Principles of Computer Engineering: Lecture 5: Source Transformation

Introduction Source transformations are useful method of circuit analysis It is theoretically possible to replace any given arbitrary linear

circuit containing any number of sources and resistances with either a Thévenin equivalent or Norton equivalent circuit

A simple source transformation allows a voltage source with series resistance to be replaced by a current source with a parallel resistance and vice versa.

Both circuits behave in the same way to external loads.a

b

Thévenin Equivalent Circuit All linear circuits can be modelled by an independent voltage

source and a series resistor The voltage of the source is the open circuit voltage of the

network across ‘a’ and ‘b’ The resistance is determined from the short circuit current

Norton Equivalent Circuit All linear circuits can be modelled by an independent current

source and a parallel resistor The current of the source is the short circuit current through ‘a’

and ‘b’ The resistance is determined from the open circuit voltage

(same as Thévenin resistance)

Example 1 Find both Thévenin and Norton equivalent circuits for the

circuit below

Example 2 (TQ5) Find both Thevenin and Norton equivalent circuits for the

circuit below

Summary Thévenin equivalent circuit Norton equivalent circuit Examples Questions

Tutorial Question 6 Solve for io using source transformations

Principles of Computer Engineering:Lab Exp 5: Source Transformation

Dr. Steve Alty

Network Simplification1. Construct circuit network2. Measure open-circuit voltage3. Measure short-circuit current4. Determine Thévenin equivalent

circuit5. Determine Norton equivalent

DC Power Supply

+15

0

b a

1k

1k

1k 4.7k

4.7k

Measured

Voc = vs

Isc = is R=Voc/Isc

Network Simplification1. Use equivalents to work out effect

of 1k resistor as a load (across ‘a’ and ‘b’) on Thévenin and 4.7 k resistor as a load on Norton.

2. Measure actual effect of resistors using DMM

3. Analyse circuit and calculate mathematically the equivalent circuits

DC Power Supply

+15

0

b a

1k

1k

1k 4.7k

4.7k

Results

Rab (kΩ) Predicted Vab (v)

Measured Vab (v)

Predicted Iab (mA)

Measured Iab (mA)

1(vs=??, R=??)

4.7(is=??, R=??)

Proof

Theoretical calculate the Voc and Isc

DC Power Supply

+15

0

b a

1k

1k

1k 4.7k

4.7k

Open circuit

Req=1 + (5.7||5.7) = 1+2.85 = 3.85 kΩIs = 15/3.85 = 3.9 mAV1k = 3.9×1 = 3.9vV5.7k = 15-3.9 = 11.1vVb0 = (1/5.7) ×11.1 = 1.95vVa0 = (4.7/5.7) ×11.1 =9.15vVab = Va0 - Vb0 = 9.15-1.95 = 7.2v = Voc

DC Power Supply

+15

0

b a

1k

1k

1k 4.7k

4.7k

Is

I1

I3

I2

I4

Iab

Short circuit

Req = 1+(4.7||1)+(1||4.7) = 1+0.825+0.825 = 2.65 kΩ

Is = 15/2.65k = 5.66 mA

V0.82k = 5.66×0.825 = 4.67 v

I1 = 4.67/1k = 4.67 mAI2 = 4.67/4.7k = 0.99 mAI3 = 4.67/4.7k = 0.99 mAI4 = 4.67/1k = 4.67 mA

I1 =Iab + I3

Iab = I1 - I3 = 4.67 – 0.99 = 3.68 mA = Isc