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CIRCUITS by Ulaby & Maharbiz All rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press REVIEW OF LECTURES 05-08 REVIEW OF LECTURES 05 08 07/23/2013 ECE225 CIRCUIT ANALYSIS

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Page 1: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

CIRCUITS by Ulaby & MaharbizAll rights reserved. Do not reproduce or distribute.

© 2013 National Technology and Science Press

REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05 08

07/23/2013 ECE225 CIRCUIT ANALYSIS

Page 2: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Contents of Midterm #2Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

L t 05 08

Midterm #2 Schecule: 10:45 13:15, July 25, Thursday

Sections 3.5-3.6; 4.3-4.7; 5.2-5.3. Lectures 05-08 Sections 3.5 3.6; 4.3 4.7; 5.2 5.3.

Office hours this week• Double office hour on Today

• No office hour on Thursday, July 25 (Midterm#2 day)

9:30am~10:30 am & 16:00~17:00

No office hour on Thursday, July 25 (Midterm#2 day)

• TA office hour remain unchanged

Page 3: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved Do not reproduce or distribute © 2013 National Technology and Science Press All rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press

Page 4: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Summary of Chapter 3 (part 3)y p (p )

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Page 5: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Thévenin’s Theorem

Linear two-terminal circuit can be replaced by an

equivalent circuit composed of a voltage

source and a series resistor

voltage across output with no load (open circuit)

inTh RR

Resistance at terminals with all Resistance at terminals with all independent circuit sources set to zero

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Page 6: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Norton’s TheoremAll rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press

Linear two-terminal circuit can be replaced by an equivalent be replaced by an equivalent circuit composed of a current source and parallel resistor

Current through output with short circuit

Resistance at terminals with all circuit sources set to zero

Page 7: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

How Do We Find Thévenin/Norton E i l t Ci it ?Equivalent Circuits ? Method 1: Open circuit/Short circuit Method 1: Open circuit/Short circuit

1. Analyze circuit to find

2. Analyze circuit to find

Note: This method is applicable to Note: This method is applicable to any circuit, whether or not it contains dependent sources.

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Page 8: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Example 3-10: Thévenin Equivalent All rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press

p q

Page 9: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

How Do We Find Thévenin/Norton E i l t Ci it ?Equivalent Circuits?Method 2: Equivalent ResistanceMethod 2: Equivalent Resistance

1. Analyze circuit to find either

or 2 ll d d b2. Deactivate all independent sources by replacing voltage sources with short circuits and current sources with open circuits.3. Simplify circuit to find equivalent resistance

Note: This method does not apply to circuits that contain dependent sources.p

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Page 10: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Example 3-11: RTh

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p Th

(Ci i h d d )Replace with SC (Circuit has no dependent sources)

Replace with OC

Page 11: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

How Do We Find Thévenin/Norton E i l t Ci it ?Equivalent Circuits?

Method 3:

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Page 12: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

ExampleExample All rights reserved. Do not

reproduce or distribute. © 2013 National Technology

and Science Press p

Page 13: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

E l ( t )

All rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press

To find

Example (cont.)

Page 14: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Power TransferAll rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press

In many situations, we want to y ,maximize power transfer to the load

Page 15: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved. Do not reproduce or distribute. © 2013 National Technology and Science Press

Page 16: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved. Do not reproduce or distribute. © 2013 National Technology

and Science Press

Page 17: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved. Do not copy or distribute. © 2013

National Technology and Science Press

Page 18: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Summary of Chapter 4y p

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Page 19: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Circuit Analysis With Ideal Op AmpsAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

y p p

Use nodal analysis as before, but with “golden rules”

N No current into op amp

No voltage drop across op amp input

Do not apply KCL at op amp output

Page 20: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Noninverting Amplifierg p

At node

021

Rv

Rvv non

spn vvv

so vR

RRv2

21

(max) = Vcc

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Page 21: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Inverting Amplifierg p

0 pn vv

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Page 22: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Example 4-2: Input Current SourceAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

p pRelate output voltage to input current sourcep

Page 23: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Summing Amplifierg p

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Page 24: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Example 4-3: p

Solution:

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© 2013 National Technology and Science Press

Page 25: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Difference AmplifierAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

p

Note negative gain of channel 1

Page 26: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Voltage FollowerAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

g

“Buffers” Sections of CircuitBuffers Sections of Circuit

depends on both input and load iresistors

is immune to input and load resistors

What is the op amp doing?

Page 27: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved. Do not copy or distribute.

© 2013 National Technology and Science Press

Page 28: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved. Do not copy or distribute.

© 2013 National Technology and Science Press

Page 29: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Summary of Chapter 5 (Parts 1)All rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

y p ( )

Page 30: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

CapacitorsAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

p

Passive element that stores energy in electric fieldPassive element that stores energy in electric field

Parallel plate capacitor

AdAC

0 1 tdtit

0oC t

For DC, capacitor looks like open circuitlike open circuit

Voltage on capacitor must be continuous (no abrupt change)abrupt change)

Page 31: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Energy Stored in Capacitorgy p

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Page 32: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Capacitor Response: Given v(t), determine i(t), p(t), and w(t)

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C = C =

Page 33: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

RC Circuits at dcAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

At dc no currents flow through capacitors: open circuits At dc no currents flow through capacitors: open circuits

Page 34: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Capacitors in Seriesp

Use KVL, current same ,through each capacitor

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Page 35: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Capacitors in Parallelp

Use KCL, voltage same , gacross each capacitor

NCCCCC 321eq

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Page 36: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Voltage Divisiong

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Page 37: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

InductorsAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

Passive element that stores energy in magnetic fieldgy g

Solenoid Wound Inductor

0o

1 tidttvL

it

t

At dc, inductor looks like a short circuit

Current through inductor Current through inductor must be continuous (no abrupt change)

AN 2

lANL 2

Page 38: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Inductor Response to p

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Page 39: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Inductors in Series

Use KVL, current is same ,through all inductors

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Page 40: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

Inductors in Parallel

V lt i ll Voltage is same across all inductors

Inductors add together in the same way resistors do

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Page 41: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

RL Circuits at dcAll rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press

At dc no voltage across inductors: short circuit At dc no voltage across inductors: short circuit

Page 42: REVIEW OF LECTURES 05-08REVIEW OF LECTURES 05-08 07/23/2013 ECE225 CIRCUIT ANALYSIS. Contents of Midterm #2 Midterm #2 Schecule: 10:45~13:15 July 25 Thursday

All rights reserved. Do not copy or distribute. © 2013 National Technology and Science Press