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

Dr. Holbert March 6, 2006 ECE201 Lect-12 1

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 2

A More Realistic Source Model

v

s (

t

) + –

R s i

(

t

) +

v

(

t

) – The Circuit The Source ECE201 Lect-12 3

I-V Relationship The I-V relationship for this source model is

v

(

t

) =

v s

(

t

) -

R s i

(

t

)

v

(

t

) ECE201 Lect-12

i

(

t

) 4

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

:

v oc

(

t

) ECE201 Lect-12 5

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

:

i sc

(

t

) ECE201 Lect-12 6

v

(

t

)

v oc

(

t

) and

i sc

(

t

)

v oc

(

t

)

i sc

(

t

)

i

(

t

) ECE201 Lect-12 7

v oc

(

t

) and

i sc

(

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 8

Equivalent Current Source

i

s (

t

)

R s i

(

t

) +

v

(

t

) – The Circuit

i s

(

t

) 

v s

(

t

)

R s

ECE201 Lect-12 9

V s

+ – Source Transformation

R s I s R s V s



R s I s I s



V s R s

ECE201 Lect-12 10

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 11

V

1 + – Averaging Circuit 1k W 1k W +

V out

– 1k W + –

V

2 How can source transformation make analysis of this circuit easier?

ECE201 Lect-12 12

V

1 Source Transformations + – 1k W 1k W +

V out

– 1k W + –

V

2 ECE201 Lect-12 13

Source Transformations

V

1 /1k W 1k W + 1k W

V out

– 1k W

V

2 /1k W Which is a single node-pair circuit that we can use current division on!

ECE201 Lect-12 14

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 15

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

I

1 +

I

2 , then new voltage is

R

(

I

1 +

I

2 ) =

RI

1 +

RI

2 =

V

1 +

V

2 ECE201 Lect-12 16

Class Example • Learning Extension E5.1

ECE201 Lect-12 17

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 18

The Summing Circuit

V

1 + – 1k W 1k W +

V out

– 1k W + –

V

2 ECE201 Lect-12 19

V

1 + – Superposition 1k W +

V ’ out

– 1k 1k W W

+

1k W 1k W 1k W + –

V ’’ out

+ –

V

2 ECE201 Lect-12 20

Use of Superposition

V out = V

’

out V

’

out V

’’

out

=

V

1 /3 =

V

2 /3 +

V

’’

out

=

V

1 /3 +

V

2 /3 ECE201 Lect-12 21

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  – Current source  short circuit.

open circuit.

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

ECE201 Lect-12 22

Problem 2mA 2k W

I

0 4mA 1k W 12V – + 2k W ECE201 Lect-12 23

2mA Source Contribution 2k W 2mA

I’

0

I’

0 = -4/3 mA 1k W ECE201 Lect-12 2k W 24

4mA Source Contribution

I’’

0 = 0 2k W 4mA

I’’

0 1k W ECE201 Lect-12 2k W 25

12V Source Contribution 2k W

I’’’

0 = -4 mA

I’’’

0 ECE201 Lect-12 1k W 12V – + 2k W 26

Final Result

I’

0 = -4/3 mA

I’’’

0

I’’

0 = 0 = -4 mA

I

0 =

I’

0 +

I’’

0 +

I’’’

0 = -16/3 mA ECE201 Lect-12 27

Superposition Procedure 1. 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 28

Class Example • Learning Extension E5.2

• Learning Extension E8.15(a) & (b) ECE201 Lect-12 29