ee70 review - · pdf fileb. reference directions “downhill: resistor ... kcl at the...
TRANSCRIPT
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