SIMPLE THOERY ABOUT CA,PROBLEMS RELATED TO CA ETC....

1. 1
GLACEVARGHESET
EC-A
ROLL NO:45
cascade amplifier

2. cascade amplifier 2
SIGNIFICANCE OF CASCADE AMPLIFIER
● The most widely used method
● Coupling a signal from one stage to the another stage and block dc
voltage from one stage to the another stage
● The signal developed across the collector
resistor of each stage is coupled into the base of the next stage
●The overall gain = product of the
individual gain
TO IMPROVE THE GAIN

3. Multistage Amplifiers
3
Multi-stage amplifiers are amplifier circuits cascaded to increased gain. We can
express gain in decibels(dB).
Two or more amplifiers can be connected to increase the gain of an ac signal.
The overall gain can be calculated by simply multiplying each gain together.
A’v = Av1Av2Av3 ……
cascade amplifier

4. Example
4
Find Vout
Av1 = 5 Av2 = 10 Av3 = 6 Avn = 8
Vin = 5 mV
cascade amplifier

5. cascade amplifier 5
A cascade amplifier is any two-port network constructed from a series
of amplifiers, where each amplifier sends its output to the input of the
next amplifier in a chain. The complication in calculating the gain
of cascaded stages is the non-ideal coupling between stages due to loading.
CONSTRUCTION

6. cascade amplifier 6

7. Cascade Connection
 small signal gain is:
by determine the voltage gain at stages 1 & stage 2
therefore
- the gain in dB
 input resistance
))(//)(//( 22121
SIN
IN
LCCmm
S
o
V
RZ
Z
RRrRgg
V
V
A
+
== π
121 //// πrRRZIN =
21 VVV AAA =
)log(20)( VdBV AA =
7
cascade amplifier

8. cascade amplifier 8
● Output resistance
assume 0=SV ,so 021 == ππ VV also
0211 == ππ VgVg mm
Therefore
LCOUT IIRRZ 2=

9. Exercise 1:
Draw the ac equivalent circuit and calculate the voltage gain, input
resistance and output resistance for the cascade BJT amplifier in
above Figure. Let the parameters are:
Ω==Ω==Ω==Ω== kRRkRRkRRkRR EECC 1,2.2,7.4,15 21214231
∞===== 0)(21 ,7.0,200,20 rVVVV ONBEQQCC ββ
9cascade amplifier

10. Solution 1 (cont.):
Ac equivalent circuit for cascade amplifier
10cascade amplifier

11. Solution 1:
DC analysis:
At Q1:
At Q2:
AC analysis:
At Q1:
At Q2:
Sg
kr
m 153.0
307.1
1
1
=
Ω=π
mAI
AI
CQ
BQ
979.3
89.19
1
1
=
= µ
Sg
kr
m 153.0
307.1
2
2
=
Ω=π
mAI
AI
CQ
BQ
979.3
89.19
2
2
=
= µ
11cascade amplifier

12. Solution 1 (cont.):
From the ac equivalent circuit:
At Q1, the voltage gain is:
Where is the o/p voltage looking to the Q1 transistor
and is the i/p resistance looking into Q2 transistor
Therefore
The voltage gain at Q1 is:
2iR
10QV
)//( 21
10
1 iCm
Q
VQ RRg
Vi
V
A −==
06.102)36.957//2.2(153.01 −=−= kAVQ
Ω=== 36.957307.1//7.4//15// 222 kkkrRR Bi π
12cascade amplifier

13. Solution 1 (cont.):
From the ac equivalent circuit:
At Q2, the voltage gain is:
Where is the i/p voltage looking into the Q2 transistor
Therefore, the voltage gain at Q2 is:
The overall gain is then,
** The large overall gain can be produced by multistage amplifiers!!
So, the main function of cascade stage is to provided the larger overall gain
2iQV
)( 2
2
0
2 Cm
iQ
VQ Rg
V
V
A −==
6.336)2.2(153.02 −=−= kAVQ
353,34)6.336)(06.102(21 =−−== VQVQV AAA
13cascade amplifier

14. Solution 1 (cont.):
From the ac equivalent circuit:
The i/p resistance is:
The o/p resistance is:
Ω=
==
36.957
307.1//7.4//15//// 121 kkkrRRRi π
Ω== kRR Co 2.22
14cascade amplifier

15. THANK YOU
15cascade amplifier