EE

70 R

evie

w

Ele

ctr

ical C

urr

ent

∫+

==

t t

tq

dt

ti

tq

dtt

dq

ti

0

)(

)(

)(

)(

)(

0

Circuit E

lem

ents

An e

lectr

ica

l cir

cuit c

onsis

ts o

f circuit e

lem

ents

such a

s v

olta

ge s

ourc

es,

resis

tances,

inducta

nces a

nd c

apacita

nces t

hat

are

conn

ecte

d in c

lose

d

path

s b

y c

ond

ucto

rs

Refe

rence D

irection

s

The v

oltag

e vab

has a

refe

rence p

ola

rity

that is

positiv

e

at

poin

t a

and n

eg

ative a

t poin

t b.

Refe

rence D

irection

s

“downhill: resistor”

“uphill: battery”

Refe

rence D

irection

s

Energ

y is t

ransfe

rred w

hen c

harg

e f

low

s t

hro

ug

h a

n

ele

ment

ha

vin

g a

voltage a

cro

ss it.

∫==

2 1

)(

)(

)(

)(

t t

dt

tp

w

ti

tv

tp

Watts

Joules

Pow

er

and E

nerg

y

•If current flows in the passive configuration the

power is given by p = vi

•If the current flows opposite to the passive

configuration, the power is given by p = -vi

Current is flowing in

the passive

configuration

Refe

rence D

irection

Dep

en

dent

Sourc

es

Resis

tors

an

d O

hm

’s L

aw

Ri

v

iRv

ab

ab==

a b

The units of resistance are Volts/Amp which are called

“ohms”. The symbol for ohms is omega: Ω

ALR

ρ=

ρis the resistivityof the material used to fabricate

the resistor. The units of resitivityare ohm-

meters (Ω

-m)

Resis

tance R

ela

ted to P

hysic

al

Para

mete

rs

Pow

er

dis

sip

ation in a

resis

tor

RvRi

vip

22=

==

Kircoh

off

’sC

urr

ent

Law

(K

CL)

•The net current entering a node is zero

•Alternatively, the sum of the currents

entering a node equals the sum of the

currents leaving a node

Kircoh

off

’sC

urr

ent

Law

(K

CL)

Series C

on

nection

cb

ai

ii

==

Kircoh

off

’sV

oltage L

aw

(K

VL)

The algebraic sum of the voltages

equals zero for any closed path (loop) in

an electrical circuit.

Kircoh

off

’sV

oltage L

aw

(K

VL)

KVL through A and B: -v

a+vb= 0 →

va= v

b

KVL through A and C: -v

a-vc= 0 →

va= -vc

Para

llel C

on

nectio

n

Equiv

ale

nt

Series R

esis

tance

eqiR

RR

Ri

iRiR

iRv

vv

v=

++

=+

+=

++

=)

(3

21

32

13

21

eqR

v

RR

Rv

Rv

Rv

Rvi

ii

i=

++

=+

+=

++

=3

21

32

13

21

11

1

Equiv

ale

nt

Para

llel R

esis

tance

Circuit A

naly

sis

usin

g

Series/P

ara

llel E

quiv

ale

nts

1.

Begin

by locating a

com

bin

ation o

f re

sis

tances that are

in s

eries o

r para

llel. O

ften the p

lace to s

tart

is

fart

hest fr

om

the s

ourc

e.

2.

Redra

w the c

ircuit w

ith the e

quiv

ale

nt

resis

tance for

the c

om

bin

ation found in

ste

p 1

.

total

32

1

11

1v

RR

R

Ri

Rv

++

==

total

32

1

22

2v

RR

R

Ri

Rv

++

==

Of the total voltage, the fraction that appears across a

given resistance in a series circuit is the ratio of the given

resistance to the total series resistance.

Voltag

e D

ivis

ion

total

21

2

1

1i

RR

R

Rvi

+=

=

total

21

1

2

2i

RR

R

Rvi

+=

=

For two resistances in parallel, the fraction of the total

current flowing in a resistance is the ratio of the other

resistance to the sum of the two resistances.

Curr

en

t D

ivis

ion

Nod

e V

oltag

e A

naly

sis

sv

v=

1

03

32

42

2

12

=−

++

−R

vv

Rv

R

vv

03

23

53

1

13

=−

++

−R

vv

Rv

R

vv

Nod

e V

oltag

e A

naly

sis

Mesh C

urr

en

t A

naly

sis

Mesh C

urr

en

t A

naly

sis

Théve

nin

Eq

uiv

ale

nt

Circuits

oc

tv

V=

Théve

nin

Eq

uiv

ale

nt

Circuits

ttsc

RVi

=

Théve

nin

Eq

uiv

ale

nt

Circuits

scoc

ivR

t=

Théve

nin

Eq

uiv

ale

nt

Circuits

Théve

nin

Eq

uiv

ale

nt

Circuits

Nort

on

Equiv

ale

nt

Circuits

scn

iI

=

Nort

on

Equiv

ale

nt

Circuits

Sourc

e T

ran

sfo

rma

tions

tL

LLL

Lt

tL

LL

Lt

tL

RR

dRdP

RR

R

VR

iP

RR

VI

=→

=

+=

=+

=0

2

2

Maxim

um

Po

wer

Tra

nsfe

r

The superposition principle s

tate

s

that th

e tota

l re

sponse is the s

um

of

the r

esponses to e

ach o

f th

e

independent sourc

es a

cting

indiv

idually

. In

equation form

, th

is is

nT

rr

rr

++

+=

L2

1

Supe

rpositio

n P

rin

cip

le

Supe

rpositio

n P

rin

cip

le

Supe

rpositio

n P

rin

cip

le

VV

vR

R

Rv

s5

15

10

5

5

21

21

=+

=+

=

Current source

open circuit

Supe

rpositio

n P

rin

cip

le

Voltage source

short circuit

VV

Vv

vv

VA

AR

R

RR

iR

iv T

seq

s

66

.11

66

.6

5

66

.6

)33

.3

)(2(

510

)5

)(10

()

2(

21

21

21

2

=+

=+

=

=Ω

=+

=+

==

Req

The input resistance R

iis the equivalent resistance see when

looking into the input term

inals of the am

plifier. R

ois the output

resistance. A

vocis the open circuit voltage gain.

Voltag

e-A

mp

lifie

r M

odel

Ideally

, an a

mplif

ier

pro

duces a

n o

utp

ut

sig

nal w

ith identical w

aveshape a

s the

input sig

nal, b

ut w

ith a

larg

er

am

plit

ude.

()() t

vA

tv

iv

o=

Voltag

e G

ain

ioi

iiA

=Li

v

ii

Lo

ioi

RRA

Rv

Rv

iiA

==

=

Curr

en

t G

ain

io PPG

=

()

Liv

iv

ii

oo

io

RRA

AA

IV

IV

PPG

2=

==

=

Pow

er

Gain

Ope

rational A

mplif

ier

Opera

tional am

plif

iers

are

alm

ost alw

ays

used w

ith n

egative feedback, in

whic

h p

art

of

the o

utp

ut sig

nal is

retu

rned to the input in

oppositio

n to the s

ourc

e s

ignal.

Sum

min

g P

oin

t C

on

str

ain

t

In a

negative feedback s

yste

m, th

e ideal op-

am

p o

utp

ut voltage a

ttain

s the v

alu

e n

eeded

to forc

e the d

iffe

rential in

put voltage a

nd input

curr

ent to

zero

. W

e c

all

this

fact th

e

summing-point constraint.

Sum

min

g P

oin

t C

on

str

ain

t

1.

Verify

that negative feedback is p

resent.

2.

Assum

e that th

e d

iffe

rential in

put

voltage

and th

e input curr

ent of th

e o

p a

mp a

re

forc

ed to z

ero

. (T

his

is the s

um

min

g-p

oin

t

constr

ain

t.)

3.

Apply

sta

ndard

circuit-a

naly

sis

princip

les,

such a

s K

irchhoff’s

law

s a

nd O

hm

’s law

, to

solv

e for

the q

uantities o

f in

tere

st.

Sum

min

g P

oin

t C

on

str

ain

t

The B

asic

Invert

er

Apply

ing the S

um

min

g P

oin

t C

onstr

ain

t

12

inRR

vvA

ov

−=

=

Invert

ing A

mplif

ier

Sum

min

g A

mplif

ier

12

in

1RR

vvA

ov

+=

=

Non

-in

vert

ing

Am

plif

iers

10

11

12=

∞+

=+

==

RR

vvA

ov

inVoltag

e F

ollo

wer

()()

()

0

0

tq

dt

ti

tq

t t

+=∫

()()

()

0

0

1t

vdt

ti

Ct

v

t t

+=

∫

Cqv

vqC

==

Cap

acitance

Cap

acitance

s in P

ara

llel

Cap

acitance

s in S

eries

32

1

11

11

CC

CC

eq

++

=

Inducta

nce

The polarity of the voltage is such as to oppose the change in

current (Lenz’s law).

Inducta

nce

Series I

nducta

nces

32

1L

LL

Leq

++

=

Para

llel In

du

cta

nces 3

21

11

LL

LLeq

++

=

Mutu

al In

ducta

nce

Fields are aiding

Fields are opposing

Magnetic flux produced by one coil links the other coil

Dis

cha

rge o

f a C

ap

acitan

ce

thro

ug

h a

Re

sis

tance

()()

0=

+R

tv

dt

tdv

CC

C()

()0

=+

tv

dt

tdv

RC

CC

KCL at the top node of the

circuit:

dt

dv

Cdt

dq

iCv

qvq

CC

==

→=

→=

i Ci R

R

tv

iC

R

)(

=

Dis

charg

e o

f a C

apacitance thro

ugh a

Resis

tance

()()

0=

+t

vdt

tdv

RC

CC

()st

CKe

tv

=

0=

+st

stKe

RCKse

()() t

vRC

dt

tdv

CC

1−

=

We need a function v

C(t)that has the same form

as it’s

derivative.

Substituting this in for v c(t)

RC

s1

−=

()RC

t

CKe

tv

−=

()RC

t

iC

eV

tv

−=

()

iC

Vv

=+

0

Dis

charg

e o

f a C

apacitance thro

ugh a

Resis

tance

Solving for s:

Substituting into v

c(t):

Initial Condition:

Full Solution:

Dis

charg

e o

f a C

apacitance thro

ugh a

Resis

tance

()RC

t

iC

eV

tv

−=

()

iC

Vv

=+

0

()RC

t

CKe

tv

−=

To find the unknown constant K, we need to use the

boundary conditions at t=0. At t=0 the capacitor is initially

charged to a voltage V

iand then discharges through the

resistor.

Dis

charg

e o

f a C

apacitance thro

ugh a

Resis

tance

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

KCL at the node that joins

the resistor and the capacitor

Current into the

capacitor:

dt

dv

Cc

Current through the

resistor:

R

Vt

vS

C−)

(

0)

()

(=

−+

R

Vt

v

dt

tdv

CS

CC

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

0)

()

(=

−+

R

Vt

v

dt

tdv

CS

CC

Rearranging:

SC

CV

tv

dt

tdv

RC

=+

)(

)(

This is a linear first-order differential equation with

contantcoefficients.

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

The boundary conditions are given by the fact that the

voltage across the capacitance cannot change

instantaneously:

0)

0()

0(=

−=

+C

Cv

v

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

stC

eK

Kt

v2

1)

(+

=

SC

CV

tv

dt

tdv

RC

=+

)(

)(

Try the solution:

Substituting into the differential equation:

Gives:

Sst

VK

eK

RCs

=+

+1

2)

1(

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

Sst

VK

eK

RCs

=+

+1

2)

1(

For equality, the coefficient of es

tmust be zero:

RC

sRCs

10

1−

=→

=+

Which gives K

1=V

S

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

RC

tS

stC

eK

Ve

KK

tv

/2

21

)(

−+

=+

=

sS

SC

VK

KV

eK

Vv

−=

→=

+=

+=

+2

20

20

)0(

Substituting in for K

1and s:

Evaluating at t=0 and rem

embering that v

C(0+)=0

RC

tS

Sst

Ce

VV

eK

Kt

v/

21

)(

−−

=+

=

Substituting in for K

2gives:

()τ

t

ss

Ce

VV

tv

−−

=

Charg

ing a

Capacitance fro

m a

DC

Sourc

e thro

ugh a

Resis

tance

DC

Ste

ady S

tate

dt

tdv

Ct

iC

C

)(

)(

=

In steady state, the voltage is constant, so the

current through the capacitor is zero, so it behaves

as an open circuit.

DC

Ste

ady S

tate

dtt

di

Lt

vL

L

)(

)(

=

In steady state, the current is constant, so the

voltage across and inductor is zero, so it behaves

as a short circuit.

The steps in determining the forced response for

RLC circuits with dc sources are:

1. Replace capacitances with open circuits.

2. Replace inductances with short circuits.

3. Solve the remaining circuit.

DC

Ste

ady S

tate

RL T

ransie

nt

Analy

sis

S

S

Vdtt

di

LR

ti

tv

Rt

iV

=+

=+

+−

)(

)(

0)

()

(

SV

dtt

di

LR

ti

=+

)(

)(

ste

KK

ti

Try

21

)(

+=

S

stV

esL

KRK

RK

=+

+)

(2

21

AV

RVK

VRK

SS

250

100

11

=Ω

==

→=

RLs

sLK

RK

−=

→=

+0

22

ste

KK

ti

Try

21

)(

+=

SV

dtt

di

LR

ti

=+

)(

)(

RL T

ransie

nt

Analy

sis

)(

)(

)(

)(

)(

)(

)(

)(

)(

tf

tx

dtt

dx

R

tv

ti

dtt

di

RL

tv

tRi

dtt

di

L

t

t

=+

=+

=+

τ

RC

and R

L C

ircuits w

ith G

enera

l

Sourc

es

First order differential

equation with constant

coefficients

Forcing

function

RC

an

d R

L C

ircuits w

ith G

en

era

l

Sourc

es

The g

enera

l solu

tion c

onsis

ts

of tw

o p

art

s.

The particular solution (also called the

forced response) is any expression that

satisfies the equation.

In order to have a solution that satisfies the

initial conditions, we must add the

complementary solution to the particular

solution.

)(

)(

)(

tf

tx

dtt

dx

=+

τ

The homogeneous equation is obtained by

setting the forcing function to zero.

The complementary solution (also called

the natural response) is obtained by

solving the homogeneous equation.

0)

()

(=

+t

xdtt

dx

τ

Integrators produce output voltages that are

proportional to the running tim

e integral of

the input voltages. In a running tim

e integral,

the upper lim

it of integration is t .

Inte

gra

tors

and D

iffe

rentia

tors

()()

dt

tv

RC

tv

t

oin

0

1∫

−=

()dt

dv

RC

tvo

in−

=

Diffe

rentiato

r C

ircuit

dtt

dv

ti

Cdtt

di

Rdt

ti

dL

s)

()

(1

)(

)( 2

2

=+

+

Differentiating with respect to tim

e:

()()

()(

)() t

vv

dt

ti

Ct

Ri

dtt

di

Ls

C

t

=+

++

∫0

1

0

Secon

d–O

rder

Cir

cuits

dtt

dv

Lt

iLC

dtt

di

LR

dt

ti

ds

)(

1)

(1

)(

)( 2

2

=+

+ LR 2=

α

LC

10=

ω

Dam

pening

coefficient

Undam

ped

resonant

frequency

dtt

dv

Lt

fs

)(

1)

(=

)(

)(

)(

2)

(2 0

2

2

tf

ti

dtt

di

dt

ti

d=

++

ωα

Define:

Forcing function

Secon

d–O

rder

Cir

cuits

()()

()()

()()

()0

22

2 02

2

2 02

2

=+

+

=+

+

ti

dtt

di

dt

ti

d

solution

ary

Complemen

t

tf

ti

dtt

di

dt

ti

d

solution

Particular

ωα

ωα

Solu

tion o

f th

e S

econd-O

rder

Equation

02

:

0)

2(

:

02

:)

(

2 02

2 02

2 02

=+

+

=+

+

=+

+

= ωα

ωα

ωα

ss

equation

stic

Chara

cteri

Ke

ssFactoring

Ke

sKe

Ke

s

Ke

ti

Try

st

stst

st

stC()

()()

02

2 02

2

=+

+t

idtt

di

dt

ti

dω

α

Solu

tion o

f th

e C

om

ple

menta

ry

Equation

0ωα

ζ=

2 0

2

1ω

αα

−+

−=

s

2 0

2

2ω

αα

−−

−=

s

Dam

pening ratio

Roots of the characteristic equation:

Solu

tion o

f th

e C

om

ple

menta

ry

Equation

1. Ove

rdamped

case (ζ> 1). If ζ> 1 (or

equivalently, if α

> ω

0), the roots of the

characteristic equation are real and distinct.

Then the complementary solution is:

()t

st

sc

eK

eK

ti

21

21

+=

In this case, we say that the circuit is

ove

rdamped.

2. Critica

lly damped

case (ζ= 1). If ζ= 1 (or

equivalently, if α

= ω

0), the roots are real

and equal. Then the complementary solution

is

()t

st

sc

teK

eK

ti

11

21

+=

In this case, we say that the circuit is

critically damped.

3. Underdamped

case (ζ< 1). Finally, if ζ

< 1

(or equivalently, if α

< ω

0), the roots are

complex. (By the term

complex, we mean that

the roots involve the square root of –1.) In other

words, the roots are of the form

:

nn

js

js

ωα

ωα

−−

=+

−=

21

and

in which j is the square root of -1 and the

natural freq

uen

cyis given by: 2

2 0α

ωω

−=

n

For complex roots, the complementary solution

is of the form

:

()(

)(

) te

Kt

eK

ti

nt

nt

cω

ωα

αsin

cos

21

−−

+=

In this case, we say that the circuit is

underdamped

.

Circuits w

ith P

ara

llel L a

nd C

∫=

++

+t

nL

ti

idt

tv

Lt

vR

dtt

dv

C

0

)(

)0(

)(

1)

(1

)(

We can replace the circuit with it’s Norton equivalent

and then analyze the circuit by writing KCL at the top

node:

dt

dv

Cdt

dq

ii

vdt

Li

RVi

t

cL

LR

==

+=

=∫ 0

)0(

1

Circuits w

ith P

ara

llel L a

nd C

dtt

di

Ct

vLC

dtt

dv

RC

dt

tv

d

dtt

di

tv

Ldtt

dv

Rdt

tv

dC

ating

differenti

ti

idt

tv

Lt

vR

dtt

dv

C

n

n

t

nL

)(

1)

(1

)(

1)

(

)(

)(

1)

(1

)(

:

)(

)0(

)(

1)

(1

)(

2

2

0

=+

+

=+

+

=+

++

∫

Circuits w

ith P

ara

llel L a

nd C

)(

)(

)(

2)

(

)(

1)

(

1

2

1

)(

1)

(1

)(

1)

(

2 0

2

0

2

tf

tv

dtt

dv

dt

tv

d

dtt

di

Ct

ffunction

Forcing

LC

freq

uen

cyreso

nant

Undamped

RC

tco

efficien

Dampen

ing

dtt

di

Ct

vLC

dtt

dv

RC

dt

tv

d

n

n

=+

+

=

=

=

=+

+

ωα

ω

α

Circuits w

ith P

ara

llel L a

nd C

LRcircuit

Series

RC

circuit

Para

llel

tf

tv

dtt

dv

dt

tv

d

22

1

)(

)(

)(

2)

(2 0

2

==

=+

+

αα

ωα

This is the same equation as we found for the series LC

circuit with the following changes for α:

LL

Lj

IV

×=

ω

o90

∠=

=L

Lj

ZL

ωω

LL

LZI

V=

Co

mple

x I

mp

eda

nces-I

nd

ucto

r

Co

mple

x I

mp

eda

nces-I

nd

ucto

r

CC

CZI

V=

o90

11

1−

∠=

=−

=C

Cj

Cj

ZC

ωω

ω

RR

RI

V=

Co

mple

x I

mp

eda

nces-C

apacitor

Co

mple

x I

mp

eda

nces-C

apacitor

RR

RI

V=

Imp

ed

ances-R

esis

tor

Imp

ed

ances-R

esis

tor

03

21

=−

+V

VV

We c

an a

pply

KV

L d

irectly to p

hasors

.

The s

um

of th

e p

hasor

voltages e

quals

zero

for

any c

losed p

ath

.

The s

um

of th

e p

hasor

curr

ents

ente

ring a

node m

ust equal th

e s

um

of th

e p

hasor

curr

ents

leavin

g.

out

inI

I=

Kirchh

off

’sLaw

s in

Phasor

Fo

rm

Pow

er

in A

C C

ircuits

ZVI

where

IZV

Z

mm

mm

=−

∠=

∠∠=

=θ

θo0

VI For θ>

0 the load is called “inductive”

since Z=jωLfor an inductor

For θ<

0 the load is “capacitive”

since Z=-j/ωC for a capacitor

Load Im

pedance in the C

om

ple

x P

lane

ZXZR

jXR

ZZ

==

+=

∠=

)sin(

)cos(

θθ

θ

Pow

er

for

a G

en

era

l Loa

d

)cos(

)cos(

θ

θ

θθ

θ

==

−=

PF

IV

Prm

srm

s

curren

tvo

ltage

Power angle:

If the phase angle for the voltage is not zero, we

define the power angle θ:

AC

Pow

er

Calc

ula

tions

()

[]

WI

VP

θcos

rms

rms

=

()

[]

VAR

IV

Qθ

sin

rms

rms

=

Average Power:

Reactive Power:

Apparent Power:

[]

VA

IV

RMS

RMS

()2

rms

rms

22

IV

QP

=+

Pow

er

Trian

gle

s

Average power

Reactive

power

Average

power

Apparent

power

The Thevenin

equivalent for an ac circuit consists of a

phasorvoltage source V

tin series with a complex

impedance Z

t

Theve

nin

Eq

uiv

ale

nt

Circuits

The Thévenin voltage is equal to the open-circuit

phasor voltage of the original circuit.

oc

VV

=t

We can find the Thévenin impedance by zeroing the

independent sources and determining the im

pedance

looking into the circuit terminals.

Theve

nin

Eq

uiv

ale

nt

Circuits

The Thévenin impedance equals the open-circuit

voltage divided by the short-circuit current.

scscoc

IV

IVt

tZ

==

Theve

nin

Eq

uiv

ale

nt

Circuits

Nort

on

Equiv

ale

nt

Circuit

The Norton equivalent for an ac circuit consists of a

phasorcurrent source I nin parallel with a complex

impedance Z

t

scII

=n

The Thevenin

equivalent of a two-terminal circuit

delivering power to a load impedance.

Maxim

um

Avera

ge P

ow

er

Tra

nsfe

r

If the load can take on any complex value,

maxim

um power transfer is attained for a load

impedance equal to the complex conjugate of the

Thévenin impedance.

If the load is required to be a pure resistance,

maxim

um power transfer is attained for a load

resistance equal to the magnitude of the

Thévenin impedance.

Maxim

um

Avera

ge P

ow

er

Tra

nsfe

r

Tra

nsfe

r F

un

ctions

The transfer function H

(f ) of the two-port filter

is defined to be the ratio of the phasor output

voltage to the phasor input voltage as a function

of frequency:

()

inout

VV=

fH

First-

Ord

er

Low

Pa

ss F

ilter

RC

ff

fj

fRC

jf

H

fCj

RfC

jfC

j

fCj

R

BB

inout

inout

in

ππ

ππ

π

π

2

1

)/

(1

1

21

1)

(

2

12

1

2

1

2

1

=+

=+

==

+

=

=

+=

VV

VI

V

VI

Half power

frequency

()

()

()

∠+

∠=

+=

BB

B

fff

ff

fj

fH

arctan

1

01

1

1

2

o

()

()2

1

1

Bff

fH

+=

()

−=

∠Bff

fH

arctan

First-

Ord

er

Low

Pa

ss F

ilter

For low frequency signals the magnitude of the transfer

function is unity and the phase is 0

° . Low frequency signals

are passed while high frequency signals are attenuated and

phase shifted.

First-

Ord

er

Low

Pa

ss F

ilter

Magnitude B

ode P

lot fo

r F

irst-

Ord

er

Low

Pass F

ilter

1.A horizontal line at zero for f < fB/10.

2.A sloping line from zero phase at fB/10 to –90°at 10f B.

3.A horizontal line at –90°for f > 10f B.

First-

Ord

er

Hig

h-P

ass F

ilter

fRC

fwhere

ff

j

ff

j

fRC

j

fRC

j

fRC

jfC

jR

R

BB

B

inout

inin

inout

π

ππ

ππ

2

1

)/

(1

)/

(

12

2

2

11

1

2

1

=+

=

+=

−=

−=

VV

VV

VV

First-

Ord

er

Hig

h-P

ass F

ilter

()

()

()

()

()

()

()

()

()

()

()

()

BB

B

B

BB

B

B

B

ff

ff

fH

ff

ff

fH

ff

ff

ff

ff

j

ff

jf

H

arctan

90

arctan90

1

arctan

1

90

1

2

2inout

−=

∠∠

=∠

+=

∠+

∠=

+=

=

o

o

o

VV

()

()

()2

1B

B ff

ff

fH

+=

()

()

Bff

fH

arctan

90

−=

∠o

First-

Ord

er

Hig

h-P

ass F

ilter

Bode P

lots

for

the F

irst-

Ord

er

Hig

h-

Pass F

ilter

()

()

()

0log

10

)log(

20

)log(

20

2=

−≈

>>

≈<<

BB

B

BB

ff

ff

fH

ff

For

ff

fH

ff

For

()

()

()

()

()

()

()

()

()

o

o

o

090

arctan

90

0log

10

)log(

20

)log(

20

1log

10

)log(

20

2

2

≈∠

>>

≈∠

<<

−=

∠

=−

≈>>

≈<<

+−

=

fH

ff

For

fH

ff

For

fff

H

ff

ff

fH

ff

For

ff

fH

ff

For

ff

ff

fH

BB

B

BB

B

BB

BB

Series R

eso

nance

LC

f

LC

fC

fL

f

reso

nance

For

fCj

RfL

jf

Zs

π

ππ

π

ππ

2

1

)2(

1

2

12

:

2

12

)(

0

2

2 00

0

=

=→

=

−+

=

For resonance the reactance of the inductor and the capacitor cancel:

Series R

eso

nance

CR

fQ

LC

ffrom

Cf

LSubstitute

R

Lf

Resistance

reso

nance

at

inductance

of

Rea

ctance

Q

ss

0

02

02

0

2

1

2

1

)(

)2(

1

2

π

πππ

=

===≡

Quality factor Q

S

Series R

eso

nant

Band

-Pass F

ilter

−+

=→

−+

==

−+

==

ff

ffjQ

ff

ffjQ

R

ff

ffjQ

R

fZ

ssR

s

sR

s

s

s

s

0

0

0

0

0

0

1

1

1

1

/

)(

VVV

IV

VV

I

()

ff

ff

jQs

sR

00

1

1

−+

=VV

Series R

eso

nant

Band

-Pass F

ilter

Para

llel R

esona

nce

()

()

fLj

fCj

RZ

pπ

π2

12

1

1

−+

= At resonance Z

Pis purely resistive:

()

LC

fL

fj

Cf

jπ

ππ

2

12

12

00

0=

→=

Para

llel R

esona

nce

CR

fQ

LC

ffrom

Cf

LSubstitute

LfR

reso

nance

at

inductance

of

Rea

ctance

Resistance

Q

PP

0

02

02

0

2

2

1

)(

)2(

1

2 ππ

ππ

=

===

≡

Quality factor Q

P

Para

llel R

esona

nce

−+

==

ff

ffjQ

RZ

P

Pout

0

0

1

II

VV

outfor constant current,

varying the frequency

Secon

d O

rde

r Lo

w-P

ass F

ilter

−+−

=

−+

−

==

−+

−

=−

+

−

=+

+=

ff

ffjQ

ff

jQ

LC

ffff

R

Lf

j

fRCj

fH

LC

ffff

R

Lf

j

fRCj

fCjfL

jR

fCj

ZZ

Z

Z

S

S

inout

inin

inC

LR

Cout

0

0

0

00

0

00

0

1

)/

(

2

12

1

2)

(

2

12

1

2

222 π

ππ

ππ

π

πππ

VV

VV

VV

()

()

()

()

()

()

()

()

()2

00

2

0

00

12

00

2

0

000

1

1

90

1

ff

ff

Q

ff

Qf

H

ff

ff

QTan

ff

ff

Q

ff

Q

ff

ff

jQ

ff

jQf

H

S

s

sS

s

s

s

−+

=

−∠

−+

−∠

=

−+

−=

=

−

o

inout

VV

Secon

d O

rde

r Lo

w-P

ass F

ilter

Secon

d O

rde

r Lo

w-P

ass F

ilter

Secon

d O

rde

r H

igh-P

ass F

ilter

At low frequency the capacitor

is an open circuit

At high frequency the

capacitor is a short and the

inductor is open

Secon

d O

rde

r B

and

-Pass F

ilter

At low frequency the capacitor

is an open circuit

At high frequency the inductor

is an open circuit

Secon

d O

rde

r B

and

-Reje

ct

Filt

er

At low frequency the

capacitor is an open

circuit

At high frequency the

inductor is an open

circuit

First-

Ord

er

Low

-Pa

ss F

ilter

ff

B

Bif

ff

if

if

ff

f

f

f

ff

f

f

ff

if

io

CR

f

ff

jRR

CfR

jRR

ZZf

H

CfR

j

RZ

R

CfR

j

R

fCj

RZ

ZZ

VVf

H

π

π

π

π

π

2

1

)/

(1

1

21

1)

(

21

21

2

111

1

)(

=

+

−=

+

−=

−=

+=

+=

+=

−=

=

A low-pass filter with a dc gain of -R

f/Ri

First-

Ord

er

Hig

h-P

ass F

ilter

ii

B

B

B

if

ii

if

ii

ii

if

ii

f

if

ff

ii

i

if

io

CR

f

ff

j

ff

j

RR

CR

fj

CR

fj

RR

CR

fj

CR

fj

Cf

jR

R

ZZf

H

RZ

Cf

jR

Z

ZZ

vvf

H

π

π

π

π

π

π

π

2

1

)/

(1

)/

(

21

2

21

2

2

1)

(

2

1

)(

=

+

−=

+

−=

+−

=

+−

=−

=

=+

=

−=

=

A high-pass filter with a high frequency gain of -R

f/Ri

φλφ

NBA

d

A

==

⋅=∫

AB

Magnetic flux passing through a surface area A:

For a constant magnetic flux density perpendicular

to the surface:

The flux linking a coil with N turns:

Flu

x L

inkage

s a

nd F

ara

day’s

Law

Faraday’s law

of magnetic induction:

dt

de

λ=

The voltage induced in a coil whenever its

flux linkages are changing. Changes occur

from:

•Magnetic field changing in tim

e

•Coil m

oving relative to m

agnetic field

Fara

day’s

Law

Lenz’s law states that the polarity of the

induced voltage is such that the voltage

would produce a current (through an

external resistance) that opposes the

original change in flux linkages.

Lenz’s

Law

Lenz’s

Law

intensity

field

Magnetic

H=

=H

Bµ

ytpermea

bili

Relative

r0

70

10

4 µµµ

πµ

=

×=

−Am

Wb

Ampère’s Law

:∑

∫=

⋅i

dl

H

Magnetic F

ield

Inte

nsity a

nd A

mpère

’s

Law

The line integral of the magnetic field

intensity around a closed path is equal to the

sum of the currents flowing through the

surface bounded by the path.

Am

père

’sLa

w

Magn

etic F

ield

Inte

nsity a

nd

Am

père

’sLa

w

iHl

dlength

lincrem

enta

the

as

direction

same

the

in

points

and

magnitude

constant

a

has

H

field

magnetic

the

If

pro

duct

dot

Hdl

d

Σ=

=•

l

lH

)cos(θ

In m

any engineering applications, we need to

compute the magnetic fields for structures that

lack sufficient symmetry for straight-forw

ard

application of Ampère’s law. Then, we use an

approxim

ate method known as magnetic-circuit

analysis.

Magn

etic C

ircuits

magnetomotive force (m

mf) of an N-turn

current-carrying coil

IN

=

F

reluctance of a path for magnetic flux

Aµl=

R

φR

F=

Analog: Voltage (emf)

Analog: Resistance

Analog: Ohm’s Law

R

INr

NI

rR

rR

Al

rA

Rl

2

21

21

12

2

22

2

µφ

µππ

µµ

ππ

===

=

=

=

==

RF

FR

Magn

etic C

ircuit f

or

Toro

idalC

oil

A M

agnetic C

ircuit w

ith R

elu

cta

nces in

Series a

nd P

ara

llel

aaa

aaa

cb

a

ab

cb

a

ba

total

c

ba

ctotal

AB

AB

RR

R

divider

curren

tR

R

R

R

Ni

RR

RR

φφ

φφ

φφφ

==

+=

+==

++

=

)(

11

1

A M

agnetic C

ircuit w

ith R

elu

cta

nces in

Series a

nd P

ara

llel

1111

11

1

iiL

λλ

==←

2222

22

2

iiL

λλ

==←

212

2

21

121

1

12

ii

ii

Mλ

λλ

λ=

==

=←

←

Self

inductance

for coil 1

Self

inductance

for coil 2

Mutual inductance between coils 1 and 2:

Mutu

al In

ducta

nce

Mutu

al In

ducta

nce 21

22

2

12

11

1

λλ

λ

λλ

λ

±=

±=

Total fluxes linking the coils:

Currents entering the dotted terminals

produce aiding fluxes

Mutu

al In

ducta

nce

Circuit E

quations f

or

Mutu

al

Inducta

nce

dt

di

Ldt

di

Mdt

de

dt

di

Mdt

di

Ldt

de

iL

Mi

Mi

iL

22

12

2

21

11

1

22

12

21

11

+±

==

±=

=

+±

=

±=

λλ

λλ

Ideal T

ransfo

rmers

()() t

vNN

tv

1

122

=

)(

)(

121

2t

iNN

ti

= ()() t

pt

p1

2=

Ideal T

ransfo

rmers

Mecha

nic

al A

nalo

g

()()

121

21

212

112

21

122

)(

)(

Fll

Ft

iNN

ti

dll

dt

vNN

tv

==

==

d1

d2

LL

ZNN

Z

2

21

11

==

′IV

Imp

ed

ance T

ransfo

rmatio

ns

Sem

icond

ucto

r D

iode

Shockle

y E

quation

breakdown

reverse

reaching

until

-for

TD

sD

Vv

Ii

<<−

≈

−

=

1exp

T

Ds

DnVv

Ii

mV

qkTVT

26

≅=

TV

for

exp

>>

≈

DT

Ds

Dv

nVv

Ii

DD

SS

vRi

V+

=

Assume V

SSand R are known. Find iDand v

D

Load-L

ine A

naly

sis

of D

iode C

ircuits

DD

SS

vRi

V+

=D

DSS

vRi

V+

=

Load-L

ine A

naly

sis

of D

iode C

ircuits

The ideal diode acts as a short

circuit for forw

ard currents

and as an open circuit with

reverse voltage applied.

i D> 0

v D< 0 →

diode is in the “on”state

v D< 0

I D = 0 →

diode is in the “off”state

Ideal D

iod

e M

odel

Assum

ed S

tate

s f

or

Analy

sis

of

Ideal-

Dio

de C

ircuits

1. Assume a state for each diode, either on (i.e., a

short circuit) or off (i.e., an open circuit). For n

diodes there are 2npossible combinations of

diode states.

2. Analyze the circuit to determine the current

through the diodes assumed to be on and the

voltage across the diodes assumed to be off.

3. Check to see if the result is consistent with the

assumed state for each diode. Current must flow

in the forw

ard direction for diodes assumed to

be on. Furthermore, the voltage across the

diodes assumed to be off m

ust be positive at the

cathode (i.e., reverse bias).

4. If the results are consistent with the assumed

states, the analysis is finished. Otherwise, return

to step 1 and choose a different combination of

diode states.

Half-W

ave R

ectifier

with R

esis

tive L

oad

The diode is on during the positive half of the cycle. The

diode is off during the negative half of the cycle.

CV

VC

TI

Qr

L=

∆=

≅∆

The charge removed from

the capacitor in one cycle:

Half-W

ave R

ectifier

with S

mooth

ing

Capacitor

r

L VTI

C2

=The capacitance required for a

full-w

ave rectifier is given by:

Full-

Wave R

ectifier

Full-

Wave R

ectifier

Clip

pe

r C

ircu

it

NP

N B

ipola

r Junction T

ransis

tor

Bia

s C

onditio

ns f

or

PN

Junction

s

The base em

itter p-n

junction of an npn

transistor is norm

ally

forw

ard biased

The base collector p-n

junction of an npn

transistor is norm

ally

reverse biased

Equations o

f O

pe

ration

−

=

1exp

TBE

ES

EVv

Ii

BC

Ei

ii

+=

From Kirchoff’scurrent law:

Equations o

f O

pe

ration

EC ii=

α

Define αas the ratio of collector current to emitter

current:

Values for αrange from 0.9 to 0.999 with 0.99 being

typical. Since:

EB

BE

BC

Ei

ii

ii

ii

01

.0

99

.0

=→

+=

+=

Most of the em

itter current comes from the collector

and very little (≈1%) from the base.

Equations o

f O

pe

ration

ααβ

−=

=1

BC ii

Define β

as the ratio of collector current to base

current:

Values for βrange from about 10 to 1,000 with a

common value being β

≈100.

BC

ii

β=

The collector current is an amplified version of the

base current.

Only a small fraction of the em

itter current flows into the base

provided that the collector-base junction is reverse biased and the

base-em

itter junction is forw

ard biased.

The base region is very thin

Co

mm

on-E

mitte

r C

hara

cte

ristics

bias

reve

rse

vv

if v

vv

v

BC

BE

CE

CE

BE

BC

→<

→>

−=

0

v BC

v CE

Co

mm

on-E

mitte

r In

put

Cha

racte

ristics

−

−

=1

exp

)1(

TBE

ES

BVv

Ii

α

Co

mm

on-E

mitte

r O

utp

ut

Cha

racte

ristics

100

==

ββ

for

ii

BC

Am

plif

ication b

y t

he B

JT

A small change in v

BEresults in a large change in iBif the

base em

itter is forw

ard biased. Provided v

CEis m

ore than a

few tenth’s of a volt, this change in iBresults in a larger

change in iCsince iC=βi

B.

Co

mm

on-E

mitte

r A

mplif

ier

Load-L

ine A

naly

sis

of a C

om

mon E

mitte

r

Am

plif

ier

(Input C

ircuit)

()()

() tv

ti

Rt

vV

BE

BB

BB

+=

+in

CE

CC

CC

vi

RV

+=

Load-L

ine A

naly

sis

of a C

om

mon E

mitte

r

Am

plif

ier

(Outp

ut C

ircuit)

As v i

n(t) goes positive, the load line moves upward and to the

right, and the value of i Bincreases. This causes the operating

point on the output to m

ove upwards, decreasing v

CE→

An

increase in v

in(t)results in a m

uch larger decrease in v

CEso that

the common emitter am

plifier is an inverting amplifier

Invert

ing A

mplif

ier

Except for reversal of current directions and

voltage polarities, the pnpBJT is almost

identical to the npnBJT.

PN

P B

ipola

r Junction T

ransis

tor

BC

E

BC

EB

EC

ii

i

ii

ii

ii

+==

−==

β

α

α

)1(

PN

P B

ipola

r Junction T

ransis

tor

NM

OS

Tra

nsis

tor

NM

OS

Tra

nsis

tor

Ope

ration in t

he C

uto

ff R

egio

n

toGS

DV

vi

≤=

for

0

By applying a positive bias

between the Gate (G

) and the body

(B), electrons are attracted to the

gate to form

a conducting n-typ

e

channel between the source and

drain. The positive charge on the

gate and the negative charge in the

channel form

a capacitor where:

Ope

ration S

lightly A

bove C

ut-

Off

DS

toGS

ox

DS

vV

vLtW

i)

(−

=µε

For sm

all values of v D

S, i D

is proportional to v

DS. The device

behaves as a resistance whose value depends on v

GS.

Ope

ration S

lightly A

bove C

ut-

Off

()

[]

22

DS

DS

toGS

Dv

vv

vC

i−

−=

2KP

LWC

=

Ope

ration in t

he T

riode R

egio

n

Vt

vt

vin

GS

4)

()

(+

=

Load-L

ine A

naly

sis

of a S

imple

NM

OS

Circuit

()() t

vt

iR

vDS

DD

DD

+=

Load-L

ine A

naly

sis

of a S

imple

NM

OS

Circuit

Load-L

ine A

naly

sis

of a S

imple

NM

OS

Circuit

CM

OS

Invert

er

Tw

o-I

nput

CM

OS

NA

ND

Gate

Tw

o-I

nput

CM

OS

NO

R G

ate