ECE 201, Section 4
Lecture 6
Prof. Peter Bermel
August 31, 2012
Recap from Friday
• Series resistors:
= = / ; currents equal
• Parallel resistors
= = / ; voltages equal
• Series-parallel circuits
– Analyzed iteratively
8/31/2012
ECE 201-3, Prof. Bermel
Dependent Sources
• Can use current or voltage to control output
current or voltage
Control type
Voltage
Output
type
8/31/2012
Current
Voltage
Current
VCVS
V=µvx
CCVS
V=IR
+
-
VCCS
I=gV
CCCS
I=βIx
ECE 201-3, Prof. Bermel
+
-
Dependent Sources Example
• What is the output voltage and current, gain,
and total power dissipated?
4Ω
+
I1
V1
20 V
+
-
12 Ω
I=V1/5
8/31/2012
ECE 201-3, Prof. Bermel
2Ω
I2
4Ω
Dependent Sources Solution
4Ω
+
I1
V1
20 V
+
-
12 Ω
Io=V1/5
2Ω
I2
4Ω
-
• Voltage division V1=15V, Io=3A, I1=2A, I2=1A
• Gain g=I2R2/(20V)=0.2 (a 7 dB attenuator)
• Power dissipated=400/16+4*3=37 W
8/31/2012
ECE 201-3, Prof. Bermel
Nodal Analysis
• General linear circuits aren’t simple
combination of series and parallel circuits
• Instead, must apply KCL and Ohm’s law to
solve for voltage at all unknown nodes
8/31/2012
ECE 201-3, Prof. Bermel
Nodal Analysis Example
• For these 5 resistors with a voltage source, solve
for the voltages and currents everywhere:
G1
I1
A
G4
B
I3
I4
G3
C
I2
G2
I5
G5
Io
Io
+ V a
8/31/2012
ECE 201-3, Prof. Bermel
Nodal Analysis Example
G1
• Using KCL:
= + = + = + = + • Using Ohm’s Law:
I1
A
G4
B
I3
G3
C
I2
G2
Io
+ V − = − + a
= − + − − + ( − ) = +
8/31/2012
ECE 201-3, Prof. Bermel
I4
I5
G5
Io
Nodal Analysis Example
G1
• Grouping by voltages:
+ + − = A
−
+ (
+ + ) = + + (
+ ) = (
+ )
• Third equation is sum of first
two and can be eliminated;
good cross-check
• Leaves 2 equations for 2
unknowns, and 8/31/2012
ECE 201-3, Prof. Bermel
I1
G4
B
I3
I4
G3
C
I2
G2
I5
G5
Io
Io
+ V a
Nodal Analysis Example
• Rewriting as matrix:
+ + −
−
= + + • Use matrix inversion formula:
1
−
=
=
det() − • Note det = − . Now check:
1
1 0
−
=
=
0 1
det() − 8/31/2012
ECE 201-3, Prof. Bermel
Nodal Analysis Example
• Using matrix inversion formula, we obtain:
1
+ + =
+ + + + − + + • If % = (0.2,0.4,0.5,0.1,0.7), = 5, then:
1 1.6 0.5 0.2
2.524
=
5 =
2.039
1.03 0.5 0.8 0.4
And % = (0.495, 1.185, 0.243, 0.252, 1.428)
Finally, check KCL is obeyed
8/31/2012
ECE 201-3, Prof. Bermel
Nodal Analysis
• Should always be able to solve problems with
2 unknowns using matrix inversion formula
• What about more than 2 unknowns?
– Adjoint method (calculate cofactor matrix, take
transpose, divide by determinant)
– Software techniques
8/31/2012
ECE 201-3, Prof. Bermel
Free, Web-Enabled Software
• SPICE on nanoHUB:
https://nanohub.org/tools/spice3f4
• Falstad circuit simulator:
http://www.falstad.com/circuit/index.html
8/31/2012
ECE 201-3, Prof. Bermel
Homework
• HW #4 solution now posted
• HW #5 due today by 4:30 pm in EE 325B
• HW #6 due Wednesday: DeCarlo & Lin, Chapter 2:
– Problem 46
– Problem 62 [In place of Pload = 100 Pin, let Pload = 10 Pin.]
– Problem 63
8/31/2012
ECE 201-3, Prof. Bermel