First, consider signal models for BJTs. A simple model for BJTs at low frequencies is the following: where node
B
corresponds to the base of the BJT, node
E
corresponds to the emitter of the BJT, the node
C
corresponds to the collector of the BJT, the resistor r
is the BJT's base-emitter resistance, the dimensionless parameter  is its current transfer ratio, and the conductance
G
is the collector-emitter conductance. This model is essentially the traditional h
-parameter model shown in the figure below with h iE
 r
and h oE

G
. The figure also shows a hybrid model, an alternative simple model that easily accommodates small capacitors that can be ignored at low frequencies but that are essential for modeling the signal behavior of the BJT at high frequencies:

Because these two models must be electrically equivalent, their parameters are related. At low frequencies where the capacitors are negligible, note that h iE
  r x
 r

and that h oE
 g
CE
. Because the size of the controlled source must be the same size in both models for them to be electrically equivalent, we see that  i
B
 g v m

. Also, i
B
 v
 r

so that
 v
  r
 g m

 r


One advantage of the hybrid model is that, with suitable parameter values, it can serve as a signal model for FETs as well as BJTs: where node
G
corresponds to the gate of the FET, node
S
corresponds to the source of the FET, and node
D
corresponds to the drain of the FET. Because the gate of the FET is essentially an open circuit at low frequencies and thus

draws negligible current, the resistance between the gate and the source is essentially infinite.
Because similar hybrid models apply to both FETs and BJTs, these notes will focus on the hybrid model rather than the h
-parameter model. Furthermore, because the BJT hybrid model is slightly more complex than the FET hybrid model, these notes will focus on the common amplifier configurations implemented with BJTs. Much of the analysis can be carried over to FET amplifiers by letting
B

G
,
E

S
,
C

D
, and then letting r

  .
At low frequencies, the hybrid models for BJTs and FETs are often drawn in a form that emphasizes their similarity:
In these forms, the resistance between the collector and the emitter and the resistance between the drain and the source are represented as resistances rather than as conductances.
Consider some BJT common emitter amplifier configurations. First consider a discrete component amplifier with the emitter resistor unbypassed:

Here is the discrete component common emitter amplifier with the emitter resistor bypassed:
Here is an IC common emitter amplifier (no bias circuitry shown) with a current source as an active load:

Here is an IC PNP   common emitter amplifier (no bias circuitry shown) with a two BJT basic current source (
Q
1
and
Q
2
) as a load:
The signal equivalent circuit at midband for each of these circuits has the same form:

For discrete component circuits, typically r R
so that o C r o
is neglected. In the
IC circuits with active loads, the resistor
R
is missing but is replaced with some
C combination of r o
and the Norton resistance (which can be very large) of the current source that serves as the active load. In either case, we let the symbol
R
represent the appropriate resistance.
C
We wish to calculate the Thevenin equivalent circuit for both the input terminals and the output terminals:
By noting that v
is the only independent source in the circuit, we conclude that th the Thevenin voltage at the input terminals is zero. Thus, the Thevenin equivalent circuit will have the following simplified form:

Note that we have defined the input to the amplifier as the base terminal of the transistor. In a discrete component amplifier, therefore, the voltage divider bias resistors are included in
R
rather than in th
R
. in
To begin the analysis necessary to find values for this equivalent circuit, define node voltages v
B
, v
E
, v
C
. In terms of the node voltages, note that v

 v
B
 v
E
.
Next, use KCL to write node equations at each of the three nodes: v
B
 v th
R th
 v
B
 v
E r


0 v
E
 v
B r

 v
E
R
E
 g m
 v
B
 v
E
 
0 g m
 v
B
 v
E
  v
C
R
C
 v
C

0
R
L
Collect terms: v
B


1

1
R th r



 v
E



1 r



 v
C
 v th
R th v
B



1
 r
 g m


 v
E


1

1
 r

R
E g m


 v
C
 

0 v
B
 
 v
E

 g m

 v
C


1

1
R
C
R
L



0
Write the equations in matrix form:









R
1 th

1
 r
 g
 m
1 r
 g m

1 r

1

1 r

R
E
 g m
 g m
0
0
1

1
R R
C L




 v v v
B
E
C
 v th

 

R th

 


0
0



 
The following MATLAB m-file solves the node equations and calculates
R in
,
  oc
, and
R out
as function of the parameters.

%BJT Common Emitter amplifier
%Define symbols syms vB vE vC vth Rth RL rpi gm RC RE vin iin Rin vout iout Rout voutoc ioutsc v A b
%Define matrix equation Av = b.
%Define the square matrix of the coefficients, A.
A=[1/Rth+1/rpi -1/rpi 0; -1/rpi-gm 1/rpi+1/RE+gm 0; gm -gm
1/RC+1/RL];
%Define the column matrix, b.
b=[vth/Rth; 0; 0];
%Solve for the node voltages, v.
v = A\b; vB = v(1); vE = v(2); vC = v(3);
%Find the input resistance, Rin.
%The input voltage is vin: vin = vB;
%The input current is iin: iin = (vth - vB)/Rth;
%The input resistance is Rin.
Rin = vin/iin;
Rin = simplify(Rin)
%Find the output voltage, vout.
vout = vC; vout = simplify(vout);
%Find the open circuit output voltage, voutoc.
voutoc = limit(vout, RL, inf)
%Find the output current, iout.
iout = vout/RL;
%Find the short circuit output current, ioutsc.
ioutsc = limit(iout, RL, 0);
%Find the output resistance, Rout.
Rout = voutoc/ioutsc;
Rout = simplify(Rout)
For
R in
, the MATLAB m-file gives:
Rin = RE+rpi+gm*rpi*RE

R in

R
E
 r

 g r R m E
 r

 
1
 m


E
Recall that   g r m

. Thus,
R in
 r

   
1

R
E
Notice that because of the multiplication effect of the current source, the emitter current is a factor   
1
 larger than the base current. This multiplied current means that the voltage across the resistor
R
is
E
  
1
 times larger than the unmultiplied base current would produce. From the perspective of the base current, therefore, the resistor
R
looks like a resistance with value
E
  
1

R
E
.
For   oc
, the MATLAB m-file gives: voutoc = -gm*RC*vth*rpi/(Rth+rpi+gm*rpi*RE+RE)
  oc

R th
 g R v r m C th

 r

 g r R m E

R
E
The negative sign indicates that this amplifier inverts its input signal. Recall that
  g r m

. Thus,
  oc
 
R th
 r



R
C
  
1

R
E v th
In an IC amplifier with no emitter resistor (bias stability being achieved with current steering, for example) or if
R
E
in a discrete component amplifier is bypassed, then
R
E

0
. In that case,
  oc
 
R th

R
C
 r
 v th
With
R

E
0
, note that the voltage gain,   oc explicitly on  and r
 v th
, of the amplifier depends
, transistor parameters whose values vary significantly from unit to unit of the same type of transistor, as well as with temperature. In IC amplifiers or discrete amplifiers with active loads, recall that the resistor
R
can
C correspond to the Norton resistance of a current source that supplies bias current

to the transistor. Because the effective resistance of such a current can be high, large voltage gains are possible.
If
R

E
0
, it can introduce enough negative feedback to make the voltage gain essentially independent of the transistor parameters:
  oc
 
 


R
C
1

R
E
1

R th
 r

 
1

R
E

1 v th
  oc
 
 


1

R
C
R
E 1

1

R th
 r

 
1

R
E v th
Because
 


1


1 and we find a limit on the magnitude of the voltage gain:
  oc v th

R
C
R
E
If 
1
and   
1

R
E
R th
 r

, then
 


1


1 and
1

1

R th
 r

 
1

R
E

1
1

1

R th
 r

 
1

R
E

1

so that
  oc
 
R
C
R
E v th
Under these conditions, therefore, the voltage gain becomes independent of the transistor parameters. Because transistor parameters can vary dramatically from unit to unit of the same type as well as with temperature, the independence of the voltage gain on transistor parameters can be a significant advantage in the manufacture of amplifiers with consistent gains by using transistors with inconsistent parameters.
For
R
, the MATLAB m-file gives: out
Rout = RC
R out

R
C
Consider some BJT common collector amplifier configurations. First consider a discrete component amplifier with a single rail power supply:

Here is a discrete component amplifier with a two-rail power supply:
Here is an IC common collector amplifier with a current source as an active load:
The signal equivalent circuit at midband for each of these circuits has the same form:

We wish to calculate the Thevenin equivalent circuit for both the input terminals and the output terminals:
By noting that v
is the only independent source in the circuit, we conclude that th the Thevenin voltage at the input terminals is zero. Thus, the Thevenin equivalent circuit will have the following simplified form:
Note that we have defined the input to the amplifier as the base terminal of the transistor. In a discrete component amplifier, therefore, the voltage divider bias resistors are included in
R
rather than in th
R
. in

To begin the analysis necessary to find values for this equivalent circuit, define node voltages v
,
B v
. In terms of the node voltages, note that
E v

 v
B
 v
E
. Next, use KCL to write node equations at both nodes: v
B
 v th
R th
 v
B
 v
E r


0 v
E
 v
B r

 g m
 v
B
 v
E
  v
E
R
E
 v
E

0
R
L
Collect terms: v
B


1

1
R r th



 v
E



1 r



 v th
R th v
B



1
 r
 g m


 v
E


1 r

 g m

1

1
R R
E L



0
Write the equations in matrix form:




1

R r th

1
 r

1 g
 m
1 r


1 r

 g m

1

1
R R
E L
 v v
B
E
 v th



R
0 th


The following MATLAB m-file solves the node equations and calculates
R
, in
  oc
, and
R
as function of the parameters. out
%BJT Common Collector amplifier
%Define symbols syms vB vE vth Rth RL rpi gm RC RE vin iin Rin vout iout Rout voutoc ioutsc v A b
%Define matrix equation Av = b.
%Define the square matrix of the coefficients, A.
A=[1/Rth+1/rpi -1/rpi; -1/rpi-gm 1/rpi+gm+1/RE+1/RL];
%Define the column matrix, b.
b=[vth/Rth; 0];
%Solve for the node voltages, v.
v = A\b;

vB = v(1); vE = v(2);
%Find the input resistance, Rin.
%The input voltage is vin: vin = vB;
%The input current is iin: iin = (vth - vB)/Rth;
%The input resistance is Rin.
Rin = vin/iin;
Rin = simplify(Rin)
%Find the output voltage, vout.
vout = vE; vout = simplify(vout);
%Find the open circuit output voltage, voutoc.
voutoc = limit(vout, RL, inf)
%Find the output current, iout.
iout = vout/RL;
%Find the short circuit output current, ioutsc.
ioutsc = limit(iout, RL, 0);
%Find the output resistance, Rout.
Rout = voutoc/ioutsc;
Rout = simplify(Rout)
For
R
, the MATLAB m-file gives: in
Rin = (RE*RL+gm*rpi*RE*RL+rpi*RL+rpi*RE)/(RL+RE)
R in

R R
E L
 r R
L
 g r R R m E L
 r R
E
R
E

R
L
Recall that   g r

. Thus,
R in

R R
E L
  
1
  r


R
L

R
E
R
E

R
L

R in
   
1

R
R R
E L

E
R
L
 r

R in
 r

   
1
 
R
E
R
L


Notice that because of the multiplication effect of the current source, the emitter current is a factor   
1
 larger than the base current. This multiplied current means that the voltage across the resistor
R R
E L
is   
1
 times larger than the unmultiplied base current would produce. From the perspective of the base current, therefore, the resistor
R R
looks like a resistance with value
E L
  
1
 
R R
E L

.
For   oc
, the MATLAB m-file gives: voutoc = vth*RE*(1+gm*rpi)/(gm*rpi*RE+rpi+RE+Rth)
  oc
 g r R m
 v R th E
E

1
 g r m


 r


R
E

R th
Recall that   g r m

. Thus,
  oc

  
1


 
1

R
E
R
E

R th
 r
 v th
  oc

1

1

R th
 r

 
1

R
E v th
Because
1

1

R th
 r

 
1

R
E

1 we find a limit on the magnitude of the voltage gain:
  oc v th

1
If   
1

R
E
R th
 r

, then
1

1

R th
 r

 
1

R
E

1

so that
  oc
 v th
Under these conditions, therefore, the voltage gain becomes unity, independent of the transistor parameters.
For
R
, the MATLAB m-file gives: out
Rout = RE/(gm*rpi*RE+rpi+RE+Rth)*(rpi+Rth)
R out


R
E
E
 r


R th

 r


R
E

R th
Recall that   g r m

. Thus,
R out
 r


R
E
 r


R th

R
E


R
E

R th
 r


R
R
E th

 r




R
 th
1


R
E
R out
 r

R
E




  r
R

1 th




R
1 th
R
E


R out

R
E

 r


R th
 
1


This form shows, consistent with the schematic diagram, that the emitter resistor
R
E
appears in parallel with the impedance,  r


R th
   
1
 , presented by the remainder of the amplifier. Notice the multiplication effect of the current source causes the emitter current to be a factor   
1
 larger than the base current, the current that flows through the series combination of resistors r


R th
. This multiplied current means that the current that flows through the series combination of resistors r


R th
in response to the emitter-base (emitter-ground) voltage appears to be   
1
 times larger than what this voltage would cause to flow through such resistance. From the perspective of the emitter voltage,

therefore, the series combination r


R th
looks like a resistance with value
 r


R th
   
1
 .
Consider some BJT common base amplifier configurations. First consider a discrete component amplifier with a two-rail power supply:
Here is an IC common base amplifier with a current source as an active load:

The signal equivalent circuit at midband for each of these circuits has the same form:

We wish to calculate the Thevenin equivalent circuit for both the input terminals and the output terminals:
By noting that v in
is the only independent source in the circuit, we conclude that the Thevenin voltage at the input terminals is zero. Thus, the Thevenin equivalent circuit will have the following simplified form:
To begin the analysis necessary to find values for this equivalent circuit, define node voltages at both nodes: v
E
, v
C
. Note that v

  v
E
. Next, use KCL to write node equations v
E r

 v
E
 v th
R th
 v
E
R
E
 g m
  v
E
 
0 g m
  v
E
  v
C
R
C
 v
C
R
L

0
Collect terms: v
E


1

1

1
 r

R th
R
E g m


 v
C v
E
  g m
  v
C


1

1
R R
C L



0
Write the equations in matrix form:
 v th
R th





1

1

1
 r

R th
R
E
 g m g m
0
1

1
R R
C L

 v v
E
C



 v
R
0 th th



The following MATLAB m-file solves the node equations and calculates
R in
,
  oc
, and
R out
as function of the parameters.
%BJT Common Base amplifier
%Define symbols syms vE vC vth Rth RL rpi gm RC RE vin iin Rin vout iout Rout voutoc ioutsc v A b
%Define matrix equation Av = b.
%Define the square matrix of the coefficients, A.
A=[1/rpi+1/Rth+1/RE+gm 0; -gm 1/RC+1/RL];
%Define the column matrix, b.
b=[vth/Rth; 0];
%Solve for the node voltages, v.
v = A\b; vE = v(1); vC = v(2);
%Find the input resistance, Rin.
%The input voltage is vin: vin = vE;
%The input current is iin: iin = (vth - vE)/Rth;
%The input resistance is Rin.
Rin = vin/iin;
Rin = simplify(Rin)
%Find the output voltage, vout.
vout = vC; vout = simplify(vout);
%Find the open circuit output voltage, voutoc.
voutoc = limit(vout, RL, inf)
%Find the output current, iout.
iout = vout/RL;
%Find the short circuit output current, ioutsc.
ioutsc = limit(iout, RL, 0);

%Find the output resistance, Rout.
Rout = voutoc/ioutsc;
Rout = simplify(Rout)
For
R
, the MATLAB m-file gives: in
Rin = rpi*RE/(RE+rpi+gm*rpi*RE)
R in

R
E
 r
 r R

E g r R m

E
Recall that   g r m

. Thus,
R in

R
E
 r R r


E


R
E r

  r R
 
E
1

R
E
R in

 r


1
R
E
 r


1

R
E
R in

R
E


 r


1


This form shows, consistent with the schematic diagram, that the emitter resistor
R
E
appears in parallel with the impedance, r

  
1
 , presented by the remainder of the amplifier. Notice the multiplication effect of the current source causes the emitter current to be a factor   
1
 larger than the base current, the current that flows through the resistor r

. This multiplied current means that the current that flows through the resistor r

in response to the emitter-base (emitterground) voltage appears to be   
1
 times larger than what this voltage would cause to flow through such resistance. From the perspective of the emitter voltage, therefore, the resistor r

looks like a resistance with value r

  
1
 .
For   oc
, the MATLAB m-file gives: voutoc = vth*rpi*RE*gm*RC/(rpi*RE+RE*Rth+Rth*rpi+gm*rpi*RE*Rth)

  oc
 
E m C r R
E

R R
E th

R r th

 g r R R m E th
Recall that   g r

. Thus,
  oc


R R
E C r R
E

R R
E th

R r th

 
R R
E th v th
  oc
 r


R
E

R th

R R
C
   
1

R R
E th v th
  oc
 
R R
E C
R
E

R th r

1
   
1

R
R R
E th
E

R th v th
  oc


R
C
R R th
R th
R
E

R r th

   
1
1
 
R R
E th
 v th
  oc


R
C
R th r

  
R R
E th

1
 
R R
E th
 v th
  oc




R
C
1 R th
R R
E th r

  
1


R
E
R th v th
  oc


R
C r R th
  r


1


R R
E th

  r


1


R R
E th v th
  oc
 g R m C
 r

 
1

R th

R R
E th
 v th
  oc
 g R m C
 r

 
1

R th
R R
E th v th
Because

 r

 
1
 R R
E th

1
R th we find a limit on the magnitude of the voltage gain:
  oc v th
 g R m C
If
R th
 r

 
1
 R
E
, then
 r

 
1

R th
R R
E th

1 so that
  oc v th
 g R m C
Under these conditions, therefore, the voltage gain depends mainly on the collector resistance and on the transconductance of the transistor.
For
R
, the MATLAB m-file gives: out
Rout = RC
R out

R
C
BJT
R in v out
R out

Config.
Common
Emitter,
R
E

0 r

   
1

R
E

 


1

R
C
R
E 1

1

R th
 

1
 r

R
E v th
 


1

R
C
R
E 1

1

R th
 

1
 r

R
E

R
C
R
E
R
C
Common
Emitter,
R
E

0 r

Common
Collector

R th

R
C
 r
 v th r

   
1
 
R R
E L

1

1

R th
 r

 
1

R
E v th
 v th
R
C
R
E

 r


R th
 
1


Common
Base
R
E


 r


1

 g R
C
 r

 
1

R th
R R
E th v th
 g R
C
R
C
The reactances of the small internal capacitances of the transistors are too large to provide paths for significant current flow at midband frequencies and hence are negligible. At frequencies higher than midband, the reactances become small enough to allow significant current flows, current flows that limit the effectiveness of the circuit at high frequencies. These capacitances, for example, set the maximum bandwidth of the circuit.
The Miller effect can multiply the value of these small capacitors and hence cause currents through them to become important at lower frequencies and thereby reduce the frequency span of midband. Specifically, the Miller effect can occur when a small capacitor connects a node whose voltage controls the value

of a dependent source and a node driven by the dependent source. Let’s consider this point in more detail.
Consider the capacitor,
C

, that connects the base and collector of a BJT:
The following circuit presents exactly the same i-v characteristics to any circuit connected between node
B '
and ground or between node
C
and ground:
Although we know from the original configuration that the current into nodes
B ' and
C
is equal but opposite, the current in the collector circuit is very much larger (roughly  times larger) than the current in the base circuit. As a consequence, the current produced by the right hand controlled source usually is negligible in comparison with the collector current. This same sized current, on the other hand, can affect the operation of the base circuit significantly. Thus, the following simpler equivalent circuit can be used to calculate the main effect of
C

, the current, which we call the Miller current, produced by
C

in the base circuit:

From this circuit, we find that the Miller current produced by
C

in the base circuit: i
C


C


B '
 v
C
 dt
The size of this current clearly depends on the relationship between v
B '
and v
C
, which varies among the transistor amplifier configurations that we have analyzed.
Consider the signal equivalent circuits, including the small internal transistor capacitances, for the basic configurations considered above.
BJT
Config.
Common
Emitter

Common
Collector
Common
Base
In the common emitter circuit, v
C
 v out
and, because r x
is relatively small, v
B '
 v in
. Using the results of earlier analysis, we can estimate the value of i
C  the case   
1

R
E
R th
 r

. For this case, v out
  
R
C
R
E
 v in
. But because
for
  
1

R
E
R th
 r

, little voltage drop appears across
R th
 r x
so that v in
 v th
.
Thus, the Miller current is approximately i
C


C


B '
 v
C


C
 dt
 in
 v out
 dt

i
C


C



 in dt
R
C
R
E v in




 i
C



 1

R
C
R
E


C
 d v in dt
In the common emitter configuration, therefore, the Miller current, i
C 
, is multiplied by approximately by the magnitude,
R R
C E
, of the voltage gain for this configuration. Some of this Miller current flows through r

and
C

and produces voltage variations that change the current produced by the dependent source, and hence, the output of the amplifier. As a consequence, the Miller effect can be strong in the common emitter configuration. Note that if the inequality
  
1

R
E
R th
 r

is not satisfied, more of the Miller current flows into the base of the BJT and the Miller effect can be even stronger.
In the common collector circuit, v
C

0
, and because r x
is relatively small, v
B '
 v in
. Using the results of earlier analysis, we can estimate the value of i
C  the case   
1

R
E
R th
 r

. For this case, v out
 v th
. But because
for
  
1

R
E
R th
 r

, little voltage drop appears across
R th
 r x
so that v in
 v th
.
Thus, the Miller current is approximately: i
C 

C


B ' dt

0
 i
C


C
 d v in dt
In the common collector configuration, therefore, the Miller capacitance,
C

, is not multiplied. Thus, the Miller effect in the common collector configuration is much smaller than in the common emitter configuration.
In the common base configuration, v
C
 v out
, and because r x
is relatively small, v

B '
0
. According to earlier analysis, v out
 g R v m C in
. Using the results of earlier

analysis, we can estimate the value of i
C 
for the case
 r

 
1
 R
E
R th
. For this case, v out
 g R v m C th
. But because
 r

 
1
 R
E
R th
, little voltage drop appears across
R th
so that v in
 v th
. Thus, the Miller current is approximately: i
C


C


B '
 v
C
 dt

C
 d

0
 dt v out
 i
C

 
C
 d
  g R v m C in
 dt i
C 
  g R m C

C
 d v in dt
In the common base configuration, therefore, the Miller current, i
C 
, is multiplied approximately by the magnitude, g R
, of the voltage gain of this configuration, m C which can be large. Because the base is essentially at ground potential (because of the small value of r x
), however, the Miller current mostly flows through r x
to ground and hence has little effect on v

and hence on the dependent source and on the output of the amplifier. As a consequence, the Miller effect affects the performance of the common base configuration much less than the common emitter configuration despite the fact that the Miller current is not small.
The cascode configuration combines the common emitter and common base configurations to obtain performance similar to the common emitter performance but with reduced Miller effect. Specifically, the collector (output) of a common emitter circuit connects to the emitter (input) of a common base circuit so that the common emitter stage sees the relatively input low resistance of the common base stage as the collector resistance of the common emitter stage. The low collector resistance for the common emitter stage reduces the Miller multiplication for this stage and hence improves it high frequency performance. In earlier analysis, we saw that then common base stage is largely immune to the

Miller effect. The cascode configuration, we will see, offers performance similar to a common base emitter amplifier but with much reduced Miller effects.
Consider some BJT cascode configurations. First consider a discrete BJT cascode amplifier with an active load and an unbypassed emitter resistor:
Here is an IC BJT folded cascode amplifier (no bias circuitry shown) with a current source as an active load:

The signal equivalent circuit at midband for each of these circuits has the same form:
We wish to calculate the Thevenin equivalent circuit for both the input terminals and the output terminals:
By noting that v
is the only independent source in the circuit, we conclude that th the Thevenin voltage at the input terminals is zero. Thus, the Thevenin equivalent circuit will have the following simplified form:

To begin the analysis necessary to find values for this equivalent circuit, define node voltages v
,
B 1 v
,
E 1 v
E 2
, v
C 2
. In terms of the node voltages, note that v

1
 v
B 1
 v
E 1
and that v

2 v
E 2
. Next, use KCL to write node equations at each of the four nodes: v
B 1
 v th
R th
 v
B 1
 v
E 1 r

1

0 v
E 1
 v
B 1 r

1
 v
E 1
R
E
 g m
 v
B 1
 v
E 1
 
0 v
E 2 r

2

0
 g m 2

0
 v
E 2
  g m 1
 v
B 1
 v
E 1
 
0 g m 2

0
 v
E 2
  v
C 2
R
C
 v
C 2
R
L

0
Collect terms: v
B 1


1

1
R th r

1


 v
E 1


 r

1
1


 v
E 2
 
 v
C 2
 
 v th
R th v
B 1



1
 r

1 g m 1


 v
E 1

 r

1

1

1
R
E g m 1


 v
E 2
 
 v
C 2
 

0 v
B 1
 
 v
E 1

 g m 1

 v
E 2


1 r

2
 g m 2


 v
C 2
 

0 v
B 1
 
 v
E 1
 
 v
E 2

 g m 2

 v
C 2


1

1
R
C
R
L



0
Write the equations in matrix form:











R
1 th

1
 r

1 g

0 m 1
1 r

1 g m 1
1 r

1

1 r

1

1
R
E
 g m 1
 g m 1
0
0
0
1 r

2
 g m 2
 g m 2
1
R
C
0
0
0
1










 v
B 1

 
 v
E 1


 v
E v
C
2
2



 v th

 

R th


0


 0


 0 

R
L
The following MATLAB m-file solves the node equations and calculates
R in
,
  out oc
, and
R out
as function of the parameters.
%BJT Cascode amplifier
%Define symbols syms vB1 vE1 vE2 vC2 vth Rth RL rpi1 rpi2 gm1 gm2 RC RE vin iin
Rin vout iout Rout voutoc ioutsc v A b
%Define matrix equation Av = b.
%Define the square matrix of the coefficients, A.
A=[1/Rth+1/rpi1 -1/rpi1 0 0; -1/rpi1-gm1 1/rpi1+1/RE+gm1 0 0; gm1
-gm1 1/rpi2+gm2 0; 0 0 -gm2 1/RC+1/RL];
%Define the column matrix, b.
b=[vth/Rth; 0; 0; 0];
%Solve for the node voltages, v.
v = A\b; vB1 = v(1); vE1 = v(2); vE2 = v(3); vC2 = v(4);
%Find the input resistance, Rin.
%The input voltage is vin: vin = vB1;
%The input current is iin: iin = (vth - vB1)/Rth;
%The input resistance is Rin.
Rin = vin/iin;
Rin = simplify(Rin)
%Find the output voltage, vout.
vout = vC2; vout = simplify(vout);

%Find the open circuit output voltage, voutoc.
voutoc = limit(vout, RL, inf)
%Find the output current, iout.
iout = vout/RL;
%Find the short circuit output current, ioutsc.
ioutsc = limit(iout, RL, 0);
%Find the output resistance, Rout.
Rout = voutoc/ioutsc;
Rout = simplify(Rout)
For
R
, the MATLAB m-file gives: in
Rin = RE+rpi1+gm1*rpi1*RE
R in

R
E
 r

1
 g r R m 1 1 E
 r

1
 
1
 g r m 1

1

R
E
Recall that  
1 g m 1 r

1
. Thus,
R in
 r

1
  
1

1

R
E
This result is the same as that obtained for the BJT common emitter amplifier analyzed earlier. As in the earlier analysis, notice that because of the multiplication effect of the current source, the emitter current in BJT
Q
1
is a factor
 
1

1
 larger than the base current. This multiplied current means that the voltage across the resistor
R
E
is  
1

1
 times larger than the unmultiplied base current would produce. From the perspective of the base current, therefore, the resistor
R
E
looks like a resistance with value  
1

1

R
E
.
For   oc
, the MATLAB m-file gives: voutoc = gm2*RC*gm1*rpi2*vth*rpi1/(Rth+Rth*gm2*rpi2+rpi1+gm1*rpi1*RE+RE+gm
2*rpi2*rpi1+gm2*rpi2*gm1*rpi1*RE+gm2*rpi2*RE)
  oc

R th

R g r th m 2

2
 g m 2
R g
C m 1 r v r
2 th

1
 r

1
 g m 1 r R
1 E

R
E
 g m 2 r r
2

1
 g m 2 r g
2 m 1 r R
1 E
 g m 2 r R
2 E

The negative sign indicates that this amplifier inverts its input signal. Recall that
 
1 g m 1 r

1
and that  
2 g r m 2

2
. Thus,
  oc

R th

R th

2
 r

1
 
1
  
2 1
R
C
R
E

R
E
 
2 r

1
  
1
R
E
 
2
R
E v th
  out oc

 
2

1

R th
  
2

1

 r

1
 
2 1
 
R
C
 
1
1

R
E
  
2

1

1

R
E v th
  oc

 
2

1
 r

1
  
1

R th

R
C
 
1

1

R
E v th
  oc
 

2

2

1

1

1

1

R
C r

1

R th

1

1

R
E

 v th
  oc
 

2

2

1

1

1

1
R
C
1
R
E



 r


1
1


R
1 th


1


 v th
  oc
 
R
C
R
E

2

2

1

1

1

1



1
 r

1

R th

1

1


1


 v th
Because

1

1

1

1

2

2

1

1 and
1


1
R th
 r


1

1

1
R
E

1 we find a limit on the magnitude of the voltage gain:

If 
1
  oc v th

R
C
R
E
1
, 
2
1
and  
1

1

R
E
R th
 r

1
, then

1

1

1

1

2

2

1

1 and so that
1


1
R th
 r


1

1

1
R
E

1
  oc
 
R
C
R
E v th
Under these conditions, therefore, the voltage gain becomes independent of the transistor parameters, with the same value as the BJT common emitter amplifier analyzed earlier.
For
R out
, the MATLAB m-file gives:
Rout = RC
R out

R
C
This result is the same as for the BJT common emitter amplifier analyzed earlier.
Thus, the cascode configuration provides the same voltage gain, input resistance, and output resistance as the common emitter configuration, but with much smaller Miller effect.
If
R
C
  and the most important resistor in the collector circuit is the collectorto-emitter resistance, r o
, the common emitter configuration has an output

resistance of r o
. Under the same circumstance, the output resistance of the common base amplifier is much higher, even larger output resistance of about 
 r o
. The cascode configuration has an r r o 2

2
  r
2 o 2
, larger because the output resistance, r o
, of the BJT is much larger than its input resistance, r

, so that r o 2 r

2
. The extraordinarily high output resistance of the cascode configuration can make the cascode configuration useful for current mirrors in CMOS ICs.