cls jeead-15-16 xii phy target-7 set-1 chapter-14

8/19/2019 Cls Jeead-15-16 Xii Phy Target-7 Set-1 Chapter-14

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SECTION - ASchool/Board Exam. Type Questions

Very Short Answer Type Questions :

1. How does the dc current gain of a transistor change, if the width of the base region is increased?

Sol. Decreases

2. Why is the conductivity of n-type semiconductor greater than that of the p -type semiconductor even when both

of these have same level of doping?

Sol. Movement of electron is easier 3. Two semiconductor materials X and Y shown in the given figure are made by doping germanium crystal with

indium and arsenic respectively. The two are joined end to end and connected to a battery as shown. Will

the junction be forward biased or reverse biased?

X Y

Sol. Reverse biased

4. Name the two factors on which electrical conductivity of a pure semiconductor at a given temperature depends.

Sol. (A) Bond strength (B) Forbidden gap

5. C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge are intrinsic semiconductors?

Sol. Larger forbidden gap in C

6. In a p -n junction, the barrier potential is 0.5 V and its width is 1 micron. What is the magnitude and direction

of the electric field of the barrier?

Sol. 5 10 5 V/m from N to P side

Solutions

Chapter 14

Semiconductor Electronics :Materials, Devices and Simple Circuits


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108 Semiconductor Electronics : Materials, Devices and Simple Circuits Solution of Assignment (Set-1)


7. Figure shows reverse bias current, under different illumination intensities I 1 , I 2 , I 3 and I 4 , for a given photo diode.Arrange the intensities in decreasing order of magnitude.

I ( A)

V

I 1

I 2I

3

I 4

Sol. I 4 > I 3 > I 2 > I 1

8. Which of the two diodes D1 and D2 in the given figures (i) forward biased (ii) reverse biased?

– 8 V D 1

D2

– 8 V

Sol. D 1 : Reverse biased

D2 : Forward biased

9. In n-type semiconductor electrons are majority carriers, even though it is electrically neutral. Why?

Sol. Because intrinsic semiconductor and the doping material, both are neutral

10. If the emitter and base regions of a transistor have same doping concentration, state how collector current willchange?

Sol. Will decrease

11. How does doping affect the conductivity of a semiconductor?

Sol. Increases

12. How does the energy gap of an intrinsic semiconductor vary when doped with a trivalent impurity?

Sol. Some allowed energy levels are produced, situated in the forbidden gap slightly above the valence band.

Short Answer Type Questions :

13. What is a solar cell? How does it work? Give its one use.

Sol. Solar cell is a p -n junction device which converts solar energy into electrical energy. When photons of light

energy h > E g fall at the junction, electron-hole pairs are generated in the depletion layer, which are collected

at the two sides of the junction, giving rise to a photo voltage between the top and bottom metal electrodes.

They are used for charging storage batteries in day time.

14. The ratio of number density of free electrons to holes, eh

n

n, for three different materials, A, B and C are equal

to one, less than one and more than one respectively. Name the type of semiconductor to which A, B and

C belong. Draw labelled energy band diagrams for three materials.


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109Solution of Assignment (Set-1) Semiconductor Electronics : Materials, Devices and Simple Circuits


Sol. A : Intrinsic semiconductor

B : P-type semiconductor

C : N-type semiconductor

For energy band diagrams refer to theory.

15. The diagram shows a piece of pure semiconductor, S in series with a variable resistor R , and a source of constant voltage V . Would you increase or decrease the value of R to keep the reading of ammeter constant,when semiconductor is heated? Give reason.

A

R

S

Sol. When semiconductor is heated, its resistance decreases and hence R has to be increased to keep the current

constant.16. Determine the currents through the resistanced ‘ R ’ of the circuits (i) and (ii), when similar diodes are connected

as shown in the figure.

30 3 V

D2

D1

30 3 V

D2

D1

Sol. (i) 3

Current30

V = 0.1 A as the diodes are forward biased.

(ii) No current will flow as diode D2 is reverse-biased.

17. With the help of a diagram, show the biasing of a light emitting diode (LED). Give its two advantages over conventional incandescent lamps.

Sol. LED is forward biased

PP

N

Advantage over incandescent lamp is that LED has less power and low operational voltage. Also, LED hasfast action and requires no warm up time.

18. An n- p -n transistor is connected in common emitter configuration in which collector voltage is 8 V. The voltagedrop across load resistance of 800 connected in the collector circuit is 0.8 V. If the current amplificationfactor is 25, determine (i) collector emitter voltage and (ii) base current

Sol. V CE + 0.8 = 8 V CE = 7.2 volt

B 0.8 / 800I25

V = 40 A.


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19. The figure shows the V -I characteristic of a semiconductor diode.

10

20

30

100

80

6040

20

0.2 0.4 0.6 0.8 1.0

100 80 60 40 20

V B

I (mA)

I ( A)

V

(i) Identify the semiconductor diode used.

(ii) Draw the circuit diagram to obtain the given characteristic of this device.

Sol. (i) Since the knee voltage is around 0.6V so it’s a silicon semiconductor.

(ii) For forward bias

V+ –

A

+

–

+ –

For reverse bias

V +–

A+

–

20. X

Y

An ac signal is fed into two circuits X and Y and the corresponding output in the two cases have the waveformsshown in the figure. Name the circuit X and Y . Also draw their detailed circuit diagrams.


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Sol. Circuit X is half wave rectifier and circuit Y is full wave rectifier.

Circuit for X Circuit for Y

Output

+

–

+

–

R

–

+

~ Half waverectified

Output

R

~

D1

D2

Full wave rectifiedoutput voltage

21. X

Y

Input ( )V i V

o (output)

The set-up, shown in the figure, can produce an a.c. output without any external input signal. Identify thecomponents X and Y of this set-up.

Sol. X : Transistor as an amplifier

Y : Feedback circuit.

22. A certain n- p -n transistor has the common emitter output characteristics as shown in the figure.

0 5 10 15 20 V

CC (volt)

678

I AC (m )

I b = 60 A

I b = 40 A

(i) Find the emitter current at V cc = 10 V and I b = 60 A.

(ii) Find at this point.

Sol. (i) I E = I b + I c = 60 A + 7mA = 7.06 mA

(ii) =c

b

I

I =7

60

mA

A 117

23. For the given combination of gates, find the values of outputs Y 1 and Y 2 in the table given below. Identify thegates G 1 and G 2 .

D

ABC

G2

G1

A B C D

0

1

0

1

0

0 Y 2

Y1

Sol. G 1 is OR gate and G 2 is AND gate.

Y1 is O and Y 2 is O.


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24. In a p -n- p transistor circuit, the collector current is 10 mA. If 90% of the holes reach the collector, find emitter and base current.

Sol.10

10mA 1.1mA90

bI

11.1mAE b c I I I

25. In a transistor the base current is changed by 10 A. This results in a change of 0.01 V in base to emitter voltage and a change of 1 mA in the collector current. Find (i) the current gain “ ac ” and (ii) Transconductance“g m”

Sol. (i)1mA

10010 A

c ac

b

I

I

(ii)1mA

0.01Vc

mb

I g

I = 10 –5 A/V

26. Two amplifiers are connected one after the other in series. The first amplifier has a voltage gain of 10 and thesecond has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output a.c. signal.

Sol. Gains get multiplied therefore output ac signals 0.01 10 20 = 2 volt.27. The inputs A and B are inverted by using two NOT gates and their outputs are fed to the NOR gate as shown

in the figure. Identify the logic gate of the complete circuit so obtained. Give its symbol and truth table.

A

B

Y

Sol. The complete circuit is equivalent to AND gate. It’s symbol is

A

B Y

Truth Table is

A B Y

0 0 0

0 1 01 0 0

1 1 1

28. In the given circuit, a voltmeter V is connected across a bulb B. What changes would occur in bulb B andvoltmeter V if the resistor R is increased in value? Give reason for your answer.

6 V

BV

R

Sol. Here emitter base junction of n-p-n transistor is forward biased with a battery and resistance R . If the valueof R is increased, then emitter current I e will decrease. Hence the collector current will also decrease. Dueto which the bulb will glow ‘less bright’.

Voltmeter V measures potential difference across bulb B. Due to decrease in collector current, the potentialdifference across bulb B decreases, hence the reading of voltmeter will decrease.


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29. In only one of the circuits given below the bulb (B) lights. Which circuit is it? Give reason for your answer.

6 VB

R B

R

Sol. In first circuit the bulb will glow, because collector is connected to +ve polarity of 6V source and it will makethe collector current to flow.

30. In the figure shown, is (i) the emitter, and (ii) the collector forward or reverse biased?

B

0V

C

E

+ 1V

– 2V

Sol. (i) emitter is reverse biased

(ii) collector is forward biased.

Long Answer Type Questions :

31. (i) What do you mean by doping? Why is it done?

(ii) Explain how an intrinsic semiconductor can be converted into (a) n-type and (b) p-tpye semiconductor.

Give one example of each and their energy band diagram.Sol. (i) Doping is a process of deliberate addition of a desirable impurity in a pure semiconductor to modify its

properties in a controlled manner. It is done to increase the conductivity of semiconductor in controlledmanner at room temperature.

(ii) n -type semiconductor is formed when a pure semiconductor of silicon (Si) or germanium (Ge) is dopedwith a controlled amount of pentavalent atoms say arsenic (As) or phosphorus (P) or antimony (Sb) or bismuth (Bi). Its energy band diagram is

Donor Energy Level

C.B

V.B

E g .

p-type semiconductor is formed when a pure semiconductor of germanium (Ge) or silicon (Si) is dopedwith controlled amount of trivalent atoms say gallium (Ga). or indium (In) or boron (B) or aluminium (Al).It’s energy band diagram is

Acceptor Energy State

C.B

V.B

E g .


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32. (i) Consider an amplifier circuit using a transistor. The output power is several times greater than the inputpower. Where does the extra power come form?

(ii) A load resistor of 2 k is connected in the collector branch of an amplifier circuit using a transistor incommon-emitter mode. The current gain = 50. The input resistance of the transistor is 0.50 k . If theinput current is changed by 50 A

(a) By what amount does the output voltage change?

(b) By what amount does the input voltage change?

(c) What is the power gain?

Sol. (i) The energy for extra power at the output is being supplied by the DC battery.

(ii) (a) Output voltage change = I i × × R L= 50 A 50 2 k = 5V

(b) Input voltage change = I i R i = 50 A 0.5 k = 25 mV

(c) Power Gain = Voltage Gain Current Gain

5V 50 10,00025mV

.

33. (i) In semiconductors, thermal collisions are responsible for taking a valance electron to the conduction band.Why does the number of conduction electrons not go on increasing with time as thermal collisionscontinuously take place?

(ii) In a p -n junction, the depletion region is 400 nm wide and an electric field of 5 × 10 5 V/m exists in it.

(a) Find the height of the potential barrier.

(b) What should be the minimum kinetic energy of a conduction electron which can diffuse from n-sideto the p -side?

Sol. (i) Presence of electrons in the conduction band saturates the further production of electron-hole pair.

(ii) (a) height of potential barrier

5 95 10 400 10 m 0.2VV

m

(b) Kinetic energy = qV

= 1.6 × 10 –19 C × 0.2 V

= 0.2 eV

34. (i) Explain why there is a very small current across the junction, when a p -n junction diode is reverse biased.Does it depend on the applied voltage?

(ii)

1

S o

l a r r a

d i a t i o n

i n t e n s

i t y

I

2 3h (in )eV

Solar spectrum is shown in the figure. Here h is Planck’s constant and is frequency. Why are siliconSi and GaAs materials preferred for manufacturing solar cells? Why CdS and CdSe are not used?


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Sol. (i) Reverse biasing current is due to minority carriers, which is small in number and hence the reverse currentis small.

Reverse current is voltage independent upto certain reverse bias voltage, known as breakdown voltage.

(ii) In the solar spectrum the maxima is near 1.5 eV. For photoexcitation h > E g. Hence, semiconductor with band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has E g ~ 1.1eV while for GaAs it is ~ 1.5 3 eV. Infact, GaAs is better (in spite of its higher band gap) than Si becauseof its relatively higher absorption coefficient.

If we choose materials like CdS or CdSe ( E g ~ 2.4eV), we can use only the high energy component of the solar energy for photoconversion and a significant part of energy will be of use.

35. (i) Emitter base and collector base of a transistor are both reverse biased. What do you expect?

(ii) Explain why the emitter base junction is forward biased and the collector base is reverse biased in atransitor.

(iii) A transistor is working in CE mode. Current gain of the transistor is 69. If emitter current is 7 mA, Findbase current.

Sol. (i) If the emitter-base junction will be reverse biased then no emitter current will be produced and transistor will not work.

(ii) When emitter base junction is forward biased then only emitter current will start. The collector is reversebiased so that the current carrier emitted by the emitter is pulled out of collector.

(iii) I E = I B + I C

or, 7mA = I B + 69I B

7mA0.1mA.

70BI

SECTION - B

Model Test Paper

Very Short Answer Type Questions :

1. In n-type semiconductor number of free electrons is greater than the number of holes. Does it have negativecharge?

Sol. No. n-tye semiconductor is neutral

2. In a transistor, doping level in base is increased slightly. How will it affect (i) Collector current and (ii) Basecurrent?

Sol. (i) Collector current will be reduced

(ii) Base current will increase

3. Draw the logic circuit of AND gate and write its truth table.

Sol. AB

Y

A B Y

0 0 0

0 1 0

1 0 01 1 1


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4. State the reason, why GaAs is most commonly used in making of a solar cell.

Sol. GA as has relatively higher absorption coefficient.

5. Write the order of energy gap of silicon.

Sol. 1.1 eV


6. Draw the circuit diagram of an illuminated photo diode in reverse bias. How is photodiode used to measurelight intensity?

Sol. h

p side n side A

By measuring the change in the conductance (or resistance) of the semiconductor, one can measure theintensity of the optical signal.

7. Name the semiconductor device that can be used to regulate an unregulated dc power supply. With the helpof I -V characteristic of this device, explain its working principle.

Sol. Zener diode.

Regulatedvoltage

( )V z

R

I L

R L U

n r e g u

l a t e d v o

l t a g e

( V )

Load

mA

Forwardbias

Volt

A

V Z

Reversebiase

When unregulated voltage exceeds V Z the extra current goes through zener diode and load gets only regulatedvoltage V Z .

8. Draw the output waveform at x , using the given input A, B for the logic circuit shown below. Also identify thegate.

A X

B

A

B

(Input)


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Sol. This is AND gate. Output wave form at X will be

X

9. How is n-type semiconductor formed? Name the major charge carriers in it. Draw the energy band diagramof an n-type semiconductors.

Sol. n -type semiconductor is formed by doping pure semiconductors (S i or G e ) with pentavalent atoms like P, Asetc. Majority carriers are electrons in it. Its energy band diagram is

Donor Energy Level

C.B

V.B

E g .

10. The potential difference across the collector of a transistor, used in common emitter mode its 1.5 V, with thecollector resistance of 3 k . Find (i) the emitter and (ii) the base current, if the dc current gain of the transistor is 50.

Sol.1.5V

3kC I = 0.5 mA

I B =C I

=0.5mA

50 = 10–2 mA

I E = I B + I C = 0.51 mA

11. Which of the diodes is (i) forward biased and (ii) reverse biased in the following circuits?

– 10 V

– 8 V

+3 V

+4 V+8 V

Sol. First two circuits are reverse biased.

Last two circuits are forward biased

12. Determine the current through 20 resistance.

20

2 V

40


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Sol. 40 resistor is reverse biased, so current will flow through only 20 .

I =2V

20 = 0.1A

13. Give two advantages of LED over conventional incandescent lamps.

Sol. (a) LED has less power and low operational voltage

(b) LED has fast action and requires no warm up time


14. A semiconductor has equal electron and hole concentrations of 2 × 10 8 m –3 . On doping with a certain impurity,the hole concentration increases to 4 × 10 10 m –3 .

(i) What type of semiconductor is obtained on doping?

(ii) Calculate the new electron concentration of the semiconductor.

(iii) How does the energy gap vary with doping?Sol. (i) P -type

(ii)

282

6 3

10

2 1010 m

4 10

i e

h

nn

n

(iii) New acceptor energy level is created close to valence band in the forbidden gap region.

15. In a transistor a change of base current by 20 A, results in a change of 0.02 V in base to emitter voltageand a change of 2 mA in the collector current. Find (i) ac , (ii) transconductance of the transistor (iii) voltagegain of the transistor when used as a common emitter amplifier with a load resistance of 5 k .

Sol. (i)2mA

10020 A

C ac

B

I

I

(ii) Transconductance2mA

0.1 A/V0.02V

C m

B

I g

V

(iii) Voltage gain = g m · R L = 0.1 5 k = 500

16. Draw the circuit diagram of a full-wave rectifier and briefly explain its working principle.

Sol. Full-wave rectifier

R L

Centre

A D

1

B D2

Centre-taptransformer

Tap


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W a v e f o r m

a t A

t

W a v e f o r m

a t B

t

O u t p u

t w a v e f o r m

( a c r o s s

) R

L

t Due to

D1

Due toD

2

Due toD

1

Due toD

2

Working : The voltages at any instant at A (input of diode D1 ) and B (input of diode D2 ) with respect to thecentre tap are out of phase with each other. Suppose the input voltage to A at any instant is positive. It isclear that, at that instant, voltage at B being out of phase will be negative as shown in figure. The diode D1gets forward biased and conducts (while D2 is not conducting). Hence, during this positive half cycle we getan output current (and a consequent output voltage across the load resistor R L) as shown in last figure. Atanother instant, when the voltage at A becomes negative then the voltage at B would be +ve. Hence the diodeD1 does not conduct but the diode D2 conducts giving an output current and output voltage (across R L) duringthe negative half cycle of the input ac . Thus, we get output voltage during the +ve as well as the –ve half of the cycle (or in other words, during the full wave). This is the working of full-wave rectifier.

Long Answer Type Questions :17. Explain the formation of the depletion region for a p -n junction. How does the width of this region change when

the junction is (i) forward-biased (ii) reverse-biased (iii) How does an increase in the doping concentration affectthe width of the depletion region?

Sol. Formation of depletion region

Suppose a p -n junction has just been formed. On the n-side, there are more electrons while the number of holes on the p-side is larger. Because of this concentration gradient, electrons from n-side will diffuse towardsthe p-side of the junction while holes from the p-side will go towards the n-side. On crossing the p-n boundary,these electrons and holes may collide with each other and recombine (or annihilate) since they have opposite

charges. These electrons/holes have come from donor or acceptor impurity atom cores. Hence, such donor or acceptor atoms will get depleted of their associated electrons or holes and subsequently will be left witha “charged ion core” in the layer near the junction boundary. Hence a layer called the “depletion layer” is formedat the junction. Note that on the n-side near the junction, there is a layer of charged donor atom cores (witheffective +ve charge) while on the p-side there are charged acceptor atom cores (with effective –ve charge).

p n

+ + + +

+ + +

+ + + +

+ + +

–

–

–

–

– – – –

– – – –

– – – –

– – – –

+

+

+

+


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In forward-bias, the thickness of depletion region decreases but in Reverse-bias the thickness increases.

The potential barrier (or field across the junction) and the depletion layer width (or junction width) depends uponthe doping concentration on the two sides. Suppose N A and N D are small. The diffusing electrons and holesacross the junction can move to reasonably large distances before suffering a collision with another hole or electron to be annihilated or recombined. Hence, junction width would be large. Obviously, the junction fieldwould be weak. On the other hand, if N

A and N

D are large the junction width would be small (and hence the

junction field would be strong). In this manner, we can obtain junctions showing different behaviour by simplychanging the doping levels.

18. (i) Draw the circuit diagram of a base-biased n- p -n transistor in CE configuration. Explain how this circuitis used to obtain the transfer characteristic ( V o – V i characteristic). How do we explain the working of atransistor as a switch using this characteristic?

(ii) The typical output characteristics ( I C – V CE ) of an n- p -n transistor in CE configuration is shown in thefigure. Calculate (a) the output resistance (b) the current amplification factor ac

C o

l l e c t o r c u r r e n

t ( ) i m A

I C

Collector to emitter voltage ( ) in voltsV CE

10

8

6

4

2

0

8.5

2 4 6 8 10 12 14 16

Base current ( ) I B

10 A

20 A

30 A

40 A

50 A

60 A

Sol. (i) Refer to text for base biased n-p-n transistor circuit and working of transistor as a switch

(ii) (a) The output resistance =CE

C

V

I

at I b = 60 A we have VCE = 2V and I C = 8mA

2resistance 250

8mA

V

(b)

8.5 7 A

60 50 AC

ac B

I

I

= 150.

cls jeead-15-16 xii phy target-7 set-1 chapter-14

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