Objectives
„ To understand the principle of operation and the advantage of the cascode amplifier.
„ To explore the advantage of the CC-CE cascade amplifier and compare it with the cascode amplifier.
„ To calculate the dominant poles and the amplifier bandwidth for a multistage amplifier.
Introduction
In the last lecture the frequency response of the differential amplifier as well as the common collector-common base (CC-CB) cascade amplifier were examined. As we saw in the last lecture the obtained high bandwidth in the CC-CB amplifier was obtained by avoiding the use of the common emitter amplifier which is know with its poor frequency response. In this lecture the frequency response of the common emitter amplifier will be enhanced by means of cascading it with other configurations. The analysis will be based on the single amplifier configurations analysis done in the previous lectures. At the end a complete cascade amplifier will be considered.
The cascode amplifier
(a) The BJT Cascode (b) The MOSFET Cascode
Figure 1: The Cascade Amplifier
• Figure 1 shows the cascode amplifier for both BJT and MOSFET transistors.
• The cascode amplifier was initially invented for the triode vacuum tubes amplifiers to overcome the Miller's capacitance effect.
• The name cascode is a shortened version of "cascaded cathode" since the anode of the first triode was feeding the cathode of the next triode.
• The basic idea of the cascode is to combine the high input resistance and large transconductance of the common emitter amplifier with the current buffer and excellent frequency performance of the common base amplifier.

• The cascode amplifier may be designed either to have the same DC gain as CE (CS) amplifier with a wider bandwidth or higher DC gain with same gain-bandwidth product. The first possibility will be examined by the following example.
Example 1:
Calculate the high 3dB frequency of the BJT cascode amplifier in Figure 1-a assuming a load resistor R
L
is connected at v o
.
Solution:
The small signal circuit of the cascode amplifier in Figure 1-a may be drawn as in Figure 2.
From which it is clear that the cascode amplifier may be treated as a common emittercommon base amplifier.
Figure 2: BJT cascode amplifier small signal circuit
To simplify the analysis we may treat the amplifier as a cascade of CE amplifier with an output resistance = r e2
=1/ g m 2
and a common base amplifier with source resistance equals the output resistance of Q
1
or r o1
.
OCTC of Q 1 with load resistor 1/ g m 2
may be obtained as (refer to lecture 5, for
Authorware this is the lecture of CE high frequency response):
R C
π 1
+ R C
μ 1
= R C
T
Where R
π o 1
= r
π 1
( r x 1
+ R s
1
= R
π o 1
⎛
⎜ C
π 1
+ C
μ 1
⎛
1 + g g m 1 m 2
+ m
1
π o 1
⎞
⎠
⎞
⎟
( ) )
As I
C 2
= I
C 1
, g m 2
= g m 1
∴ R C
T 1
≅ R
π o 1
(
C
π 1
+
, gain of the first stage is unity. Assuming g m 2
R
π o 1
>>1,
2 C
μ 1
)
OCTC of Q 2 with source resistor r o1
may be obtained as (refer to lecture 6, for Authorware this is the lecture of CB high frequency response):
R
π o 2
C
π 2
+ R
μ o 2
C
μ 2
= r x 2
⎝
⎜
⎜
⎛
1 +
C
π m
2 g r o 1
⎜
⎛
⎝
1 + r o 1 r x 2
⎟
⎞
⎠
+ C
μ 2
⎜
⎛
⎝
1 +
1 + m L g r o 1
⎟
⎞
⎠ ⎠
⎟
⎟
⎞
+
μ L

Assuming that
∴ R
π o 2
C
π 2
+ r o1
>> R
L
and
R
μ o 2
C
μ 2
≅ ⎜
⎛
⎝
C g
μ f
= g m r o
>>1:
π m
2
2
+ C
μ 2
( r x 2
+ R
L
Assuming matched devices,
∴ ω
H
=
⎛
⎝
( ( x
+ R s
) )
+
1 g m
1
⎞
⎠
C
π
+
( r x
+ R
L
+ 2
(
)
⎟
⎞
⎠
( x
+ R s
) ) )
C
μ g
To compare the obtained bandwidth of the cascode with the case of CE amplifier let us assume the same parameters as in Example 4 in lecture 5 (for Authorware this is the lecture of CE high frequency response) m
= 66.4
mS r
π
= 1.5
k Ω r o
= 60.2
k Ω r
X
= 250 Ω
C
μ
= 0.5pF
R
L
= R ′
L
= 3.86k
Ω
C
π
= 20.6pF
R
π o 1
= R th 2
= 0.64
k Ω
ω
H
=
⎛
⎝
R
π o 1
+ g
1 m
⎞
⎠
1
C
π
+ ( r x
+ R
L
+ 2 R
π o 1
) C
μ
= 61.77M rad/sec
The high frequency obtained for the common amplifier in the example was 10.26Mrad/sec.
The CD-CS, CC-CE, and CD-CE cascade Amplifier
V
CC
V
DD
V
CC
I
2
I
2
Q
1
Q
1 v o v o
Q
2
Q
2
I
1
I
1
Figure 3: (a) CD–CS amplifier. (b) CC–CE amplifier. (c) CD–CE amplifier.
• As discussed in the last section the cascode configuration was invented to reduce the effect of Miller's capacitance associated with the CE (CS) amplifier to increase its bandwidth. The same result may be obtained with different configuration by cascading a CC (CD) amplifier with the CE (CS) amplifier as shown in Figure 3.
• Although, both configuration reduce the Miller's capacitance. The difference between the cascode and the CC-CE or (CD-CS) may be summarized as:

• In the cascode we isolate the load resistance from the CE collector by means of low input resistance of the CB stage
• In the CC-CE the large Miller capacitor sees a small resistance from the output of the CC which isolate the effect of the source resistance
Example 2:
Calculate the high 3dB frequency of the BJT CC-CE amplifier in Figure 3-b assuming a source with a source resistor R sig
is connected at the input.
Solution:
The small signal circuit of the CC-CE amplifier in Figure 3-b may be drawn as in Figure 4.
R sig
Q
1 v o
V sig Q
2
R
L
Figure 4: BJT cascade amplifier small signal circuit
The common collector amplifier has a load resistance equals the input resistance of Q
2
or
(r
π 2
+ r x2
) , and The CE amplifier has a source resistance equals the output resistance of Q
1 or (R sig
+r x1
+ r
π 1
)/ β .
OCTC of Q 1 with load resistor r
π 2
may be obtained as (refer to lecture 6, for Authorware this is the lecture of CC high frequency response):
R
π o 1
C
π 1
+ R
μ o 1
C
μ 1
= ( R
S
+ r x 1
+ r
π 2
+ r x 2
)
1 +
C
π 1
1 π 2
+
(
( R sig
+ r x 1
) R in 1
)
C
μ 1
R in 1
= r
π 1
(1 β
1
)( r
π 2
+ r x 2
)
Assuming β >> 1, R
R
π o 1
C
π 1
+
μ μ sig
R o 1
C
1
<<R in1
≅ ( R sig
+ x
)
μ 1
OCTC of Q 2 with source resistor R o1
may be obtained as (refer to lecture 5, for
Authorware this is the lecture of CE high frequency response):
R
π o 2
C
π 2
+ R
μ o 2
C
μ 2
=
( r
π 2
( r x 2
+ R o 1
) )
⎝
⎜
⎜
⎛
C
π 2
+ C
μ 2
⎝
⎜
⎜
⎛
1 + m L
+ (
R
L r
π 2
( r x 2
+ R o 1
) )
⎠
⎟
⎟
⎞
⎠
⎟
⎟
⎞
R o 1
=
R sig
+ r x 1
+ r
π 1
1 + β
Assuming matched devices

∴ ω
H
=
⎛
⎜ r
π
⎛
⎝ r x
+
R sig r r
π
1 + β
⎞
⎠
⎞
⎟ C
π
+ C
μ
1
⎛
⎜ ( 1 + g R
L
)
⎛
⎜ r
π
⎛
⎝ r x
+
R sig
1
+ + r
π
+ β
⎞
⎠
⎞
⎟ + R
L
+ R sig
+ r x
⎞
⎟
To compare the obtained bandwidth of the CC-CE with the case of CE amplifier let us assume the same parameters as in Example 4 in lecture 5 (for Authorware this is the lecture of CE high frequency response) g m
= 66.4
mS r
π
= 1.5
k Ω r o
= 60.2
k Ω r
X
= 250 Ω
C
μ
= 0.5pF
R
L
= R ′
L
= 3.86k
Ω
C
π
= 20.6pF
R sig
= Ω
ω
H
= 26.8M rad/sec
The high frequency obtained for the common amplifier in the example was 10.26Mrad/sec.
Multistage Amplifier
Multistage amplifiers are utilized either to have a high overall voltage gain or to enhance the amplifier characteristics by increasing the input resistance, decreasing the output resistance, or increasing the bandwidth as studied in the previous two sections by using cascode or CC-CE cascade.
As a final example let us consider multistage amplifier using CS amplifier as an input stage to provide high input impedance and CC as a final stage to provide low output impedance with most of the gain provided by a CE amplifier.
Example 3:
Use open-circuit and short-circuit time constant methods to estimate upper and lower cutoff frequencies and bandwidth of the multistage amplifier shown in Figure 5. The transistors' parameters are as follows:
M
1
has fT1 =200 MHz, λ =0.02V
-1 , K n
=12.5mA/V 2 , and CGD1 =1.0 pF .
Q
2
has fT2 =500 MHz, β
Q
3
has fT3 =500 MHz, β o2
=150, C o3
=80, C
μ2 =1.0 pF, rx2 =250 Ω, and V
A2
=80V .
μ 3
=1.0 pF,and rx3 =250 Ω, V
A3
=100V .
The transistors are biased at the following operating points:
M
Q
2
1
is biased at (4mA, 8.72).
is biased at (1.5mA, 2.7).
Q
3
is biased at (2mA, 5.4).

Figure 5: Multistage Amplifier
Solution:
Using the given operating points given, the small signal transistors' parameters may be calculated a follows: g m 1
= 2
1
(1 + λ V
DS 1 mS r o 1
=
λ I
1
D 1
= 12.5
k Ω
C
GS 1
= g
2 π f m 1
T g m 2 r
π 2
=
=
I
C
V
T
2
1 + β o g m 2
2
=
1
=
− C
GD 1
40.
I
C 2
2.52
k
=
= p
60 mS
Ω r o 2
=
V
I
C
A
2
2 =
80
1.5
= 53.3
k Ω
C
π 2
= g
2 π f m 2
T g m 3 r
π 3
=
=
I
V
C
T
3 =
1 + β o 3 g m 3
2
− C
40.
I
C
μ 2
3
= 1.01
k
= 18.1pF
=
Ω
80 mS r o 3
=
V
I
A 3
C 3
=
100
2
= 50 k Ω
C
π 3
= g
2 π m 3 f
T 3
− C
μ 3
= 24.5pF

Coupling and bypass capacitors determine the low-frequency response.
SCTC for each of the six independent coupling and bypass capacitors has to be determined.
R
1 S
= R
I
+ ( R R in 1
) 10k Ω + 1M Ω ∞ = 1.01M
Ω
R
2 S
= R
S 1
R
3 S
=
1 g m 1
(
R
D 1
R
O 1
= 200 Ω
R
B 2
R in 2
1
0.0108S
R th 2
= R
B 2
R
D 1 r o 1
= 571 Ω
= 63.3
Ω
R
D 1 r o 1
R
B 2 r
π 2
)
= 2.79k
Ω
R
4 S
= R
E
R
5 S
2
R th 2
+ r
π
β o 2
+ 1
2
=
=
(
(
R
C 2
R
O
R
C 2 r o 2
2
)
+
(
= 20.2
Ω
R
B 3
R in
R
B 3
( r
π 3
3
+
)
( β o
= 18.66k
Ω
3
R th 3
= R
B 3
R
C 2 r o 2
= 3.99k
Ω
+ 1)( R
E 3
R
L
)
) )
R
6 S
= R
L
+ R
E 3
R th 3
+ r
π
β o 3
+ 1
3 = 311 Ω
ω
L
≅ i n
∑
= 1
1
= 3154rad/s
R C i f
L
=
ω
2
L
π
= 502Hz
Device capacitances affect high-frequency response.
At high frequencies, the coupling and bypass capacitors are short circuits. Small signal model for the multi-stage amplifier is shown in Figure 6.
Figure 6: Equivalent small signal circuit of the multi-stage amplifier at high frequencies
OCTC for each of the two capacitors associated with each transistor has to be determined.

For M1,
R
L 1
= R
I 12 r
π 2
= 483 Ω
R C
T 1
= R th
⎛
⎜ C
GS 1
+ C
GD 1
⎛
1 +
= 3.99k
Ω m
=
For Q2, r
π o 2
=
R
L 2
=
( r
π 2
( R th 2
+ r x 2
) 619 Ω
R
I 23
R in 3
)
=
(
R
I 23
( r
π 3
π o T 2
= r
π o 2
⎛
⎜ C
π 2
+ C
μ 2
⎛
1 + m
+ ( β o 3
L 2
+ 1)( R
E 3
R
L
)
) )
= 3.54k
Ω
+
R
L r
π o
2
2
⎞
⎠
⎞
⎟ =
For Q3,
R th 3
= R
I 23 r o 2
L 1
+
R
L 1
R th
⎞
⎠
⎞
⎟
R
EE
=
E 3
= 232 Ω
R
π 3 O
C
π 3
+ R
μ 3 O
C
μ 3
=
⎛
⎝ r
π 3
ω
H
≅ i m
∑
= 1
1
R C i
= f
H
=
ω
B.W =
2
H
π
= 544kHz f
H
− f
L
≅ f
H
×
=
544kHz
( R th 3
+ r x 3
+ R
1 + m EE
EE
) ⎞
C
π 3
+ ( R th 3
+ x
)
μ 3
=