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 14.1, 14.4
and Notes
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ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Source
Amplifier
Load Resistance
Amplifier Output
Resistance
Amplifier Input
Resistance
Source Input
Resistance
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 Circuit Voltage Gain can be found by applying 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.
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ECE 3040 - Dr. Alan Doolittle
Amplifier Configurations
Current Amplifier: Current input and Current output
The controlled source is a 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)
•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 (for a current amplifier) for maximum current gain.
•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)
•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.
•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)
•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
possible for maximum transresistance gain.
•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 a test input voltage and measure the input current, Rin=vt/it
Or
Apply a test input current and measure the input 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
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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, rπ, 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 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, 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 b.) Apply a test current to the output and measure the output
voltage, Rout= vt/it
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Can be modeled as a current amplifier,
IC=βIB, or a transconductance amplifier ,
iC=KvBE
Output
Resistance
Input Resistance
Output
Resistance
Input Resistance
Common Emitter and Common Source
Modeled as transconductance
amplifier, iDS=KvGS
Overall Amplifier Configuration
•Emitter/Source is neither an input nor an output
•Input is between base-emitter or gate-source
•Output is between collector-emitter and drain-source
•Is a transconductance amplifier (see small signal models we have
used in previous examples)
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ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Emitter and Common Source
Portion due
to transistor
rπ
Portion due to Portion due to
transistor bias circuitry
gmvπ
or gmvgs
rο
Rc
Rout
Rin
Portion due to
bias circuitry
rπ 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 || rπ 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 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 Drain/Common Collector
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector and Common Drain
DC Circuit
D
C
G
B
E
S
Collector (or Drain) is neither an input or output
Input is Base (or Gate)
Output is Emitter (or Source)
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector
DC Circuit converted to AC Equivalent (reduced)
DC
Circuit
Note the extra
ground due to C2
AC
Circuit
ib B
C
rπ
βib
ro
E
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector
DC Circuit converted to AC Equivalent (reduced)
AC
Circuit
ib B
C
rπ
βib
ro
E
ib B
AC Circuit
(reduced)
C
βib
rπ
E (β+1)ib
vo
vo = (β o + 1)ib (R 4 || ro || R7 )
vth = ib (Rth + rπ + (β o + 1)(R 4 || ro || R7 ))
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Collector
AC Voltage Gain
vo = (β o + 1)ib (R 4 || ro || R7 )
vth = ib (Rth + rπ + (β o + 1)(R 4 || ro || R7 ))
vth = vs
Av =
R 2 || R1
R 2 || R1 + Rs

(β o + 1)ib (R 4 || ro || R7 )
vo vth  R 2 || R1 


= 
vth vs  R 2 || R1 + Rs  ib (Rth + rπ + (β o + 1)(R 4 || ro || R7 )) 

(β o + 1)RL
 R 2 || R1 
 where R L = (R 4 || ro || R7 )
Av = 

 R 2 || R1 + Rs  (Rth + rπ + (β o + 1)RL ) 
gm
multiplying numerator and denominator by
(β o + 1)






 R 2 || R1 
g m RL
Av = 



g m rπ  
 R 2 || R1 + Rs    Rth


  g m  (β + 1) + RL  + (β + 1)  
o
o










 R 2 || R1 
g m RL

Av = 



 R 2 || R1 + Rs    Rth



  g m  (β + 1) + RL  + α 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
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 Circuit converted to AC Equivalent (reduced)
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 || R 7 )
vth = vGS + g m vGS (R 4 || ro || R 7 ) = vGS (1 + g m (R 4 || ro || R 7 ))
vth = v s
R 2 || R1
R 2 || R1 + Rs
vo vGS vth  R 2 || R1  g m (R 4 || ro || R 7 ) 


Av =
= 
vGS vth v s  R 2 || R1 + Rs  (1 + g m (R 4 || ro || R 7 )) 
 R 2 || R1  g m R L 
 where R L = (R 4 || ro || R 7 )

Av = 
 R 2 || R1 + Rs  (1 + g m R L ) 
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
=
= rπ + (β o + 1)RL
i
Resistance in the emitter circuit is “multiplied”
by transistor to increase the input resistance
Rin , MOSFET
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vx vx
=
=
=∞
i
0
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

vx
vx
v
vx
vx  vx
 +
=
− β oi + x =
− β o  −
RL rπ + Rth
ro rπ + Rth
 rπ + Rth  RL
v
1
where RL = ro || R 4
= x =
1
1
ix
+
rπ + Rth RL
(β o + 1)
ix = −i − β 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, apply a test voltage to the output and measure the
output current, Rout=vx/i
i x = − g m vGS +
Rout ,MOSFET =
v
vx
= 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 Voltage Gain
•Very High Input Impedance
•Low Output Impedance
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These properties make the
CC/CD configuration very
good for impedance
transformation, I.E.
“buffering” high
impedances to low
impedances. 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
(βo+1)ith
i
Ai , BJT =
i
= βo +1
ith
Ai , MOSFET =
i
=∞
ith
Note: since R7 was originally defined as the load, the current gain should actually be (β+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
Base (or Gate) is neither an input or output
Input is Emitter (or Source)
Output is Collector (or Drain)
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base
DC Circuit converted to AC Equivalent (reduced)
DC
Circuit
AC
Circuit
Note: Jaeger let’s ro go to infinity which
makes the math dramatically easier
C
B
rπ
gmvπ
ro
vo
E
Georgia Tech
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base AC Equivalent (reduced)
C
B
vπ
AC
Circuit
rπ
gmvπ
ro
vo
E
vπ
AC
Circuit
(reduced)
rπ
gmvπ
ro
gmvπ
ro
vo
rπ
vπ
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
gmvπ
iB
i E = iC + i B = g m vπ + i + i B
 v + vo
i E = g m vπ +  π
 ro
vth = −i E Rth − i B rπ
  vπ
 + 
  rπ

 v + vo
− vth =  g m vπ +  π
 ro

also,



  vπ
 + 
  rπ
iE
vπ

v 
  Rth +  π rπ


 rπ 
1

 gm +

 v + vo  
ro
  R L so, vo = −vπ 
vo = −iC RL = − g m vπ +  π


RL
 ro  

 1+
ro




1 




 gm +  
r





o 
v
v
R
−
π
π
L




RL  


+
1


 
r


v

o 
 +  π  R + v

− vth =  g m vπ + 
π
  r   th
ro
π


























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rπ
Vth = vs
R4
= 0.8667Vs
R 4 + Rs
Rth = Rs || R 4 = 1.73K


R
 L


ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Voltage Gain


1



 gm +
ro




 vπ − vπ 
RL


 1+
ro


− vth =  g m vπ + 
ro











1

 gm +
ro
v v v
Av = o π th = −

RL
vπ vth v s
 1+
ro



R
 L




R
 L







 +  vπ
 r
  π










  Rth + vπ






⇒

 
1
  gm +

ro
 

1−




R
  1+ L

ro

− vth = vπ  g m +  

ro











 R4





 
1
 R 4 + Rs

  gm +  


ro  

 


1−
RL 







RL

  1+
 


ro    1  
 g +  
+    Rth + 1


 m 
r
ro
  π 
























−1


R
 L







 +  1
 r
  π















  Rth + 1












Thus,



for ro → ∞ and R 4 >> Rs
Av =
vo vπ vth
g m RL
=
vπ vth v s

 Rth
 g m Rth + 
 rπ

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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
 π  

g m RL
=
 
1 

 + 1
R
g
+
 th  m
r
π 
 

g m RL
=
 
g m rπ
1 gm  

 + 1
g
R
+
 th  m
g m rπ rπ g m  
 
g m RL
=
  gm
 
(g m rπ + 1) + 1
 Rth 
 
  g m rπ
g m RL
=
⇒ L arg e β o
  gm
 
(β 0 + 1) + 1
 Rth 
β
 
  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
gmvπ
ro
vo
rπ
-ix=iE
iB
vπ=−vx
vx
Input Resistance: With the load resistance attached apply a test input
voltage and measure the input current, Rin=vx/i
From before,
 v + vo
i E = g m vπ +  π
 ro
i x = −i E
and





R
 L


v x = −vπ
 v − vo
i x = g m v x +  x
 ro
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 v
 +  π
  rπ
1

 gm +
ro
and , vo = −vπ 

RL
 1+
ro

  vx
 +  r
  π



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 
  π 






 
1 
  gm +  


ro  
 


1−
R


RL  L 


  1+
 


ro    1 
 
i x = v x  g m +  
 +  r 

ro

  π 

















v
1
Rin , BJT = x =
ix 
 
1 
  gm +  

ro  
 

1−
R


RL  L 

  1+
 

r
o


  +  1
g +
m

 r

ro

  π















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Base Input Resistance
Letting ro → ∞
Rin , BJT =
vx
=
ix
Rin , BJT =
rπ
g m rπ + 1
=













=
1
1
g m + 
 rπ



rπ β o
βo +1 βo
αo
gm
≈
1
gm
Input Resistance is very small!
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Output Resistance
vπ
gmvπ
rπ
iC
iB
β o ib
rπ
Replace RL by a voltage
source, vx
ro
vr
vπ
ro
ix
vx
iE
Rth
ve
Result follows exactly after discussion in Jaeger, pages 668-670, and 683-684.
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ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Output Resistance
v x = vr + ve
= (i x − β o ib )ro + v e
β o ib
rπ
but ,
v e = i x [rπ || Rth ] = i x
ib = − i x
Rth
rπ + Rth
iC
iB
vr
ro
vπ
rπ Rth
rπ + Rth
ix
vx
iE
Rth
ve
thus,

Rth 
rR
ro + i x π th
v x = i x ro − β o  − i x
rπ + Rth 
rπ + Rth

Even Larger
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 Rth 
vx
rR
 ro + π th
= ro + β o 
ix
rπ + Rth
 rπ + Rth 
Very Large
∴ Rout ,BJT =
Output Resistance is HUGE!
ECE 3040 - Dr. Alan Doolittle
Transistor Amplifier Configurations
Common Base Current Gain
il
= αo ≈ 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:
β → ∞ ⇒ α →1
o
o
rπ → ∞
Av , BJT =
g m RL
→

 Rth  
 + 1
g
R
+
 m th 
r
 π  

Rin , BJT =
1
1
g m + 
 rπ



g m RL
g m Rth + 1
Av , MOSFET =
→ Rin , MOSFET =
1
gm
 Rth 
 Rth 
r R
r R
ro + π th = ro + g m rπ 
ro + π th
Rout , BJT = ro + β o 
rπ + Rth
rπ + Rth
 rπ + Rth 
 rπ + Rth 
Ai , BJT = α o ≈ 1 →
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→ Rout , MOSFET = ro (1 + g m Rth ) + Rth
Ai , BJT = α 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 Voltage Gain
amplifiers are normally
used instead of Common
•Very Low Input Impedance
Base/Gate. The input and
•Very High Output Impedance
output impedances are
useful for current
amplifiers but the current
gain is at best unity. Thus
a current buffer is one
useful application for the
Common Base/Gate
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ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
You can combine or Cascade configurations to produce “High Performance” amplifiers
with High input impedance, low output impedance and huge voltage gains.
vo vth ,input v1 v 2 vo
=
Av =
v s vth ,input v1 v 2
vs
v1
v2
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CC provides Low
output Impedance,
no gain, but
maintains positive
gain from
previous stage
CE provides High
Input Impedance,
high gain, and
corrects the
negative gain
from previous
stage
CS provides High
Input Impedance,
Moderately high
negative gain
vo
ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
For AC-Coupled amplifiers (capacitors between 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), 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 AC-Coupled amplifiers (capacitors between stages), the AC solution
reduces to three circuits, each of which has a load dependent on the input resistance of
the next stage!)
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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 || rπ 2 )
v1
We just found this!
v3
= − g m 2 (ro 2 || R12 || (rπ 3 + (β + 1)(ro 3 || RL 3 )))
v2
v3 = vo + v4
 1

+ g m 3 (ro 3 || RL 3 )
vo = v4 
 rπ 3

 1

+ g m 3 (ro 3 || RL 3 )
vo = (v3 − vo )
 rπ 3

 1


+ g m 3 (ro 3 || RL 3 )
r
vo

=  π3
v3
 1

1 + 
+ g m 3 (ro 3 || RL 3 )
 rπ 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 || rπ 2 ) )(− g m 2 (ro 2 || R12 || (rπ 3 + (β + 1)RL 3 )) )  π 3

vs vs vth v1 v2 v3
 1 +  1 + g m 3 (ro 3 || RL 3 ) 


 r


  π3
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ECE 3040 - Dr. Alan Doolittle
Multistage Amplifier Configurations
AC-Coupled amplifiers (capacitors between stages), have one major 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)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