Lecture 27
Amplifier Configurations
Reading:
CE/CS: Jaeger 13.6, 13.9, 13.10, 13.11
CC/CD: Jaeger 14.1, 14.3
CB/CG: Jaeger 14
14.1,
1 14
14.4
4
and Notes
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ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Source
Amplifier
Load R
Resistance
Ampliffier Output
Ressistance
Ampliifier Input
Ressistance
Sourrce Input
Ressistance
Voltage Amplifier: Voltage input and Voltage output
Load
•Any signal source has a finite “source resistance”, RS .
•The amplifier is often asked to drive current into a load of finite
impedance, RL (examples: 8 ohm speaker, 50 ohm transmission line, etc…)
The controlled source is a Voltage-controlled-Voltage Source
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Av=Open
O
Ci
Circuit
i Voltage
Vl
Gain
G i can be
b found
f
d by
b applying
l i a
voltage source with Rs=0, and measuring the open circuit output
voltage(no load or RL=infinity)
ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Why is the input and output resistance important?
•Only the voltage vin is amplified to Avvin.
•Since Rs and Rin form a voltage divider that determines vin, you want Rin as
large as possible (for a voltage amplifier) for maximum voltage gain.
•Since RL and Rout form a voltage divider that determines vout, you want Rout
as small as possible (for a voltage amplifier) for maximum voltage gain.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Current Amplifier: Current input and Current output
The controlled source is a Current
Current-controlled-Current
controlled Current Source
Ai=Short Circuit Current Gain can be found by applying a current
source with Rs= infinity, and measuring the short circuit output
current (No Load or RL=0)
0)
•Only the current iin is amplified to Aiiin.
•Since Rs and Rin form a current divider that determines iin, you want Rin as small as
possible
ibl (for
(f a currentt amplifier)
lifi ) for
f maximum
i
currentt gain.
i
•Since RL and Rout form a current divider that determines iout, you want Rout as large
as possible (for a current amplifier) for maximum current gain.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Transconductance Amplifier: Voltage input and Current output
The controlled source is a Voltage-controlled-Current Source
Gm=Transconductance Gain can be found by applying a voltage
source with Rs=0, and measuring the short circuit output current
(No Load or RL=0)
0)
•Only the voltage vin is amplified to iout=Gmvin.
•Since Rs and Rin form a voltage divider that determines vin, you want Rin as large as
possible for maximum transconductance gain.
gain
•Since RL and Rout form a current divider that determines iout, you want Rout as large
as possible for maximum transconductance gain.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Transresistance Amplifier: Current input and Voltage output
The controlled source is a Current-controlled-Voltage Source
Rm=Transresistance Gain can be found by applying a current
source with Rs=infinity, and measuring the open circuit output
voltage (RL=infinity)
infinity)
•Only the current iin is amplified to vout=Rmiin
•Since Rs and Rin form a current divider that determines iin, you want Rin as small as
ppossible for maximum transresistance gain.
g
•Since RL and Rout form a voltage divider that determines vout, you want Rout as small
as possible for maximum transresistance gain.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Input Resistance
With the load resistance attached…
Apply
l a test input
i
voltage
l
andd measure the
h input
i
current, Rin=vt/it
Or
Apply a test input current and measure the input voltage,
voltage Rin= vt/it
Output Resistance
With all input voltage sources shorted and all input current sources opened…
Apply a test voltage to the output and measure the output current, Rout=vt/it
Or
Apply a test current to the output and measure the output voltage, Rout= vt/it
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Final Summary of Transistor Amplifier Analysis
1.) a.) Determine DC operating point. Make sure the transistors are biased into
active mode ((forward active for BJTs and Saturation for MOSFET. Do not
confuse the two terms as saturation means a completely different thing for a
BJT)and b.) calculate small signal parameters gm, r, ro etc…
2.) Convert to the AC only model.
•DC Voltage sources are replaced with shorts to ground
•DC Current sources are replaced with open circuits
•Large capacitors are replaced with short circuits
•Large
Large inductors are replaced with open circuits
3.) Use a Thevenin circuit where necessary on each leg of transistor
4.) Replace transistor with small signal model
5.) Simplify the circuit as much as necessary and solve for gain.
6.) Solve for Input Resistance: With the load resistance attached… a.) Apply a test input
voltage and measure the input current, Rin=vt/it or b.) Apply a test input current and
measure the input voltage
voltage, Rin= vt/it
7.) Solve for Output Resistance: With all input voltage sources shorted and all input
current sources opened… a.) Apply a test voltage to the output and measure the output
current Rout=vt/it or bb.)) Apply a test current to the output and measure the output
current,
voltage, Rout= vt/it
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Can be modeled as a current amplifier,
IC=IB, or a transconductance amplifier ,
iC=KvBE
Outputt
Resistan
nce
Input R
Resistance
Outputt
Resistan
nce
Input R
Resistance
Common Emitter and Common Source
Modeled as transconductance
amplifier, iDS=KvGS
Overall
O
ll Amplifier
A lifi Configuration
C fi
ti
•Emitter/Source is neither an input nor an output
•Input is between base-emitter or gate-source
•Output
O t t is
i between
b t
collector-emitter
ll t
itt andd drain-source
d i
•Is a transconductance amplifier (see small signal models we have
used in previous examples)
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Emitter and Common Source
Portion due
to transistor
r
Portion due to Portion due to
transistor bias circuitry
gmv
or gmvgs
r
Rc
Rout
Rin
Portion due to
bias circuitry
r is replaced with an open
circuit for the MOSFET case
Previously, we have analyzed voltage gain. Now let us look at the
amplifier input and output resistance (these are small signal
parameters):
Rin  R 2 || R1 || r for the BJT or
Rin  R 2 || R1 for the MOSFET
Rout  ro || Rc for the BJT or MOSFET
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Summary of Common Emitter and Common Source Characteristics
•Very
Very Large Voltage Gain
•Inverting Voltage Gain (due to –gmro)
•High Input Impedance
These properties
make the CE/CS
configuration very
good for high gain
stages of amplifiers.
•High Output Impedance
Now let us consider the other two configurations of
transistor amplifiers:
•Common Gate/Common Base
•Common
C
D
Drain/Common
i /C
Collector
C ll t
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ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector and Common Drain
DC Circuit
Ci it
D
C
G
B
E
S
Collector (or Drain) is neither an input or output
Input is Base (or Gate)
p is Emitter ((or Source))
Output
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector
DC Ci
Circuit
it converted
t d to
t AC Equivalent
E i l t (reduced)
( d d)
DC
Circuit
Note the extra
ground due to C2
AC
Circuit
ib B
C
r
ib
ro
E
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector
DC Ci
Circuit
it converted
t d to
t AC Equivalent
E i l t (reduced)
( d d)
AC
Circuit
ib B
C
r
ib

ro
E
ib B
AC Circuit
(reduced)
C
ib
r
i
b
E 
vo
vo   o  1ib R 4 || ro || R 7 
vth  ib Rth  r   o  1R 4 || ro || R7 
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector
AC V
Voltage
lt
G
Gain
i
vo   o  1ib R 4 || ro || R 7 
vth  ib Rth  r   o  1R 4 || ro || R 7 
vthh  vs
Av 
R 2 || R1
R 2 || R1  Rs

 o  1ib R 4 || ro || R7 
vo vth  R 2 || R1 


 
vth vs  R 2 || R1  Rs  ib Rth  r   o  1R 4 || ro || R 7  

 R 2 || R1 
 o  1RL
 where

Av  
h R L  R 4 || ro || R 7 
 R 2 || R1  Rs  Rth  r   o  1RL  
gm
multiplying numerator and denominator by
 o  1






 R 2 || R1 
g m RL

Av  


g m r  
 R 2 || R1  Rs    Rth


  g m    1  RL     1  
o
o










 R 2 || R1 
g m RL

Av  



 R 2 || R1  Rs    Rth



  g m    1  RL    o  


  o
But for R2 || R1  Rs and g m R L  1
Av 
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 
g m RL
1 V
V
1  g m RL
Gain is positive and ~1
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Drain Conversion from DC to AC Equivalent Circuit
DC
Ci i
Circuit
G
AC
Circuit
D
gmvGS
ro
R7
S
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Emitter and Common Source
DC Ci
Circuit
it converted
t d to
t AC Equivalent
E i l t (reduced)
( d d)
G
AC
Circuit
D
gmvGS
ro
R7
S
G
D
AC Circuit
(reduced)
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Drain AC Voltage Gain
vo  g m vGS R 4 || ro || R7 
vth  vGS  g m vGS R 4 || ro || R7   vGS 1  g m R 4 || ro || R7 
vth  v s
R 2 || R1
R 2 || R1  Rs
vo vGS vth  R 2 || R1  g m R 4 || ro || R7  


 
Av 
vGS vth v s  R 2 || R1  Rs  1  g m R 4 || ro || R7  
 R 2 || R1  g m RL 
 where

Av  
h R L  R 4 || ro || R7 
 R 2 || R1  Rs  1  g m RL  
But for R2 || R1  Rs and g m R L  1
 
g m RL
1 V
Av 
V
1  g m RL
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Gain is positive and ~1
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector/Drain Input Resistance
Input Resistance: With the load resistance attached apply a test input
voltage and measure the input current, Rin=vx/i (where i=ix)
Rin , BJT
vx

 r   o  1R L
i
Resistance in the emitter circuit is “multiplied”
by transistor to increase the input resistance
Rin , MOSFET
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vx vx



0
i
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector Output Resistance
ix
RL
Output Resistance: With all input voltage sources shorted and all input
current sources opened, apply a test voltage to the output and measure the
output current, Rout=vx/ix

v
vx
vx  vx
vx
vx
 

  oi  x 
  o  
RL r  Rth
ro r  Rth
 r  Rth  RL
v
1
where RL  ro || R 4
 x 
1
1
ix

r  Rth RL
 o  1
ix  i   oi 
Rout , BJT
Two resistors in parallel: RL, and Resistance in the base circuit is “multiplied” by
transistor to decrease the output resistance
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Drain Output Resistance
gmvGS
ix
RL
Output Resistance: With all input voltage sources shorted and all input
current sources opened,
opened apply a test voltage to the output and measure the
output current, Rout=vx/i
i x   g m vGS 
Rout ,MOSFET 
vx
v
 gmvx  x
ro
RL
vx
1

1
ix
gm 
RL
where RL  ro || R 4
Two resistors in parallel: RL and inverse transconductance
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Summary of Common Collector and Common Drain Characteristics
•Unity Voltage Gain
•Non-Inverting
Non Inverting Voltage Gain
•Very High Input Impedance
•Low
L Output
O t t Impedance
I
d
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These properties
Th
ti make
k th
the
CC/CD configuration very
good for impedance
transformation I.E.
transformation,
IE
“buffering” high
impedances to low
impedances.
peda ces. CC/C
CC/CD
configurations are good
for output stages of
amplifiers due to their
very low output
impedance, I.E., very little
voltage drop in the output
resistance of the amp.
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector/Drain Current Gain
ith
Rth
vth
(o+1)ith
i
Ai , BJT 
i
 o 1
ith
Ai , MOSFET 
i

ith
Note: since R7 was originally defined as the load, the current gain should actually be (+1) (R4||ro)/(R4||ro+R7)
using a current divider.
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base and Common Gate
DC Circuit
Ci it
Base (or Gate) is neither an input or output
Input is Emitter (or Source)
p is Collector (or
( Drain))
Output
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base
DC Ci
Circuit
it converted
t d to
t AC Equivalent
E i l t (reduced)
( d d)
DC
Circuit
AC
Circuit
Note: Jaeger let’s ro go to infinity which
makes the math dramatically easier
C
B
r
gmv
ro
vo
E
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base AC Equivalent (reduced)
C
B
v
AC
Circuit
r
gmv
ro
vo
E
v
AC
Circuit
( d d)
(reduced)
r
gmv
ro
gmv
ro
vo
r
v
R4
Vth  vs
 0.8667Vs
R 4  Rs
Rth  Rs || R 4  1.73K
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Voltage Gain
iC
vo
i
ro
gmv
iB
i E  iC  i B  g m v  i  i B
 v  vo
i E  g m v   
 ro
vth  i E Rth  i B r
  v
  
  r

 v  vo
 vth   g m v   
 ro

also,



  v
  
  r

 v  vo
vo  iC RL   g m v   
 ro

iE
v
v 

  Rth    r

 r 

Vth  vs
R4
 0.8667Vs
R 4  Rs
Rth  Rs || R 4  1.73K

  R L




1



 gm 
ro



v
v






RL


 1
ro


 vth   g m v  
ro











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r


R
 L


1

 gm 
ro
so, vo  v 

RL
 1
ro






   v
 r
  










  Rth  v








R
 L


ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Voltage Gain


1



 gm 
ro




 v  v 
RL


 1
ro


 vth   g m v  
ro











1

 gm 
v v v
ro
Av  o  th  

RL
v vth v s
 1
ro



R
 L




R
 L







   v
 r
  










  Rth  v








 
1
  gm 

ro
 

1




R
  1 L

ro

 vth  v  g m   

ro











 R4




 

1
 R 4  Rs

  gm   


ro  

 


1
RL 







RL

  1
 


ro    1  
 g   
    Rth  1

 m 

ro
r

   























1


R
 L







   1
 r
  















  Rth  1












Thus,



for ro   and R 4  Rs
Av 
vo v vth
g m RL

v vth v s

 Rth
 g m Rth  
 r

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 
  1
 
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Voltage Gain
Av 
Av
Av
Av
Av
Av
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g m RL

 Rth  
  1
 g m Rth  
r
   

g m RL

 
1 

  1

R
g
 th  m
r
 
 

g m RL

 
g m r
1 gm  

  1

R
g
 th  m
g m r r g m  
 
g m RL

  gm
 
g m r  1  1
 Rth 
 
  g m r
g m RL

 L arg e  o
  gm
 
 0  1  1
 Rth 

 
  o
g m RL

Gain
1  Rth g m
is positive and can be large
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Input Resistance
iC
i
gmv
ro
vo
r
-ix=iE
iB
vvvx
vx
Input Resistance: With the load resistance attached apply a test input
voltage
l
andd measure the
h input
i
current, Rin=vx/i
From before,
 v  vo
i E  g m v   
 ro
i x  i E
and





R
 L


v x  v
 v  vo
i x  g m v x   x
 ro
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  v
  
  r
1

 gm 
ro
d , vo  v 
and

RL
 1
ro

  vx
  
  r



1

 gm 
ro
and , vo  v x 

RL
 1
ro



R
 L


ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations

1


 gm 
ro


v
v

x
x


RL

 1
ro

i x  g m v x  
ro







R
 L


 Common




   v x 
 r 
   






 
1 
  gm   


ro  
 


R
1 
RL  L 


  1
 

ro    1 



 
ix  vx g m  
   r 

ro

   

















v
1
Rin , BJT  x 
ix 
 
1 
  gm   

ro  
 

1
R


RL  L 

  1
 

r
o


    1
g 
m

 r

ro

  














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Base Input Resistance
Letting ro  
Rin , BJT 
vx

ix
Rin , BJT 
r
g m r  1















1
1
g m  
 r



r  o
o 1 o
o
gm

1
gm
Input Resistance is very small!
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Output Resistance
v
gmv
r
iC
iB
oib
r
Replace RL by a voltage
source, vx
ro
vr
v
ro
ix
vx
iE
Rth
ve
Result follows exactly after discussion in Jaeger,
Jaeger pages 668-670,
668-670 and 683-684.
683-684
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ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Output Resistance
v x  v r  ve
 i x   o ib ro  ve
oib
r
but ,
v e  i x r || Rth   i x
ib  i x
Rth
r  Rth
iC
iB
vr
ro
v
r Rth
r  Rth
ix
vx
iE
Rth
ve
thus,

Rth 
rR
ro  i x  th
v x  i x ro   o   i x
r  Rth
r  Rth 

Even Larrger
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 Rth 
vx
rR
ro   th
 ro   o 
ix
r  Rth
 r  Rth 
Very Larrge
 Rout ,BJT 
Output Resistance is HUGE!
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Current Gain
il
 o  1
Ai 
ith
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ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Gate Solution
The Common Gate solution can be found by recognizing
that the following translations can be made in our small
signal model:
     1
o
o
r  
Av , BJT 
g m RL


 Rth  
  1
g
R

 m th 
r
   

Rin , BJT 
1
1
g m  
 r



g m RL
g m Rth  1
Av , MOSFET 
 Rin , MOSFET 
1
gm
 Rth 
 Rth 
r R
r R
ro   th  ro  g m r 
ro   th
Rout , BJT  ro   o 
r  Rth
r  Rth
 r  Rth 
 r  Rth 
Ai , BJT   o  1 
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 Rout , MOSFET  ro 1  g m Rth   Rth
Ai , BJT   o  1
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Summary of Common Base and Common Gate Characteristics
The input and output
impedances are the
opposite of what is
typically needed for a
•High Voltage Gain
voltage amplifier. Thus,
Common Emitter/Source
•Non-Inverting
Non Inverting Voltage Gain
amplifiers are normally
used instead of Common
•Very Low Input Impedance
Base/Gate. The input and
•Very
V Hi
High
hO
Output
t t Impedance
I
d
output impedances are
useful for current
amplifiers but the current
gain
i is
i att best
b t unity.
it Thus
Th
a current buffer is one
useful application for the
Common Base/Gate
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
You can combine or Cascade configurations to produce “High
High Performance”
Performance amplifiers
with High input impedance, low output impedance and huge voltage gains.
vo vth ,input v1 v 2 vo
Av 

vs
v s vth ,input v1 v 2
v1
v2
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des Low
CC provid
output Im
mpedance,
no gain, but
maintainss positive
gain from
previous sstage
CE provid
des High
Input Imp
pedance,
high gain, and
corrects th
he
negative g
gain
from prev
vious
stage
CS provid
des High
Input Imp
pedance,
Moderatelly high
negative g
gain
vo
ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
For AC-Coupled amplifiers (capacitors between stages),
stages) the DC solution reduces to three
parallel and independent circuits!
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ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
For AC-Coupled amplifiers (capacitors between stages),
stages) the AC solution reduces to three
circuits, each of which has a load dependent on the input resistance of the next stage!
Continued….
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ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
Continued….(For
Continued
(For AC-Coupled amplifiers (capacitors between stages),
stages) the AC solution
reduces to three circuits, each of which has a load dependent on the input resistance of
the next stage!)
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
Continued...
vth
RG

vs
RS  RG
v1
1
vth
v3
  g m 2 ( ro 2 || R12 || RinQ 4 )
v2
v2
  g m1 ( ro1 || R11 || r 2 )
v1
We just found this!
v3
  g m 2 ( ro 2 || R12 || r 3    1RL 3 )
v2
Note: an exact solution would have RL3||ro3
v 3  vo  v 4
 1

vo  v 4 
 g m 3 ( ro 3 || RL 3 )
 r 3


 1
vo  v3  vo 
 g m 3 ( ro 3 || RL 3 )

 r 3
 1


 g m 3 ( ro 3 || RL 3 )
r
vo

  3
v3

 1
1  
 g m 3 ( ro 3 || RL 3 )

 r 3
  1


 
 g m 3 ( ro 3 || RL 3 ) 
 r

v
v v v v v

Av  o  th 1 2 3 o  (1) g m1 ( ro1 || R11 || r 2 )  g m 2 ( ro 2 || R12 || r 3    1RL 3 )    3

v s v s vth v1 v 2 v3
 1   1  g m 3 ( ro 3 || RL 3 ) 

r



 3


Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
AC-Coupled amplifiers (capacitors between stages),
stages) have one major limitation.
limitation They do
not amplify low frequencies or DC voltages. To accomplish this, we must DC-Couple
the stages as shown. Note: for this to be a DC coupled amp, C1 and C6 should also be
replaced as shorts.
Since the bias here is usually ~(2/3)V
(2/3)Vcc (Vcc =15V in this example) , it is easier to use a
PNP for the second stage so that VEB+IERE2 ~ (2/3)Vcc
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ECE 3040 - Dr. Alan Doolittle