TUTORIAL NO. – 1

Q.1 Explain drift and diffusion of charge carriers in semiconductors. Derive an expression for the electron current due to drift and diffusion.

Q.2 Distinguish between the followings: (i) Intrinsic and extrinsic semiconductor (ii) Majority and minority charge carriers (iii) P and N type semi conductor (iv) Forward and reverse biasing of P-N junction

Q.3a) For what voltage will the reverse current in a p-n junction germanium diode reach 90% of its saturation values at room temperature? b)What is the ratio of the current of a forward bias of 0.05 V to the current for the same magnitude of the reverse bias.c) If the reverse saturation current is 10 A, calculate the forward currents for voltages of 0.1, 0.2 and 0.3 V respectively

Q.4 Using the fact that a silicon diode has I0=10-14A at 25ºC and that is increases by 15% perº C rise in temperature. Find the value of I0 at 125ºC. (Ans. : 1.17 x 10-8 A)

Q.5 Assuming that the diodes in the circuit given below is ideal, utilize thevein’s theorem to simplify the circuit shown in figure and find the values of the labelled currents and voltages.

Q.6 For the circuit shown below, both the diodes are identical conducting 10mA at 0.7V and 100mA at 0.8V. find the value of R for which V= 50mV.

15V

10kΩ

20kΩ 20kΩ

IV

+50mV-

10mA

I2

D1

D2R

+V-

I1+vD1

-

+vD2

-

Q.7 For the circuit given, find the output voltage vo for the cases a) V 1 = V2 = 5V b) V1 = 5V, V2 = 0V c) V1 = V2 = 5V

Q.8 (a) Calculate the anticipated factor by which the reverse saturation current of a germanium diode is multiplied when the temperature increased from 25 to 80ºC.

(b) Repeat part(a) for a silicon diode over the range 25 to 150ºC.

Q.9 An ideal germanium p-n junction diode has at a room temperature of 125ºC a reverse saturation current of 30 µA. At a temperature of 125ºC find the dynamic resistance for a

0.2V bias in (a) the forward direction (b) the reverse direction.

Q.10 The zero-voltage barrier height at an alloy germanium p-n junction is 0.2 V. The concentration NA of acceptor atoms in the p-side is much smaller than the concentration of donor atoms in the n- material, and NA = 3 x 1020 atoms/m3. Calculate the width of the depletion layer for an applied reverse voltage of (a) 10V and (b) 0.1 V and (c) for a forward bias of 0.1 V.

5V

4.7kΩ300ΩV1

300ΩV2

V0

D1

D2

TUTORIAL NO. – 2Q.1 A half wave rectifier uses a diode with an equivalent forward resistance of 0.3If the

input a.c. voltage is 10V (rms) and the load resistance of 20. Calculate Idc and Irms in the load.

Q.2 A half wave rectifier uses a diode with a forward resistance of 100 If the input a.c. voltage is 220 V (rms) and the load resistance of 2 k. Determine (i) Imax, Idc and Irms. (ii) Peak Inverse voltage when diode is ideal (ii) Load output voltage (iv) d.c. output power and a.c. input power (v) ripple factor (vi) TUF (vii) Rectification efficiency.

Q.3(a) What is the ripple factor for the ripple of 2v on avg. of 50 v. (Ans : r = )

(b) In a power supply the d.c. output voltage drops from 44V with no load to 42 V at full load. Calculate the % voltage regulation.

Q.4 A single phase full-wave rectifier uses two diodes, the internal resistance each being 20The transformer r.m.s. secondary voltage form centre tap each end of secondary is

50 V and RL= 980. Find

(a) The mean load current (b) Rms load current (c) Output efficiency

Ans. : 45mA, 50mA, 79.58Q.5 What is the ripple 2V on average of 50V?Q.6 A full wave rectifier has a peak output voltage of 25V at 50 Hz and feed a resistive load of

1K. The filter used is shunt capacitor one with 20 F . determine (i) dc load current (ii) dc out put voltage (iii) ripple voltage (iv) ripple factor

Q.7 Give output of following clipper circuits when input to all circuits is a sinusoidal wave of peak voltage Vm.

RVi VO

––

+ +

(a)

RVi VO

––

+ +

RVi VO

––

+ +

(a)

RVi VO

––

+ +

(b)

IV

RVi VO

––

+ +

(b)

IV

Q.8 Give output of following clipper circuits when input to all circuits is a sinusoidal wave of peak voltage Vm

Q.9 Give output of following clamper circuit a square wave input having maximum voltage Vm.

R

VVO

––

+ +

(c)

Vi

R

VVO

––

+ +

(c)

Vi

R

V

VO

––

+ +

(d)

Vi

R

V

VO

––

+ +

(d)

Vi

R

V1

VO

––

+ +

(e)

Vi

V2

R

V1

VO

––

+ +

(e)

Vi

V2

RVi VO

––

+ +

(a)

(

CR

Vi VO

––

+ +

(a)

(

C RVi VO

––

+ +

(b)

(

CR

Vi VO

––

+ +

(b)

(

C

RVi VO

––

+ +

(c)

(

C

—

RVi VO

––

+ +

(c)

(

C

—

RVi VO

––

+ +

(d)

(

C

—

RVi VO

––

+ +

(d)

(

C

—

Q.10 Give output of following clamper circuit a square wave input having maximum voltage Vm.

RVi VO

––

+ +

(f)

(

C

—V1

RVi VO

––

+ +

(f)

(

C

—V1

RVi VO

––

+ +

(e)

(C

—V1

RVi VO

––

+ +

(e)

(C

—V1

Q. 1) TUTORIAL NO. – 3Q1. Draw the V-I characteristics of Zener diode and also explain avalanche breakdown and zener

breakdown. Q.2 List the application of an LED. In what respect is an led different from an ordinary PN

junction diode?Q.3 Show that Zener diode can be used as a voltage regulator.Q.4 What is Schottky diode? Why it is also called hot carrier diode? How does it differ in

construction from a normal P-N junction? Give its working, characteristics and application.

Q.5 For the zener diode network, determine VL, VR, IZ and PZ. Ans.: (a) 1z = 0A

PZ = 0WVR = 7.27 V

VL =8.73V(b) VL = 10V

VR= 6 VIL=3.33 mA

IR = 6mAIZ=1.67 mA

Pz = 26.7 mW

Q.6 Repeat above question with RL=3K

Q.7 Determine VL, VR, IZ and PZ

Vi=16V

VR

1 K IZ

RL=1.2K

PZM=30mW

Vz=10V

Vi=16V

VR

1 K IZ

RL=1.2K

PZM=30mW

Vz=10V

VR

1 K IZ

RL=1.2K

PZM=30mW

Vz=10V

(Ans. : (a) 8.73V, 7.27V, oA, oW (b) 10V, 6V, 2.67 mA, 26.7 mW) Q.8 Repeat above question with RL= 3 k

Q.9 (a) Find range of RL and IL that will maintain VL at 10V. b) Determine maximum voltage rating of diode.

(Ans. : 250 -1.25 K mW)

Q.10 Determine range of Vi that will maintain zener diode in on state.

IZ

1 K

+ VR —

16V

Vz =10V, Pzm = 30 mu)

RL

1.2 K

VL

IZ

1 K

+ VR —

16V

Vz =10V, Pzm = 30 mu)

RL

1.2 K

VL

IZ

1 k

Vz =10V, Izm = 32 mA

RL

VL

IL

Vi=50V

-

+ +

–

IZ

1 k

Vz =10V, Izm = 32 mA

RL

VL

IL

Vi=50V

-

+ +

–

(Ans. : 23.67 V – 36.87V)

IZ

220 IR

Vi

–

+

RL

IL

1.2 K

VZ= 20 V, I2m = 60 mA

+

–

VL

IZ

220 IR

Vi

–

+

RL

IL

1.2 K

VZ= 20 V, I2m = 60 mA

+

–

VL

TUTORIAL NO. – 4Q.1 a) Define α and β of a transistor and derive the relationship between them. b) Differentiate between ICO and ICBO. What is the effect of temperature on ICBO.

Q.2 Determine the d.c. bias voltage VCE and the current Ic for the voltage divider configuration.Ans. : VCE = 12.22 V

Q.3 Determine the quiescent levels of ICQ and VCEQ for the network shown below Ans.: ICQ = 1.07 mA

VCEQ = 3.69 V

22 V

RC=10 K

=140

RE=1.5 K

R1= 39 K

R2= 3.9 K

I/P

22 V

RC=10 K

=140

RE=1.5 K

R1= 39 K

R2= 3.9 K

I/P

Vcc = 10V

Rc = 4.7 K

RE = 1.2 K

= 90

250 K

RB

Vi

Vcc = 10V

Rc = 4.7 K

RE = 1.2 K

= 90

250 K

RB

Vi

Q.4 For the network shown in problem (2) Determine

(a) ICQ & VCEQ2 Assume : VCC = 20 V Ans. : ICQ = 1.86 mA, (b) Find VB, Vc, VE & VBC Rc = 4.7 K, RB = 680 K

RE = 0, β = 120VCEQ = 11.26 V,VB = 0.78V, VC = 11.26V, VE = 0, VBC = -10.56V

Q.5 Determine Vc &VB for the network. Ans. : VC = - 4.448 V

VB= -8.3V

Q.6 Determine VCEQ and IE for the network Ans. : VCEQ= 11.68 V

IE = 4.16 mA

1.2 K

100 K

Vi

RB

VEE = -9V

= 45

Vo

1.2 K

100 K

Vi

RB

VEE = -9V

= 45

Vo

240 K

Vi

RB

VEE = -20V

= 90

Vo240 K

Vi

RB

VEE = -20V

= 90

Vo

Q.7 Determine Vc & VB for the network .Ans. : VC = 8.53 V

VB= -11.59 V

Q.8 a) In the circuits shown fig (a), determine whether or not the transistor is in saturation and find IB and IC.

b) Repeat with 2k emitter resistance added as given in fig (b).

2.2 K

Vi

VCC = 20V

= 120

Vo

8.2 K 2.7 K

1.8 K2.2 K

Vi

VCC = 20V

= 120

Vo

8.2 K 2.7 K

1.8 K

Hint. : Find IB by calculation and final (IB)min = or . If (IB)calculated > (IB) min (IB calculate =

0.084 mA, IBmin = 0.033 mA) then transistor will be in saturation

(b) Ic = 1.71 mA, IB = 0.0171 mA, Vqb = 0.72

Q.9 Determine the following for the fixed bias configuration. (a) IBQ, ICQ Ans.:IBQ = 47.08 A(b) VCEQ ICQ = 2.35 mA(c) VB & Vc VCEQ = 6.83 V

VB = 0.7V, VC = 6.83V(d) VBC VBC = - 6.13 V

3 K

RB= 50 K

5V

hfe = 100

10 V

(a)

3 K

RB= 50 K

5V

hfe = 100

10 V

(a)

3 K

50 K

5V

= 100

10 V

(b)

RE = 2K

3 K

50 K

5V

= 100

10 V

(b)

RE = 2K

=50 VCE

+

-

Rc = 2.2 K240 KRB

O/p

I/p

IB

Ic

Vcc= +10V

=50 VCE

+

-

Rc = 2.2 K240 KRB

O/p

I/p

IB

Ic

Vcc= +10V

Q.910 Repeat the above problem when a resistance RE=1 Kis attached between emitter and ground.

TUTORIAL NO. – 5

Q.1 What is the need for biasing the transistor? Briefly explain the reason for keeping the operating point of a transistor as fixed.

Q.2 (a)What is thermal runaway? How can it be avoided? (b) Define stability factor.

Q.3 A base resistor biasing circuit shown in fig 1. Determine (i) IC and VCE neglect VBE and β = 60 (ii) if RB is changed to 200 k

Q.4 Calculate collector current and collector to emitter voltage for the circuit shown in fig. (2) Q.5 Calculate the coordinates of operating point in fixed biasing circuit shown in fig(3).

Given RB=120K, RC=1K. If transistor is replaced by another having β = 150, what will be new coordinates of operating point.

Q.6 It is desired to set the operating point using feedback resistor method of biasing at I C = 1mA, VCE = 8V. What will be value of RC and RB, VCC = 12V, VBE = 0.3 V and β= 100.

(i) What will be the new operating point if β changes to 150 assuming all other values of the circuit to be same.

Q.7 Calculate the value of RB in fig.(4) so that operating point is fixed at IC = 6.4 mA and VCE = 3V; β = 80.

Q.8 A voltage divider biasing circuit is shown in fig. (5). Determine emitter current, collector emitter voltage and collector voltage.

Q.9 A transistor biased by potential divider and emitter resistance biasing has its zero operating point fixed at 2 mA, 6 V. If VCC = 15 V, RE=1 k, R2 = 10 k and VBE = 0.3 V. Find value of RC and R1.

Q.10 In a CE germanium transistor amplifier circuit the bias is provided by self bias. The various parameters are VCC = 16V, RC = 3k, RE = 2k, R1 = 56 k, R2 = 20 k and α = 0.985. Determine (a) the coordinates of the operating points (b) the stability factor S.

12 VVCC

Rc

Figure - 3

RB

=60

VCC

10 V

Rc

Figure - 4

RB

=80

250

+

–

RE 5000

100 F

12 VVCC

Rc

Figure - 3

RB

=60

12 VVCC

Rc

Figure - 3

RB

=60

VCC

Rc

Figure - 3

RB

=60

VCC

10 V

Rc

Figure - 4

RB

=80

250

+

–

RE 5000

100 F

VCC

10 V

Rc

Figure - 4

RB

=80

250

+

–

RE 5000

VCC

10 V

Rc

Figure - 4

RB

=80

250

+

–

RE 5000

100 F

VCC

18 V

Rc

Figure - 5

R1

RE

2K

1 kR2

2 k

7 k

RE

1 k

VCC

+15V

Rc

Figure - 6

R1

R2

10 k

VCC

18 V

Rc

Figure - 5

R1

RE

2K

1 kR2

2 k

7 k

VCC

18 V

Rc

Figure - 5

R1

RE

2K

1 kR2

2 k

7 k

RE

1 k

VCC

+15V

Rc

Figure - 6

R1

R2

10 k

RE

1 k

VCC

+15V

Rc

Figure - 6

R1

R2

10 k

3 KRc

VCC

Figure - 1

300 KRB

+6 V +12 V

VBB

2 K

VCC

6 V

Rc

Figure - 2

200 KRB

=50

3 KRc

VCC

Figure - 1

300 KRB

+6 V +12 V

VBB

3 KRc

VCC

Figure - 1

300 KRB

+6 V +12 V

VBB

2 K

VCC

6 V

Rc

Figure - 2

200 KRB

=50

Rc

Figure - 2

200 KRB

=50

TUTORIAL NO. – 6

Q. 1) For the circuit shown, prove that the stability factor S is given by

Q. 2) In the biasing with feedback resistor method, a silicon transistor with feedback resistor is used. The operating point is at 7V, 1mA and Vcc = 12V, Assume β= 100. Determine (a) the value of RB (b) stability factor and (c) what will be the new operating point if β= 50 with all other circuit values are same.

Ans. : RB = 630 K , S=56.5(C) Ic = 0.642 mA, VCE = 8.79 V

+

Vcc

Vin

RB

RC

Vout

VBE –

IB

+

Vcc

Vin

RB

RC

Vout

VBE –

IB

+

Vcc

Vin

RB

RC

Vout

VBE –

IB

Ic

+

Vcc

Vin

RB

RC

Vout

VBE –

IB

Ic

Q. 3) In a CE germanium transistor amplifier circuit, the bias is provided by self bias, i.e. emitter resistor and potential divider arrangement, the various parameters are Vcc = 16V, Rc = 3K, RE=2K, R1=56 K, R2=20K and α = 0.985. Determine (a) co-ordinates of the operating points (b) the stability factor S.

Ans. : (a)Ic= 1.73 mA, VCE=7.35V (b) S=7.537

Q. 4) A voltage divider bias circuit is designed to establish the Q-point at VCE = 12V, Ic = 2mA and stability factor S< 5.1. If Vcc = 24V, VBE = 0.7 V, β=50 and Rc = 47K. Determine the values of RE, R1, R2. (Ans. : RE = 1.3 K, R1 = 6.4K, R2=6.5K)

Q. 5) For the two – battery transistor circuit shown, prove that the stabilization factors is given by –

Q. 6) Determine the stability factor S for the circuit shown.

Rc

RB

B

RE

+– V1

+

– V2

Rc

RB

B

RE

+– V1

+

– V2

R1

R2 Ro

B

Vcc

RcR1

R2 Ro

B

Vcc

Rc

Q.7 The transistor amplifier shown uses a transistor whose parameters are given (refer

book). Calculate AI = , Av, Avs, Ro, Ri.

RS

Va

RL

10 K

VS

Ii

Ri

R2

10 K

R1

100 K 5 K

Ie

–

+

Vo

RS

Va

RL

10 K

VS

Ii

Ri

R2

10 K

R1

100 K 5 K

Ie

–

+RS

Va

RL

10 K

VS

Ii

Ri

R2

10 K

R1

100 K 5 K

Ie

–

+

Vo

Q.8 For amplifier show calculate Ri, , Av, Avs and , = (for the transistor parameter

refer book)

Q.9 For the circuit shown is a two stage amplifier circuit CE – CC configuration. The transistor parameters at corresponding Q-point are

hie=2 hfe= 50 hre=6x10– 4 hoe = 25A/Vhic=2 hfc = -51 hrc = 1 hoc = 25A/V

Vcc

RL

VS

10 KI2

–

+

Vo

R1

200 KRS

10 K

Ri

+

–

iR

Vcc

RL

VS

10 KI2

–

+

Vo

R1

200 KRS

10 K

Ri

+

–

iR

Find the input, output impedances and individual, as well as overall, voltage and current gains.

Ai2= 45.3 Ri2 = 228.5 K, Av2 = 0.991Ai1= 45.5 Ri1 = 1.87 K, Av1 = 16.6

Avs = -75.3

TUTORIAL NO. – 7

Vcc

VS

–

+

Vo

RC1

+

–

5 K

Q1

RS

1 K

Q2

Rc2

5 K

Vcc

VS

–

+

Vo

RC1

+

–

5 K

Q1

RS

1 K

Q2

Rc2

5 K

Q. 1) (a) In terms of h-parameters and the source resistance, derive the equation for output admittance. (b) Find (i) Avs in terms of Av (ii) AIS in terms of AI.(c) Derive the expression for Av in terms of AI. (d) In terms of the h-parameter and the load impedance, derive the expression for (a) A I

and Ri. Q.2 Derive the equation for voltage gain, current gain, input impedance and out put

impedance for a BJT using h-parameter model for (a) CE configuration (b) CB configuration (c) CC configuration

Q.3. Draw the AC equivalent of a CE amplifier with fixed bias using h-parameter model. Q.4. Justify the validity of approximate hybrid model applicable in low frequency region.Q.5 What is Miller’s Theorem. Why it is used in hybrid models.Q.6 Which of the configurations (CB, CE, CC) has the (a) highest Ri, (b) Lowest Ri (c) Highest Ro (d) Lowest Ro (e) Lowest Av (f) highest Av. (g) Lowest AI (h) Highest AI.

(b) Draw the circuit of an emitter follower. List its three most important characteristics. Q.7 For the figure shown is connected as common emitter amplifier. If RL = 10K. and Rs =

1K . Find the various gains and input and output impedance.

Q.8 (a) Draw the equivalent circuit for CE & CC configurations subjected to the restriction that RL = 0, Show that the input impedance of the two circuits are same. (b) Draw the circuits for the CE & CC configurations subjected to the restriction that the input is open circuited. Show that output impedances of two are same.

Q.9 For any single transistor amplifier prove that RI = hi /(1-hr Av)

Two – port active

Network

(Transistor)

V1 V2

ZinYo

Rs I1 I2

IL

RLVs–

+ Two – port active

Network

(Transistor)

V1 V2

ZinYo

Rs I1 I2

IL

RLVs–

+

TUTORIAL NO. – 8

Q.1 Why a field effect transistor is called so? Also explain why BJT is bipolar device while FET is unipolar device.

Q.2 Define and explain the parameters transconductance gm, drain resistance rd and amplification factor of a JFET. Establish the relation between them.

Q.3 Explain how the transconductance of a JFET varies with drain current and gate voltage.Q.4 Show that if a JFET is operated at sufficiently low drain voltage, it behaves as a

resistance R given approximately by

Q. 5 Given IDSS = 9 mA and VP = -3.5 V, determine ID when(a) VGS = 0V(b) VGS = -2V(c) VGS = -3.5V(d) VGS = -5V

Q.6 Determine the value of Rs required to self bias an N channel JFET with IDSS = 50 mA, Vp = -10 V and VGSQ = 5 V.

(Ans.: Rs = 400 )

Q.7 In an N channel JFET biased by voltage divider method, determine value of R S to give operating point ID = 4mA & VDS = 8V. Given VDD = 25 V, Rg1 = 1.2 M. Rg2 = 0.6 IDSS = 12 mA, Vp = -4 V.

(Ans.: 2.5 K ) Q.8 For JFET biased in self bias configuration. Determine VDS and VGS.

Given VDD = 25 V, RD = 3 K , RS = 400 and ID = 2mA

(Ans.: 18.2 V, -0.8 V) Q.9 A common source FET amplifier uses load resistance RD = 100 K and an unbypassed

resistor RS is the source circuit. The FET has drain resistance rd = 200 k and = 20. Compute the voltage gain and output impedance Ro, for the following values of Rs (i) 2 k (ii) 10 k and (iii) 20 k

(Ans.: -5. 847, -3.921, -2.777)

Q.10 A CD amplifier uses FET having rd = 300 k and = 15. Compute (a) the output impedance and (b) the voltage gain for the following values of load resistance Rs (i) 100 k (ii) 300 K

(Ans.: 7895, 0.882)

TUTORIAL NO. – 9

Q.1 What do you understand by feedback in amplifiers? Explain the terms feedback factor and open loop gain.

Q.2 What are the advantages and disadvantages of positive and negative feedback.

Q.3 How does negative feed back reduce distortion in an amplifier? The distortion in an amplifier is found to be 3% when the feed back ratio of negative feed back amplifier is 0.04. When the feed back is removed the distortion becomes 15%. Find the open loop gain and closed loop gain.

Q.4 prove that the voltage –series feed back with RS =0, AIf = AI

Q. 5 An amplifier with open loop voltage gain AV = 1000± 100 is available. It is necessary to have an amplifier whose voltage gain varies by no more than ± 0.1 percent.(a) find the reverse transmission factor β of the feed back network used.(b) find the gain with feed back.

Q.6 Explain the characteristics of Negative feed back amplifiers.

Q.7 Calculate the gain of negative feed back amplifier having A = -2000 and β = -1/10.

Q.8 Calculate the gain, input and output impedances of a voltage series feedback amplifier having A = -300, Ri = 1.5 kΩ, Ro = 50 kΩ and β = -1/15.

Q.9 Calculate the gain with and without feed back for an FET amplifier for circuit values R 1 = 800 kΩ, R2 = 200 kΩ, Ro = 40 kΩ, RD = 8 kΩ and gm = 5000µS.

Q.10 An FET phase shift oscillator having gm = 6000 µS, rd = 36 kΩ and feed back resistor R = 12kΩ is to operate at 2.5 kHz. Select C for specified oscillator operation.

TUTORIAL NO. – 10

Q.1 How are amplifier classified based on the biasing condition?Q.2 Explain the following type of distortion in amplifiers

(i) harmonic distortion (ii) frequency distortion (iii) phase distortion

Q.3 Draw the circuit diagram of a push pull amplifier and explain its working.

Q.4 Prove that the efficiency of class B amplifier is 78.5%.Q. 5 For a class B amplifier providing a 20 V peak signal to a 16Ω load (speaker) and a power

supply of VCC = 30V, determine the input power , output power and the circuit efficiency.

Q.6 Calculate the harmonic distortion components for an output signal having fundamentalamplitude of 2.1 V second harmonic amplitude of 0.3 V third harmonic component of

0.1 V and the fourth harmonic component of 0.05V.

Q.7 For distortion reading of D2 = 0.15, D3 = 0.01 and D4 = 0.05 with I1 = 3.3 A and Rc = 4Ω, calculate the total harmonic distortion fundamental power component and total power.

Q.8 Calculate the gain, input and output impedances of a voltage series feedback amplifier having A = -300, Ri = 1.5 kΩ, Ro = 50 kΩ and β = -1/15.

Q.9 Calculate the gain with and without feed back for an FET amplifier for circuit values R 1 = 800 kΩ, R2 = 200 kΩ, Ro = 40 kΩ, RD = 8 kΩ and gm = 5000µS.

Q.10 An FET phase shift oscillator having gm = 6000 µS, rd = 36 kΩ and feed back resistor R = 12kΩ is to operate at 2.5 kHz. Select C for specified oscillator operation.