lecture16(20-3-11)
TRANSCRIPT
7/27/2019 Lecture16(20-3-11)
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EE 332
DEVICES AND CIRCUITS II
Lecture 16
Coupling & Bypass Capacitor Design
BYPASS AND COUPLING CAPACITOR DESIGN
/CE AMPLIFIERS
R1
C1
RE
Q 1
vo
vi
R2
Rs
C3
C2
RL
RC
VCC
C1 and C2 are coupling capacitors C3 is bypassing capacitor
BYPASS AND COUPLING CAPACITOR DESIGN
/CE AMPLIFIERS
C1
RE
Q 1
vo
vi
RB
Rs
C2
RL
RC
Replace C3 with short circuits by assuming its large
B
C
E
BYPASS AND COUPLING CAPACITOR
DESIGN/INTRODUCTION
Since impedance of a capacitor increases with
decreasing frequency, coupling and bypass capacitors reduce amplifier gain at low frequencies.
To choose capacitor values, short‐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
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BYPASS AND COUPLING CAPACITOR DESIGN
/CE COUPLING
CAPACITORS
Consider C1 (with vi=0): using short‐circuit time constant
technique, look at each capacitor on its own each time, and
short other capacitors
Note that C1 forms a series loop with Rs+Ri
To neglect C1, its impedance must be
much smaller than (Rs+Ri)
1
( )1
11
( )
/ /
s i
s i
i B b
R R C
C R R
R R r
ω
ω
∴ << +
⇒ >>+
=
Where ω is the lowest frequency to be amplified.
C1
Ri
Rs
BYPASS AND COUPLING CAPACITOR DESIGN
/CE BYPASSING CAPACITOR
Replace C1 and C2 with short circuits by assuming both are large
C3
RE
Q 1vo
vi
RB
Rs
RLR
C
B
C
E
BYPASS AND COUPLING CAPACITOR DESIGN
/CE BYPASSING CAPACITOR
1( / / )
3
13
( / / )
E e
E e
R r C
C R r
ω
ω
∴ <<
⇒ >>
Note that C3 forms a parallel loop with RE and re
To neglect
C3,
its
impedance
must
be
much smaller than (RE//re)
Where ω is the lowest frequency to be amplified.
C3
re
RE
Consider C3 (with vi=0): using short‐circuit time constant
technique, look at each capacitor on its own each time, and
short other capacitors
BYPASS AND COUPLING CAPACITOR DESIGN
/CE COUPLING
CAPACITORS
C2
RL
Ro
1( )
2
12
( )
/ /
o L
o L
o C c
R R C
C R R
R R r
ω
ω
∴ << +
⇒ >>+
=
Consider C2 (with vi=0): using short‐circuit time constant
technique, look at each capacitor on its own each time, and
short other capacitors
Note that C2 forms a series loop with RL+Ro
To neglect C2, its impedance must be
much smaller than (Rs+Ri)
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COUPLING AND BYPASS CAPACITOR DESIGN
/C‐E AMPLIFIERS
(EXAMPLE)
Problem: Choose values of coupling and bypass capacitors.
Given data: f = 1000Hz, values of all resistors and input and
output resistances for C‐E amplifiers.
, ,,
1,2
70 0.7
70 , 15
BE ON
A CC
V V
Q V V V V
β = =⎧ ⎫
= ⎨ ⎬= =⎩ ⎭
1 2
300 , 160 , 22 ,
3 , 2 , 100 ,
C
E s L
R k R k R k
R k R k R k
Ω Ω Ω
Ω Ω Ω
= = =⎧ ⎫⎨ ⎬
= = =⎩ ⎭
C1
RE
Q 1
vi
R1
Rs
C2
RL
RC
R2
C3
VCC
COUPLING AND BYPASS CAPACITOR DESIGN
/C‐E AMPLIFIERS (EXAMPLE)
For C‐E amplifier:
13( / / )
1; / /
E e
CE
e E e e
m
C R r
r R r r g
ω>>
≅ ≅
C
1
RE
Q 1
vi
R1
Rs
C2
RL
RCR2
VCC
C3
Solution
Solution
COUPLING AND BYPASS CAPACITOR DESIGN
/CC AMPLIFIERS
C1
RE
Q 1
vo
viRB
Rs
C2
RL
RC
C1
RE
Q 1
viR1
Rs
C2
RL
RCR2
VCC
COUPLING AND BYPASS CAPACITOR DESIGN
/C‐
E
AMPLIFIERS
(EXAMPLE)
11
( )
/ /
, ( 1) 300
77.2
11 1.6
2000 (2 77.2 )
s i
CE
i B b
CE
b E
i
C R R
R R r
an d r r R k
R k
C n F
k k
π
ω
β
π
Ω
Ω
>>+
=
= + + ≅
⇒ =
∴ >> =
× +
For C‐E amplifier:
12
( )
/ /
12 1.3
2000 (22 100 )
o L
CE
o C c C
C R R
R R r R
C n F k k
ω
π
>>+
= ≅
∴ >> =× +
C1
RE
Q 1
vi
R1
Rs
C2
RL
RCR
2
C3
VCC
Solution
Solution
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BYPASS AND COUPLING CAPACITOR DESIGN
/CC COUPLING
CAPACITORS
Consider C1 (with vi=0): using short‐circuit time constant
technique, look at each capacitor on its own each time, and
short other capacitors
Note that C1 forms a series loop with Rs+Ri
To neglect C1, its impedance must be
much smaller than (Rs+Ri)
1( )
1
11
( )
/ /
s i
s i
i B b
R R C
C R R
R R r
ω
ω
∴ << +
⇒ >>+
=
C1
Ri
Rs
COUPLING AND BYPASS CAPACITOR DESIGN
/CB AMPLIFIERS
C2
RE
Q 1
vo
RB Rs
C1
RL
C2
RE
Q 1
vi
R1
RsC1
RL
RCR2
VCC
RC
vi
BYPASS AND COUPLING CAPACITOR DESIGN
/CB COUPLING CAPACITORS
Consider C1 (with vi=0): using short‐circuit time constant
technique, look at each capacitor on its own each time, and
short other capacitors
Note that C1 forms a series loop with Rs+Ri
Hence, to
neglect
C1,
its
impedance
must be much smaller than (Rs+Ri)
1( )
1
11
( )
s i
s i
R R C
C R R
ω
ω
/ /i E e
R R r
∴ << +
⇒ >>+
C1
Ri
Rs
BYPASS AND COUPLING CAPACITOR DESIGN
/CC COUPLING
CAPACITORS
C2
RL
Ro
Hence, to neglect C2, its impedance
must be much smaller than (Ro+RL)
1
( )2
12
( )
/ /
o L
o L
o E e
R R C
C R R
R R r
ω
ω
∴ << +
⇒ >>+
=
Consider C2 (with vi=0): using short‐circuit time constant
technique, look at each capacitor on its own each time, and
short other capacitors
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BYPASS
AND
COUPLING
CAPACITOR
DESIGN/CB COUPLING CAPACITORS
C2
RL
Ro
Consider C2
To neglect C2, its impedance must be
much smaller than (Ro+RL)
1( )
21
2( )
/ /
o L
o L
o C c
R R
C
C R R
R R r
ω
ω
∴ << +
⇒ >>+
=
Where ω is the lowest frequency to be amplified.
End of Lecture 16
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