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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


– 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)


= – 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)


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


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


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


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.


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.


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)


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.


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).


(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


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


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.


(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.


Fig. P. 7.24.1(a)

Soln : Given : hfe = 100, hie = 1.1 k hre = hoe = 0

Step 1 : Draw 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


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


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.


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


Soln. :


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.


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)



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.



Soln. :


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 :


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.


AI2 = ( 1 + hfe ) = 1 + 50 = 51


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


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.


AI1 = 1 + hfe

1 + hoe RL1 =

1 + 50

1 + [ ]2.5 10–5 55.67 103 = 21.32


Ri1 = hie + AI1 RL1 = 1.1 k + ( 21.32 55.67 103 ) Ri1 = 1.187 M


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)


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 :


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



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.


AI2 = ( 1 + hfe ) = 101


Ri2 = hie + ( 1 + hfe ) RL2

Ri2 = 1 k + [101 900 ] = 91.9 k


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.


( 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


Ri1 = hie + AI1 RL1 = 1 k + (30.6 + 91.9 k) = 2.813 M



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


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


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


hfe = – AV hie

RC = –

(– 150) 1.43

2.2 = 97.5 ...Ans.


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.


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


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.


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


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.


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).


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


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


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.

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