Review of Circuit Analysis
Common Symbols
Independent Sources
Battery
• Constant voltage
• DC voltage source
+-
Voltage Source
• Can be constant or time varying
• +/- indicates polarity
Current Source
• Can be constant or time varying
• Arrow denotes direction of current flow
Dependent Sources
• Depend upon the voltage/current measured at some location in the circuit to provide its value
• Often a linear proportionality term, but can be non-linear
+-
Dependent voltage source
Dependent current source
Reference Voltages
Ground
• Reference potential
• 0V
Vcc / Vdd
• Supply voltage
• Typically highest potential
• Often called a supply “rail”
1
Ohm’s Law
V = IR
R
V1
V2
I
(V1 − V2 ) = IR
I=
V1 − V2
R
More generally, as applied to complex impedances, Z
Z
V1
V2
I
(V1 − V2 ) = IZ
I=
V1 − V2
Z
Resistors in Series
R1
R2
Rser = R1 + R2
=
100Ω
Ex.
Rser =
500Ω
=
2
Resistors in Parallel
R1
Rpar = R1//R2
=
R2
1
1
1
=
+
R par R1 R2
R par =
R1 R2
= R1 R2
R1 + R2
Ex.
300Ω
Rpar =
=
150Ω
The parallel combination will always be less than the smaller of the two resistors
Useful cases
• Parallel combination of 2 identical resistors
•
Case where one resistor is much larger than the other
→ Equivalent resistance looks mostly like the smaller of the two resistors
3
Kirchoff’s Voltage Law (KVL)
• “Loop Equations”
• Sum of all voltage drops around any loop equals zero
∑V
∑V
x
up
=0
=∑ Vdown
Ex. Find the current flowing through R2 and the voltage across R2.
R1
Vin
+-
R2
4
Kirchoff’s Current Law (KCL)
• “Node Equations”
• Sum of all currents entering [exiting] a node equals zero
∑I = 0
∑ I =∑ I
x
in
out
Ex. Find the current flowing through R2.
Iin
R1
R2
5
Superposition
• Only valid for linear circuits
• For the case where there are multiple voltage/current sources
• The voltage [current] at any node [branch] of a circuit can be found by adding the sum of the
contributions of each source, while all others are turned off
• Method
o Keep only one source on. Turn all others off. Solve for the desired voltage/current.
Vx
+-
Ix
o
o
o
o
OFF
+-
OFF
0V
=
Short Circuit
0A
=
Open Circuit
Turn first source off. Turn next source on. Solve for the desired voltage/current.
…
Repeat for all N sources
Desired voltage/current is the sum of each case
Ex.
R2=10kΩ
R1=1kΩ
+
V1=1V
+-
Vx=?
+-
V2=5V
-
6
Thevenin / Norton Equivalent Circuits
• “Black box” representation of a circuit
• Exactly the same when viewed from the terminals
• Internal circuitry may be different
Thevenin
Norton
How to determine values – Vth, IN, Rth=RN
Rth = RN
Turn off all sources (V → short circuit; I → open circuit)
Solve for the equivalent resistance
Vth
Disconnect (i.e. open circuit) the circuit at the location of interest
Solve for the “open-circuit voltage” at this location
IN
Short circuit the location of interest
Calculate the current that flows through this short (i.e. “short-circuit current”)
Ex. Find the Thevenin and Norton equivalent circuits to the left of the amplifier.
R1
Vin
+-
R2
R3
Amplifier
7
8
Source Transformation
• Thevenin and Norton versions are equivalent
• You can switch between the two representations without issue
Vx = I x Rx
where
Ix =
Vx
Rx
9
Measure Input Impedance
•
•
•
•
Let the output voltage float
Apply a test current source at the input, Itest
Measure the change in the input voltage,
Vtest
Rin =
Vtest
I test
Vtest
Let Vout float
Itest
Ex. Find the input resistance of the following circuit.
R1
+
Vin
-
R2
+
Vx
-
0.01Vx
R3
+
Vout
-
10
Measure Output Impedance
• Turn off the input voltage
• Apply a test current source at the output, Itest
• Measure the change in the output voltage, Vtest
•
Rout =
Vtest
I test
Ex. Find the output resistance of the following circuit.
R1
+
Vin
-
R2
+
Vx
-
0.01Vx
R3
+
Vout
-
11
Capacitors
• Store energy in the form of an electric field
• Typically parallel plates separated by a non-conductive material
I =C
V =
dV
dt
1 t
I ( x )dx + Vinit
C ∫0
Capacitors in parallel → add like resistors in series
C1
Cpar = C1 + C2
Capacitors in series → like resistors in parallel
C1
C2
=
Cser = C1 // C2
=
C2
Impedance of capacitors is frequency dependent (sinusoidal signals)
As ω → 0
The capacitor acts like an open circuit
As ω → ∞
The capacitor acts like a short circuit
Complex Impedance
ZC =
1
=
j ωC
Easier method of using complex impedances
→ Use Laplace transforms…
dV 

Laplace I = C

dt 

I = sCV
Solve Ohm’s Law
Z CL =
1
sC
Where s = σ + jω is the complex frequency
If you replace the capacitance with its complex impedance, you can treat it like a resistor
12
Inductors
• Store energy in the form of an magnetic field
• Typically coiled wire/conductive material
V =L
I=
dI
dt
1 t
V ( x )dx + I init
L ∫0
Combine like resistors (series/parallel)
Impedance of inductors is frequency dependent
As ω → 0
The inductor acts like a short circuit
As ω → ∞
The inductor acts like an open circuit
Complex Impedance
Z L = j ωL =
Using Laplace Notation
dI 

Laplace V = L 
dt 

V = sLI
∴ Z L = sL
Where s = σ + jω is the complex frequency
Treat complex impedance like a resistor and follow normal circuit analysis
13
Time-Domain Response of Circuits with Energy-Storage Elements
Ex. Find Vout(t) for the following circuit for t > 0. The switch closes at time t = 0
(
)
Let the capacitor have no initial charge stored on it → Vout t = 0 − = 0V
Let the input signal be a constant voltage
t=0
R
Vin
+-
C
+
Vout
-
14