Lecture 10 coupling and bypass design

EE 322
El t i II
El
ec
t
ron
i
cs
II
Lecture 10
Coupling and Bypass Capacitor Design
8/18/2013 1
Coupling and Bypass Capacitor Design
 Since impedance of a capacitor increases with decreasing
f li d b it d lifi
f
requency, coup
li
ng an
d

b
ypass capac
it
ors re
d
uce amp
lifi
er
gain at low frequencies.
 To choose capacitor values, shor

t
-circuit time constant
method is used: each capacitor is considered separately
with all other capacitors replaced by short circuits.
 To neglect a capacitor, the magnitude of capacitive
impedance must be much smaller than the equivalent
resistance appearing at its terminals.
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 2
Coupling and Bypass Capacitor Design: C-E
and C
S Amplifiers
and

C
-
S

Amplifiers
Consider C
1
(with v
i
=0): using
short-circuit time constant
technique, look at each
capacitor on its own each

capacitor

on

its

own

each

time, and short other
capacitors
TltC
it i d
T
o neg
l
ec
t

C
1
,
it
s
i
mpe
d
ance
must be much smaller than

(R
I
+R
in
)
C
C
in
RR
C





101
1
1
1
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 3




in
R

I
R
C
in
R
I
R
C






11
Coupling and Bypass Capacitor Design: C-E
and C
S Amplifiers
and

C
-
S

Amplifiers
R
iB
R
R
For C-E amplifier,

R
iB
B
R
in
R

For C-S amplifier,
R
iG
RR

G
in
For coupling capacitor C
1
,




in
R
I
R
C
in
R
I
R

C





10
1
1
1
 is chosen to be lowest frequency
for which midband operation is
needed in given application.
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
needed

in

given

application.
Chap 14 - 4
Coupling and Bypass Capacitor Design: C-E
and C
S Amplifiers
and

C
-
S

Amplifiers
For C-E amplifier,
R
iC
R
R
C
3
R
iC
C
R
ou
t
R

For C-S amplifier,
R
iD
R
D
out
R

iD
D

out
For coupling capacitor C
3
,
C
3




out
RR
C
out
RR
C




3
10
3
3
1
3

 is chosen to be lowest frequency
for which midband operation is
needed in given application

Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
needed

in

given

application
.
Chap 14 - 5
Coupling and Bypass Capacitor Design: C-E
and C
-
S Amplifiers (contd.)
and

C
S

Amplifiers

(contd.)
In this case, we can neglect impedances
fi
C
d
C

fi d h
o
f
capac
i
tors
C
1
an
d

C
3
,
fi
n
d
t
h
e
equivalent resistance looking up into
emitter or source of amplifier.























m
g
E
RR
C
1
4
1
2

C
2







m
g








C
1
1
2


















m
g
S
RR
1
4

C
2
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 6
Coupling and Bypass Capacitor Design: C-E
and C-S Am
p
lifiers
(
Exam
p
le
)
p(p)
 Problem: Choose values of coupling and bypass capacitors.

 Given data:
f
= 1000Hz
,
values of all resistors and in
p
ut and out
p
ut
f
,pp
resistances for C-E.
?
R
iB
B
R
in
R
?
?
1
C
C


?
1
?
1

1





C
in
R
I
R
C



?
)
/
1
(
1
2







R

R
C


?
2
)
/
1
(
4








C
m
g
E
R
R



?
3

?
1
3




C
C
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill


3
3
3

out
RR

Chap 14 - 9
Coupling and Bypass Capacitor Design: C-E
and C-S Am
p
lifiers
(
Exam
p

le
)
p(p)
kΩ1.78
R
iB
B
R
in
R
F
02
0
1 99nF
1



C
C
F
02
.
0
1
1
.
99nF
1
















C
in
R
I
R
C


nF2.67
)
/
1
(
1
2








R
R
C


F0.68
2
)
/
1
(
4










C
m

g
E
R
R


F015.0
3
nF31.1
1
3




C
C


3
3
3



out
RR
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design

McGraw-Hill
Chap 14 - 10
Coupling and Bypass Capacitor Design: C-E
and C-S Am
p
lifiers
(
Exam
p
le
)
p(p)
 Problem: Choose values of coupling and bypass capacitors.
 Given data: f = 1000Hz, values of all resistors and input and output
resistances for C
-
S amplifier
resistances

for

C
S

amplifier
.
kΩ892


G

R
in
R
1
p
F1800
1
178pF
1
1













C
in
R
I
R
C




nF3.55
1
2




C


F0.56
2
)/1(
6
2










C
m
g

S
RR
F
015
0
F
31
1
1
C
C


F
015
.
0
3
n
F
31
.
1
3
1
3








C
out
RR
C
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 11
Coupling and Bypass Capacitor Design: C-C
and C
-
D Amplifiers
and

C
-
D

Amplifiers
For C-C amplifier,
R
iB
B
R
in
R 

For C-D amplifier,
R
iG
G
R
in
R 
For coupling capacitor C
1
,










in
R
I
R
C

1
1
Jaeger/Blalock
7/1/03

Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 12
Coupling and Bypass Capacitor Design: C-C
and C
-
D Amplifiers
and

C
-
D

Amplifiers
For C-C amplifier,
C
R
iE
R
out
R
6

For C-D amplifier,
C
3
R
iS
R
out

R
6

For coupling capacitor C
3
,


C

1
C
3


out
RR
C


3
3

Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 13
Coupling and Bypass Capacitor Design:
C

-
CandC
-
D Amplifiers (Example)
C
C

and

C
D

Amplifiers

(Example)
 Problem: Choose values of coupling and bypass capacitors.

Given data
:
f
= 1000Hz values of all resistors and input and output

Given

data
:

f

=

1000Hz
,
values

of

all

resistors

and

input

and

output

resistances for both C-E and C-S amplifiers.
 Analysis:
For C
C amplifier:
For C-D amplifier:
For

C
-
C

amplifier:
kΩ5.95
R
iB
B
R
in
R
kΩ892


G
R
in
R
89pF
1

C
 120
6
R
iE
R
out
R
 k74.1
6
R
iS

R
out
R
pF
8200
1
816pF
1
1













C
in
R
I
R
C

pF1000

1
89pF
1












C
in
R
I
R
C

6
pF
8200
1

C



pF
8200
pF795
3
1
3





C
out
RR
C



pF
8200
3
pF782
3
1
3






C
out
RR
C

Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
pF
8200
3


C
pF
8200
3

C
Chap 14 - 14
Coupling and Bypass Capacitor Design:
C
-
BandC
-
G Amplifiers
C
B

and

C
G

Amplifiers
For C-B amplifier,
R
iE
R
in
R
6

For C-G amplifier,
R
iS
R
in
R
6

For coupling capacitor
C
For

coupling

capacitor

C
1
,



C
1
1










in
R
I
R
C

1
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill

Chap 14 - 15
Coupling and Bypass Capacitor Design:
C
-
BandC
-
G Amplifiers
C
B

and

C
G

Amplifiers
For C-B amplifier,
C
3
R
iC
C
R
out
R 
For C-G amplifier,
3
R
iD
D

R
out
R 
C
3
For coupling capacitor C
3
,


R
R
C 
1
3


out
R
R

3
3

Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 16
Coupling and Bypass Capacitor Design:

C
-
BandC
-
G Amplifiers (contd.)
C
B

and

C
G

Amplifiers

(contd.)
In this case, we can neglect impedances
fi
C
d
C
hfidh
o
f
capac
i
tors
C
1
an

d

C
3
, t
h
e
fi
n
d
t
h
e
equivalent resistance looking up into
emitter or source of amplifier.
C
2
1
)
6
)(1(
2
1
RR
R
CG
e
q
I
RR

o
rRR
R
CB
eq










C
2
2
1
q
R
CGCB
eq
C
,
1
2


R

eq

C
2
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 17
Coupling and Bypass Capacitor Design: C-B
and C
-
G Amplifiers (Example)
and

C
G

Amplifiers

(Example)
 Problem: Choose values of coupling and bypass capacitors.

Given data
:
f
= 1000Hz values of all resistors and input and output

Given

data
:

f

=

1000Hz
,
values

of

all

resistors

and

input

and

output

resistances for both C-E and C-S amplifiers.
 Analysis:
For C-B amplifier:
Ω100Ω102kΩ13
6


R
iE
R
in
R
kΩ9.21MΩ93.3kΩ22 
R
iC
C
R
out
R
F82.0
1
75.8nF
1
1














C
in
R
I
R
C


F0.027nF38.2
1
2





C


F015.0
3
nF31.1
1
3




 C

R
R
C


)
6
)(1(
21
2












I
RR
o
rRR
Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill



3


out
R
R
Chap 14 - 18
Coupling and Bypass Capacitor Design: C-B and
C-G Amplifiers (Example contd.)
For C
G amplifier:
For

C
-
G

amplifier:
kΩ74.1kΩ04.2kΩ12
6

R
iS
R
in
R
k
Ω

9
20
k
Ω
410
k
Ω
22
R
R
R
F42.0
1
42.6nF
1
1











C
R
R

C
k
Ω
9
.
20
k
Ω
410
k
Ω
22



R
iD
D
R
out
R








in

R
I
R
pF1800pF178
2
1
1
2









RR
C



F015.0
3
nF31.1
3
1
3





 C
out
RR
C
2
1




Jaeger/Blalock
7/1/03
Microelectronic Circuit Design
McGraw-Hill
Chap 14 - 19
End of Lecture 10