ece2262 electric circuits chapter 5: circuit...

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ECE2262 Electric Circuits

Chapter 5: Circuit Theorems

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Equivalence

Linearity

Superposition

Thevenin’s and Norton’s Theorems

Maximum Power Transfer

Analysis of Circuits Using Circuit Theorems

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5. 1 Equivalence

Two circuits are equivalent if they have the same i-v characteristics at a specified pair of terminals

Our aim is to simplify analysis replacing complicated sub-circuits by simpler equivalent circuits

4

V1 V2

I1

I2

?

?

5

Source Transformation A source transformation allows a voltage source in series with a resistor to be replaced by a current source in parallel with the same resistor or vice versa.

a

b

a

b

RL RL!

• The current in RL • The current in RL

iL =vs

R + RL

iL =R

R + RL

is

These circuits are equivalent if these resistor currents are the same

vsR + RL

= RR + RL

is ! is =vsR

vs = Ris

6

The resistance in parallel with the voltage source

The resistance in series with the current source

These circuits are equivalent with respect to terminals a,b since they produce the same voltage and current in any resistor RL inserted between nodes a,b .

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Example Find the power P6V = 6V ! i6V

We could use the node-voltage method with 3 node-voltage equations

Let’s use the source transformation strategy by reducing the circuit in a way that preserves the identity of the branch containing the 6 V source.

!

8

Step 1

Step 2 20! || 5! = 4! and 4!" 8A = 32V

6 + 4 + 10 = 20! in series with 32 V source ! 32V20!

= 1.6A

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Step 3

Step 4 30! || 20! = 12! and 12!"1.6A = 19.2V

The current in the direction of the voltage drop across the 6 V source is

i6V = 19.2 ! 616

= 0.825A ! P6V = 6V ! i6V = 4.95 W (absorbing)

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Example Find v0 and P250V , P8A

Remove the resistors 125 ! and 10 ! , they do not influence the formula for

v0 ! equivalent circuit

source transformation

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25 ||100 || 20 = 10 !

v0 = 2A!10" = 20V

12

P250V ! need I250V

I250V = 250125

+ 250 ! v025

= 250125

+ 250 ! 2025

= 11.2

P250V = 250 !11.2 = 2800 W (delivering)

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P8A ! need V8A

v0 !10 " 8 !V8A = 0 ! V8A = 20 ! 80 = !60V

P8A =V8A ! 8A = "480 W (delivering)

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5. 2 Linearity = additivity + homogeneity

v1v2 v3

ix

2ix

2!is

vs

4!

1!

The node-voltage method v1 : v1 = vs

v2 : v2 ! v14

+ 2 ixv1!v32

! + is = 0 ! 34v1 +

14v2 ! v3 = !is

v3 : v31+ v3 ! v1

2! is = 0 ! ! 1

2v1 +

32v3 = is

15

vs

is

1 0 03/ 4 1 / 4 !1!1 / 2 0 3 / 2

"

#

$$$

%

&

'''

v1v2v3

"

#

$$$

%

&

'''=

vs!isis

"

#

$$$

%

&

'''

v1v2v3

"

#

$$$

%

&

'''=

1 0 0!5 / 3 4 / 3 01/ 3 0 2 / 3

"

#

$$$

%

&

'''

vs!isis

"

#

$$$

%

&

'''

4 8 / 3

x! y ! Ay = x ! y = A!1x

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Additivity: If x1! y1 and x2 ! y2 then x1 + x2 ! y1 + y2 Homogeneity: If x! y then

!x!!y for any number !

vs

is

!v

!

!

vs, is( ) = 4V, 2A( ) v = 3V

"8V ,"4A( ) v =

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Additivity: If x1! y1 and x2 ! y2 then x1 + x2 ! y1 + y2

Homogeneity: If x! y then !x!!y for any number ! ! Linearity: x1! y1 and x2 ! y2 then

!1x1 +! 2x2 !!1y1 +! 2y2

for any numbers !1 , ! 2

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Example

• Let I0 = 1mA , find the corresponding I • The true value of I = 6mA

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I2I3

I0 = 1mA

• CD: I0 =66 + 3

I2 =23I2 ! I2 =

32I0 =

32mA

• 3 || 6 + 2 = 4k! ! VI = I2 ! 4 = 6V • I3 =VI4 + 8

= 12

mA

• I = I2 + I3 =32+ 12= 2mA

Hence for the assumed I0 = 1mA we have I = 2mA , then by the linearity if I = 6mA = 3! 2 ! 3!1= 3mA ! I0

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Example: Circuits 1 and 2 below are identical except for the voltage sources. Assuming that I1 = 5A then the value of I2 is ?

I2+

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5. 3 Superposition

Superposition Principle:

Let f1,... fN be set of independent source strengths (vS ,iS ) in a linear resistive circuit. Then, any electrical response y in the circuit v,i( ) can be expressed as y = k1 f1 + k2 f2 + ....+ kN fN ,where k1,..,kN are constants coefficients, unique for each response

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

The principle of superposition allows us to reduce a complicated multi- source problem to several simple problems. Each problem contains only a single independent source.

= 0 = 0

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Example

v1v2 v3

ix

2ix

2!is

vs

4!

1!

• If vs = 0 (short circuit) ! v1' = 0 , v2

' = ! 43is , v3

' = 23is

• If is = 0 (open circuit) ! v1'' = vs , v2

'' = ! 53vs , v3

'' = 13vs

• It is clear that v1 = v1' + v1

'' , v2 = v2' + v2

'' , v3 = v3' + v3

''

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Example 5.3 Find V0

(a) Inactivate the 3V source ! Response to 2 mA

CD: I0 =1+ 21+ 2 + 6

! 2m = 23mA ! V0

' = I0 ! 6k = 4V

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(b) Inactivate the 2mA source ! Response to 3 V

V0'' = 6

6 + 2 +1! 3V = 2 V

By the superposition principle: V0 =V0' +V0

'' = 4 + 2 = 6 V.

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Example 5.4 Find V0

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(a) Inactivate the 2mA source

• 2 + 6( ) || 4 = 8 / 3k ! VD: V1 =8 / 38 / 3+ 2

! 6V = 247V

• VD: V0' = 66 + 2

!V1 = 187V

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(b) Inactivate the 6 V source

• 2 || 4 = 4 / 3k ! V0'' = 2 + 4

3!"#

$%& || 6 ' 2m = 30

7V

By the superposition principle: V0 =V0' +V0

'' = 187+ 307

= 487

! 6.86V

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Example Find v0 : circuit with dependent sources

(a) Response to the 10 V source

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• v!' = "0.4 # v!

'( )#10 ! v!' = 0 ! 0.4 ! v"

' = 0

• v0' = 205 + 20

!10 = 8V

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(b) Response to the 5 A source

• KCL at a : v0''

5+ v0

''

20! 0.4 " v#

'' = 0 ! 5v0'' ! 8v"

'' = 0

• KCL at b: 0.4 ! v"'' + vb # 2i"

''

10# 5 = 0

! 4v!

'' + vb " 2i!'' = 50 ! v!

'' = vb " 2i!'' & i!

'' = "v0'' / 5

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• Hence we have 5v0

'' ! 8v"'' = 0 and 4v!

'' + vb " 2i!'' = 50 ! v!

'' = vb " 2i!'' & i!

'' = "v0'' / 5

! 9v0

'' ! 40vb = 02v0

'' + 5vb = 50

"#$%

! v0'' = 16V & vb = 18 / 5V

• By the superposition principle: v0' + v0

'' = 8 +16 = 24 V

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Example Superposition Applied to Op-Amp Circuits

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• Contribution of V1

This is a basic inverting circuit: V01 = ! R2R1V1

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• Contribution of V2

This is a basic non-inverting circuit: V02 = 1+ R2R1

!"#

$%&V2

Principle of Superposition: V0 =V01 +V02 = ! R2R1V1 + 1+ R2

R1

"#$

%&'V2

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Warning: Power does not obey superposition

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5. 4 Thevenin’s and Norton’s Theorems

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Linear Circuit

A BLOAD

Any Circuit

i

v+!

a

b

!VTh + i " RTh + v = 0

v = VTh ! i "RTh

• the load draws current i and results in voltage v •

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Thevenin Equivalent of Circuit A

BLOAD

Any Circuit

i

v+!

a

b

v = VTh ! i "RTh

A

• The values of VTh (Thevenin voltage) and RTh may be either positive or negative

• RTh - the Thevenin resistance is a quantity in a mathematical model - it is not a physical resistor

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BLOAD

Any Circuit

i

v+!

a

b

v = VTh ! i "RTh

A

• How to calculate the Thevenin Voltage VTh from v ?

Assume i = 0 (open circuit circuit - no external load) ! v = VTh

VTh = voc

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Norton Equivalent of Circuit A

BLOAD

Any Circuit

i

v+!

a

b

v = VTh ! i "RThIN

IN =VThRTh

ARTh

• How to calculate the Norton Current IN from v?

v = VTh ! i "RTh !

vRTh

= VThRThIN!

! i

Assume v = 0 (short circuit) ! IN = isc

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RTh - The Thevenin Resistance 1. The most general way for obtaining RTh is to use

RTh =VThIN

,

where • VTh = voc - open circuit voltage

• IN = isc - short circuit current

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2. The Thevenin resistance RTh can be determined directly by a source suppression method without finding the Thevenin voltage and Norton current. This applies directly to circuits that contain only independent sources. This is a result of the linearity property of the circuit.

RTh =VThIN

= VTh !!IN !!

! - scaling applied to independent sources

! ! 0 - the Thevenin resistance remains unchanged even in the limit case when all independent sources are suppressed to zero.

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(1) Replace all independent voltage sources in the circuit by short circuits and all independent current sources by open circuits

a

b

Independent SourcesDeactivated

RThA

(2) If the remaining circuit contains no dependent sources, then RTh is the equivalent resistance, which can be determined by using series/parallel resistor combinations.

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A. Thevenin/Norton Equivalent for Circuits with Independent Sources

Example Find a Thevenin/Norton Equivalent

1. Open circuit voltage at a ! b : vab = v1 ! the voltage across the 3 A source

KCL : v1 ! 255

+ v120

! 3 = 0 ! v1 = 32V ! voc = 32V

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2. Short circuit current at a ! b :

KCL: v2 ! 255

+ v220

! 3+ v224= 0 ! v2 = 16 V

isc =164

= 4A ! RTh =vocisc

= 324

= 8 !

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3. The Thevenin Equivalent

4. The Norton Equivalent

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5. The Thevenin resistance ! Direct Method

RTh

RTh = 4 + 20 || 5 = 8!

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6. The circuit with load

24!

24!

V24! = 2424 + 8

" 32 = 24V

I24! = 24V24!

= 1A ! P24! = 12 " 24 = 24W

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Example 5.5

1. Open circuit voltage

Voc = 3+V1 , V1 = 2 !10"3( )! 3k = 6V ! Voc = 9V

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2. Short circuit current

Isc = 2m + I1

= 2m + 3V3k

= 2m +1m = 3mA

RTh =VocIsc

= 9V3m

= 3k

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3. RTh

RTh = 3k! !

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4. The circuit with load

V0 =66 + 3

! 9 = 6V ! P6k! = 62

6k= 6mW

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Example 5.6 Find V0

12V

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1. Open circuit voltage and RTh

Voc1 =V6k =66 + 3

!12 = 8V

RTh = 2 + 3 || 6 = 4k!

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2. Circuit with load

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3. Second Iteration

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• Open circuit voltage and RTh

Voc2 = 8 + 2m ! 4k = 16V

RTh = 4k!

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4. Circuit with load

16V

V0 =8

8 + 4 + 4!16 = 8V

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Example 5.7 Find V0

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1. Open circuit voltage and RTh

• Mesh-Current Method: !6 + 4k " I1 + 2k " I1 ! I2( ) = 0I2 = 2m

#$%

! I1 = 5 / 3 mA

• KVL: Voc = 4k ! I1 + 2k ! I2 "Voc =203+ 4 = 32

3V

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• RTh

RTh = 2 + 2 || 4 =103k!

Note: Isc =VocRTh

= 32 / 310 / 3

= 3.2mA

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2. Thevenin Equivalent with Load

Thevenin

V0 =6

6 + 103

! 323

= 487

V = 6.857...V

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B. Thevenin and Norton Equivalent for Circuits with Dependent Sources

Valid and Invalid Partitions

We cannot split the dependent source and its controlling variable when we break the circuit to find the equivalent Thevenin/Norton circuits

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B1 Thevenin/Norton Equivalent for Circuits with Only Dependent Sources: Test Source Approach

I0

RTh =1VI0

RTh =V01mA

V0

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Example 5.8 Find RTh

I0

RTh =1VI0

• Voc = ?

• Thevenin equivalent ?

71

I0 = I1 + I2 + I3 • KVL (big loop): !V1 !Vx +1 = 0 ! V1 = 1!Vx

• KCL (at V1 ): V11k

+ V1 ! 2Vx2k

+ V1 !11k

= 0 ! Vx =37V

• I0 = I1 + I2 + I3 ! I1 =Vx1k

= 37mA , I2 =

1! 2Vx1k

= 17mA , I3 =

12k

= 12mA

! I0 =1514

mA ! RTh =1 VI0

= 1415

k!

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Example 5. 9 Find RTh

• Voc = ?

• Thevenin equivalent ?

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V1

RTh =V2

1 mAV2

• KCL: V1 : V1 ! 2000Ix

2k+ V11k

+ V1 !V23k

= 0 ! Ix =V11k

V2 : V2 !V13k

+ V22k

!1m = 0

! V2 =107V ! RTh =

V2

1 mA= 10

7k!

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B2 Thevenin/Norton Equivalent for Circuits with Both Independent and Dependent Sources

Example 5.10 Find V0

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supernodeVoc +12

1. Open circuit voltage

• KCL at the supernode: Voc +12( )+ 2000Ix'

1k+ Voc +12

2k+ Voc2kIx'!

= 0

Since Ix' = Voc

2k ! Voc = !6V !VTh = "6V

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2. Short circuit current and RTh

!

Isc = ! 121|| 2k

= ! 122 / 3k

= !18mA

RTh =VocIsc

= !6V!18mA

= 13k!

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3. Circuit with Load

Thevenin

Vo =1

1+1+ 13

! "6( ) = "187V = - 2.57…V

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Example Find Thevenin Equivalent

1. Open circuit voltage VTh = vab = v25! = v

ix = 0( ) ! VTh = !20i( )" 25 = !500i !

i = 5 ! 3v2k

= 5 ! 3VTh2k

! VTh = !5V

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2. Short circuit current and RTh

! v = 0 !

isc = !20i , i = 52k

= 2.5mA ! isc = !20 " 2.5 = !50mA

RTh =VThisc

= !5!50m

= 100"

Thevenin

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Find the Thevenin Equivalent RTh Using a Test Source The equivalent method is to first deactivate all independent sources and then apply either a test voltage source or a test current source The Thevenin resistance RTh is calculated as

• RTh =VtestI0

• RTh =V0Itest

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Example

Deactivate the independent sources and excite the circuit by a test source

Note: Use, e.g., vT = 1V

82

RTh =vTiT

• iT = vT25

+ 20i

• i = !3vT2k

= ! 32vTmA

! • iT = vT25

+ 20 ! " 32vTmA

#$%

&'( = = vT

25! 602000

vT

! • iTvT

= 125

! 6200

= 1100

! RTh =vTiT

= 100!

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5. 5 Maximum Power Transfer

• Circuit analysis plays an important role in the analysis of systems designed to transfer power from a source to a load.

• The efficiency of the power transfer: power utility systems are a good example of this type because they are concerned with the generation, transmission, and distribution of large quantities of electric power. If a power utility system is inefficient, a large percentage of the power generated is lost in the transmission and distribution processes, and thus wasted.

•Power transferred: communication and instrumentation systems are good examples because in the transmission of information, or data, via electric signals, the power available at the transmitter or detector is limited. Thus, transmitting as much of this power as possible to the receiver, or load, is desirable. In such applications the amount of power being transferred is small, so the efficiency of transfer is not a primary concern.

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• We consider maximum power transfer in systems that can be modeled by a purely resistive circuit.

VTh,RTh( )

We wish to determine the value of of RL that permits maximum power delivery to RL for the given circuit represented by the Thevenin Equivalent

86

• Represent the Resistive Network by its Thevenin Equivalent

Pload = i2RL =

VThRTh + RL

!"#

$%&

2

RL 'maxRL

87

Pload =VTh

RTh + RL

!"#

$%&

2

RL

Pload

RL

88

Maximum is reached at the point where dPloaddRL

= 0 ! Pload =VTh

RTh + RL

"#$

%&'

2

RL

dPloaddRL

=VTh2 ddRL

RL

RTh + RL( )2!"#

$#

%&#

'#

= VTh2 RTh + RL( )2 ! 2RL RTh + RL( )

RTh + RL( )4"

#$$

%

&''

= 0

89

Condition for Maximum Power Transfer

RL = RTh

The Maximum Power Delivered to RL

Pload =VTh

RTh + RL

!"#

$%&

2

RL !RL=RTh! Pmax =

VTh2RTh

!"#

$%&

2

RTh

Pmax =

VTh2

4RTh

90

Example Find RL that results in max. power and the corresponding max. power that can be delivered to RL .

• VTh =150

150 + 30! 360 = 300 V • RTh = 150 || 30 = 25!

91

iL ! maxvL !max

• Pload =VTh

RTh + RL

!"#

$%&

2

RL =300

25 + RL

!"#

$%&

2

RL

Pload

RL

! RL = 25! for the maximum power transfer with

Pmax =30025 + 25

!"#

$%&

2

25 = 900W

92


93

1. RTh

RTh = 4k + 3k || 6k = 6k This is the resistance for maximum power transfer

If we wish to find the value of the power that can be transferred then we need the Thevenin voltage !

94

2. Voc

* loop 1: I1 = 2mA * loop 2: 3k I2 ! I1( )+ 6kI2 + 3V = 0 ! I2 =13mA

* KVL: Voc = 6kI2 + 4kI1 ! Voc = 10V

* Pmax =VTh2

4RTh =

102

4 ! 6k= 256mW

95


96

1. Open circuit voltage Voc

Voc ! 2000Ix'

• KCL at the supernode:

Voc ! 2000Ix

'

3k +1k+ Voc2k

! 4m = 0 and Ix' = Voc

2k ! Voc = 8V

97

2. Short circuit current Isc and RTh

• Ix

'' = 0 ! dependent source is zero ! Isc = 4mA

• RTh =8V4mA

= 2k! <-------- Obtain by the test source approach

98

3. Circuit with Load

6

• Pload ==8V

6k + RL

!"#

$%&

2

RL is maximized by RL = 6k!

• The maximum power transfer: Pmax =8V

6k + 6k!"#

$%&

2

6k = 83mW

99

Example Plot Vout , I , Pin , Pout and Pout / Pin as a function of R2

iL ! maxvL !max

PL !max

100

* Vout =R2

R1 + R2Vin ; * I = Vin

R1 + R2

* Pin = I !Vin = Vin2

R1 + R2 * Pout = I !Vout =

VinR1 + R2

!"#

$%&

2

R2

* efficiency = PoutPin

= R2

R1 + R2 =

R2 / R11+ R2 / R1

! !1+!

101

10.5

102

* As R2! then Vout ! Vin = 5 V * As R2! then I!

* small R2 ! Vout is small , large R2 ! I is small ! * maximum power transfer (R2 / R1 = 1) does not correspond to max. efficiency

* At R2 / R1 = 1 ! efficiency = PoutPin

= 0.5 (50%)

* The fact that the eff. is higher for R2 > R1 is due to the fact that a higher percentage of the source power is transferred to the load (more $ for MH), but the value of the load power is lower since the total circuit resistance goes up maximum power transfer! 2R1 < R1 + R2

ece2262 electric circuits chapter 5: circuit...

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