+ - Can be constant or time varying +/- indicates polarity - - PDF document

1

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”

2 Ohm’s Law

IR V =

V1 V2 R I ( )

R V V I IR V V

2 1 2 1

− = = −

More generally, as applied to complex impedances, Z

V1 V2 Z I ( )

Z V V I IZ V V

2 1 2 1

− = = −

Resistors in Series

R1 R2 = Rser = R1 + R2

Ex.

100Ω = Rser = 500Ω

3 Resistors in Parallel

R1 R2 = Rpar = R1//R2

2 1

1 1 1 R R Rpar + =

2 1 2 1 2 1

R R R R R R Rpar = + =

Ex.

300Ω 150Ω = Rpar =

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

4 Kirchoff’s Voltage Law (KVL)

• “Loop Equations”
• Sum of all voltage drops around any loop equals zero

∑

= 0

x

V

∑ ∑

=

down up

V V

• Ex. Find the current flowing through R2 and the voltage across R2.

+

• R1

R2 Vin

5 Kirchoff’s Current Law (KCL)

• “Node Equations”
• Sum of all currents entering [exiting] a node equals zero

∑

= 0

x

I

∑ ∑

=

• ut

in

I I

• Ex. Find the current flowing through R2.

R1 R2 Iin

6 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
• Keep only one source on. Turn all others off. Solve for the desired voltage/current.

+

• OFF

+

• Vx

0V = Short Circuit Ix OFF 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.

+

• R1=1kΩ

V1=1V R2=10kΩ

+

• V2=5V

+

• Vx=?

7 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 R3 R2 Amplifier

8

9 Source Transformation

• Thevenin and Norton versions are equivalent
• You can switch between the two representations without issue

where

x x x x x x

R V I R I V = =

10 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

• test

test in

I V R =

Itest Let Vout float Vtest

• Ex. Find the input resistance of the following circuit.

R2 R1 R3 0.01Vx + Vout

• +

Vx

• +

Vin

11 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
• test

test

• ut

I V R =

• Ex. Find the output resistance of the following circuit.

R2 R1 R3 0.01Vx + Vout

• +

Vx

• +

Vin

12 Capacitors

• Store energy in the form of an electric field
• Typically parallel plates separated by a non-conductive material

dt dV C I =

( )

init t

V dx x I C V + =

∫

1

Capacitors in parallel → add like resistors in series

C1 C2 = Cpar = C1 + C2

Capacitors in series → like resistors in parallel

C1 C2 = Cser = C1 // C2

Impedance of capacitors is frequency dependent (sinusoidal signals) As

→ ω

The capacitor acts like an open circuit As

∞ → ω

The capacitor acts like a short circuit Complex Impedance

= = C j ZC ω 1

Easier method of using complex impedances → Use Laplace transforms…

sCV I dt dV C I Laplace =       =

Solve Ohm’s Law

sC ZCL 1 =

Where

ω σ j s + =

is the complex frequency If you replace the capacitance with its complex impedance, you can treat it like a resistor

13 Inductors

• Store energy in the form of an magnetic field
• Typically coiled wire/conductive material

dt dI L V =

( )

init t

I dx x V L I + = ∫ 1

Combine like resistors (series/parallel) Impedance of inductors is frequency dependent As

→ ω

The inductor acts like a short circuit As

∞ → ω

The inductor acts like an open circuit Complex Impedance

= = L j Z L ω

Using Laplace Notation

sL Z sLI V dt dI L V Laplace

L =

∴ =       =

Where

ω σ j s + =

is the complex frequency Treat complex impedance like a resistor and follow normal circuit analysis

14 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 →

( )

V t Vout = =

−

Let the input signal be a constant voltage

+

• R

C Vin + Vout

• t = 0