓𝒄𝒖𝒊𝒕
, where 𝑨∞ 𝒗𝒊 and 𝒊𝒐,𝒔𝒉𝒐𝒓𝒕_𝒄𝒊𝒓𝒄𝒖𝒊𝒕 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 𝒊𝒊 = 𝟎 and 𝒗𝒊 = 𝟎)
and applying a voltage signal 𝒗𝒙 to the output of the
amplifier. If one denotes the current drawn from 𝒗𝒙 as 𝒊𝒙
𝒗
(note that it is opposite in direction of 𝒊𝒐 ), then 𝑹𝒐 = 𝒊 𝒙.
𝒙
These techniques are conceptually correct for a theoretical
calculations, but it is not employed in actual practice to
characterize 𝑹𝒐 .
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 amplifier’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
𝒊𝒐
𝑹𝒔
𝑹𝒐
= 𝑨𝒊𝒔
∙
𝒊𝒔
𝑹 𝒔 + 𝑹𝒊 𝑹𝒐 + 𝑹𝑳
𝑹𝒐
𝑹 𝒐 + 𝑹𝑳
𝒗𝒔
𝒊𝒊 = 𝒊𝒔 =
𝑹 𝒔 + 𝑹𝒊
𝒊𝒐 = 𝑨𝒊𝒔 𝒊𝒊
𝑅𝑠
𝑅𝐿
𝑣𝑠
𝒊𝒔
𝟏
𝒊𝒐
𝑹𝒐
𝑹𝒔 + 𝑹𝒊
𝑹𝒐
𝑹𝒔 + 𝑹𝒊
=
= 𝑨𝒊𝒔 𝒊𝒊
∙
=
= 𝑨𝒊𝒔
∙
𝒗𝒔 𝑹𝒔
𝒊𝒔
𝑹𝒐 + 𝑹𝑳
𝒗𝒔
𝑹𝒐 + 𝑹𝑳
𝑹𝒔
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 𝑹𝒔 be connected to the input
and a load resistance 𝑹𝑳 be connected to the output. Show that the overall voltage gain is
given by
𝒗𝒐
𝑹𝒊
𝑹 𝒐 𝑹𝑳
=
𝑮
∙
𝒎 𝑹 +𝑹 𝑹 +𝑹 .
𝒗
𝒔
𝒊
𝒔
𝒐
𝑳
𝑹 𝑹
𝒗𝒐 = 𝑨𝒊𝒔 𝒊𝒊 𝑹 𝒐+𝑹𝑳 and 𝒗𝒔 = 𝒊𝒊 𝑹𝒔 + 𝑹𝒊
𝒐
𝑳
𝒗𝒐
𝑹𝒐 𝑹𝑳
𝟏
= 𝑨𝒊𝒔 𝒊𝒊
∙
=
𝒗𝒔
𝑹𝒐 + 𝑹𝑳 𝒊𝒊 𝑹𝒔 + 𝑹𝒊
= 𝑨𝒊𝒔
𝑅𝑠
𝑅𝐿
𝑣𝑠
𝑨𝒊𝒔 𝒊𝒐 𝟏
𝒊𝒐
𝑹𝒐 𝑹𝑳
𝟏
𝑹𝒊 𝑨𝒊𝒔 𝑹𝒐 𝑹𝑳
𝑹𝒊
= ∙
= = 𝑮𝒎 =
∙
∙
=
∙
∙
=
𝑹𝒊
𝒊 𝒔 𝑹 𝒊 𝒗𝒊
𝑹𝒐 + 𝑹𝑳 𝑹𝒔 + 𝑹𝒊 𝑹𝒊
𝑹𝒊 𝑹𝒐 + 𝑹𝑳 𝑹𝒔 + 𝑹𝒊
= 𝑮𝒎 ∙
𝑹𝒐 𝑹𝑳
𝑹𝒊
∙
𝑹𝒐 + 𝑹 𝑳 𝑹𝒔 + 𝑹𝒊
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 𝒊𝒔 having a resistance 𝑹𝒔 , and
let the output be connected to a load resistance 𝑹𝑳 . Show that the overall gain is given by
𝒗𝒐
𝑹𝒔
𝑹𝑳
= 𝑹𝒎
∙
.
𝒊𝒔
𝑹𝒔 + 𝑹𝒊 𝑹𝑳 + 𝑹𝒐
𝒊𝒔
The voltage division gives
𝑹𝒔
𝑹𝑳 , 𝒗𝒐
𝒗𝒐
𝑹𝑳
𝑹𝒎 𝑹 𝑳
=
⟹ 𝒗𝒐 = 𝒊𝒊
𝑹𝒎 𝒊𝒊 𝑹𝒐 + 𝑹 𝑳
𝑹𝒐 + 𝑹 𝑳
The current division gives
𝑹𝒔
𝑹𝒔 + 𝑹𝒊
𝒊𝒊 = 𝒊𝒔
⟹ 𝒊𝒔 = 𝒊𝒊
.
𝑹𝒔 + 𝑹𝒊
𝑹𝒔
Then
𝒗𝒐
𝒊𝒔
= 𝒊𝒊
𝑹𝒎 𝑹𝑳
𝑹𝒐 +𝑹𝑳
∙
𝑹𝒔
𝒊𝒊 𝑹𝒔 +𝑹𝒊
= 𝑹𝒎
𝑹𝑳
𝑹𝒐 +𝑹𝑳
∙
𝑹𝒔
.
𝑹𝒔 +𝑹𝒊
27
E1.21. Find the input resistance between terminals B (base) and G (ground) in the circuit
shown in Fig. The voltage 𝒗𝒙 is a test voltage (base – ground) with the input resistance
𝒗
𝑹𝒊𝒏 = 𝑹𝑩𝑮 defined as 𝑹𝒊𝒏 = 𝒊 𝒙 .
𝒙
𝒊𝒆
From KCL one may get that
𝒊𝒆 = 𝒊𝒃 + 𝜷𝒊𝒃 ⟹ 𝒊𝒆 = 𝒊𝒃 𝟏 + 𝜷
From KVL one may get that
𝒗𝒙 = 𝒊𝒃 𝒓𝝅 + 𝒊𝒆 𝑹𝒆 ⟹ 𝒊𝒙 = 𝒊𝒃 ⟹ 𝒗𝒙 = 𝒊𝒃 𝒓𝝅 + 𝒊𝒃 𝟏 + 𝜷 𝑹𝒆 = 𝒊𝒃 𝒓𝝅 + 𝟏 + 𝜷 𝑹𝒆
Thus, 𝑹𝒊𝒏 =
𝒗𝒙
𝒊𝒙
=
𝒗𝒙
𝒊𝒃
=
𝒊𝒃 𝒓𝝅 + 𝟏+𝜷 𝑹𝒆
𝒊𝒃
= 𝒓𝝅 + 𝟏 + 𝜷 𝑹𝒆 .
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
• 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.
• Its resistivity falls as its temperature rises; metals behave in the opposite way.
• 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.
• 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.
• 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
• A semiconductor device is an electronic component that relies on the electronic
properties of a semiconductor material for its function.
• 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.
• 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
• 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.
• 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 “hole”, 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 “hole”. This process may repeat itself, with the
result in an electric current of holes. As temperature increases, more covalent bonds are
broken and electron–hole 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
“recombination”. The recombination
rate is proportional to the number
of free electrons and holes, which in
turn is determined by the thermal
“generation” rate.
4
• A thermal equilibrium is a state of intrinsic s/c crystal that is featuring by the
recombination rate equal to the generation rate
• In thermal equilibrium, the concentration of free electrons 𝒏 is equal to the
concentration of holes 𝒑,
𝒏 = 𝒑 = 𝒏𝒊
where 𝒏𝒊 = 𝑩𝑻𝟑/𝟐 𝒆−𝑬𝒈 /𝟐𝒌𝑻 is the number of free electrons and holes in a unit
volume (𝒄𝒎𝟑 ) of intrinsic silicon at a given temperature 𝑻 𝐊 ; 𝑩 = 𝟕. 𝟑 ×
𝟑
−𝟐
𝟏𝟓
−𝟑
𝟏𝟎 [𝒄𝒎 𝑲 ] is a material-dependent parameter that is given for silicon; 𝑬𝒈 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 𝑬𝒈 =1.12 [eV] ; and 𝒌 = 𝟖. 𝟔𝟐 × 𝟏𝟎−𝟓 𝒆𝑽/𝑲 is Boltzmann’s constant.
• It is useful to express the product of the hole and free-electron concentration as
𝒑𝒏 = 𝒏𝟐𝒊 ,
where for the intrinsic silicon at the room temperature, 𝒏𝒊 ≈ 𝟏. 𝟓 × 𝟏𝟎𝟏𝟎 /𝒄𝒎𝟑 .
5
Energy Bands, Band Gap
(K. Kano, “Semiconductor Devices”)
Conductance
band
Valence
band
The band gap is temperature dependent 𝒎𝐚𝐭𝐞𝐫𝐢𝐚𝐥 𝐩𝐚𝐫𝐚𝐦𝐞𝐭𝐞𝐫 𝑬𝒈 𝑻 = 𝑬𝒈 𝟎
𝜶𝑻𝟐
− 𝑻+𝜷
Silicon: 𝑬𝒈 𝟎 = 𝟏. 𝟏𝟕 𝒆𝑽, 𝜶 = 𝟒. 𝟕𝟑 × 𝟏𝟎−𝟒 , 𝜷 = 𝟔𝟑𝟔, 𝑬𝒈 𝟑𝟎𝟎𝑲 = 𝟏. 𝟏𝟐𝟓 𝒆𝑽.
6
[𝒆𝑽],
Doped (Extrinsic) Semiconductors
• 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.
• To increase the concentration of free electrons, 𝒏, silicon is doped with an element
with a valence of 5, such as phosphorus. The resulting doped silicon is then said to
be of 𝒏 type.
• To increase the concentration of holes, 𝒑, 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
p–n junction
Two electrically neutral
crystals separated by a
parallel air gap.
Two crystals are connected
by p–n junction:
• Electron
and
hole
concentration
are
plotted with blue and
red lines, respectively.
• The gray regions are
charge-neutral.
The
light-red
zone
is
positively charged.
• The light-blue zone is
negatively charged.
• 𝑬 is inner electric field
of the depleted zone.
11
The inner field 𝑬- 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 𝑽𝒃𝒊 across the p-n
junction is given by
𝑽𝒃𝒊 = 𝑽𝑻 𝒍𝒏
𝑵𝑨 𝑵𝑫
𝒏𝟐𝒊
≈ 𝟎. 𝟕 [𝑽],
where 𝑵𝑨 ~𝟏𝟎𝟏𝟕 /𝒄𝒎𝟑 and 𝑵𝑫 ~𝟏𝟎𝟏𝟓 /𝒄𝒎𝟑
are the doping concentrations of the 𝒑 side
and 𝒏 side of the junction, respectively;
𝒏𝒊 ≈ 𝟏. 𝟓 × 𝟏𝟎𝟏𝟎 /𝒄𝒎𝟑 is concentration of
free electrons 𝒏 and holes in intrinsic s/c;
and 𝑽𝑻 = 𝒌𝑻Τ𝒒 = 𝟐𝟓. 𝟗 𝒎𝑽 (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
• The Diffusion Current 𝑰𝑫 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 𝑰𝑫 current is from the p side to the n side.
• The Drift Current 𝑰𝑺 . 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 𝑰𝑺 , whose direction is
from the n side to the p side of the junction.
• 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:
𝑰𝑫 = 𝑰𝑺 .
15
𝑰𝑫
𝑰𝑺
metal
metal
• When the p-n junction terminals are left open-circuited, the voltage measured
between them will be zero. That is, the voltage 𝑽𝒃𝒊 across the depletion region
does not appear between the junction terminals. This is because similar built-in
voltages arise at the metal–semiconductor 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.
16
• 𝑵𝑨 > 𝑵𝑫 is a common situation in practice;
• The depletion region extends in both the p and n sides and equal amounts of
charge exist on both sides 𝑸+ = 𝑸− ;
• Since 𝑵𝑨 > 𝑵𝑫 , the width of the depletion layer will not be the same on the two
sides. Rather, to uncover the same amount of charge, …
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