The cascade amplifier configuration consists of a common emitter stage followed by a common base stage. This provides good high frequency performance. The input resistance and current gain are similar to a single CE stage, while the output resistance is like a CB stage. Cascade amplifiers are used in tuned RF amplifiers and as wideband amplifiers due to their isolation between input and output. Analysis shows the overall voltage gain is approximately equal to the product of the individual stage gains, while the input resistance is the hie of the first stage and the output resistance is the Rc of the second stage.

2. The cascade configuration
consists of a common emitter
configuration and a common
base configuration.
The CB stage provides a good
high frequency operation.

3. A CE-CB cascade
connection is as shown in
the fig. :

4. Features of a Cascade
Amplifier :
 The input resistance and current gain of the
cascade configuration are equal to the
corresponding value of a single stage CE
amplifier.
 The output resistance is equal to that of a CB
configuration.
 The millers capacitor shunting the CE input
stage is very small.

5. Applications of Cascade
Amplifier :
• In the tuned RF amplifiers in T.V.
circuits.
• It is also used as wideband amplifier.
• The isolation provided between input
and output by the cascade amplifiers
is very high.

6. Step 1 : Analysis of stage 2 :
1. Current gain A12 = – hfb / (1 + hobRL )
Neglecting hob we get A12 ~ – hib
2. Input resistance Ri2 = hib+hrbA12RL
Neglecting hrb we get, Ri2 = hib
3. Voltage gain Av2 = A12 x ( RL2 / Ri2 )
= – hfb x (RL2 / hib )
But hfb ~ – hfe /(1+hfe) and hib = hie / (1+ hfe)
So,
Av = hfe / (1+hfe) X (RL(1+hfe) / hie)
= hfeRL / hie

7. Step 2 : Analysis of stage 1:
1.Load resistance of stage 1, Rl1 = Ri2= hie/(1+hfe)
2.Current gain A11 = – hfe
3.Input resistance Ril = hie
4.Voltage gain, Av1=A11 X (RL1/Ril)
= -hfe X (hie / (1+hfe) hie)
So,
Av1 = -hfe/(1+hfe)

8. Step 3 : Overall voltage gain :
Av = Av1 X Av2
= -[ hfe / (1+hfe) ] X [ (hfeRL ) / hie ]
Since hfe ~ (1+hfe)
So,
Av ~ -(hfeRL) / hie
Step 4 : Overall input resistance :
Ri = Ril = hie
and R’i = hie || Ri3

9. Step 5 : Overall output resistance :
Ro = Ro2 = Rc
Step 6 : Overall current gain :
The overall current gain,
AI = AI1 X AI2
= - hfe X – hfb
= -hfe X (hfe / (1+hfe))
~ -hfe

10. Conclusions :
The cascade configuration has :
 Low input resistance.
 Moderate to high voltage and current
gains.
 High output resistance.

11. Thank you
- Rakesh Makhija.