Unit - 2
Multistage Amplifiers
Q1) For the circuit shown in below figure find input impendence RI if hie = hre = hoe = 0 and hFe is the same of each of transistor.
A1)
Given hie = hre = hoe = 0 and
hFe1 = hFe2 = hFe3 = …... = hFeN = hFe
Check for approximation: hoeRLN’ 0.1
As hoe is zero so all transistor is in valid approximation
Now, Ain = 1 + hFe, RiN = hie + (1 + hFe) RLN
Or RiN = 0 + (1 + hFe) Re
= (1 + hFe) Re;
1 + hFe,
= hie + (1 + hFe)
Re
Re
1 + hFe,
= hie + (1 + hFe)
Re
Hence, QN → (1 + hFe) Re
QN-1 → Re
QN-2 → Re
Similarly, Q[N-(n-1)] → Re
Ri1 = Re (Ans)
Also, AI = AI1 × AI2 × ……. × AIN
Hence, AI = (1 + hFe) N
Q2) Determine VONS for the cascade amplifier shown in figure
Use hie = 1 KΩ, hFe = 100, hre = 0, hoe = 0
A2)
Given hie = 1 KΩ, hFe = 100, hre = 0, hoe = 0
RC1 = RC2 = 5 KΩ
Q1 → CE
Q2 → CE
The equivalent circuit can be drawn as –
Analysis of Q2 [CE]: Ri2 = 5KΩ hoeRL2 = 0 (approximate analysis)
AI2 = = -hfe = -100
Ri2 = hie = 1KΩ
AV2 = =
AV2 =
Analysis of Q1 [CE]: RL1 = RC1 || Ri2
= 5 K || 1 K = 0.8333 K
hoeRL1 = 0 (approximate analysis)
AI1 = hFe = -100
Ri1 = hie = 1KΩ
AV1 = =
AV = AV1 × AV2
AV = (-500) × (-83.33) = 41650 (Ans)
Q3) Find the voltage gain AVS of the amplifier shown.
Assume hie = 1000Ω, hre = 10-4, hFe = 50, hoe = 10-4 A/V
A3) Given: hie = 1000Ω, hre = 10-4, hFe = 50, hoe = 10-4 A/V, RL = 5 KΩ
Q1 → CE, Q2 → CC
For analysis of amplifier, we consider DC supply voltage as ground and modified circuit is given as –
Analysis of Q2 [CC]: hoeRL’ = 10-4 × 5 × 103 = 0.5 > 0.1
approximate analysis is not valid, therefore exact analysis.
AI2 = = = = 34
Ri2 = hie + AI2RL’ = 1 + 34 × 5 = 171 KΩ
AV2 = = 0.994
Analysis of Q1 [CE]:
RL1 = 10 || Ri2 = 9.447 KΩ
hoeRL1 = 10-4 × 9.447 × 103 = 0.9447 > 0.1
so exact analysis
AI1 = = -25.71
Ri1 = hie + hreAI1RL1 = 1 + 10-4 × (-25.71) × 9.447 = 975.71 Ω
AV1 = =
AV =
But Vi2 = VO1
AV = AV1 × AV2
AV = (-249.11) × (0.994) = -247.615
AVS-247.615 × 0.66101 = -163.675 (Ans.)
AI =
25.71
AI2 =
IO = Ie2, Ib1 = Ii
AI = (-1) × (25.71) ×
-0.05524
AI = -25.71 × 34 × 0.05524
= -48.288 (Ans)
Q4) For the circuit shown, compute AI, AV, AVS, RI and RO’
A4) Given: Re = 0.1 KΩ, RL = 3 KΩ RS = 10 KΩ
Q1 → CC, Q2 → CE
AC analysis –
Analysis of Q2 [CE]:
hoe (Re + RL) = (0.1 + 3) = 0.0775 < 0.1
approximate analysis
AI2 = = -hFe = -50
Ri2 = hie + (1 + hFe) Re = 1.1 + 51 × 0.1 = 6.2 KΩ
AV2 = = -24.193
Analysis of Q1 [CE]:
hoe (Re + RL’)
RL’ = Ri2 || 8 K= 3.492 K
(0.5 + 3.492) = 0.0998 < 0.1
So approximate analysis
AI1 = = -50
Ri1 = hie + (1 + hFe) Re = 1.1 + 51 × 0.5 = 26.6 KΩ (Ans)
AV1 = =
Hence,
AV = AV1 × AV2 = 158.8 (Ans)
AVS
=
AVS = 158.8 × 0.7261 = 115.41 (Ans)
AI =
AI1 = 50
AI2 =
AI = (-1) × (50) ×
-0.5633
AI = 50 × 50 × 0.5633
= 1408.25 (Ans)
RO1 = ∞, RS2 = 8 || ∞ = 8 KΩ
RO2 =∞, RO’ = ∞ || 3 K = 3 K (Ans)
Q5) Calculate AI, AV, AVS and Ri
A5) Given: RS = 1 KΩ, RL = 5 KΩ
Amplifier configuration: cascade amplifier [CE – CB]
Analysis of Q2 [CB]:
hoeRL 0.1
2.45 × 10-3 < 0.1 which is less than 0.1
Approximate analysis:
AI2 = = -hFb = 0.98
Ri2 = hib = 21.6 Ω
AV2 = = 226.85
Analysis of Q1 [CE]:
RL1 = Ri2 = 21.6 Ω
Checking approximation:
hoeRL1 0.1
= 5.4 × 10-4 < 0.1, which is less than 0.1
So approximate analysis
AI1 = = -50
Ri1 = hie = 1.1 KΩ (Ans)
AV1 = =
Hence,
AV = AV1 × AV2 = -222.723 (Ans)
AVS = 0.5238 × (-222.732) = -116.66 (Ans)
[ since can be given as ]
AVS = 158.8 × 0.7261 = 115.41 (Ans)
AI = AI1 × AI2 = -50 × 0.98 = -49 (Ans)
Q6) For the circuit shown in figure. Calculate the values of Ri, AI, AV and Ro. Assume the following h-parameters for both the transistors. Hie = 1.1 KΩ, hFe = 50, hre = 2.5 × 10-4, hoe = 25 μA/V
A6) Given: hie = 1.1 KΩ, hFe = 50, hre = 2.5 × 10-4,
hoe = 25 μA/V, RS = 1 KΩ, RE = 1 Ω
Amplifier configuration:
Q1 → CC, Q2 → CC
Analysis of second stage:
Checking approximation: hoeRc2 0.1
Let us first analyse the second stage. For this stage,
Analysis of Q2 [CE]:
hoeRL2 = 1 × 103 × 25 × 10-6 = 25 × 10-3 [ RC = RE = 1]
which is less than 0.1
approximate analysis:
a) Current gain of second stage (AI2):
AI2 1+ hFe = 1 + 50 = 51 ----------(1)
b) Input resistance
Ri2 = hie + (1 + hFe) RE = 1.1 K + (51 × 1 K)
Ri2 = 52.1 KΩ -----------(2)
c) Voltage gain (AV2):
AV2 = = 1 - 0.978 -----------(3)
Analysis of the first stage:
The load resistance for the first stage is the input resistance of the second stage Ri2
hoeRi2 = 25 × 10-6 × 52.1 × 10-3 = 1.3
As hoeRi2 > 0.1, we cannot use the approximate analysis. Hence, we will have to use the exact analysis.
a) Current gain (AI1):
We can write the exact equation for the current gain as,
AI1 =
Substituting the values, we get,
AI1 = = 22.41 ----------(4)
b) Input resistance (RI1):
Ri1 = hie + AI1Ri2 = 1.1 K + (22.41 × 52.1 K) = 1.169 MΩ ----------(5)
c) Voltage gain (AV1):
AV1 = = 1 - 0.999 Ω ----------(6)
d) Output resistance (RO1):
RO1 = Ω ------------(7)
e) Output resistance of second stage (RO2):
RO2 =
Substituting the values, we get,
RO2 = = 22.38 Ω -------------(8)
Combine the result of analysis of stage 1&2:
Overall voltage gain (AV): AV = AV1 × AV2
= 0.978 × 0.999
= 0.977 Ans.
Overall current gain (AI): AI = AI1 × AI2
= 22.41 × 51 = 1142.9 Ans.
Overall input resistance (Ri) = Ri1 = 1.169 MΩ Ans.
Overall output resistance (RO) = RO2 = 22.38 Ω Ans.
Q7) For the two-stage cascade, shown calculate AI, AV, Ri and RO’.
A7) Given: RB = R1 = 100 KΩ, RL = 4 KΩ Re = 0.1 K
Amplifier configuration: Q1 → CE, Q2 → CC with Re
AC analysis: For a.c. analysis of amplifier we consider the d.c. supply as ground and apply Miller’s theorem across feedback resistance (RB = 100 KΩ) then modified circuit is given as:
Analysis of Q2 [CC]:
Check for approximation, hoeR1 0.1
hoeRL = × 4 = 0.1
Hence, valid approximation
Approximate analysis:
Current gain:
AI2 = = 1 + hFe = 1 + 50 = 51
Input Impendence:
Ri2 = hie + (1 + hFe) RL = 1.1 + 51 × 4 = 205.1 KΩ
Voltage gain:
AV2 = = 0.995
Analysis of Q1 [CE with Re]:
hoe (Re + RL’)
RL’ = || 2 || Ri2
|AV1| >> 1 for CE, << 1, 1
RL1 = 100 || 2 || 205.1 = 1.942 KΩ
Check for approximation hoe (RL’ + Re) 0.1
= 0.051 < 0.1
Hence, valid approximation
Approximate analysis:
Current gain:
AI1 = = -50
Ri1 = hie + (1 + hFe) Re = 1.1 + 51 × 0.1 = 6.2 KΩ
Voltage gain:
AV1 = =
AV = AV1 × AV2 = -15.661 × 0.995
= - 15.583 (Ans)
Input Impendence:
Ri = || Ri1 = 6.03 K || 6.2 K = 3.05 KΩ (Ans)
Now, RO’ = source resistance for Q2 & RO1 = ∞
RB’ = ∞ || || 2K = 1.958 KΩ
From the fig. (a):
RO’ = RO2 || 4 K = 0.060 || 4 = 0.059 KΩ (Ans)
Overall current gain:
AI =
And
AI1 = 50
AI2 =
AI = (-1) × (51) ×
From fig. (b):
From fig. (c):
AI = 51 × (- 0.009) × 50 × 0.492 = -11.29 Ans.
Q8) For the circuit shown, find AV, AVS, AI = , Ri and RO
A8) Given RC = 100 KΩ, Ri = 470 KΩ, R1 = 40 KΩ, R2 = 100 KΩ
Amplifier configuration: Q1 → CC, Q2 → CC with Re
AC analysis: For a.c. analysis of amplifier we consider the d.c. supply voltage as ground and all capacitor as short circuit and apply Miller’s theorem across 470K feedback resistance then modified circuit is given as:
Let us assume AV = 0.96 [Since in CC-CC configuration AV 1]
R2’ = 1K || 40K || 100K || 100K ||
= 1K || 40K || 50K || 11280K = 0.956 KΩ
Analysis of Q2 [CC]:
Check for approximation, hoeRL’ 0.1
0.0239 0.1, which is less than 0.1
Approximate analysis:
AI2 = = 1 + hFe = 51
Ri2 = hie + (1 + hFe) RL’ = 1.1 + 51 × 0.956 = 49.856 KΩ
AV2 = = 0.9779
Analysis of Q1 [CC]:
RL1 = Ri = 49.856 KΩ
Checking approximation:
hoeRL1 0.1
which is greater than 0.1
Exact analysis:
AI1 =
Ri1 = hie + AI1RI1 = 1.1 + 22.70299 × 49.856
= 1132.98 KΩ Ans.
AV1 = =
AV = AV1 × AV2 = 0.976 Ans.
R = 19583.33 KΩ
From Fig. (a):
Ri = R || Ri = 19583.33 K || 1132.98 K = 1071.017 KΩ
AVS = = 0.976 × = 0.8927 Ans.
AI =
AI
AI =
AI = 1047.30
RS1 = 100 K || 19583.33 K = 99.491 KΩ
RO1 = , VO1 = hOe + 0.532 mʊ
RO1 = 1.8796 KΩ
RS2 = RO1 = 1.8796 KΩ
RO2 = KΩ
RO’ = RO2 || RL’ = 0.0584 || 0.956 = 55 Ω Ans.
Q9) For the two-stage cascade shown, find AI, AV, Ri and RO’.
A9) Given: RB = 100 KΩ, Re = 0.05 KΩ, RL = 2 KΩ
Amplifier configuration: Q2 → CE with Re, Q1 → CE with Re
AC analysis: For a.c. analysis of amplifier we consider the d.c. supply as ground and apply Miller’s theorem across (RB = 100 KΩ) then modified circuit is given as:
Analysis of Q2 [CE with Re]:
Check for approximation,
hoe (Re+RL) 0.1
hoe (Re + RL) = × [0.5 + 2] = 0.05125 < 0.1
Approximate analysis:
AI2 = = - hFe = -50
Ri2 = hie + (1 + hFe) Re = 1.1 + 51 × 0.05 = 3.65 KΩ
AV2 = = -27.397
Analysis of Q1 [CE with Re]:
RL’ = || 10 || Ri2
Assume |AV1| = 1 for CE,
<< 1, 1
RL’ = 100 || 10 || 3.65 = 2.604 KΩ
Check for approximation hoe (RL’ + Re) 0.1
hoe (RL’ + Re) = = 0.078 < 0.1
Approximate analysis:
AI1 = = -50
Ri1 = hie + (1 + hFe) Re = 1.1 + 51 × 0.5 = 26.6 KΩ
AV1 = =
AV = AV1 × AV2 = (-4.895) × (-27.397)
= 134.108 (Ans)
Ri = || Ri1 = 16.964 || 26.6 = 10.358 KΩ (Ans)
AI =
AI1 = 50
AI2 =
AI = (-1) × (50) ×
From fig. (b),
From fig. (d):
AI = 50 × 0.71 × 50 × 0.389 = 690.475 Ans.
RO2 = ∞, RO’ = 2 || ∞ = 2 KΩ (Ans)
Q10) Three cascaded stages have an overall upper 3dB frequency of 16kHz and a lower 3dB frequency of 25 HZ. What is the value of FL and FH of each stage? Assume that all the stages are identical Also calculate the bandwidth of each stage
A10) Given: FL(n) = 25 HZ, FH(n) = 16 KHZ n = 3
Lower 3dB frequency, FL for each stage:
Therefore FL(n)
FL = FL (n) ×
= 25 × = 12.75 HZ
Upper 3dB frequency FH for each stage:
Therefore FH(n) = FH
FH=
FH= = 31.38 KHZ
Bandwidth of each stage: BW = FH – FL =31.38 KHZ – 12.75HZ
= 31.37 KHZ