Download Small Signal * Low Frequency Transistor amplifier Circuits

UNIT-IV
Small Signal – Low Frequency
Transistor amplifier Circuits
contents
Transistor as an Amplifier
Analysis of Transistor Amplifier Circuits using
h – parameters
Miller’s theorem and it’s dual
Simplified CE and CC hybrid model
The CE amplifier with emitter resistance
Darlington pair
Analysis of Single Stage Amplifiers
Hybrid Parameter Model
Ii
Io
Linear Two port
Device
Vi
Ii
1
Vi
1'
Vo
Io
hi
hrVo
hfIi
ho
2
Vo
2'
Vi  h11Ii  h12Vo  hi I i  hrVo
I o  h21Ii  h22Vo  h f I i  hoVo
h-Parameters
Vi
h11 
Ii
Io
h21 
Ii
Vo  0
Vi
h12 
Vo
Ii  0
Vo  0
Io
h22 
Vo
Ii  0
h11 = hi = Input Resistance
h12 = hr = Reverse Transfer Voltage Ratio
h21 = hf = Forward Transfer Current Ratio
h22 = ho = Output Admittance
Use of h – parameters to describe a transistor have the
following advantages.
h – parameters are real numbers up to radio
frequencies
They are easy to measure
They can be determined from the transistor static
characteristics curves.
They are convenient to use in circuit analysis and
design.
Easily convert able from one configuration to other
Readily supplied by manufactories
Analysis of CE amplifier using hybrid model
hre
VS
vce
-

For finding h parameters quantities Vbe and ic are taken
independent variables

For finding voltage gain, current gain, input impedance,
output impedance h parameter equations are
vbe = hie ib + hre vce --------------------------------(1)
ic = hfe ib + hoe vce---------------------------------- (2)
 Putting vce =0 in above equations we will get hie and hfe ,
ib =0 in above equations we will get hre and hoe
 current gain
output current
input current
iL
Ai 
ib
Ai 
Apply current divider rule to the output circuit
1
ho e
1

ho e
h f eib
iL  
RL
iL 
 h f eib
1  ho e RL
from  eqn
 h fe
Ai 
1  ho e RL
To find input resistance
vs
Ri 
ib
Applying KVL to input circuit
Vs = hie ib + hre vce
Vs = ib hie + hre iL RL
(vce = iLRL)
Vs = ib hie + hre Ai ib RL ( iL =Ai ib )
Substituting in equation (2)
Ri = hie + hre Ai RL
Ri
ib hie
C
+
hfe ib
RL
hre vce
VS
-
1/hoe
E
vce
-
To find voltage gain
Av
=
output voltage vce

inputvoltage
vs
i R
Av  L L
ib Ri
since
RL
Av  Ai
Ri
Ri 
vs
ib
To find output resistance
ib
hie
+
RO 
vce
iC
hre vce
hfe ib
1/hoe
RL
vce
 Applying KC L to
the output circuit.
iC = hfe ib + i1
iC = hfe ib + vce hoe
Applying KVL to input circuit
- ( hie ib + hre vce) =0
 hrevce
ib 
hie
substituting for ib in equation
RO1
-
RO
iC  h fe (
 hrevce
)  vce hoe
hie
substituting in equation
RO 
1
 h fe hr e
hie
 hoe
To find output resistance with RL
RO1 = RO||RL
(Since RL is in parallel with the voltage source, total
output resistance is the parallel combination of R L and RO)
PROBLEM
1.A common emitter amplifier has the following h- parameters. hie =1KΩ, hre = 10-4,
hfe =100, hoe = 12µmho Find current gain, Voltage gain, Ri, Ro, power gain.
Take RL = 2KΩ. Also find output power take vS = 500 mV ( rms).
Ai 
 hf e
1  hoe R L
 100
1  (12  10 6  2  10 3
Ai  97.656
Ai 
Ri  hi e  hr e Ai R L
Ri  (1  10 3 )  (10  4  97.656  2  10 3 )
Ri  980.468
Av  Ai
RL
Ri
2  10 3
Av  97.656 
980.468
Av  199.2
RO 
1
 h f e h re
hie
 hoe
R O  0.5M  500 K
RO
1
 R O || R L 
RO
1

RO
1
RO R L
RO  R L
500  2
500  2
 1.992 K
AP  AV  Ai
AP  199.2  97.656
AP  19.453  10 3
2
v
PO  ce
RL
( Av .VS ) 2
PO 
RL
( 1.99.2) 2 (500  10  3 ) 2
PO 
2  103
PO  4.96Watt
 MILLER’S THEOREM
Miller’s theorem states that when a resistance ( or capacitance ) is connected across input and
output terminals, the same can be replaced by two independent resistances ( or capacitances )
connected one across the input terminals and the other across output terminals.
These are called Miller equivalent resistances ( or capacitances).
PROOF:
Applying Ohm’s law
VS  VO
R
V
Let A  O
VS
i
         (1)
i
RMI
VS
Then VO  AVS
Substituting in (1)
VS  AVS
R
V (1  A)
 S
R
V
 S
 R 


1

A


i 

R
VS  0
         (2)
RMI
+
RS
RL
+
vS1
-
VO
Amplifier
A
-
 comparing equations (1) and (2), the same current i will flow through a resistance R
when it is between input terminal and ground. Therefore, the KCL at the input terminal is
not affected by replacing the resisitance R by R is called Miller equivalent resistance
on the input side.
MI
.
.
MI
.
To find Miller equivalent resistance on the output side.
V0  VS
             (3)
R
V
Let A  O
then VO  AVS
VS
i
i
 1
VO
VO 1  
 A 
A 
R
R
VO 
VO
 RA 


 A 1 
V 0
i O
- - - - - - - - - - - (4);
RMO
R
i
 A 1 
VO 

 A 
R

+
RS
RA
RMO 
A 1
RL
+
vS1
VO
Amplifier
-
A
-
RMO
is the Miller equivalent resistance on the output side
 The same current I will flow through the resistance RMO if R is replaced by RMO
which is connected across output terminal and ground.
 Replacing R by Miller equivalent resistances, the circuit is as follows.
+
+
i
RS
RMI
Amplifier
A
+
i
RMO
+
RL
Vo
vS1
-
-
-
-
 Dual of Miller’s Theorem
 Consider a general amplifier as shown below. In this amplifier, the resistance R is common to the
input and output circuits do not have a common resistance.
The purpose of the following analysis is to remove the inter dependence of input and output
circuits. So that either input circuits or output circuits can be solved independently
Applying KVL to input circuit
VS  i1 R S  Vi  (i1  i2 ) R  0
But i2 = -iL
VS  i1 RS  Vi  (i1  i L ) R  0
RS
+
VS  i1 RS  Vi  (i1  Ai i1 ) R  0
+
VS  i1 RS  Vi  i1 (1  Ai ) R  0
+
Amplifier
+
VS  i1 RS  Vi  RMI i1  0
Vi
VS
RMI  (1  Ai ) R
vO1
-
(1  Ai ) R
-
is the resistance which when connected in series with the
input circuit instead of R will not affect the input circuit.
i1
-
R
i2
Repeating the above analysis for the output
circuit.
|
-
VO  VO  (i1  i 2 ) R  0
But i 2   i L
VO  VO  (i1  i L ) R  0
|
|
iL
 iL ) R  0
Ai
VO  VO  i L (
|
RL
vO
From the above equation,
VO  VO  (
iL
1
1) R  0
Ai
VO  VO  i L (1 
|
1
)R  0
Ai
VO  VO  i L R MO  0
|
 A 1 
 R
R MO   i
 Ai 
From the equation, the resistance
RMO
 when connected in the output circuit alone does
not alter the KVL equation.
Thus, without affecting the electrical nature of the circuit, the circuit can be redrawn a
RMI = R( 1-Ai)
RS
RMO = R( Ai – 1)
+
iAi
L
i1
Amplifier
RL
VO
-
Approximate h-parameter equivalent circuit.
hre is the reverse voltage gain or reciprocal of voltage gain. For an amplifier, voltage gain is high
therefore hre is very low for approximation take hre Vce is zero or voltage source is replaced by
short circuit
In h parameter equivalent
is open circuit
1
is
hoe
internal resistance of the current source hfe ib for ideal condition it is infinity so it
Darlington pair
•
Allow for much greater gain in a circuit
•
β = β1 * β2