Circuit Theorems

Dr. Mustafa Kemal Uyguroğlu

Circuit Theorems Overview

Introduction Linearity Superpositions Source Transformation Thévenin and Norton Equivalents Maximum Power Transfer

INTRODUCTION

A largecomplex circuits

Simplifycircuit analysis

Circuit Theorems

‧Thevenin’s theorem Norton theorem‧‧Circuit linearity Superposition‧‧source transformation 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 v1 = i1R and v2 = i2R

then applying (i1 + i2)

v = (i1 + i2) R = i1R + i2R = v1 + v2

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 v1 = i1R then K v1 =K i1R

Linearity Property

A linear circuit is one whose output is linearly related (or directly proportional) to its input.

Suppose vs = 10 V gives i = 2 A. According to the linearity principle, vs = 5 V will give i = 1 A.

vV0I0

i

Linearity Property - Example

i0

Solve for v0 and i0 as a function of Vs

Linearity Property – Example (continued)

Linearity Property - Example

1 A

+

5 V

-

+ 3 V -

+

8V

-

2 A

3 A

+ 6 V -

+

14 V

-

2 A

5 A

This shows that assuming I0 = 1 A gives Is = 5 A; the actual source current of 15 A will give I0 = 3 A as the actual value.

Ladder Circuit

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. 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 current2. Repeat step 1 for each of the other independent sources.3. Find the total contribution by adding algebraically all the

contributions due to the independent sources. Dependent sources are left intact.

Superposition - Problem

2k1k

2k12V

I0

2mA

4mA– +

2mA Source Contribution

2k1k

2k

I’0

2mA

I’0 = -4/3 mA

4mA Source Contribution

2k1k

2k

I’’0

4mA

I’’0 = 0

12V Source Contribution

2k1k

2k12V

I’’’0

– +

I’’’0 = -4 mA

Final Result

I’0 = -4/3 mAI’’0 = 0I’’’0 = -4 mA

I0 = 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 = 0i2(2) + 5 = i1(2) + 2v2Ohm's law: i(1) = v2

v2(2) + i2(2) +5 = 0 => i2 = -(5+2v2)/2i2(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 vs in series with a resistor R by a current source is in parallel with a resistor R, or vice versa

Source Transformation

RviRiv s

sss or

Source Transformation

sss IRV s

ss R

VI

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 vo in the circuit in Fig 4.17.

Example 4.6

Fig 4.18

Example 4.6

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

and

A4.0)2(82

2

i

V2.3)4.0(88 ivo

Example 4.7

Find vx in Fig.4.20 using source transformation

Example 4.7

Applying KVL around the loop in Fig 4.21(b) gives (4.7.1)Appling KVL to the loop containing only the 3V voltage source, the resistor, and vx yields (4.7.2)

01853 xvi

ivvi xx 3013

01853 xvi

Example 4.7

Substituting this into Eq.(4.7.1), we obtain

Alternatively thus

A5.403515 ii

A5.40184 iviv xx

V5.73 ivx

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)

Circuit with independent sources

RTh

Voc

Thevenin equivalent circuit

+–

No Independent Sources

Circuit without independent sources

RTh

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 voc

– Find RTh

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, VOC.

Norton: Compute the short circuit current, ISC.

For case 3(b) both VOC=0 and ISC=0 [so skip step 2]

Thevenin/Norton Analysis

3. Compute the Thevenin equivalent resistance, RTh

(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

RTh = VTest/Itest

(c) If there are both independent and dependent sources, then compute RTh from VOC/ISC.

Thevenin/Norton Analysis

4. Thevenin: Replace circuit with VOC in series with RTh

Norton: Replace circuit with ISC in parallel with RTh

Note: for 3(b) the equivalent network is merely RTh , that is, no voltage (or current) source.

Only steps 2 & 4 differ from Thevenin & Norton!