ECA

1. Unit-1
Analysis And Design of Small
Signal Low Frequency BJT
Amplifiers

2. Classification Of Amplifiers
1. According to frequency capabilities.
2. According to coupling methods.
• AF Amplifier are used to amplify the signals lying in the audio
range ( i.e. 20 Hz to 20 kHz )
• RF amplifiers are used to amplify signals having very high
frequency.
Amplifiers are classified as audio amplifiers , radio frequency
amplifiers
• R-C coupled amplifiers,
• Transformer coupled amplifiers
• Direct Coupled

3. Classification Of Amplifiers
3. According to use.
• Amplify the input voltage, if possible with minimal current at the
output.
• The power gain of the voltage amplifier is low.
• The main application is to strengthen the signal to make it less
affected by noise and attenuation.
• Ideal voltage amplifier have infinite input impedance & zero output
impedance.
a. Voltage amplifiers
• Amplify the input power, if possible with minimal change in the
output voltage
• Power amp. are used in devices which require a large power across
the loads.
b. Power amplifiers

4. • In multi stage amplifiers, power amplification is made in the
final stages
 Audio amplifiers and RF amplifiers use it to deliver
sufficient power the load.
 Servo motor controllers use power it to drive the motors.

5. Classification Of Amplifiers
Voltage amplifiers Power amplifiers
current gain low high
Voltage gain high low
Heat dissipation low high
cooling mechanism not required required
Transistor Size Small Large to dissipate heat
Base Width small Wide to handle higher
current
Beta Usually high >100 Low usually < 20

6. Distortion in amplifiers
Linear waveform distortion
If the gain of an amplifier has a different magnitude for the
various frequency components of the input signal, a form of
distortion known as amplitude distortion .
Fig a)input waveform Fig b)output waveform
Fig: Linear amplitude distortion

7. Phase Distortion
•The phase shift between input and output waveforms
depend up on the signals of different frequencies.
•If the phase shift of an amplifier is not proportional to
frequency, phase distortion occurs.
Fig a): Amplifier A has no phase shift Fig b): Amplifier B has linear phase versus
frequency

8. Fig c) Amplifier C has phase that is not proportional to frequency
Fig: Effect of amplifier phase response

9. Requirements for Distortion less Amplification:
To avoid linear waveform distortion, an amplifier should have
constant gain magnitude and a phase response that is linear versus
frequency for the range of frequencies contained in the input signal.

10. Common Emitter amplifier
1. One of the primary uses of a transistor is to amplify ac
signals.
2. This could be an audio signal or perhaps some high
frequency radio signal.
3. It has to be able to do this without distorting the original
input.
4. Most transistors amplifiers are designed to operate in the
linear region
5. The common-emitter amplifier has high voltage and
current gain

11. Definition of Small Signal
• ac input signal voltages and currents are in the order ±10% of
Q-point voltages and currents.
• Eg : If dc current is 10mA, the ac current (peak to peak) is less
than 0.1 mA.
Purpose of components:
• R1, R2, RE – form the biasing circuit & stabilization.
• C1(Coupling capacitor) – couple the signal to the base.
• C2(Bypass capacitor) - use in parallel with RE to provide low
reactance path to amplified ac signal
• C3(Coupling capacitor)- couple the signal to the next stage

12. How BJT Amplifies?(Working)
B
C I
I 

1. When a weak ac signal is given to the base, a small ac
base current start flowing.
2. Due to BJT, a much larger ac current flow through RC
3. Therefore a large voltage appear across the collector
circuit.
4. There is 1800 phase shift between input and output.

13. SMALL-SIGNAL LOW-FREQUENCY
OPERATION OF TRANSISTORS
Hybrid Parameters and Two-Port Network:
•For the hybrid equivalent model to be described, the parameters are defined at
an operating point that may or may not give an actual picture of the operating
condition of the amplifier.
•The quantities hie , hre , hfe and hoe are called the hybrid parameters and are the
components of a small-signal equivalent circuit..
Two-port system representation (Black
model realisation)

14. Hybrid Parameter Model
hi
hrVo
ho
hfIi
Vi
Ii
2
2'
Io
Vo
1
1'
11 12
21 22
i i o i i r o
o i o f i o o
V h I h V h I h V
I h I h V h I h V
   
   
Linear Two
port Device
Vi
Ii Io
Vo

15. 11 12
21 22
0 0
0 0
i i
o i
i o
o o
o i
i o
V V
h h
V I
I V
I I
h h
V I
I V
 
 
 
 
h-Parameters
h11 = hi = Input Resistance
h12 = hr = Reverse Transfer Voltage Ratio
h21 = hf = Forward Transfer Current Ratio
h22 = ho = Output Admittance

16. Hybrid Equivalent Model
The hybrid parameters: hie, hre, hfe, hoe are developed and used to
model the transistor. These parameters can be found in a
specification sheet for a transistor.

17. Determination of h parameters
0V
V
o
i
12
0V
V
i
i
11
o
12
i
11
i
o
o
V
V
h
I
V
h
V
h
I
h
V






0A
I
o
o
22
0V
V
o
i
21
o
o
22
i
21
O
o
o
V
I
h
I
I
h
,
0V
V
Solving
V
h
I
h
I







Forward Current gain
Output conductance
Input impedance
Reverse transfer
Voltage gain

18. General h-Parameters for any
Transistor Configuration
hi = input resistance
hr = reverse transfer voltage ratio (Vi/Vo)
hf = forward transfer current ratio (Io/Ii)
ho = output conductance

19. EXPRESSIONS OF CURRENT GAIN,
INPUT RESISTANCE, VOLTAGE
GAIN AND OUTPUT RESISTANCE
The h-parameter equivalent circuit of a transistor amplifier
having a voltage source Vg , with its input resistance Rg
connected to the input terminals and a load resistance RL
connected to the output terminals.
h-Parameter equivalent circuit of a transistor

20. EXPRESSIONS OF CURRENT GAIN,
INPUT RESISTANCE, VOLTAGE
GAIN AND OUTPUT RESISTANCE
Current Gain (AI)
Input Resistance (RI)

21. EXPRESSIONS OF CURRENT GAIN,
INPUT RESISTANCE, VOLTAGE
GAIN AND OUTPUT RESISTANCE
Voltage Gain: Voltage gain or voltage amplification is defined
as the ratio of the output voltage V2 to the input voltage V1.
Where,
Output Resistance (RO):

22. Common emitter hybrid equivalent circuit

23. Simplified common emitter hybrid model
•Figure shows the CE amplifier equivalent circuit in terms of h-parameters
•Since1 / hoe in parallel with RL is approximately equal to RL
•If 1 / hoe >> RL then hoe may be neglected. Under these conditions.
•Ic = hfe IB .
•hre vc = hre Ic RL = hre hfe Ib RL .

24. • Since h feh re » 0.01, this voltage may be neglected in
comparison with h ic Ib drop across h ie provided RL is not very
large.
• If load resistance RL is small than hoe and hre can be
neglected.
• Output impedance seems to be infinite.
• When Vs = 0, and an external voltage is applied at the output
we fined Ib = 0, I C = 0. True value depends upon RS and lies
between 40 K and 80K.
• Similarly same calculations for CC and CB can be done.

25. CE amplifier with an Emitter resistor
The voltage gain of a CE stage depends upon hfe.
This transistor parameter depends upon temperature, aging and
the operating point.
Moreover, hfe may vary widely from device to device, even for
same type of transistor.
To stabilize voltage gain A V of each stage, it should be
independent of hfe.

26. • A simple and effective way is to connect an emitter resistor
Re as shown in fig.
• The resistor provides negative feedback and provide
stabilization.
• Without RE, when β increases or decreases -> ICQ and VCEQ
also vary, thus Q-point will be shifted and makes the circuit
unstable.
• By adding RE, there will be not much shift in Q-point is
stabilized even with variation of β. Moreover, the voltage gain
is less dependent on transistor current gain in ac analysis.

27. Common-emitter circuit with
emitter resistor
R1
R2
RC
CC
vs
VCC
vo
RE
Emitter resistor

28. Thevenin circuit analysis
RC
VCC
RTH
VTH
RE
• Apply KVL around B-E loop,
0
)
1
(
0
)
(
)
(









E
BQ
on
BE
TH
BQ
TH
E
EQ
on
BE
TH
BQ
TH
R
I
V
R
I
V
R
I
V
R
I
V

  )
(
)
1
( on
BE
TH
E
TH
BQ V
V
R
R
I 


 
E
TH
on
BE
TH
BQ
R
R
V
V
I
)
1
(
)
(






29. Thevenin circuit analysis
• We will get collector current as:
• Apply KVL around C-E loop to find VCEQ,
0



 E
EQ
CEQ
C
CQ
CC R
I
V
R
I
V













E
TH
on
BE
TH
BQ
CQ
R
R
V
V
I
I
)
1
(
)
(



 
E
C
BQ
CC
E
BQ
C
BQ
CC
E
EQ
C
CQ
CC
CEQ
R
R
I
V
R
I
R
I
V
R
I
R
I
V
V
)
1
(
)
1
(
















30. Ac analysis
small-signal equivalent circuit-

31. Small-signal hybrid-π parameters
• The ac output voltage is: (if we consider equivalent
circuit with current gain β)
• Input voltage equation:
• Input resistance looking into the base of BJT, Rib:
• Input resistance to the amplifier is:
C
b
o R
I
V )
(


E
b
b
b
in R
I
I
r
I
V )
( 
 


E
b
in
ib R
r
I
V
R )
1
( 
 



ib
i R
R
R
R 2
1


32. Small-signal hybrid-π parameters
• By voltage divider, we get relate Vin and Vs:
• Small-signal voltage gain is then:
• If Ri>>RS and if (1+β)RE >> rπ, voltage gain is:
s
S
i
i
in V
R
R
R
V 























S
i
i
E
C
s
o
v
R
R
R
R
r
R
V
V
A
)
1
( 


E
C
E
C
s
o
v
R
R
R
R
V
V
A






)
1
( 

Exact value
Approximate value

33. •In the low-frequency region of the single-stage BJT it is the RC
combinations formed by the network capacitors CC , CE , and
the network resistive parameters that determine the cut off
frequencies
Low frequency response of BJT
Amplifier
Fig: Common Emitter amplifier

34. Gain falls of due to the
effects of CC and CE
Cc and CE short cct & Cπ
and Cμ open cct.
Gain almost constant.
Midband
Gain falls of due to the
effects of C and C
Fig: Frequency response of BJT Amplifier

35. Basic concepts-Low Frequency Range-
• At low frequency range, the gain falloff due to coupling
capacitors and bypass capacitors.
• As signal frequency  , the XC  - no longer behave as short
circuits.
• Coupling capacitors C1 and C3 limit the passage of very low
frequencies. Emitter bypass C2 capacitor will have high
reactance to low frequencies as well, limiting the gain. Also
the capacitance causes a phase shift of the signal.

36. Low frequency response of BJT Amplifier
• Assume CC and CE are short-ckt, midrange Av is:
 
L
C
o
S
m
s
o
v R
R
r
R
r
R
R
r
R
R
g
V
V
A














2
1
2
1

37. Low frequency amplifier response
• At the low frequency ac equivalent circuit of a capacitor
coupled amplifier, we can see there are three RC circuits that
will limit low frequency response. The input at the base, the
output at the collector, and the emitter.
Input RC Circuit
Output RC Circuit
Bypass RC Circuit

38. Low frequency amplifier response
-Input RC Circuit
• As frequency decreases, XC1
increases  less base voltage
due to more voltage drop across
C1. Thus, voltage gain is
reduced.
• A critical point occurs when o/p
voltage is 70.7% of its midrange
value. This condition occur
when XC1=Rin.
in
C
in
in
base V
X
R
R
V










2
1
2
in
in
in
in
in
in
in
in
base V
.
V
R
R
V
R
R
R
V 707
0
2
2
2




















dB
)
.
log(
V
V
log
in
base
3
707
0
20
20 











39. Low frequency amplifier response
-Input RC Circuit
• The frequency at which the gain is down by 3 dB is called the
lower critical frequency (fcl). This frequency can be
determined by the formula below.
• fcl also known as lower cutoff frequency, lower corner
frequency or lower break frequency.
in
)
input
(
cl
C
R
C
f
X 


1
1
2
1
1
1
2
1
2
1
C
)
R
R
(
@
C
R
f
in
S
in
)
input
(
cl




Resistance of input
source taken into
account

 r
R
R
R
@
R i
in 2
1
R1S

40. Low frequency amplifier response
-Input RC Circuit
• The decrease in voltage gain with frequency is called the roll-
off.
• A ten times change in frequency is called a decade.
• For each ten times reduction in frequency below fc, there is a
20 dB reduction in voltage gain.
• The attenuation measured in dB at each decade is dB/decade.

41. Low frequency amplifier response-Input
RC Circuit
• This typical plot of dB Av vs
frequency is called Bode
plot.
• From Bode plot, it is flat
(0dB) down to critical
frequency, at which point
gain drop at -20dB/ decade.
Above fc are the midrange
frequencies.
• Sometimes roll-off is
expressed in dB/octave,
which is a doubling or
halving of the frequency.
Midrange
Low frequency

42. Low frequency amplifier response
-Input RC Circuit
• Input RC circuit also causes an increasing in phase shift
through amplifier.
• At midrange, phase shift is approximately zero since XC1=0Ω.
• At lower frequencies, higher values of XC1 cause a phase shift
and o/p voltage of RC circuit leads i/p voltage.
• Phase shift in input RC circuit is:
• At critical frequency, XC1=Rin, so:









 
in
C
R
X
tan 1
1
0
1
1
45
1 










 

)
(
tan
R
R
tan
in
in

43. Low frequency amplifier response
-Input RC Circuit
• As phase shift approach 90ο,
frequency approaches zero.
• The input RC circuit causes base
voltage to lead input voltage
below midrange by an amount
equal to circuit phase angle, θ.

44. Low frequency amplifier response cont..
-Output RC Circuit
• The output RC circuit affects the response similarly to the
input RC circuit. The formula below is used to determine the
cutoff frequency of the output circuit.
3
2
1
C
)
R
r
R
(
f
L
o
C
)
output
(
cl



R2S

45. Low frequency amplifier response
-Output RC Circuit
• Phase angle in o/p RC circuit is:
• θ≈0o for midrange frequencies and approaches 90o as
frequency approaches zero (XC3 ∞).
• At critical frequency, phase shift =45o.










 
L
C
C
R
R
X
tan 3
1

46. Low frequency amplifier response
-Bypass RC Circuit
• At low frequencies, XC2 is in parallel with RE creates an
impedance that reduces the voltage gain.

47. Low frequency amplifier response
-Bypass RC Circuit
R1||R2
R3S
RE
RoutCC
Rth
Replacing C1
and C2 by
short circuits
Rs
Input source
is short-circuit
2
1 R
R
R
R S
th 
OUTCC
E
S R
R
R 
3














1
1
1
th
b
b
th
b
e
e
e
OUTCC
R
r
I
)
(
I
)
R
r
(
I
)
(
V
I
V
R



 
1
3
th
E
S
R
r
R
R
RC||RL

48. Low frequency amplifier response
-Bypass RC Circuit
• The lower cutoff frequency for bypass RC circuit is:
2
2
3
1
2
1
2
1
C
R
r
R
C
R
f
th
E
S
)
bypass
(
cl

















49. C-E Amplifier with Emitter Bypass
Capacitor
CE provides a short
circuit to ground for
the ac signals

50. • By include RE, it provide stability of Q-point.
• If RE is too high +++> small-signal voltage gain will be
reduced severely. (see Av equation)
• Thus, RE is split to RE1 & RE2 and the second resistor is
bypassed with “emitter bypass capacitor”. CE provides a
short circuit to ground for ac signal.
• So, only RE1 is a part of ac equivalent circuit.
• For dc stability: RE=RE1+RE2
• For ac gain stability: RE=RE1 since CE will short RE2 to
ground.

51. Example based on CE amplifier
• Given β=100, VBE=0.7V and VA=100V. Determine: (a)
small-signal voltage gain (b) input resistance seen by
the signal source, Rin and the output resistance looking
back into the output terminal, Ro.

52. AC Load Line Analysis
• Dc load line -> a way of visualizing r/ship between Q-
point and transistor characteristic.
• When capacitor included in cct, a new effective load line
 ac load line exist.
• Ac load line -> visualizing r/ship between small-signal
response and transistor characteristic.
• Ac operating region is on ac load line.

53. Ac load line cont..

54. Ac load line cont..
• For Dc load line:
• Apply KVL around collector-emitter loop,
• But
• Substitute and rearrange both equations:
• If β>>1, then we can approximate
Dc load line
equation

55. Cont..
• For ac analysis, apply KVL around collector-emitter loop,
• Assume ic ≈ ie,
• The slope is given by:
• The slope of ac load differ from dc load line  RE2 is not
included in the equivalent circuit. Small-signal C-E voltage
and collector current response are functions of resistor RC
and RE1.
0
1 

 E
e
ce
C
c R
i
v
R
i
)
( 1
E
C
c
ce R
R
i
v 


1
1
E
C R
R
Slope




56. Dc and ac load lines for CE circuit

57. AC load line cont..
v

+ VCE
0
+ IC
ICQ
VCEQ
Q
CC
V
E
C
CC
R
R
V

i

L
C
CEQ
R
R
V
i 

)
( L
C
CQ R
R
I
v 


)
( L
C
CQ
CEQ
off
cut R
R
I
V
v 



L
C
CEQ
CQ
csat
R
R
V
I
i 


58. Maximum symmetrical
swing
• When symmetrical sinusoidal signal applied to i/p of
amplifier, symmetrical sinusoidal signal generated at o/p.
• Use ac load line to determine the maximum output
symmetrical swing.
• If output exceed limit, a portion of o/p signal will be
clipped and signal distortion occur.

59. 1. draw the ac load line
IC
VCE
ac load line
0
2. add the Q point
Q
3. add ib~ vin
4. add reference lines
5. sketch ic
6. sketch vce

60. Saturation & Cut- Off Regions
RESTRICT MAXIMUM UNDISTORTED SIGNAL

61. Maximum undistorted signal

62. Bias (ICQ) Below Load Line Centre
62
IC
ac load line
0
Q
VCE
max
05
.
0 C
I
CQ
I
peak
C
I 


ICmax
)
(
)
max
05
.
0
( L
R
C
R
C
I
CQ
I
peak
v 



)
(
)
max
05
.
0
(
2 L
R
C
R
C
I
CQ
I
peak
peak
v 





ICQ
VCEQ

63. Bias (ICQ) Above Load LineCentre
IC
ac load line
0
Q
VCE
CQ
I
C
I
peak
C
I 

 max
95
.
0
ICQ
)
(
)
max
95
.
0
( L
R
C
R
CQ
I
C
I
peak
v 



)
(
)
max
95
.
0
(
2 L
R
C
R
CQ
I
C
I
peak
peak
v 





VCEQ
ICmax

64. Saturation Distortion
IC
ac load line
0
Q
VCE
ICQ

65. Cut-Off Distortion
IC
ac load line
0
Q
VCE
ICQ

66. 66
IC
ac load line
0 VCE
ICQ
Q
Clipping

67. Effect of coupling and bypass capacitors
67
Midband range
Gain falls of due to the
effects of CC and CE
Gain falls of due to the effects
of stray capacitance and
transistor capacitance effects
L
H
BW f
f
f 


68. • Frequency response of an amplifier is the graph of its gain versus the
frequency.
• Cutoff frequencies : the frequencies at which the voltage gain equals
0.707 of its maximum value.
• Mid band : the band of frequencies between 10fL and 0.1fH where the
voltage gain is maximum. The region where coupling & bypass
capacitors act as short circuits and the stray capacitance and transistor
capacitance effects act as open circuits.
• Bandwidth : the band between upper and lower cutoff frequencies
• Outside the mid band, the voltage gain can be determined by these
equations:
 2
1 /
1 f
f
A
A mid


 2
2
/
1 f
f
A
A mid


Below mid band Above mid band

69. Lower & Upper Critical frequency
• Critical frequency a.k.a the cutoff frequency
• The frequency at which output power drops by 3 dB.
[in real number, 0.5 of it’s midrange value.
• An output voltage drop of 3dB represents about a
0.707 drop from the midrange value in real number.
• Power is often measured in units of dBm. This is
decibels with reference to 1mW of power. [0 dBm =
1mW], where
.
dBm
0
mW
1
mW
1
log
10 






69

70. Gain & frequencies
• Gain-bandwidth product : constant value of the product
of the voltage gain and the bandwidth.
• Unity-gain frequency : the frequency at which the
amplifier’s gain is 1
BW
A
f mid
T 
70

71. Fig: Common Emitter RC Coupled amplifier
Design of Single stage RC Coupled amplifier

72. The design of a single stage RC coupled amplifier is shown
below.
The nominal vale of collector current Ic and hfe can be
obtained from the datasheet of the transistor.
Design of Re and Ce
Let voltage across Re;
VRe = 10%Vcc ………….(1)
Voltage across Rc;
VRc = 40% Vcc. ……………..(2)
The remaining 50% will drop across the collector-emitter .
From (1) and (2)
Rc =0.4 (Vcc/Ic) and Re = 01(Vcc/Ic).

73. Design of R1 and R2.
Base current
Ib = Ic/hfe.
Let Ic ≈ Ie .
Let current through R1;
IR1 = 10Ib.
Also voltage across R2 ;
VR2 must be equal to Vbe + VRe.
From this VR2 can be found.
There fore VR1 = Vcc-VR2.
Since VR1 ,VR2 and IR1 are found we can find R1 and R2 using
the following equations.
R1 = VR1/IR1 and R2 = VR2/IR1.

74. Finding Ce.
Impedance of emitter by-pass capacitor should be one by tenth
of Re.
i.e., XCe = 1/10 (Re) .
Also XCe = 1/2∏FCe.
F can be selected to be 100Hz.
From this Ce can be found.
Finding Cin.
Impedance of the input capacitor(Cin) should be one by tenth of
the transistors input impedance (Rin).
i.e., XCin = 1/10 (Rin)
Rin = R1 parallel R2 parallel (1 + (hfe re))
re = 25mV/Ie.
Xcin = 1/2∏FCin.
From this Cin can be found.

75. Finding Cout.
Impedance of the output capacitor (Cout) must be one by tenth
of the circuit’s
output resistance (Rout).
i.e, XCout = 1/10 (Rout).
Rout = Rc.
XCout = 1/ 2∏FCout.
From this Cout can be found.
Setting the gain.
Introducing a suitable load resistor RL across the transistor’s
collector and ground will set the gain. This is not shown
in Fig1.
Expression for the voltage gain (Av) of a common emitter
transistor amplifier is as follows.
Av = -(rc/re)
re = 25mV/Ie
and rc = Rc parallel RL
From this RL can be found

76. PART-II
Multistage amplifiers

77. 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

78. 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 ……

79. Example
Find Vout
Av1 = 5 Av2 = 10 Av3 = 6 Avn = 8
Vin = 5 mV

80. Decibel (dB)
• A logarithmic measurement of the ration of power or voltage
• Power gain is expressed in dB by the formula:
where ap is the actual power gain, Pout/Pin
Voltage gain is expressed by
• If av is greater than 1, the dB is +ve, and if av is less than 1, the dB gain
is –ve value & usually called attenuation
v
dB
V a
A log
20
)
( 
P
P a
A log
10

80

81. 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
)
(
1
)
(
2
/
1
1
)
(
1
/
1
2
)
(
2
BW
1
2
1
2
T
C
T
C
n
C
T
C
n
C
T
C
f
f
f
f
f
f







82. Different coupling schemes used in amplifiers
 Multistage amplifier configuration:
Rc1 Rc2
RB
Vi Q1
Q2
R2
RL
R1
Vo
Vi Q1
Q2
R3
R1 R2
Vi
Vo
R1
Vi
Vo
T
Cascade /RC coupling
Cascode
Transformer coupling
Darlington/Direct coupling
Q1
Q2
Q1
Q2

83. 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
Fig: A two stage amplifier in a cascade
connection
Fig: Small signal equivalent circuit of cascade configuration

84. i) Cascade Connection (cont.)
 small signal gain is:
by determine the voltage gain at stages 1 & stage 2
therefore
- the gain in dB
 input resistance
Output resistance
- assume ,so also
Therefore
)
)(
//
)(
//
( 2
2
1
2
1
S
i
i
L
C
C
m
m
S
o
V
R
R
R
R
R
r
R
g
g
V
V
A


 
1
2
1 //
// 
r
R
R
Ris 
0

S
V 0
2
1 
 
 V
V 0
2
1
1 
 
 V
g
V
g m
m
2
0 C
R
R 
2
1 V
V
V A
A
A 
)
log(
20
)
( V
dB
V A
A 

85. 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:











 k
R
R
k
R
R
k
R
R
k
R
R E
E
C
C 1
,
2
.
2
,
7
.
4
,
15 2
1
2
1
4
2
3
1





 0
)
(
2
1 ,
7
.
0
,
200
,
20 r
V
V
V
V ON
BE
Q
Q
CC 


86. Solution 1 (cont.)
2
1
1
// R
R
RB

1

r


1

V


2

V 2

r
2
C
R
1
C
R
1
1 
V
gm
2
2 
V
gm
4
3
2
// R
R
RB

i
V
0
V
1
B 2
B
1
E
1
C
2
E
2
C
i
R
0
R
Ac equivalent circuit for cascade amplifier

87. Solution 1
DC analysis:
At Q1:
At Q2:
AC analysis:
At Q1:
At Q2:
S
g
k
r
m 153
.
0
307
.
1
1
1




Why the Q-point values same for both
Q1 & Q2 ?
mA
I
A
I
CQ
BQ
979
.
3
89
.
19
1
1

 
S
g
k
r
m 153
.
0
307
.
1
2
2




mA
I
A
I
CQ
BQ
979
.
3
89
.
19
2
2

 

88. 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:
2
i
R
1
0Q
V
)
//
( 2
1
1
0
1 i
C
m
Q
VQ R
R
g
Vi
V
A 


06
.
102
)
36
.
957
//
2
.
2
(
153
.
0
1 


 k
AVQ



 36
.
957
307
.
1
//
7
.
4
//
15
// 2
2
2 k
k
k
r
R
R B
i 

89. 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
2
iQ
V
)
( 2
2
0
2 C
m
iQ
VQ R
g
V
V
A 


6
.
336
)
2
.
2
(
153
.
0
2 


 k
AVQ
353
,
34
)
6
.
336
)(
06
.
102
(
2
1 



 VQ
VQ
V A
A
A

90. Solution 1 (cont.)
From the ac equivalent circuit:
The i/p resistance is:
The o/p resistance is:




36
.
957
307
.
1
//
7
.
4
//
15
//
// 1
2
1 k
k
k
r
R
R
Ri 


 k
R
R C
o 2
.
2
2

91. 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
Fig(a): Cascode amplifier Fig(b):Ac equivalent circuit

92. ii) Cascode Connection (cont.)
From the small equivalent circuit, since the capacitors act as short circuit,
by KCL equation at E2:
solving for voltage
Where
the output voltage is
or
2
2
2
2
1
1 


 V
g
r
V
V
g m
m 

2

V
 
S
m V
g
r
V 1
2
2
2
1 












2
2
2 
 r
gm

)
//
)(
( 2
2 L
C
m
o R
R
V
g
V 


 
S
L
C
m
m
o V
R
R
r
g
g
V )
//
1 2
2
2
1 













93. ii) Cascode Connection (cont.)
Therefore the small signal voltage gain:
From above equation shows that:
So, the cascode gain is the approximately
* The gain same as a single-stage C-E amplifier
 
L
C
m
m
S
V R
R
r
g
g
V
V
A //
1 2
2
2
1
0














 
L
C
m
V R
R
g
A //
1


1
1
1 2
2
2
2
2 










 



r
gm

94. 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
So, the current gain
If
The Darlington connection
provides a current gain of
2
1

 
D
1
 2



 
 2
1
2

 
D
Figure : Darlington transistor

95. iii) Darlington Connection (cont.)
 Figure shown a Darlington configuration
Fig(a): Darlington pair connection fig(b):AC equivalent
circuit

96. iii) Darlington Connection (cont.)
 The small current gain :
Since
Therefore
Then,
The o/p current is:
The overall gain is:
* The overall small-signal current gain = the product of the individual
current gains
i
i
m
m I
I
r
g
V
g 1
1
1
1
1 

 

2
1
2 )
( 
  r
I
I
V i
i 

i
i
m
m I
I
V
g
V
g
I )
1
( 1
2
1
2
2
1
1
0 



 




2
1
1
2
1
0
)
1
( 



 




i
i
I
I
A
i
o
i I
I
A /

1
1 
 r
I
V i


97. iii) Darlington Connection (cont.)
 The input resistance:
Known that:
So, the i/p resistance is:
The base of Q2 is connected to the emitter of Q1, which means that the i/p resistance to
Q2 is multiplied by the factor , as we saw in circuits with emitter
resistor.
So, we can write: and
Therefore
The i/p resistance is then approximately
*The i/p resistance tends to be large because of the multiplication
2
1
1
2
1 )
1
( 


  r
I
r
I
V
V
V i
i
i 




2
1
1 )
1
( 
  r
r
Ri 


1
1
CQ
T
I
V
r

 
2
2
1

CQ
CQ
I
I 
2
1
2
2
1
1 
 

 r
I
V
r
CQ
T










2
1
2 
 r
Ri 
)
1
( 1


