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Basic Electronics (GTU) 7-1 Transistor at Low Frequencies
Chapter 7 : Transistor at Low Frequencies
Section 7.17 :
Ex. 7.17.2 : For the amplifier shown in the Fig. P. 7.17.2(a), find voltage gain Vo/VS, without using
Miller's theorem. Transistor h-parameters are hie = 1 k, hfe = 50, hre = hoe = 0. Verify the
answer using Miller’s theorem. .Page No. 7-50.
(a) (b) AC equivalent circuit
Fig. P. 7.17.2
Soln. :
Method I : Gain without using Miller's theorem
Step 1 : Draw the AC equivalent as shown in Fig. P. 7.17.2(b).
Step 2 : The load resistance RL = 10 k
Since hoe RL = 0, use simplified model for transistor.
Fig. P. 7.17.2(c) shows the simplified CE hybrid equivalent circuit of the given amplifier.
Fig. P. 7.17.2(c) : Simplified h-parameter equivalent circuit
Step 3 : Calculate gain Vo / VS :
1. By applying KCL at the base terminal of the transistor we get,
Ii = If + Ib
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Basic Electronics (GTU) 7-2 Transistor at Low Frequencies
– Ii + If + Ib = 0
Vi – VS
10 k +
Vi – Vo
200 k +
Vi
hie = 0 Since hie = 1 k
Vi
1
10 k +
1
200 k +
1
1k =
VS
10 k +
Vo
200 k
Vi = 90.49 10– 3
VS + 4.52 10– 3
Vo ... (1)
2. By applying KCL at node „C‟, the collector we obtain,
If = hfe Ib + IL
But If = Vi – Vo
200 k and IL =
Vo
10 k
hfe Ib + Vo – Vi
200 k +
Vo
10 k = 0 But, Ib =
Vi
hie
hfe
hie Vi +
Vo – Vi
200 k +
Vo
10 k = 0 Since hfe = 50 and hie = 1 k
Vi
50
1 k –
1
200 k = – Vo
1
200 k +
1
10 k
49.995 10– 3
Vi = – 0.105 10– 3
Vo ...(2)
Eliminating Vi from Equations (1) and (2), we get,
Vo
VS = – 13.66 ...Ans.
Answer : AVS = – 13.66
Method II : Gain using the Miller's Theorem
The hybrid equivalent circuit using Miller's theorem is shown in Fig. P. 7.17.2(d).
R1 and R2 are Miller resistances. Their values are given by,
R1 = Rf
1 – AV and R2 =
AV Rf
AV – 1 Rf = 100 k
Fig. P. 7.17.2(d) : Hybrid equivalent circuit using Miller's theorem
Referring to Fig. P. 7.17.2(d) we get,
Vo = Voltage across (R2 || RL) = – hfe Ib (R2 || RL)
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Basic Electronics (GTU) 7-3 Transistor at Low Frequencies
= – hfe Ib (Rf || RL) ... since R2 Rf
and Vi = hie Ib
AV = Vo / Vi = – hfe Ib (Rf || RL)
hie Ib =
– hfe (Rf || RL)
hie
= – 50 (200 k || 10 k)
1 k =
– 50 (200 10)
(200 + 10)
AV = – 476
Input resistance Ri = hie || R1
But R1 = Rf / (1 – AV) = 200 k
1 + 476 = 419.3
Ri = 1 k || (419.3) = 295.43
AVS =
Ri
RS + Ri
AV = 295.45
10000 + 295.45 – 476
AVS = – 13.66 ...Ans.
Note that the gain obtained using Miller's theorem is exactly same as that obtained without using
Millers theorem.
Ex. 7.17.3 : The transistor used in the circuit of Fig. P. 7.17.3(a) has the following parameters :
hie = 500 , hre = 2.4 10– 4
, hfe = 60 and hoe = 1/40 k. Calculate :
1. Vo / VS 2. Ri 3. R
o .
Assume all capacitors to be very large. .Page No. 7-50.
Fig. P. 7.17.3(a)
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Basic Electronics (GTU) 7-4 Transistor at Low Frequencies
Soln. :
Step 1 : AC equivalent circuit and h-parameter equivalent circuit :
The AC equivalent circuit is shown in Fig. P. 7.17.3(b).
Fig. P. 7.17.3(b) : AC equivalent circuit
Draw the h-parameter equivalent circuit :
The h-parameter equivalent circuit is as shown in Fig. P. 7.17.3(c).
(c) h-parameter equivalent circuit (d)
Fig. P. 7.17.3
We have drawn an approximate h-parameter equivalent circuit in Fig. P. 7.17.3(c). This is
because,
hoe RL =
1
40 k (RC || RL || 60 k) =
1
40k (4 k || 5 k || 60 k)
hoe RL = 0.05352 which is less than 0.1.
Step 2 : Input resistance Ri :
Ri = hie || 120 k = 500 || 120 k = 498 ...Ans.
Step 3 : Output resistance Ro :
Ro = 60 k || 5 k || 4 k = 60 k ||
5 4
9 = 60 k || (2.22 k) = 2141 ...Ans.
Step 4 : Voltage gain (Vo / VS) :
Vo
VS =
Vo
Vi
Vi
VS ...(1)
But Vo = – hfe Ib [ 60 k || 5 k || 4 k ] and Vi = hie Ib
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Basic Electronics (GTU) 7-5 Transistor at Low Frequencies
Vo
Vi =
– hfe Ib [ 60 k || 5 k || 4 k ]
hie Ib =
– hfe ( 60 k || 5 k || 4 k )
hie
Vo
Vi =
– 60 2141
500 = – 256.92 ...(2)
And referring to Fig. P. 7.17.3 (d) we can write that,
Vi = R
i
Ri + RS
VS Vi
VS =
498
498 + 1000 = 0.332 ...(3)
Substitute Equations (2) and (3) into Equation (1) to get,
Vo
VS = – 256.92 0.332 = – 85.41 ...Ans.
Section 7.18 :
Ex. 7.18.2 : A CE amplifier with bypassed RE using a PNP transistor is shown in Fig. P. 7.18.2(a).
Calculate AI, AIS, Ri, Ri , AV, AVS, Ro and Ro using approximate analysis. Assume
hie = 1.1 k and hfe = 50. .Page No. 7-58.
Fig. P. 7.18.2(a) : Given circuit
Soln. : Type of analysis : Approximate ...given
Step 1 : Draw the hybrid equivalent circuit :
The hybrid equivalent circuit is as shown in Fig. P. 7.18.2(b).
Fig. P. 7.18.2(b) : Hybrid equivalent circuit
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Basic Electronics (GTU) 7-6 Transistor at Low Frequencies
Note the changed current directions due to use of PNP transistor.
Step 2 : Calculate AI and AIS :
AI = IL / Ib = Ic / – Ib = – hfe = – 50
Note that the negative sign is introduced in order to compensate for the reversed direction of Ib as
compared to the standard direction.
AIS = IL
IS =
IL
Ic
Ic
Ib
Ib
IS
But IL = RC
( RC + RL ) Ic
IL
Ic =
RC
RC + RL
Ic / Ib = hfe and Ib = – RB
RB + Ri IS,
Ib
IS =
– RB
(RB + Ri )
AIS = RC
( RC + RL ) hfe
– RB
(RB + Ri )
AIS = – hfe RC RB
( RC + RL ) ( RB + Ri ) ...(1)
RB = R1 || R2 = 100 k || 10 k = 9.09 k, Ri = hie = 1.1 k
AIS = – 50 5 k 9.09 k
(5 k + 10 k) (9.09 k + 1.1 k) = – 14.867 ...Ans.
Step 3 : Calculate Ri and Ri :
Looking at Fig. P. 7.18.2(b),
Ri = hie = 1.1 k ...Ans.
and Ri = Ri || RB = 1.1 k || 9.09 k
Ri = 0.9812 k or 981.2 ...Ans.
Step 4 : Calculate AV and AVS :
AV = Vo / Vb
But Vo = Voltage across ( RC || RL )
AV = hfe Ib ( RC || RL ) ...(see Fig. P. 7.18.2(d))
and Vb = – Ib Ri
= – Ib hie ...(see Fig. P. 7.18.2(c))
(c) Calculation of Vb (d) Calculation of Vo
Fig. P. 7.18.2
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Basic Electronics (GTU) 7-7 Transistor at Low Frequencies
AV = hfe Ib (RC || RL)
– Ib hie =
– hfe (RC || RL)
hie
AVS = – 50 (5 k || 10 k)
1.1 k =
– 50 3.33
1.1 = – 151.51 ...Ans.
AVS =
Ri
( Ri + RS )
AV = 0.9812
(0.9812 + 10) – 151.51 = – 13.54 ...Ans.
Step 5 : Calculate Ro and Ro :
Yo = hoe – hfe hre
RS + hie
Note : We can use all the standard expressions for PNP transistor as well.
As this is approximate analysis, hoe and hre are to be neglected.
Yo = 0
Ro = 1/ Yo =
Ro = Ro || (RC || RL ) = || (RC || RL )
= RC || RL = 5 k || 10 k = 3.33 k ...Ans.
Section 7.19 :
Ex. 7.19.2 : The transistor in Fig. P. 7.19.2(a) has hie = 2.5 k and hFE hfe = 200. Calculate :
(a) Quiescent values of IC and VCE. (b) Small signal gain Vo / VS.
Assume all capacitors to be arbitrarily large. .Page No. 7-67
(a) Given circuit (b) DC equivalent circuit
Fig. P. 7.19.2
Soln. : We will solve this example in two parts.
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Basic Electronics (GTU) 7-8 Transistor at Low Frequencies
Part I : DC Analysis
Step 1 : Draw the DC equivalent circuit :
By assuming all the capacitors to be open circuit, we can draw the DC equivalent circuit as
shown in Fig. P. 7.19.2(b).
Step 2 : Calculate ICQ :
Refer to the dc equivalent circuit shown in Fig. P. 7.19.2(b). All the capacitors are replaced by
open circuit.
Apply KVL to the collector circuit to write,
3 = 1.5 k IC + VCE + (0.6 + 0.15) k IE – 3
Substitute IE = IC + IB to get,
6 = 1.5 k IC + VCE + 0.75 k IC + 0.75 k IB
6 = 2.25 k IC + VCE + 0.75 k IB ...(1)
Now consider the base loop of Fig. P. 7.19.2(b) and apply KVL to write,
3 = 38 k I1 + VBE + (0.75 k) IE – 3
6 = 38 k I1 + VBE + 0.75 k IB + 0.75 k IC ...(2)
Also, I1 = I2 + IB and I2 = VbR2
I2 =
0.75 k IB + 0.75 k IC + VBE
24.4 k ...(3)
Substitute I1 = I2 + IB and substitute the expression for I2 into Equation (2) to get,
6 = 38 k
0.75 k IB + 0.75 k IC + VBE
24.4 k + 38 k IB + 0.75 k IB + 0.75 k IC ...(4)
Substitute VBE = 0.7 V
6 = 1.16 k IB + 1.16 k IC + 1.09 + 0.75 k IB + 0.75 k IC + 38 k IB
4.9 = 39.91 k IB + 1.91 k IC ...(5)
Now substitute IB = ICh
FE
4.9 = 39.91 k IC
hFE + 1.91 k IC
Substitute hFE = 200 to get,
4.9 = 199.55 IC + 1.91 IC
IC = 2.43 mA ...Ans.
Thus the quiescent value of IC = 2.43 mA.
Step 3 : To obtain the value of VCEQ :
IB = ICQ
hFE =
2.43 10– 3
200 = 12.2 A
Substitute the values of IB and IC into Equation (1) to get,
6 = (2.25 2.43) + VCE + 0.75 103 12.2 10
– 6
VCEQ = 0.5234 Volt ...Ans.
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Basic Electronics (GTU) 7-9 Transistor at Low Frequencies
Part II : AC Analysis
Step 4 : To obtain the small signal gain :
The ac equivalent circuit of the given amplifier is as shown in Fig. P. 7.19.2(c) and the
h-parameter equivalent circuit is as shown in Fig. P. 7.19.2(d).
Fig. P. 7.19.2 (c) : AC equivalent circuit
Fig. P. 7.19.2 (d) : h-parameter equivalent circuit
Refer to Fig. P. 7.19.2(d).
RB = (R1 || R2) = 38 24.4
(38 + 24.4) = 14.85 k
Voltage gain AVS = Vo
VS =
Vo
Vb
Vb
VS ...(6)
But Vo = IL RC = – IC 1.5 k ...(7)
Vb = R
i
(Ri + RS )
VS
where, Ri = RB || [hie + (1 + hfe ) RE ]
= 14.85 k || [2.5 + (201 0.6)] k = 14.85 123.1
(14.85 + 123.1) k
Ri = 13.25 k
Vb = 13.25
13.25 + 0.3 VS
Vb
VS = 0.978 ...(8)
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Basic Electronics (GTU) 7-10 Transistor at Low Frequencies
Also, Vb = Ib hie + (1 + hfe ) Ib RE = Ib [hie + (1 + hfe ) RE]
Vb = Ib [2.5 k + (201 0.6 k )] = 123.1 k Ib ...(9)
Substitute Equations (7), (8) and (9) into Equation (6) to get,
AVS = – 1.5 k Ic
123.1 k Ib 0.978
But Ic
Ib = hfe = 200
AVS = – 1.5 k hfe
123.1 k 0.978
Voltage gain AVS = – 1.5 200 0.978
123.1 = – 2.4 ...Ans.
Section 7.20 :
Ex. 7.20.2 : For the amplifier shown in Fig. P. 7.20.2(a) calculate all the important parameters.
.Page No. 7-70.
Fig. P. 7.20.2(a) : Given circuit
Soln. :
Step 1 : Draw the ac equivalent circuit :
The ac equivalent circuits are shown in Figs. P. 7.20.2(b) and (c).
Step 2 : Draw the hybrid equivalent circuit :
R
L = RE || RL = 3k || 5k = 1.875 k
hoc R
L = 25 10– 6
1.875 103 = 0.04687
As hoc R
L < 0.1 let us neglect 1/ hoc from the equivalent circuit. The hybrid equivalent circuit is
shown in Fig. P. 7.20.2(d).
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Basic Electronics (GTU) 7-11 Transistor at Low Frequencies
(b) AC equivalent circuit (c) Simplified ac equivalent
Fig. P. 7.20.2
Method of analysis :
We can use any one of the following methods :
1. Analysis using standard formulae 2. Analysis using direct method
For this example we will use the second method.
Fig. P. 7.20.2(d) shows the simplified hybrid equivalent for the CC configuration. Note that the parameter hoc has been deleted from the circuit.
Fig. P. 7.20.2(d) : Hybrid equivalent circuit
RB = R1 || R2 = 300k || 100k = 300k 100k
400k = 75 k
Current gain :
AI = I
L
Ib =
IeIb
= (1 + hfe)
Input resistance :
Ri = VbIb
= hic Ib + hrc Ve
Ib
But Ve = Vo = Ie (RE || RL) = hfc Ib (RE || RL) and hrc = 1
Ri = [ hic + hfc (RE || RL ) ] Ib
Ib
Ri = hic + hfc (RE || RL ) = 1.2k + 101 (3k || 5k)
Ri = 1.2k + (101 1.875k) = 190.575 k ...Ans.
And R
i = RB || Ri = 75k || 190.575k
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Basic Electronics (GTU) 7-12 Transistor at Low Frequencies
R
i = 75 190.575
(75 + 190.575) k = 53.82 k ...Ans.
Voltage gain :
AV = Vo / Vb = Ie (RE || RL )
hic Ib + hrc Ve
But Ve = Vo = Ie (Re || RL) and hrc = 1
AV = Ie (RE || RL)
hic Ib + Ie (RE || RL)
But Ib = Ie / hfc
AV = Ie (RE || RL )
Iehfc
hic + Ie (RE || RL)
= Ie (RE || RL )
Ie
hic
hfc + (RE || RL)
AV = (3k || 5k)
1.2k
101 + (3k || 5k)
= 0.9937 ...Ans.
Voltage gain (Overall) : AVS = R
L
RS + R
i
AV = 53.82
10 + 53.82 0.9937 = 0.8379 ...Ans.
Current gain AIS :
AIS = ILIS
= ILIe
IeIb
IbIS
But IL
Ie =
RERE + RL
... using current division principle
But Ie / Ib = hfc and Ib / IS = Ri
Ri + RB ...using current division
AIS = RE
RE + RL hfc
RiRi + RB
= 3k
3k + 5k 101
190.575k
190.575k + 75k = 27.18 ...Ans.
Output Resistance Ro :
Set VS = 0 Ib = 0 hfc Ib = 0 hence the current source is equivalent to an open circuit.
Refer Fig. P. 7.20.2(e) to obtain Ro and R
o .
Ro =
and R
o = Ro || (RE || RL ) = RE || RL = 3k || 5k = 1.875 k …Ans.
Fig. P. 7.20.2(e) : Equivalent circuit to obtain Ro and R
o
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Basic Electronics (GTU) 7-13 Transistor at Low Frequencies
Section 7.23 :
Ex. 7.23.1 : For the two stage amplifier circuit shown in Fig. P. 7.23.1(a) calculate
1. Ri 2. Ro 3. AV 4. AVS. If both the transistors are identical, with hie = 1k, hfe = 50,
hre and hoe = 0. All the capacitors are assumed to be large. .Page No. 7-79.
Fig. P. 7.23.1(a)
Soln. :
Steps to be followed :
Step 1 : Analyze the second stage.
Step 2 : Analyze the first stage.
Step 3 : Calculate the overall parameter values.
Step 1 : Analysis of the second stage :
The second stage of the given circuit is a CE amplifier with emitter resistance bypassed. As
hoe = hre = 0 the approximate analysis is to be carried out :
(a) Current gain ( AI2 ) : AI2 = – hfe = – 50 ...Ans.
(b) Input resistance ( Ri2 ) : Ri2 = hie + hre AI2 RL2
But hre = 0, Ri2 = hie = 1k ...Ans.
The input resistance Ri2
= Ri2 || ( R1 || R2 ) = hie || ( R1 || R2 )
= 1 k || ( 47k || 4.7k ) = 810.3
(c) Voltage gain (AV2) : AV2 = AI2
RL2
Ri2, But R
L2 = RC2 = 3 k
AV2 = – 50 3
1 = – 150 ...Ans.
Step 2 : Analysis of the first stage :
(a) Current gain ( AI1 ) : AI1 = – hfe = – 50 ...Ans.
(b) Input resistance ( Ri1 ) : Ri1 = hie = 1 k ...Ans.
(c) Overall input resistance ( Ri1
) : Ri1
= hie || ( R1 || R2 ) = 1k || ( 100k || 10k ) = 900 ...Ans.
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Basic Electronics (GTU) 7-14 Transistor at Low Frequencies
(d) Voltage gain of first stage ( AV1 ) : AV1 = AI1 R
L1
Ri1
But R L1
= RC1 || Ri2
and AI1 = – hfe
AV1 =
–hfe [ RC1 || Ri2
]
hie
Substituting the values, AV1 = – 50 12 k || 0.81 k
1 k = – 50
0.759 k
1 k = – 37.95 ...Ans.
Step 3 : Calculate the overall parameters :
Overall voltage gain ( AV ) : AV = AV1 AV2 = – 37.95 – 150 = 5692.7 ...Ans.
Overall voltage gain in dB : AV dB = 20 log10 ( 5692.7 ) = 75.1 dB ...Ans.
Overall voltage gain ( AVS ) : AVS = Vo
Vi
Vi
VS = AV
Vi
VS
Referring to Fig. P. 7.23.1(b), Vi
VS =
Ri1
( Ri1
+ RS)
AVS = AV
Ri1
( Ri1
+ RS) Fig. P. 7.23.1(b)
Substituting the values we get, AVS = 5692.7 900
(900 + 1000) = 2696.5 ...Ans.
Output resistance of first stage ( Ro1 ) :
Yo1 = hoe – hre hfe
RS1 + hie
But as hoe = hre = 0
Yo1 = 0 and Ro1 = 1
Yo1 =
Output resistance of second stage (Ro2) :
Yo2 = hoe – hre hfe
RS2 + hie
Again Yo2 = 0 and Ro2 =
Overall output resistance ( Ro ) :
Ro = Ro2 || RL2 = || 3 k = 3 k ...Ans.
Section 7.24 :
Ex. 7.24.1 : The transistors shown in Fig. P. 7.24.1(a) are identical with hie = 1.1 k, hfe = 100 and
negligible hre and hoe. Calculate : 1. A I = io
is 2. AV =
Vo
Vs .Page No. 7-82.
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Basic Electronics (GTU) 7-15 Transistor at Low Frequencies
Fig. P. 7.24.1(a)
Soln : Given : hfe = 100, hie = 1.1 k hre = hoe = 0
Step 1 : Draw AC equivalent circuit :
Fig. P. 7.24.1(b) : AC equivalent circuit
Step 2 : Analysis of stage 2 :
1. AI2 = – hfb = hfe
1 + hfe =
100
101 = 0.99
2. Ri2 = hib = hie
1 + hfe =
1100
101 = 10.89
3. AV2 = AI2 RL2
Ri2 = + 0.99
4000
10.89 = + 363.6
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Basic Electronics (GTU) 7-16 Transistor at Low Frequencies
Step 3 : Analysis of stage 1 :
1. Load resistance RL1 = Ri2 = hib = 10.89
2. Current gain AI1 = – hfe = – 100
3. Input resistance Ri1 = hie + (1 + hfe) RE 1 = 1.1 k + (101 0.1 k) = 11.2 k
4. AV1 = AI1 RL1
Ri1 = – hfe
hi b
Ri1 = – 100
10.89
11.2 103 = – 0.09723.
Step 4 : Overall current gain :
AIS = Io
IS =
Io
Ie2
Ie2
Ib1
Ib1
IS = AI2 AI1
( R1 || R2 )
Ri1 + ( R1 || R2 )
= 0.99 – 100 ( 150 k || 9 k )
11.2 k + ( 150 k || 9 k )
But 150 k || 9 k = 8.49 k AIS = – 99 8.49
( 11.2 + 8.49 ) = – 42.687 ... Ans.
Step 5 : Overall voltage gain :
AV = AV1 AV2 = – 0.09723 363.6 = – 35.35
AVS =
Ri1
RS + Ri1
AV
But Ri1
= Ri1 || ( R1 || R2 ) = 11.2k || 8.49 k = 4.829 k
AVS = – 4.829
0.5 + 4.829 – 35.35 = 32.03 ...Ans.
Section 7.25 :
Ex. 7.25.1 : Calculate : Zi , Zo , AVS = Vo / VS and AIS = IL / IS for the circuit shown in Fig. P. 7.25.1.
Assume both transistors to be identical with hie = 1.1 k, hfe = 50 and negligible hre and
hoe. .Page No. 7-86.
Fig. P. 7.25.1 : Given circuit
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Basic Electronics (GTU) 7-17 Transistor at Low Frequencies
Soln. :
The given circuit is a CE-CC cascade configuration.
Since the values of hoe and hre are not given, we have to perform the approximate analysis.
Analysis of stage II :
1. Current gain AI2 = – hfe = – [– ( 1 + hfe )] = 51
2. Input resistance Ri2 = hic + AI2 hrc RL2
But hic = hie and hrc = 1
Ri2 = 1.1 k + (51 10k) = 511.1 k
3. Voltage gain AV2 = ( 1 + hfe ) RL2
Ri2 =
51 10
511.1 = 0.9978
Analysis of stage I :
1. Current gain AI1 = – hfe = – 50
2. Input resistance Ri1 = hie = 1.1 k
3. Voltage gain AV1 = – hfe RL1
Ri1
But RL1 = RC1 || Ri2 = 5 k || 511.1 k = 4.95 k
AV1 = – 50 4.95
1.1 = – 225
4. Output resistance Ro1, Yo1 = hoe – hfe hre
hie + RC1 = 0 Ro1 =
Overall parameters :
1. Voltage gain AV = AV1 AV2 = – 225 0.9978 = – 224.5
2. Input resistance Ri = Ri1 = 1.1 k ...Ans.
3. AVS = AV Ri
Ri + RS =
– 224.5 1.1 k
1.1 k + 0.5 k = – 154.34 ...Ans.
4. Output resistance Ro, Yo2 = hoc – hfc hrc
hic + RS = hoe +
( 1 + hfe ) 1
hfe + RS
But hoe = 0 and RS = Ro1 || RC1 = || RC1 = RC1
Yo2 = 1 + hfe
hie + RC1
Ro2 = hie + RC1
( 1 + hfe ) =
1.1 k + 5 k
51
Ro2 = 119.6
Ro = Ro2 || RE2 = 119.6 || 10 k = 118.2 ...Ans.
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Basic Electronics (GTU) 7-18 Transistor at Low Frequencies
Section 7.28 :
Ex. 7.28.1 : Calculate the dc bias voltages and currents for the circuit shown in Fig. P. 7.28.1.
.Page No. 7-92.
Fig. P. 7.28.1
Soln. :
1. Base current ( IB ) :
Substituting the values into Equation (7.28.5) we get,
IB = 15 – 1.4
3 M + ( 5000 330 ) = 2.92 A ...Ans.
2. Emitter current ( IE ) : IE = D IB = 14.62 mA ...Ans.
3. Emitter voltage ( VE ) : VE = IE RE = 14.62 10 – 3
330 = 4.82 volt ...Ans.
4. Base voltage ( VB ) : VB = VE + VBE = 4.82 + 1.4 = 6.22 volts ...Ans.
Ex. 7.28.2 : For the circuit shown in Fig. P. 7.28.2 calculate the values of Ri , AI , AV and Ro. Assume
the following h parameters for both the transistors : hie = 1.1 k, hfe = 50,
hre = 2.5 10 – 4
, hoe = 25 A / V. .Page No. 7-98.
Fig. P. 7.28.2
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Basic Electronics (GTU) 7-19 Transistor at Low Frequencies
Soln. :
Steps to be followed :
Step 1 : In a two stage amplifier always analyze the second stage first and then carry out the
analysis of the first stage.
Step 2 : Analyze the second stage to obtain the current gain AI2, input resistance Ri2 and voltage gain
AV2.
Step 3 : Analyze the first stage to obtain current gain AI1, input resistance Ri and voltage gain AV1.
Step 4 : Combine the results obtained in steps 2 and 3 to obtain the overall results.
Step 1 : Analysis of the second stage :
Let us first analyze the second stage. For this stage,
RL hoe = 1 103 25 10
– 6 = 25 10
– 3 [∵ RL = RE = 1 k ]
As RL hoe < 0.1 we can use the approximate analysis.
(a) Current gain of second stage ( AI2 ) :
AI2 ( 1 + hfe ) = 1 + 50 = 51 ...(1)
(b) Input resistance ( Ri2 ) :
Ri2 = hie + ( 1 + hfe ) RE = 1.1 k + ( 51 1 k)
Ri2 = 52.1 k ...(2)
(c) Voltage gain ( AV2 ) :
Referring to Equation (7.28.24),
AV2 = 1 – hie
Ri2 = 1 –
1.1
52.1 = 0.978 ...(3)
Step 2 : Analysis of the first stage :
The load resistance for the first stage is the input resistance of second stage Ri2.
hoe RL = hoe Ri2 = 25 10 – 6
52.1 103 = 1.3
As hoe RL > 0.1, we cannot use the approximate analysis. Hence we will have to use the exact
analysis.
(a) Current gain ( AI1 ) :
Referring to Equation (7.28.12(a)), we can write the exact equation for the current gain as,
AI1 = 1 + hfe
1 + hoe ( 1 + hfe ) RE
Substituting the values we get, AI1 = 1 + 50
1 + ( 25 10 – 6
51 1 103
) = 22.41 ...(4)
(b) Input resistance (Ri1) :
Referring to Equation (7.28.16), we get,
Ri1 = hie + AI1 Ri2 = 1.1 k + (22.41 52.1 k ) = 1.169 M ...(5)
-
Basic Electronics (GTU) 7-20 Transistor at Low Frequencies
(c) Voltage gain ( AV1 ) :
AV1 = 1 – hie
Ri1 = 1 –
1.1
1169 = 0.999 ...(6)
(d) Output resistance ( Ro1 ) :
We can write, Ro1 = hie + RS
1 + hfe =
1100 + 1000
51 = 41.17 ...(7)
(e) Output resistance of second stage ( Ro2) :
Referring to Equation (7.28.35) we can write,
Ro2 = RS + hie
( 1 + hfe )2 +
hie
(1 + hfe)
Substituting the values we get,
Ro2 = 1000 + 1100
( 51 )2 +
1100
( 51 ) = 22.38 ...(8)
Step 3 : Combine the results of analysis of stage 1 and 2 :
Overall voltage gain ( AV ) :
AV = AV1 AV2 = 0.999 0.978 = 0.977 ...Ans.
Overall current gain ( AI ) :
AI = AI1 AI2 = 22.41 51 = 1142.9 ...Ans.
Overall input resistance :
Ri = Ri1 = 1.169 M ...Ans.
Overall output resistance :
Ro = Ro2 = 22.38 ...Ans.
Ex. 7.28.3 : Fig. P. 7.28.3(a) shows bootstrapped Darlington pair with identical transistors. Calculate :
1. AV 2. AVS 3. Ri 4. Ro
Transistor data : hie = 1.1 k , hfe = 50, hre = 2.4 10–4
, hoe = 2.5 10–5
A/V
.Page No. 7-106.
Fig. P. 7.28.3(a) : Given circuit
-
Basic Electronics (GTU) 7-21 Transistor at Low Frequencies
Soln. :
Steps to be followed :
Step 1 : Draw the ac equivalent circuit of the amplifier.
Step 2 : Analyze second stage to obtain the current gain, input resistance, voltage gain and output
resistance.
Step 3 : Analyze the first stage to obtain the current gain, input resistance, voltage gain and output
resistance.
Step 4 : Calculate the values of voltage gain, input resistance and output resistance for the overall
configuration.
Step 1 : AC equivalent circuit :
Fig. P. 7.28.3(b) : AC equivalent circuit
The ac equivalent circuit is as shown in Fig. P. 7.28.3 (b). Note that “ R3 ” has been split into
Miller equivalent resistances RM1 and RM2. ( The 68 k resistor is actually the resistor R3 ).
Step 2 : Analysis of second stage :
The value of the product hoe RL for the second stage is given by,
hoe RL2 = 2.5 10–5
( 1.2 k || 10 k || 470 k )
Note that RL2 = 1.2 k || 10 k || 470 k = 1.07 k without considering RM2 as it is very large and
have no effect on the value of RL2.
hoe RL2 = 2.5 10–5
1.07 103 = 0.02675
As hoe RL2 < 0.1 we will use the approximate analysis for the second stage.
(a) Current gain ( AI2 ) :
AI2 = ( 1 + hfe ) = 1 + 50 = 51
(b) Input resistance ( Ri2 ) :
Ri2 = hie + ( 1 + hfe ) RL2 = 1.1 k + ( 51 1.07 k ) = 55.67 k
(c) Voltage gain ( AV ) :
AV2 = 1 – hie
Ri2 = 1 –
1.1
55.67 = 0.98
-
Basic Electronics (GTU) 7-22 Transistor at Low Frequencies
Step 3 : Analysis of first stage :
For the first stage, RL1 = Ri2 = 55.67 k
hoe RL1 = 2.5 10–5
55.67 103 = 1.39
As hoe RL1 > 0.1 we have to use the exact analysis for the first stage.
(a) Current gain ( AI1 ) :
AI1 = 1 + hfe
1 + hoe RL1 =
1 + 50
1 + [ ]2.5 10–5 55.67 103 = 21.32
(b) Input resistance ( Ri1 ) :
Ri1 = hie + AI1 RL1 = 1.1 k + ( 21.32 55.67 103 ) Ri1 = 1.187 M
(c) Voltage gain ( AV1 ) :
AV1 = 1 – hie
Ri1 = 1 –
1.1 103
1.187 106 = 0.999
Step 4 : Overall voltage gain ( AV ) :
AV = AV1 AV2 = 0.999 0.98 = 0.979 ...Ans.
Step 5 : Miller equivalent resistances :
RM1 = R3
1 – AV =
68 k
1 – 0.979 = 3.238 M
Step 6 : Overall input resistance ( Ri ) :
Ri = Ri1 || RM1 = ( 1.187 M || 3.238 M )
Ri = 0.8685 M = 868.5 k ...Ans.
Step 7 : Overall voltage gain ( AVS ) :
AVS = Vo
Vi
Vi
VS
AVS = AV Vi
VS
From Fig. P. 7.28.3(c), Vi
VS =
Ri
( )Ri + RS
AVS = AV R
i
( )Ri + RS Fig. P. 7.28.3(c)
-
Basic Electronics (GTU) 7-23 Transistor at Low Frequencies
Substituting the values we get, AVS = 0.979 868.5
( )868.5 + 220 = 0.7978 ...Ans.
Step 8 : Output resistance ( Ro1 ) :
Ro1 = RS1 + hie
( ) 1 + hfe
Here, RS1 = RM1 || RS = 3.238 M || 220 k = 206 k
Ro1 = (206 + 1.1) k
51 = 4.06 k
Step 9 : Output resistance ( Ro2 ) :
Ro2 = RS2 + hie
( )1 + hfe
Now, RS2 = Ro1 = 4.06 k
Ro2 = ( )4.06 + 1.1 k
51 = 0.101 k or 101
Ro2
= Ro2 || RL2 = 101 || 1.07 = 92.28 ...Ans.
Ex. 7.28.4 : Fig. P. 7.28.4(a) shows bootstrap darlington pair with identical transistors. Calculate AV,
AVS, Ri , Ro . The transistor parameters are hie = 1 k, hre = 2.4 10
–4, hfe = 100 and
hoe = 2.5 10–5
A/V. .Page No. 7-106
Fig. P. 7.28.4(a)
Soln. :
Step 1 : To draw the ac equivalent circuit :
Step 2 : Analysis of the second stage :
The effective load resistance RL2 for the second stage can be obtained from Fig. P. 7.28.4(b) as,
RL2 = 1 k || 10 k || 100 k = 0.9 k or 900
-
Basic Electronics (GTU) 7-24 Transistor at Low Frequencies
Fig. P. 7.28.4(b) : AC equivalent circuit
For second stage hoe RL2 = 2.5 10–5
900 = 0.0225
As hoe RL2 < 0.1 we are going to use the approximate analysis.
(a) Current gain ( AI2 ) :
AI2 = ( 1 + hfe ) = 101
(b) Input resistance ( Ri2 ) :
Ri2 = hie + ( 1 + hfe ) RL2
Ri2 = 1 k + [101 900 ] = 91.9 k
(c) Voltage gain ( AV2 ) :
AV2 = 1 – hie
Ri2
Substituting the values we get, AV2 = 1 – 1 k
91.9 k = 0.989
1. Analysis of the first stage
:
For the first stage RL1 = Ri2 = 91.9 k
hoe RL1 = 2.5 10–5
91.9 103 = 2.29
As hoe RL > 0.1 we must use the exact analysis for the first stage.
(a) Current gain ( AI1 ) :
( AI1 ) = 1 + hfe
1 + hoe RL1
where, RL1 = Ri2 = 91.9 k
AI1 = 1 + 100
1 + ( )2.5 10–5 91.9 103 = 30.6
(b) Input resistance ( Ri1 ) :
Ri1 = hie + AI1 RL1 = 1 k + (30.6 + 91.9 k) = 2.813 M
-
Basic Electronics (GTU) 7-25 Transistor at Low Frequencies
(c) Voltage gain ( AV1 ) :
AV1 = 1 – hie
Ri1 = 1 –
1 k
2.813 M = 0.999
2. Overall voltage gain :
AV = AV1 AV2 = 0.999 0.989 = 0.988
3. Miller equivalent resistance :
RM1 = R3
1 – AV =
50 k
1 – 0.988 = 4.17 M
4. Overall input resistance ( Ri ) :
Ri = RM1 || Ri1
= 4.17 M || 2.813 M = 1.68 M ...Ans.
5. Overall voltage gain ( AVS ) :
AVS = Vo
Vi
Vi
VS = AV
Ri
Ri + RS
Substituting the values we get,
AVS = 0.988 1.68
( ) 1.68 + 0.01 = 0.982 ...Ans.
6. Output resistance ( Ro1 ) :
Ro1
= RS1 + hie
( ) 1 + hfe Fig. P. 7.28.4(c)
Here, RS = RM1 || RS = 4.17 M || 10 M = 9.98 k.
Ro1
= ( ) 9.98 + 1 k
101 = 108.7
7. Output resistance ( Ro2 ) :
Ro2 = RS2 + hie
( ) 1 + hfe
Here, RS2 = Ro1 = 108.7
Ro2 = 108.7 + 1000
( ) 101 = 10.98
Output resistance Ro2
= Ro2 || RL2 = 10.98 || 900 = 10.83
-
Basic Electronics (GTU) 7-26 Transistor at Low Frequencies
Section 7.29 :
Ex. 7.29.1 : For the transistor amplifier shown in Fig. P. 7.29.1, find hie and hfe if AV = – 150 and
Ri = 1 k assume hre = hoe = 0. .Page No. 7-107.
Fig. P. 7.29.1
Soln. :
This is a common emitter amplifier with emitter resistance RE bypassed through CE. From the
Fig. P. 7.29.1 we can write the values of different components as,
RC = 2.2 k RB = R1 || R2 = R1 R2
R1 + R2 =
5 10
15 = 3.33 k
1. To obtain hie :
Ri = RB || hie ...(1)
Ri = RB hie
RB + hie Ri RB + R
i hie = RB hie
Ri RB = (RB – Ri ) hie hie =
Ri RB
(RB – Ri )
Substituting the values we get, hie = 1 3.33
(3.33 – 1) = 1.43 k ...Ans.
2. To obtain hfe :
From Table 7.18.1 we can write the expression for AV as :
AV = – hfe (RC || RL)
hie ...(2)
In this example RL = RC || RL = RC
AV = – hfe RC
hie
Substituting the values we get,
hfe = – AV hie
RC = –
(– 150) 1.43
2.2 = 97.5 ...Ans.
-
Basic Electronics (GTU) 7-27 Transistor at Low Frequencies
Ex. 7.29.2 : In the laboratory, measurement made on the circuit shown in the Fig. P. 7.29.2(a) and
voltage gain VoVS
was found to be 135 and input resistance Rin = 800 . Neglecting the
parameters hre and hoe find the approximate values of hfe and hie. .Page No. 7-107.
(a) (b) AC equivalent circuit
Fig. P. 7.29.2
Soln. :
To calculate hie :
From the ac equivalent circuit of Fig. P. 7.29.2(b),
we have, Ri = hie
Ri = Ri || RB, where RB = R1 || R2 = 2k
800 = hie || 2k
800 = 2k hie
hie + 2000 hie = 1333 ...Ans.
To calculate hfe :
We have voltage gain,
AV = AVS = VoVS
= AI R
L
Ri
But AI = – hfe – 135 = – hfe 2 k
hie
135 = – hfe 2 k
hie = – hfe
2k
1333 hfe = 90 ...Ans.
Ex. 7.29.3 : Transistor used in the amplifier shown in Fig. P. 7.29.3(a) has the following parameters :
hie = 1.5 k, hfe = 75, hre = hoe = 0. Find :
1. AV1 = Vo1/VS 2. AV2 = Vo2/VS
3. AIS = IL/IS 4. Ri. .Page No. 7-107.
-
Basic Electronics (GTU) 7-28 Transistor at Low Frequencies
Fig. P. 7.29.3(a)
Soln. : h parameter model :
Fig. P. 7.29.3(b)
hfe = 75, hie = 1.5 k
1. AV1
= Vol
VS , Vol = – hfe Ib (10 k || 9 k)
Vi = Ib 1.5 k + (1 + hfe) Ib 1.5 k
Vol
Vi =
– hfe (10 k || 9 k)
1.5 k + (1 + hfe) 1.5 k =
– 355.26
115.5 = – 3.076
AV1
= Vol
Vi
ViVS
But Vi = VS (90 k || 10 k)
1 k + (90 k || 10 k) =
9 k VS10 k
= 0.9 VS
AV1
= – 3.076 0.9 = – 2.7684 ...Ans.
– ve sign indicates 180 phase shift in output.
2. AV2
= Vo2
VS =
Vo2Vi
Vi
VS
-
Basic Electronics (GTU) 7-29 Transistor at Low Frequencies
But Vo2 = (1 + hfe) Ib 1.5 k
and Vi = 1.5 k Ib + 1.5 k (1 + hfe) Ib
Vo2Vi
= (1 + hfe) Ib 1.5 k
1.5 k Ib [1 + (1 + hfe)] =
1 + hfe
2 + hfe =
1 + 75
2 + 75
Vo2Vi
= 0.987
We have already obtained the value of Vi
VS = 0.9
AV2
= 0.987 0.9 = 0.8883 ...Ans.
3. AIS = IL
IS =
IL
IC
IC
Ib
Ib
IS
IL
IC =
– 10 k
10 k + 9 k = – 0.5,
ICIb
= + hfe = + 75
Ib
IS =
9 k
9 k + Ri = 0.072
where, Ri = hie + (1 + hfe) RE = 1.5 k + (76 1.5 k) = 115.5 k
Substituting the values
AIS = – 2.7 ...Ans.
4. Ri = VBE
Ib = 1.5 k + 1.5 k (1 + hfe)
Ri = 115.5 k and Ri = Ri || 9 k
Ri = 8.35 k ...Ans.
Ex. 7.29.4 : The transistor amplifier shown in Fig. P. 7.29.4
uses a BJT with the following parameters :
hie = 1.1 k, hfe = 50, hre = 2.5 10– 4
,
hoe = 24 A / V. Calculate :
1. AI = IoIi 2. AV 3. AVS
4. Ro 5. Ri. .Page No. 7-108.
Soln. :
1. hoe RL = 24 10 – 6
5 103 = 0.12. Fig. P. 7.29.4 : Given circuit
Hence we have to perform the exact analysis.
2. Current gain AI = – hfe
1 + hoe RL =
– 50
1 + (24 10 – 6
5 103) = – 41.66 ...Ans.
3. Input resistance Ri = hie + hre AI RL = 1.1 k + (2.5 10 – 4
– 41.66 5 103) = 1048
Ri = Ri || R1 || R2 = 1048 || (100 k || 10 k) = 1048 || 9090.9 = 939.67 ...Ans.
-
Basic Electronics (GTU) 7-30 Transistor at Low Frequencies
4. Voltage gain AV = AI RL
Ri = – 41.66
5 103
1048 = – 198.76 ...Ans.
5. Voltage gain AVS = AV Ri
Ri + rS =
– 198.76 939.67
939.67 + 10,000 = – 17 ...Ans.
6. Output resistance :
We know that Yo = hoe – hfe hre
rS + hie = 24 10
– 6 –
50 2.5 10 – 4
10 103 + 1.1 10
3 = 2.29 10 – 5
Ro = 1 / Yo = 1 / 2.29 10 – 5
= 43.668 k ...Ans.
and Ro = Ro || RL = 43.668 k || 5k = 4.486 k ...Ans.
Ex. 7.29.5 : For the amplifier shown in Fig. P. 7.29.5 assume hie = 1 k and hfe = 100.
Also hoe = hre = 0.
Find AI, AV, AVS, AIS, Ri and Ro . Assume all capacitors to be short circuit.
.Page No. 7-108.
Fig. P. 7.29.5
Soln. :
Type of amplifier : Common emitter with bypassed RE
Analysis : Approximate
Step 1 : Draw h-parameter equivalent circuit :
h-parameter equivalent circuit is shown in Fig. P. 7.29.5(a).
Fig. P. 7.29.5(a) : h-parameter equivalent circuit
-
Basic Electronics (GTU) 7-31 Transistor at Low Frequencies
RB = R1 || R2 = 100 k || 10 k = 9.09 k
R L = RC || RL = 4.7 k || 10 k = 3.2 k
Step 2 : Analysis :
1. AI = – hfe = – 100
2. Ri = hie = 1 k
3. Ri = Ri || RB = 1 k || 9.09 k = 0.9008 k or 900.8
4. AV = AI R
L
Ri = – 100
3.2
1 = – 320
5. AVS = AV R
i
Ri + RS
= – 320 0.9008
0.9008 + 10 = – 26.44
6. Ro = 1 / Yo and Yo = hoe – hfe hre
RS + hie = 0 Ro =
7. Ro = Ro || RL = R
L = 3.2 k
Ex. 7.29.6 : The transistor amplifier, shown in Fig. P. 7.29.6, uses a transistor having h-parameters as
hie = 1.1 k, hfe = 50, hre = 2.5 10 – 4
and hoe = 25 A /V.
Calculate Io/Ii, AV, AVS, Ro , Ri .Page No. 7-109.
Fig. P. 7.29.6
Soln. :
Given : hie = 1.1 k, hfe = 50, hre = 2.5 10 – 4
, hoe = 25 A / V
To find : AI, AV, AVS, Ri , R
o
Type of analysis :
hoe RL = 25 10 – 6 6.2 10
3 = 0.155
Since hoe RL > 0.1 exact analysis should be carried out.
Step 1 : Current gain :
AI = IoIi
= – hfe
1 + hoe RL =
– 50
1 + 0.155 = – 43.29 ...Ans.
-
Basic Electronics (GTU) 7-32 Transistor at Low Frequencies
Step 2 : Input resistance :
Ri = hie + hre AI RL = 1100 + (2.5 10 – 4
– 43.29 6.2 103) = 1033
Ri = R1 || R2 || Ri = 110 k || 12 k || 1.033 k
= 10.82 k || 1.033 k = 0.9429 k or 942.9 ...Ans.
Step 3 : Voltage gain :
AV = AI RLRi
= – 43.29 6.2
1.033 = – 259.8 ...Ans.
AVS = AV R
i
Ri + RS
= – 259.8 0.9429
0.9429 + 1 = – 126 ...Ans.
Step 4 : Output resistance :
Yo = hoe – hfe hre
(RS + hie) = 25 10
– 6 –
50 2.5 10 – 4
1000 + 1100 = 1.9047 10
– 5
Ro = 1
Yo = 52500 or 52.5 k
and Ro = Ro || RC = 52.5 k || 6.2 k = 5.545 k ...Ans.
Ex. 7.29.7 : The transistor amplifier shown in Fig. P. 7.29.7(a) uses a transistor with hie = 1.1 k,
hfe = 50, hre = 2.5 10– 4
and hoe = 25 10 – 6
A / V. Calculate AI = I0 / Ii, AV, AVS, Ro , Ri .
.Page No. 7-109.
Fig. P. 7.29.7(a)
Soln. :
Type of amplifier : Common emitter with bypassed RE
Analysis : Approximate
Step 1 : Draw h-parameter equivalent circuit :
h-parameter equivalent circuit is shown in Fig. P. 7.29.7(b).
-
Basic Electronics (GTU) 7-33 Transistor at Low Frequencies
Fig. P. 7.29.7(b)
RB = 100 k || 10 k = 9.09 k RL = RC = 4.7 k
Step 2 : Analysis :
1. AI = – hfe = – 50
2. Ri = hie = 1.1k
3. Ri = Ri || RB = 1.1 k || 9.09 k = 0.981 k or 981.35
4. AV = AI R
L
Ri = – 50
4.7 k
1.1 k = – 213.63
5. AVS = AV R
i
Ri + RS
= – 213.63 0.981
0.981 + 10 = – 19.08
6. Ro = 1
Yo and Yo = hoe –
hfe hre
RS + hie = 25 10
– 6 –
50 2.5 10 – 4
10 k + 1.1 k = 23.87 10
– 6 mho
Ro = 1
23.87 10 – 6 = 41.88 k
7. Ro = Ro || RL = 41.88 k || 4.7 k = 3.59 k
Ex. 7.29.8 : For the amplifier circuit in Fig. P. 7.29.8, calculate : AVS = VoVS
, AIS = IoIS
, Ri , Ro .
For the BJT hie = 1.1 k, hfe = 50, hre = 2.4 10 – 4
, hoe = 25 A/V. .Page No. 7-110.
Fig. P. 7.29.8 : Given circuit
-
Basic Electronics (GTU) 7-34 Transistor at Low Frequencies
Soln. :
Load Resistance :
R L = RC || RL = 2k || 2k = 1k
hoe RL = 25 10
– 6 1 10
3 = 25 10
– 3 = 0.025
As hoe RL < 0.1 we will use the approximate analysis.
Step 1 : Calculate Ai :
Ai = – hfe = – 50
Step 2 : Calculate Ri, Ri :
Ri = hie = 1.1 k
Ri = hie || R1 || R2 = 1.1 k || (100 k || 10k) = 1.1 k || (9.09 k) = 0.9812 k
Step 3 : Calculate AiS :
AiS = IoIS
= IoIb
IbIS
= Ai hie
hie + RB
But RB = R1 || R2 = 9.09 k AiS = – 50 1.1
1.1 + 9.09 = – 5.397
Step 4 : Calculate AVS :
AV = Ai R
L
Ri = – 50
1
1.1 = – 45.45
AVS = AV
Ri
Ri + RS
= – 45.45 0.9812
(0.9812 + 2.2) = – 14 ...Ans.
Step 5 : Calculate Ro :
Ro = RL = RC || RL = 1 k ...Ans.
Ex. 7.29.9 : A transistor used in the amplifier circuit shown in Fig. P. 7.29.9 has h parameters
hie = 800 , hoe = 50 10 – 6
mho and hfe = 55. Calculate :
1. Voltage and power of circuit. 2. Express voltage and power gain in dB.
3. Voltage and power gain if hoe is neglected. .Page No. 7-110.
Fig. P. 7.29.9
-
Basic Electronics (GTU) 7-35 Transistor at Low Frequencies
Soln. :
Part 1 : AV and AP by taking hoe into account :
1. Current gain AI = – hfe
1 + hoe RL =
– 55
1 + (50 10 – 6
2 103) = – 50
2. Voltage gain AV = AI
RL
Ri
But Ri = hie + hr Ai RL = hie = 800
AV = – 50 2 10
3
800 = – 125 ...Ans.
3. Power gain AP = AV AI = – 125 – 50 = 6250 ...Ans.
Part 2 : AV and AP in dB :
AV dB = 20 log10 (125) = 41.94 dB ...Ans.
AP dB = 10 log10 (6250) = 37.96 dB ...Ans.
Part 3 : AV and AP neglecting hoe :
AI – hfe = – 55
Ri hie = 800
AV = AI
RL
Ri = – 55
2000
800 = – 137.5
AP = AV Ai = – 137.5 – 55 = 7562.5 ...Ans.