ECE201Lect-14

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ECE201 Lect-14 1

Norton's Theorem (5.3, 8.8)

Dr. Holbert

March 20, 2006

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ECE201 Lect-14 2

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 requiresessentially the same process as finding a

Thevenin equivalent circuit.

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Independent Sources

Circuit with one ormore independent

sources

RTh

Norton equivalentcircuit

I sc

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No Independent Sources

Circuit withoutindependent sources

RTh

Norton equivalentcircuit

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Finding the Norton Equivalent

• Circuits with independent sources:

– Find V oc

and I sc

– Compute RTh

• Circuits without independent sources:

– Apply a test voltage (current) source – Find resulting current (voltage)

– Compute RTh

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Example: Strain Gauge

• Strain is the amount of deformation of a

body due to an applied force-it is defined as

the fractional change in length.

• Strain can be positive (tensile) or negative

(compressive).

• One type of strain gauge is made of a foilgrid on a thin backing.

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A Strain Gauge

• The strain gauge’s resistance varies as a

function of the strain:

D R =

GF e R

• e is the strain, R is the nominal resistance,

GF is the Gauge Factor

Backing

Foil

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Typical values

• Measured strain values are typically fairly

small-usually less than 10-3.

• GF is usually close to 2.

• Typical values for R are 120W, 350W, and

1000W.

• A typical change in resistance is

D R = 2•10-3•120W = 0.24W

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Measuring Small Changes in R

• To measure such small changes in

resistance, the strain gauge is placed in a

Wheatstone bridge circuit.

• The bridge circuit uses an excitation voltage

source and produces a voltage that depends

on D R.

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The Bridge Circuit

R+D R

V ex

R

R

R

+ – V out

+ –

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Norton Equivalent for Any e

e

e

64

2

R

V I ex sc

e

e

21

64

4

R RTh

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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

) bothV OC =0 and

I SC =0 [so skip step 2]

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3. Compute the Thevenin equivalent resistance, RTh

(or

impedance, ZTh

).

(a) If there are only independent sources, then short

circuit all the voltage sources and open circuit the currentsources (just like superposition).

(b) If there are only dependent sources, then must use a

test voltage or current source in order to calculate

RTh (or ZTh) = V Test / I test (c) If there are both independent and dependent sources,

then compute RTh (or Z

Th) from V OC / I SC .

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4. Thevenin: Replace circuit with V OC

in series with RTh

, ZTh

.

Norton: Replace circuit with I SC

in parallel with RTh

, ZTh

.

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

Only steps 2 & 4 differ from Thevenin & Norton!

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Class Examples

• Learning Extension E5.4

• Learning Extension E8.15(c)

ECE201Lect-14

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