EE 201 source transformations – 1

Consider the two circuits below. In particular, look at the current and voltage of RL in each circuit. Using any of the techniques we seen so far, it is easy to find iRL and vRL for each case.

Source transformations

Interesting: From the point of view of the resistor RL, the series combination of the voltage source and resistor RS gives the exact same result as the parallel combination of current source and resistor RP.

+–

RS

RL–

+vRLVS

iRL

5 V

2 k!

1 k!RP RL

–

+vRLIS

iRL

2.5 mA 1 k!2 k!

vRL =RL

RL + RSVS

=1 kΩ

1 kΩ + 2 kΩ (5 V) = 1.67 V

iRL = 1.67 mA, P = 2.78 mW

iRL =1RL

1RL

+ 1RP

IS

=1

1 kΩ1

1 kΩ + 12 kΩ

(2.5 mA) = 1.67mA

vRL = 1.67 V, P = 2.78 mW

EE 201 source transformations – 2

The series combination seems to behave identically to the parallel combination. This suggests that we may, in the right circumstances, replace one configuration for the other. Making this switch is known as a source transformation.

VS

RS

IS

RPIS = VS/RS

RP = RS

IS

RP

VS

RS

VS = ISRP

RS = RP

EE 201 source transformations – 3

These are perfectly viable substitutions. From the point of view of whatever circuitry is attached to the two terminals, the result will be exactly the same for the two source configurations.

This is an example of bigger, more important idea known as the Thevenin equivalent of circuit. We will introduce this later, and make extensive use of it in discussing amplifiers, etc.

If you are trying calculate something about the two components (VS-RS or IP-RP), you cannot transform them. In transforming them, you lose the ability to calculate a specific property.

EE 201 source transformations – 4

25 Ω

50 Ω 200 Ω 100 Ω50 V

100 Ω200 Ω50 Ω25 Ω2 A

Then use a current divider:

Example 1

Find iR4 in the circuit at right.

Use a source transformation to put everything in parallel.

IS = VS/R1 = 50 V/25 ! = 2A.

+–

R1

R2 R3 R4VSiR4

R1 R2 R3 R4ISiR4

iR4 =1R4

1R1 + 1

R2 + 1R3 + 1

R4IS

=1

100 Ω1

25 Ω + 150 Ω + 1

200 Ω + 1100 Ω

(2 A) = 0.267 A Easy.

EE 201 source transformations – 5

Example 2

Find iR2 in the circuit at right.

5 kΩ

7.5 kΩ20 V 2 mA

Transform the voltage source / resistor combo. IST = VS/R1 = 20 V/5 Ω = 4 mA.

4 mA

6 mA

Combine the two current sources, IP = IST + IS = 6 mA…

…and use the current divider then once again.

+–

VS IS

R1

R2iR2

5 kΩR1 R2IST

iR2

IS7.5 kΩ 2 mA

5 kΩ 7.5 kΩR1 R2IP

iR2

iR2 =1R2

1R1 + 1

R2IP

=1

7.5 kΩ1

7.5 kΩ + 15 kΩ

(6 mA) = 2.4 mA

EE 201 source transformations – 6

Example 3

VS+– IS

R1

R2iR1

(An incorrect application)

Find iR1 in the circuit at left.

40 V

1 kΩ

1 kΩ 15 mA

The previous example worked nicely so use the same method. Transform VS & R1, and use current divider with the total current.

ISR1 R2

iR1ISTL5� =

�5�

�5� + �

5�(,67 + ,6)

40 mA 1 kΩ 1 kΩ 15 mA =�

�Nż�

�Nż + ��Nż

(��P$) = ��.�P$

It seems nice, but it is wrong because you cannot transform the component for which you are trying to find voltage or current. (To see that it is wrong, insert iR1 = 27.5 mA in the original circuit and show that there are serious inconsistencies with the currents and voltages.)

EE 201 source transformations – 7

Example 3 (Redo it correctly.)

VS+– IS

R1

R2iR1 Find iR1 in the circuit at left.

Transform IS & R2.

VS+–

R1 R2

iR1 VST+–

40 V

1 kΩ

1 kΩ 15 mA

40 V

1 kΩ 1 kΩ

15 V

Writing a KVL loop equation and solving for iR1 gives

L5� =96 � 9675� + 5�

=��9� ��9�Nż + �Nż = ��.�P$This is the correct answer.

EE 201 source transformations – 8

Example 4

VS1 +–

+–

VS2 ISR1

R2

R3

R4

R5Find vR5 in the circuit at left.

120 V 5 Ω36 A

60 V20 Ω

6 Ω

1.6 Ω

8 Ω vR5

+

–

Transform two voltage sources to current sources. (Pay attention to polarity.)

ISIST1 IST2R1 R2 R3

R4R5

36 A6 A 12 A

20 Ω 5 Ω

6 Ω

1.6 Ω

8 Ω vR5

+

–

EE 201 source transformations – 9

Example 4 (cont.)Add the parallel current sources into one. Combine the parallel resistors into one.

Ieq ReqR4

R5

1.6 Ω

8 Ω vR5

+

–

Ieq = 6A – 12 A + 36 A = 30 A.

Req = 20 Ω || 5 Ω || 6 Ω = 2.4 Ω.

30 A 2.4 Ω

Transform Ieq & Req:

+–Veqt

Req R4R5

72 V

2.4 Ω 1.6 Ω

8 Ω vR5

+

–

Use voltage divider:

Y5� =5�

5HT + 5� + 5�9HTW

=�ż

�.�ż + �.�ż + �ż (��9)

= ��9