ECE201 Lect-12 1

Equivalence/Linearity (5.1);Superposition (5.2, 8.8)

Dr. Holbert

March 6, 2006

ECE201 Lect-12 2

Equivalent Sources

• An ideal current source has the voltage necessary to provide its rated current.

• An ideal voltage source supplies the current necessary to provide its rated voltage.

• A real voltage source cannot supply arbitrarily large amounts of current.

• A real current source cannot have an arbitrarily large terminal voltage.

ECE201 Lect-12 3

A More Realistic Source Model

vs(t)

Rs The

Circuit

The Source

i(t)

+

–

v(t)+–

ECE201 Lect-12 4

I-V Relationship

The I-V relationship for this source model is

v(t) = vs(t) - Rs i(t)v(t)

i(t)

ECE201 Lect-12 5

Open Circuit Voltage

• If the current flowing from a source is zero, then the source is connected to an open circuit.

• The voltage at the source terminals with i(t) equal to zero is called the open circuit voltage:

voc(t)

ECE201 Lect-12 6

Short Circuit Current

• If the voltage across the source terminals is zero, then the source is connected to a short circuit.

• The current that flows when v(t) equals zero is called the short circuit current:

isc(t)

ECE201 Lect-12 7

voc(t) and isc(t)

v(t)

i(t)

voc(t)

isc(t)

ECE201 Lect-12 8

voc(t) and isc(t)

• Since the open circuit voltage and the short circuit current determine where the I-V line crosses both axes, they completely define the line.

• Any circuit that has the same I-V characteristics is an equivalent circuit.

ECE201 Lect-12 9

Equivalent Current Source

is(t) Rs

The

Circuit

i(t)

+

–

v(t)

s

ss R

tvti

)()(

ECE201 Lect-12 10

Source Transformation

Vs

Rs

Is Rs

sss IRV s

ss R

VI

+–

ECE201 Lect-12 11

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.

ECE201 Lect-12 12

Averaging Circuit

How can source transformation make analysis of this circuit easier?

+

–

Vout

1k

1k

1k

V1 V2

+–

+–

ECE201 Lect-12 13

Source Transformations

+

–

Vout

1k

1k

1k

V1 V2

+–

+–

ECE201 Lect-12 14

Source Transformations

+

–

Vout

1k1kV1 /1k

1kV2 /1k

Which is a single node-pair circuit that we can use current division on!

ECE201 Lect-12 15

Linearity

Linearity leads to many useful properties of circuits:

– Superposition: the effect of each source can be considered separately.

– Equivalent circuits: any linear network can be represented by an equivalent source and resistance (Thevenin’s and Norton’s theorems).

ECE201 Lect-12 16

Linearity

• More important as a concept than as an analysis methodology, but allows addition and scaling of current/voltage values

• Use a resistor as for example (V = R I):– If current is KI, then new voltage is

R (KI) = KV

– If current is I1 + I2, then new voltage is

R(I1 + I2) = RI1 + RI2 = V1 + V2

ECE201 Lect-12 17

Class Example

• Learning Extension E5.1

ECE201 Lect-12 18

Superposition

“In any linear circuit containing multiple independent sources, the current or voltage at any point in the circuit may be calculated as the algebraic sum of the individual contributions of each source acting alone.”

ECE201 Lect-12 19

The Summing Circuit

+

–

Vout

1k

1k

1k

V1 V2

+–

+–

ECE201 Lect-12 20

Superposition

+

–

V’out

1k

1k

1k

V1

+

–

V’’out

1k

1k

1k

V2++–

+–

ECE201 Lect-12 21

Use of Superposition

V’out = V1/3

V’’out = V2/3

Vout = V’out + V’’out = V1/3 + V2/3

ECE201 Lect-12 22

How to Apply Superposition

• To find the contribution due to an individual independent source, zero out the other independent sources in the circuit.

– Voltage source short circuit.

– Current source open circuit.

• Solve the resulting circuit using your favorite technique(s).

ECE201 Lect-12 23

Problem

2k1k

2k12V

I0

2mA

4mA

– +

ECE201 Lect-12 24

2mA Source Contribution

2k1k

2k

I’0

2mA

I’0 = -4/3 mA

ECE201 Lect-12 25

4mA Source Contribution

2k1k

2k

I’’0

4mA

I’’0 = 0

ECE201 Lect-12 26

12V Source Contribution

2k1k

2k12V

I’’’0

– +

I’’’0 = -4 mA

ECE201 Lect-12 27

Final Result

I’0 = -4/3 mA

I’’0 = 0

I’’’0 = -4 mA

I0 = I’0+ I’’0+ I’’’0 = -16/3 mA

ECE201 Lect-12 28

Superposition Procedure1. For each independent voltage and current source (repeat the

following):

a) Replace the other independent voltage sources with a short circuit (i.e., V = 0).

b) Replace the other independent current sources with an open circuit (i.e., I = 0).

Note: Dependent sources are not changed!

c) Calculate the contribution of this particular voltage or current source to the desired output parameter.

2. Algebraically sum the individual contributions (current and/or voltage) from each independent source.

ECE201 Lect-12 29

Class Example

• Learning Extension E5.2

• Learning Extension E8.15(a) & (b)