survival skills for circuit analysis

Survival Skills forCircuit Analysis

What you need to know from ECE 109Phyllis R. Nelson

[email protected]

Professor, Department of Electrical and Computer Engineering

California State Polytechnic University, Pomona

P. R. Nelson Fall 2010 WhatToKnow —- – p. 1/46

mailto:[email protected]

Basic CircuitConcepts

All circuits can be analyzed by manymethods.




Some methods have specific advantages forcalculating a particular result in a specificcircuit.





Some methods yield useful intuition aboutcircuit behavior.





Some methods yield useful intuition aboutcircuit behavior.

No single analysis method will be thebest for all possible circuits!


Kirchhoff’s Laws

These two methods are the most general tools ofcircuit analysis.

Sign errors in writing Kirchhoff’s laws are themost common error I observe in student work inhigher-level classes.

Sign errors are not trivial errors!


Kirchhoff’s current lawis equivalent to stating that electrical chargecannot be stored, created, or destroyed locally ina conductor.

The algebraic sum of all currentsentering a node is zero.

Alternate statements:

The algebraic sum of the currents leaving anode is zero.

The total current entering a node is equal tothe total current leaving the node.


Kirchhoff’s voltage law

is derived by considering the forces on anelectrical charge that travels around a closedloop in a circuit and has the same starting andending velocity. The total force on the chargemust be zero, so the total change in electricalpotential energy (voltage) around the loop mustbe zero.

The algebraic sum of the voltagedrops around any closed loop in a

circuit is zero.


Solving K*L equations

Kirchhoff’s laws applied directly result in∑

i

Ii = 0 (KCL)

∑

i

Vi = 0 (KVL)


Solving K*L . . .

Solving KCL means finding the node voltages.

Solving KVL means finding the branch currents.

⇒ need relationships between currentand voltage for each circuit element.

These relationships describe the operation of thecircuit elements.


Ohm’s law+ V −

I V = IR

Ohm’s law is an example of a current-voltage(I-V ) relation.

To use K*L, you must know the I-V relationfor every circuit element.

I-V relations include definition of therelationship between the direction of thecurrent and the sign of the voltage.


Series resistors,voltage divider

R1

+ V1−

R2

+ V2−

Ri

+ Vi−

Rn

+ Vn−

−VT+I

VT = I Req

Prove that these equations are true:

Req =n∑

j=1

Rj Vi =Ri

∑nj=1 Rj

VT


VT =n∑

j=1

Vj =n∑

j=1

I Rj = I

n∑

j=1

Rj = I Req

⇒ Req =n∑

j=1

Rj

Vi = I Ri =

(

VT

Req

)

Ri =

(

Ri∑n

j=1 Rj

)

VT


Parallel resistors,current divider

−

V

+

I

R1 I1 R2 I2 V = I Req

Prove that these equations are true:

R−1eq =

2∑

i=1

R−1i Req =

R1R2

R1 + R2Ix =

(

Ry

Rx + Ry

)

I

Which can be extended to more resistors?P. R. Nelson Fall 2010 WhatToKnow —- – p. 11/46

I = I1 + I2 =V

R1+

V

R2= V

(

2∑

i=1

R−1i

)

=V

Req

⇒ R−1eq =

2∑

i=1

R−1i

Req =1

R1+

1

R2=

R1R2

R1 + R2


Let x = 1 and y = 2. Then

I1 =V

R1=

I Req

R1=

(

R1R2

R1+R2

)

R1I =

(

R2

R1 + R2

)

I

If x = 2 and y = 1 then

I2 =

(

R1

R1 + R2

)

I


For n resistors in parallel,

I =n∑

j=1

Ij =n∑

j=1

V

Rj

= V

(

n∑

j=1

R−1j

)

=V

Req

⇒ R−1eq =

n∑

j=1

R−1j

This is the only formula for parallel resistors thatextends easily to more than two resistors.


Illegal circuits

The following configurations are not allowed:

a short-circuited voltage source

an open-circuited current source

Why?


KCL Example

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.


KCL Example

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.

Hint: Since KCL is applied at nodes, how manyKCL equations will be required?


KCL Example - 1

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.


KCL Example - 1

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.

1. Since KCL is applied at nodes, count thenodes.


KCL Example - 1

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.

1. Since KCL is applied at nodes, count thenodes. (4)


KCL Example - 2, 3

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.


KCL Example - 2, 3

−+Vx

R1

R2

R3

R4 Iy

Find VR2and VR4

.

2. Choose one node as ground. Choose forconvenience!

3. Label node voltages.


−+Vx

R1

R2

R3

R4 Iy

V2 V4


−+Vx

R1

R2

R3

R4 Iy

V2 V4

Both VR2and VR4

are connected to ground.

The voltage source is connected to ground⇒ one less KCL equation.


KCL Example - 4

−+Vx

R1

R2

R3

R4 Iy

V2 V4

2 unknown node voltages ⇒ 2 KCL equations.

4. Define currents for all elements connected tonodes “2” and “4”.


KCL Example - Eqn’s

−+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:



−+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:I1 − I2 − I3 = 0



−+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:I1 − I2 − I3 = 0

KCL at 4:



−+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

KCL at 2:I1 − I2 − I3 = 0

KCL at 4:I3 − I4 − Iy = 0


KCL Example - I-V’s

−+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

Write unknown I’s in terms of node V’s:


KCL Example - I-V’s

−+Vx

R1

I1R2 I2

R3

I3R4 I4 Iy

V2 V4

Write unknown I’s in terms of node V’s:

I1 =Vx − V2

R1I2 =

V2

R2

I3 =V2 − V4

R3I4 =

V4

R4P. R. Nelson Fall 2010 WhatToKnow —- – p. 22/46

KCL Example -Substitution

Vx − V2

R1−

V2

R2−

V2 − V4

R3= 0

V2 − V4

R3−

V4

R4− Iy = 0

Check! 2 equations in 2 unknowns

This pair of equations can be solved, but it’smessy unless resistor values are given.


KVL Example

−+VA

R1

R2

R3

I3

IB R4

Find I3.


KVL Example

−+VA

R1

R2

R3

I3

IB R4

Find I3.

Hint: Since KVL is applied around closed loops,how many KVL equations will be required?


KVL Example - 1

−+VA

R1

R2

R3

I3

IB R4

Find I3.


KVL Example - 1

−+VA

R1

R2

R3

I3

IB R4

Find I3.

1. Count the number of “windowpanes” in thecircuit.


KVL Example - 1

−+VA

R1

R2

R3

I3

IB R4

Find I3.

1. Count the number of “windowpanes” in thecircuit. (3)


KVL Example - 2, 3

−+VA

R1

R2

R3

I3

IB R4

Find I3.


KVL Example - 2, 3

−+VA

R1

R2

R3

I3

IB R4

Find I3.

2. Choose one closed loop for eachwindowpane. Choose for convenience!

3. Choose a direction for each loop.P. R. Nelson Fall 2010 WhatToKnow —- – p. 26/46

KVL Example

−+VA

R1

R2

R3

I3

IB R4

Only one loop passes through R3, ensuring thatthe unknown can be used as a “loop current.”


KVL Example - 4, 5

−+VA

R1

R2

R3

I3

IB R4

4. Define loop current labels for each loop. Becareful not to give different currents the samelabel!

5. Define voltages for all branch elements.P. R. Nelson Fall 2010 WhatToKnow —- – p. 28/46

KVL Example - I & V

−+VA

R1

R2

R3

IB R4I1 I3 I4

−+VA

R1

R2

R3

IB R4

+ V1 − + V3 −

V2

−

+V4

−


KVL Example - Eqn’s

KVL for loop 1:



KVL for loop 1:

−VA + V1 + V2 = 0



KVL for loop 1:

−VA + V1 + V2 = 0

KVL for loop 3:



KVL for loop 1:

−VA + V1 + V2 = 0

KVL for loop 3:

−V2 + V3 + V4 = 0



KVL for loop 1:

−VA + V1 + V2 = 0

KVL for loop 3:

−V2 + V3 + V4 = 0

KVL for loop 4: ??!?



KVL for loop 1:

−VA + V1 + V2 = 0

KVL for loop 3:

−V2 + V3 + V4 = 0

KVL for loop 4: ??!?−V4 + V4 = 0 is a tautology . . .



KVL for loop 1:

−VA + V1 + V2 = 0

KVL for loop 3:

−V2 + V3 + V4 = 0

KVL for loop 4: ??!?−V4 + V4 = 0 is a tautology . . .

. . . but the current source current is

IB = I4 − I3


KVL Example - I-V’s

−+VA

R1

R2

R3

IB R4I1 I3 I4

+ V1 − + V3 −

V2

−

+V4

−

Write unknown V’s in terms of loop I’s:


KVL Example - I-V’s

−+VA

R1

R2

R3

IB R4I1 I3 I4

+ V1 − + V3 −

V2

−

+V4

−

Write unknown V’s in terms of loop I’s:

V1 = I1 R1 V2 = (I1 − I3) R2

V3 = I3 R3 V4 = I4 R4


KVL Example -Substitution

−VA + R1 I1 + (I1 − I3) R2 = 0

− (I1 − I3) R2 + I3 R3 + I4 R4 = 0

I4 − I3 = IB

Check! 3 equations in 3 unknowns

Collect terms, then solve this set ofequations by Kramer’s rule.


KVL Example - Matrix

R1 + R2 −R2 0

−R2 R2 + R3 R4

0 −1 1

I1

I3

I4

=

VA

0

IB

∆ =

R1 + R2 −R2 0

−R2 R2 + R3 R4

0 −1 1

= (R1 + R2) (R2 + R3) + (R1 + R2) R4 − R22

= R1R2 + R1R3 + R1R4 + R2R3 + R2R4


∆ I3 =

R1 + R2 VA 0

−R2 0 R4

0 IB 1

= R2VA − (R1 + R2) R4VA

I3 =R2 VA − (R1 + R2) R4 IB

R1R2 + R1R3 + R1R4 + R2R3 + R2R4


Equivalent circuits

Equivalent circuits enable us to treat a portion ofa circuit like a “black box” in that we can give itthe simplest possible model.

There are two possible models that account forboth power sources and impedance:

a voltage source in series with a resistor

a current source in parallel with a resistor

Equivalent circuits can be used tosimplify a circuit before solving

for an unknown!P. R. Nelson Fall 2010 WhatToKnow —- – p. 35/46

Thèvenin equivalent

−+Voc

Req

I

−

V+ V = Voc − Req I

I =Voc

Req

−V

Req

If the terminals are connected to an open circuit,I = 0 and V = Voc.

If the terminals are shorted, V = 0 andI = Voc/Req = Isc.


Norton equivalent

Isc Req

−

V+

I

I = Isc −V

Req

V = Req Isc − Req I

If the terminals are shorted, V = 0 and I = Isc.

If the terminals are connected to an open circuit,I = 0 and V = Req Isc.


Graphicalrepresentation

All possible combinations of pairs of voltage andcurrent values for either equivalent circuit modelcan be represented as the line through the points(V = 0, ISC) and (VOC , 0).

I

V

ISC

VOC


Superposition

A voltage or current in a linear circuit is thesuperposition (sum) of the results for eachsource alone.

Superposition can be used along with paralleland series resistors, voltage and current dividers,and Thèvenin and Norton equivalent circuits tosimplify the analysis of a circuit.

Superposition often gives solutionsin a mathematical form that enhancesintuition.


SuperpositionExample

−+Vx

R1

R2 I2 R3 Iy

Find the current I2 using superposition.

Solve for the current I2 with

Iy = 0 so that Vx is the only source

Vx = 0 so that Iy is the only source

then add the results.P. R. Nelson Fall 2010 WhatToKnow —- – p. 40/46

SuperpositionExample - V

Set Iy = 0 and solve for the current I2V .

−+Vx

R1

R2 I2V R3


SuperpositionExample - V

Set Iy = 0 and solve for the current I2V .

−+Vx

R1

R2 I2V R3

I2V =

(

R3

R1R2 + R1R3 + R2R3

)

Vx


SuperpositionExample - I, Result

Set Vx = 0and solve forthe current I2I .

R1

R2 I2I R3 Iy




R1

R2 I2I R3 Iy

I2I = −

(

R1R3

R1R2 + R1R3 + R2R3

)

Iy




R1

R2 I2I R3 Iy

I2I = −

(

R1R3

R1R2 + R1R3 + R2R3

)

Iy

I2 =

(

R3

R1R2 + R1R3 + R2R3

)

(Vx − R1Iy)


Dependent Sources

Circuits containing dependent sources are solvedby the same methods used for other circuits.

There is one extra step, because the final resultcannot contain the dependent variable!

Not eliminating the dependentvariable from the final result isanother of the most commonerrors I see in student work!


Dependent SourceExample

−+Vx

R1

R2

+V2

−

+−

BV2

R3

+V3

−

Find V3.


Dependent SourceExample

−+Vx

R1

R2

+V2

−

+−

BV2

R3

+V3

−

Find V3.

Hint: The answer may only containVx, R1, R2, R3, and B.


Dependent SourceExample - Eqn’s

−+Vx

R1

R2

+V2

−

+−

BV2

R3

+V3

−

Find V3.



−+Vx

R1

R2

+V2

−

+−

BV2

R3

+V3

−

Find V3.

V2 − Vx

R1+

V2

R2+

V3

R3= 0



−+Vx

R1

R2

+V2

−

+−

BV2

R3

+V3

−

Find V3.

V2 − Vx

R1+

V2

R2+

V3

R3= 0

−V2 + B V2 + V3 = 0


Dependent SourceExample - Result

−+Vx

R1

R2

+V2

−

+−

BV2

R3

+V3

−

Find V3.

V3 =Vx

R1

[

R1+R2

(1−B)R1R2

+ 1R3

]


survival skills for circuit analysis

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