EGR 260 – Circuit Analysis
Lecture #9
Reading Assignment: Sections 4.1-4.9 in Electric Circuits, 9th Edition by Nilsson
Mesh Equations (continued)
Dependent Sources:
The key to using mesh analysis on circuits with dependent sources is to redefine
the control variables in terms of mesh currents.
Example: Use mesh equations to determine V1 and i1 in the circuit shown below.
3V1
_+
3
6
8
+
i1
5
4
V1
6A
_
1
EGR 260 – Circuit Analysis
Lecture #9
Supermesh
If a circuit contains an internal current source, a supermesh is required in order to
perform mesh analysis. A supermesh is the new, larger mesh that is created by
removing the internal current source. A new mesh current is not added. The
supermesh simply shows the path for a KVL equation around the supermesh.
Example:
1) Note that the following circuit has an internal current source, so a supermesh is
required.
6
50V
+
_
IA
8
4
+
V1
_
IB
4A
IC
5
2
Lecture #9
EGR 260 – Circuit Analysis
2) The supermesh is the new, larger mesh created by removing the current source
(as shown on the following page).
Supermesh
6
50V
+
_
8
4
IA
+
V1
_
IB
5
IC
3) Note that the supermesh defines a path for a KVL equation. No new mesh
current is defined.
4) Also note that the internal current source can be used to form a relationship
between currents IB and IC. In general, this is referred to as the supermesh
relationship.
6
50V
+
_
IA
8
4
+
V1
_
IB
4A
IC
5
Supermesh
relationship:
IB - I C = 4
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Lecture #9
EGR 260 – Circuit Analysis
Example: Analyze the circuit shown below using mesh equations. Use the results
to find the voltage V1.
8
6
50V
+
_
4
+
V1
4A
5
_
4
Lecture #9
EGR 260 – Circuit Analysis
Example: Analyze the circuit shown below using mesh equations. Use the results
to find the voltage V1 and the current i1. Is a supermesh required? Also discuss
systems of units.
48 V
_+
_
5k
V1
3 mA
4k
+
8k
i1
10k
+
_
75 V
5
Lecture #9
EGR 260 – Circuit Analysis
Reading Assignment: Chapter 5 in Electric Circuits, 7th Edition by Nilsson
Chapter 5 – Operational Amplifiers
Note: We are temporarily skipping the remaining sections of Chapter 4. We will
cover them after completing Chapter 5.
Operational Amplifier - An operational amplifier (op amp) is a high gain differential
amplifier with nearly ideal external characteristics. Internally the op amp is
constructed using many transistors.
-
V
+VDC
I-
V
+
V+
I+
_
Io
VO
+
-VDC
Note: Sometimes the supply voltage
connections are not shown
Terminology:
V+ = non-inverting input voltage
V- = inverting input voltage
Vo = output voltage
Io = output current
I+ = non-inverting input current
I- = inverting input current
VDC = positive and negative DC
supply voltages used to power the
op amp (typically 5V to 30V)
V = V+ - V- = difference voltage
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Lecture #9
EGR 260 – Circuit Analysis
Reading Assignment: Chapter 5 in Electric Circuits, 6th Edition by Nilsson
Operational Amplifiers
U1
3
+
V+
8-pin package (3D view)
7
8-pin package pinout
OS2
2
-
V-
OUT
OS1
5
6
1
4
uA 741
uA741 symbol in PSPICE
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Lecture #9
EGR 260 – Circuit Analysis
Typical Operational Amplifier Schematic
8
EGR 260 – Circuit Analysis
Lecture #9
Open-loop versus closed-loop operation:
Open-loop:
• Relatively rare
• Op amp specifications may be important
A OL = open-loop gain
_
V
(typical value: A OL = 100,000)
V
+
+
V
VO
+
V = differential input voltage
In general, Vo = A OL  V  or A OL =
Vo
V
Closed-loop:
• Most commonly used
• Some sort of feedback from output to input exists
• The input voltage, Vin, is defined according to the application
Feedback
A CL = closed-loop gain
_
Vin
VO
A CL =
Vo
Vin
+
9
Lecture #9
EGR 260 – Circuit Analysis
An op amp circuit can be easily analyzed using the following ideal assumptions.
Ideal op-amp assumptions:
 Assume that V = 0, so V+ = V Assume the input resistance is infinite, so I+ = I- = 0
 Realize the all voltages defined above are node voltages w.r.t. a common ground (as
illustrated below)
Illustration: Draw an op amp showing a common negative terminal for all node
voltages.
10
Lecture #9
EGR 260 – Circuit Analysis
Example: Determine an expression for Vo in the inverting amplifier shown below.
Illustrate the results using both DC and AC inputs.
R2
_
Vin
R1
+
Vo
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Lecture #9
EGR 260 – Circuit Analysis
Load Connections: Vo is typically determined independent of the load (the circuit
connected to the output). Once Vo has been determined, it essentially acts like a
voltage source to the load.
Load Current: Io is the output current for an op amp. It can be found using KCL.
Example: Find Vo , I1, V2, and Io below.
12k
_
2V
Vo
4k
+
Io
3k
1k
I1
6k
+
9k
V2
_
12
Lecture #9
EGR 260 – Circuit Analysis
Example: Determine an expression for Vo in the non-inverting amplifier shown below.
R2
_
R1
Vo
Vin
+
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Lecture #9
EGR 260 – Circuit Analysis
Example: Determine Vo in the inverting summing amplifier shown below.
RF
V1
_
R1
V2
+
Vo
R2
V3
R3
14
Lecture #9
EGR 260 – Circuit Analysis
Example: Determine Vo in the non-inverting summing amplifier shown below.
RF
_
R
+
Vo
V1
R1
V2
R2
V3
R3
15
EGR 260 – Circuit Analysis
Lecture #9
Example: Determine VL in the circuit shown below.
3k
_
2k
12 V
+
_
12k
4k
+
2k
+
3k
4k
VL
_
16