Circuit Theorems

Dr. Mustafa Kemal Uyguro ğlu

Circuit Theorems Overview

 Introduction  Linearity  Superpositions   Source Transformation Th évenin and Norton Equivalents  Maximum Power Transfer

INTRODUCTION

A large complex circuits Simplify circuit analysis Circuit Theorems

‧

Thevenin’s theorem

‧

Circuit linearity

‧

source transformation

‧ ‧ ‧

Norton theorem Superposition max. power transfer

Linearity Property

A linear element or circuit satisfies the properties of 

Additivity :

requires that the response to a sum of inputs is the sum of the responses to each input applied separately.

If

v

1 =

i

1 R and

v

2 =

i

2 R then applying (

i

1 +

i

2 )

v =

(

i

1 +

i

2 ) R =

i

1 R

+ i

2 R =

v

1 +

v

2

Linearity Property



Homogeneity:

If you

multiply the input

(i.e.

current

) by some

constant K

, then the

output response

(

voltage

) is

scaled by the same constant

. If

v

1 =

i

1 R then K

v

1 = K

i

1 R

Linearity Property

 A linear circuit is one whose output is linearly related (or directly proportional) to its input. i I 0 Suppose v s = 10 V gives

i

the linearity principle, v s = 2 A. According to = 5 V will give

i

= 1 A.

V 0 v

Linearity Property - Example

i 0

Solve for

v

0 and

i

0 as a function of

V

s

Linearity Property – Example

(continued)

Linearity Property - Example

Ladder Circuit

3 A 5 A 2 A + 14 V + 6 V 2 A + 8V + 3 V 1 A + 5 V This shows that assuming

I

0 current of 15 A will give

I

0 = 1 A gives

I

s = 5 A; the actual source = 3 A as the actual value.

Superposition

 The

superposition

principle states that the voltage across (or current through) an element in a linear circuit is the algebraic sum of the voltages across (or currents through) that element due to each independent source acting alone.

Steps to apply superposition principle

 1.

2.

3.

Turn off all independent sources except one source. Find the output (voltage or current) due to that active source using nodal or mesh analysis.

  Turn off voltages sources = short voltage sources; make it equal to zero voltage Turn off current sources = open current sources; make it equal to zero current Repeat step 1 for each of the other independent sources.

Find the total contribution by adding algebraically all the contributions due to the independent sources.

Dependent sources are left intact .

Superposition - Problem

2mA 2k W

I

0 4mA 1k W 12V – + 2k W

2mA Source Contribution

2mA 2k W 1k W

I’

0

I’

0 = -4/3 mA 2k W

4mA Source Contribution

2k W 4mA 1k W

I’’

0

I’’

0 = 0 2k W

12V Source Contribution

2k W 1k W

I’’’

0

I’’’

0 = -4 mA 12V – + 2k W

Final Result

I’

0

I’’

0 = -4/3 mA = 0

I’’’

0 = -4 mA

I

0 =

I’

0 +

I’’

0 +

I’’’

0 = -16/3 mA

Example

 find v using superposition

one independent source at a time, dependent source remains

KCL: i = i1 + i2 Ohm's law: i = v1 / 1 = v1 KVL: 5 = i (1 + 1) + i2(2) KVL: 5 = i(1 + 1) + i1(2) + 2v1 10 = i(4) + (i1+i2)(2) + 2v1 10 = v1(4) + v1(2) + 2v1 v1 =

10/8

V

Consider the other independent source

KCL: i = i1 + i2 KVL: i(1 + 1) + i2(2) + 5 = 0 i2(2) + 5 = i1(2) + 2v2 Ohm's law: i(1) = v2 v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2 i2(2) + 5 = i1(2) + 2v2 -2v2 = (i - i2)(2) + 2v2 -2v2 = [v2 + (5+2v2)/2](2) + 2v2 -4v2 = 2v2 + 5 +2v2 -8v2 = 5 =>

v2 = - 5/8

V from superposition: v = -5/8 + 10/8

v = 5/8 V

Source Transformation

 A source transformation is the process of replacing a voltage source

v s

resistor

R

by a current source in series with a

i s

in parallel with a resistor

R

, or vice versa

Source Transformation

v s



i s R

or

i s



v s R

Source Transformation

V s



R s I s I s



V s R s

Source Transformation

 Equivalent sources can be used to simplify the analysis of some circuits.

 A voltage source in series with a resistor is transformed into a current source in parallel with a resistor.

 A current source in parallel with a resistor is transformed into a voltage source in series with a resistor.

Example 4.6

 Use source transformation to find

v o

circuit in Fig 4.17.

in the

Example 4.6

Fig 4.18

Example 4.6

we use current division in Fig.4.18(c) to get

i

 2 2  8 ( 2 )  0 .

4 A and

v o

 8

i

 8 ( 0 .

4 )  3 .

2 V

Example 4.7

 Find

v x

in Fig.4.20 using source transformation

Example 4.7

 3  5

i



v x

 18  0  Applying KVL around the loop in Fig 4.21(b) gives  3  5

i



v

 18  0 (4.7.1)

x

 Appling KVL to the loop containing only the 3V voltage source, the resistor, and

vx

yields  3  1

i



v x

 0 

v x

 3 

i

(4.7.2)

Example 4.7

Substituting this into Eq.(4.7.1), we obtain 15  Alternatively 5

i

 3  0 

i

  4 .

5 A 

v x

 4

i



v x

 18  0 

i

  4 .

5 A thus

v x

 3 

i

 7 .

5 V

Thevenin’s Theorem

  Any circuit with sources (dependent and/or independent) and resistors can be replaced by an equivalent circuit containing a single voltage source and a single resistor.

Thevenin’s theorem implies that we can replace arbitrarily complicated networks with simple networks for purposes of analysis.

Implications

 We use Thevenin’s theorem to justify the concept of input and output resistance for amplifier circuits.

 We model transducers as equivalent sources and resistances.

 We model stereo speakers as an equivalent resistance.

Independent Sources (Thevenin)

R Th V oc

+ – Circuit with independent sources Thevenin equivalent circuit

No Independent Sources

Circuit without independent sources

R Th

Thevenin equivalent circuit

Introduction

 Any Thevenin equivalent circuit is in turn equivalent to a current source in parallel with a resistor [source transformation].

 A current source in parallel with a resistor is called a Norton equivalent circuit.

 Finding a Norton equivalent circuit requires essentially the same process as finding a Thevenin equivalent circuit.

Computing Thevenin Equivalent

 Basic steps to determining Thevenin equivalent are – – Find

v oc

Find

R Th

Thevenin/Norton Analysis

1. Pick a good breaking point in the circuit (cannot split a dependent source and its control variable). 2.

Thevenin

: Compute the open circuit voltage,

V OC

.

Norton

: Compute the short circuit current,

I SC

.

For case 3(

b

) both

V OC

=0 and

I SC

=0 [so skip step 2]

Thevenin/Norton Analysis

3. Compute the Thevenin equivalent resistance,

R Th

(

a

) If there are only independent sources, then short circuit all the voltage sources and open circuit the current sources (just like superposition). (

b

) If there are only dependent sources, then must use a test voltage or current source in order to calculate

R Th

=

V Test

/

I test

(

c

) If there are both independent and dependent sources, then compute

R Th

from

V OC

/

I SC

.

Thevenin/Norton Analysis

4.

Thevenin

: Replace circuit with

V OC

in series with

R Th

Norton

: Replace circuit with

I SC

in parallel with

R Th

Note: for 3(

b

) the equivalent network is merely

R Th ,

that is, no voltage (or current) source.

Only steps 2 & 4 differ from Thevenin & Norton!