Download multistage amplifiers

MULTISTAGE
AMPLIFIERS
Multistage Amplifiers
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 ……
Introduction
 Many applications cannot be handle with single-
transistor amplifiers in order to meet the specification
of a given amplification factor, input resistance and
output resistance
 As a solution – transistor amplifier circuits can be
connected in series or cascaded amplifiers
 This can be done either to increase the overall small-
signal voltage gain or provide an overall voltage gain
greater than 1 with a very low output resistance
Multistage Amplifiers
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 ……
Example
Find Vout
Vin = 5 mV
Av1 = 5
Av2 = 10
Av3 = 6
Avn = 8
Multistage Amplifier Cutoff
Frequencies and Bandwidth
 When amplifiers having equal cutoff frequencies are cascaded,
the cutoff frequencies and bandwidth of the multistage circuit are
found using
f C 2 (T )  f C 2 21/ n  1
f C1(T ) 
f C1
21 / n  1
BW  f C 2 (T )  f C1(T )
6
Introduction (cont.)
 Multistage amplifier configuration:
RL
Rc1
RB
R1
Rc2
Vo
Q2
Q2
R2
Vi
Q1
Vi
Q1
R3
Cascade /RC coupling
Vo
T
Cascode
Q2
R1
R2
R1
Vo
Vi
Q1
Q2
Vi
Transformer coupling
Q1
Darlington/Direct coupling
i) Cascade Connection
-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
**refer page 219
i) Cascade Connection (cont.)
 small signal gain is:
by determine the voltage gain at stages 1 & stage 2
AV  AV 1 AV 2
therefore
Vo
Ri
AV 
 g m1 g m 2 ( RC1 // r 2 )( RC 2 // RL )(
)
VS
Ri  RS
- the gain in dB
AV ( dB)  20 log( AV )
 input resistance
Ris  R1 // R2 // r 1
Output resistance
- assume VS  0
,so
Therefore
R0  RC 2
V 1  V 2  0
also
gm1V 1  gmV 2  0
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:
VCC  20V ,  Q1   Q 2  200,VBE (ON )  0.7V , r0  
R1  R3  15k, R2  R4  4.7k, RC1  RC 2  2.2k, RE1  RE 2  1k
Solution 1 (cont.):
Ri
Vi
C1
B1
 R1 // R2
V 1
r 1
g m1V 1

E1
C2

RC1

RB1
B2
RB 2
V 2
RC 2
r 2

 R3 // R4
g m 2V 2
E2
Ac equivalent circuit for cascade amplifier
R0
V0
Solution 1:
DC analysis:
At Q1:
I BQ1  19.89A
I CQ1  3.979mA
At Q2:
I BQ 2  19.89A
I CQ 2  3.979mA
AC analysis:
At Q1:
r 1  1.307k
g m1  0.153S
At Q2:
r 2  1.307 k
g m 2  0.153S
Why the Q-point values same for both
Q1 & Q2 ?
Solution 1 (cont.):
From the ac equivalent circuit:
At Q1, the voltage gain is:
AVQ1 
Where
and
V0Q1
Ri 2
V0Q1
Vi
  g m ( RC1 // Ri 2 )
is the o/p voltage looking to the Q1 transistor
is the i/p resistance looking into Q2 transistor
Therefore
Ri 2  RB 2 // r 2  15k // 4.7k // 1.307k  957.36
The voltage gain at Q1 is:
AVQ1  0.153(2.2k // 957.36)  102.06
Solution 1 (cont.):
From the ac equivalent circuit:
At Q2, the voltage gain is:
AVQ2
V0

  g m ( RC 2 )
ViQ2
Where ViQ 2 is the i/p voltage looking into the Q2 transistor
Therefore, the voltage gain at Q2 is:
AVQ 2  0.153(2.2k )  336.6
The overall gain is then,
AV  AVQ1 AVQ 2  (102.06)( 336.6)  34,353
** The large overall gain can be produced by multistage amplifiers!!
So, the main function of cascade stage is to provided the larger overall gain
Solution 1 (cont.):
From the ac equivalent circuit:
The i/p resistance is:
Ri  R1 // R2 // r 1  15k // 4.7k // 1.307k
 957.36
The o/p resistance is:
Ro  RC 2  2.2k
ii) Cascode Connection
-A cascode connection has one transistor on top of (in series with) another
-The i/p into a C-E amp. (Q1) is, which drives a C-B amp. (Q2)
-The o/p signal current of Q1 is the i/p signal of Q2
-The advantage: provide a high i/p impedance with low voltage gain to
ensure the i/p Miller capacitance is at a min. with the C-B stage providing good
high freq. operation
**refer page 223
ii) Cascode Connection (cont.)
From the small equivalent circuit, since the capacitors act as short circuit,
by KCL equation at E2:
V 2
 g m 2V 2
r 2
solving for voltage V 2
g m1V 1 
 r 
V 2    2 g m1VS 
 1  2 
Where
 2  gm 2r 2
the output voltage is
Vo  ( g m 2V 2 )( RC // RL )
or
 r 2 
RC // RL )VS 
Vo   g m1 g m 2 
 1  2 
ii) Cascode Connection (cont.)
Therefore the small signal voltage gain:
 r 2 
V0
RC // RL 
AV 
  g m1 g m 2 
VS
 1  2 
From above equation shows that:
 r 2 
2
 
g m 2 
1
 1  2  1  2
So, the cascode gain is the approximately
AV   gm1 RC // RL 
* The gain same as a single-stage C-E amplifier
iii) Darlington Connection
-The main feature is that the composite transistor acts as a single unit with a
current gain that is the product of the current gains of the individual transistors
-Provide high current gain than a single BJT
-The connection is made using two separate transistors having current gains of
 and 
1
2
So, the current gain
 D  1  2
If
1   2  
The Darlington connection
provides a current gain of
D   2
Figure 1: Darlington transistor
iii) Darlington Connection (cont.)
 Figure shown a Darlington configuration
refer page 222
iii) Darlington Connection (cont.)
 The small current gain Ai  I o / I i :
Since
V 1  I i r 1
Therefore
g m1V 1  g m1r 1 I i  1 I i
Then,
V 2  ( I i  1 I i )r 2
The o/p current is:
I 0  g m1V 1  g m 2V 2  1 I i   2 (1  1 ) I i
The overall gain is:
I0
Ai 
 1   2 (1  1 )  1  2
Ii
** The overall small-signal current gain = the product of the individual
current gains
iii) Darlington Connection (cont.)
 The input resistance:
Known that:
Vi  V 1  V 2  I i r 1  I i (1  1 )r 2
So, the i/p resistance is:
Ri  r 1  (1  1 )r 2
The base of Q2 is connnected to the emitter of Q1, which means that the i/p
resistance to Q2 is multiplied by the factor (1  1 ) , as we saw in
circuits with emitter resistor.
So, we can write:
Therefore
r 
1VT
  2VT
r 1  1 
I
 CQ 2
I CQ1
and
I CQ1 
I CQ 2
2

  1 r 2


The i/p resistance is then approximately
Ri  21r 2
**The i/p resistance tends to be large because of the

multiplication