Target Publications Pvt. Ltd. Chapter 19: Semiconductors

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Hints to Problems for Practice

1. Resistance in forward bias

Rf = VI

∆∆

= 3

0.012 10−×

= 5 Ω

2. Rr = VI

∆∆

= 6

6.8 4(13 6) 10−

−− ×

= 2.87

× 106 = 4 × 105Ω

3. Rr = VI

∆∆

= ( )62

13 8I 40 10−

−− ×

∴ I2 – (40 × 10–6) = 3

5250 10×

= 0.02×10−3

∴ I2 = (20 + 40) × 10−6 = 60 µA 4. Rs = 150 + 50 = 200Ω Rp = 200Ω // 200Ω = 100Ω

I = p

E 0.7R− = 6 0.7

100− = 5.3× 10–2 A

= 53 mA

5. Rd = VI

∆∆

= 3

0.0110 10−×

= 1 Ω 6. I = I1 + I2

I1 = Z

L

VR

= 5100

= 120

A = 50 mA

I = ZE VR−

= 10 580− = 1

16A = 62.5 mA

∴ I2 = I – I1 = 12.5 mA 7. For half wave rectifier, finput = foutput = 50 Hz For full wave rectifier, 2 finput = foutput = 120 Hz 8.

Voltage drop across resistor VR = E − VZ = 6 – 2 = 4 V Current through series resistor

IR = VR / R = 4200

= 20 mA.

The smallest value of load resistance for which stabilisation occurs is that which allows a negligible (≈ 0) zener current.

∴ Current thought the load resistance, IL = IR = 20 mA.

∴ Smallest Value of RL = Z

Z

VI

= 3

620 10−×

= 300 Ω 9. 0 c

(I ) = 1.412 mA

IC = ( )0 cI

2 ≈ 1 mA [∵ Irms = 0I

2]

β = C

B

II

= 3

6

1020 10

−

−× = 50

10. Ic = 90% IE

∴ IE = C100I90

= 10 109× ≈ 11 mA

11. β = C

B

II

= 3

6

3 10100 10

−

−

××

= 30

IE = IC + IB = (3 + 0.1) mA = 3.1 mA

α = 1β+β

= 3031

= 0.97

12. For C.E. transistor

βa.c. = C

B

II

∆∆

= 3

6

1050 10

−

−× = 20

∆IE = ∆IC + ∆IB = (1000 + 50)10−6 = 1050 × 10−6 A = 1050 µA

αa.c. = a.c.

a.c.1β+β

= 2021

13. βa.c. = C

B

II

∆∆

= 3

6

3 1020 10

−

−

××

= 150

19 SEMICONDUCTORS

E

RVZ

Target Publications Pvt. Ltd. Std. XII Sci.: Physics Numericals

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14. For C.E transistor:

i. βa.c. = C

B

II

∆∆

= 3

6

30 1030 10

−

−

××

= 100

ii. Ri = BE

B

VI

∆∆

= 3

6

30 1030 10

−

−

××

= 1000Ω

iii. gm = a.c.

iRβ

= 1001000

= 0.1

iv. VA = o

i

RR

× βa.c.

= 3

3

5 1010×

× 100 = 500 15. For C. B. transistor,

Current gain α = 1β+β

= 7576

α = C

E

II

⇒ IC = α IE = 7576

× 5

= 4.93 mA

16. α = 1β+β

= 100101

= 0.99

17. VA = o

i

RR

β = 350002050

× 50 = 853.66

18. βa.c. = ( )( )

o a.c.

i a.c.

VV

∴ ( )o a.c.V = 80 × 0.4 = 32 mV

19. α = C

E

II

∴ IE = 20.98

= 2.0408 mA

IB = IE – IC = 0.0408 mA = 40.8 µA 20. For a C. E. amplifier,

VA = o

i

RR

β = 52

× 20 = 50

Hints to Multiple Choice Questions

1. D2 in reverse bias will have infinite resistance and will not conduct

R = Resistance of D1 + 70 Ω + 200 Ω = 300Ω.

I = VR

= 12300

= 0.04 A

2. R = VI

= ( )D3

e.m.f . V6 10−

−×

= 3

4 0.46 10−

−×

= 600 Ω 3. Diode being in reverse biased mode will not

conduct so A2 will read 0.

I1 = 5V20Ω

= 0.25 A so A1 will read 0.25 A

4. VB > VA hence, p type of diode should be

connected towards B and n-type towards A.

I = B A

0

V V 1 ( 6)R R 50 200

− − − −=

+ + = 0.02 A = 20 mA

5. E = hν

ν = 19

34

2 1.6 106.62 10

−

−

× ××

= 4.8 × 1014 Hz

ν ≈ 5 × 1014 Hz. 6. Potential difference across resistor is V = −1 − (−3) = 2 V

∴ I = V 2VR 100

=Ω

= 20 mA.

7. Reverse resistance = 6

V 1I 0.5 10−

∆=

∆ ×=2×106 Ω.

8. Forward resistance = 3

V 0.7 0.6I (15 5) 10−

∆ −=

∆ − ×

= 3

0.110 10−×

= 10 Ω 9. R = effective

3

V 12 0.6=I 2×10−

− = 3

11.42 10−×

= 5.7×103 Ω 10. Band gap = 2.5 eV

∴ Corresponding wavelength λ = g

hcE

34 8

19

6.63 10 3 102.5 1.6 10

−

−

× × ×× ×

= 4960 Å

Nearest option is (C). 11. As nenh = 2

in

∴ ne = ( )216 32

i22 3

h

10 mnn 5 10 m

−

−=×

= 2 × 109 m−3

12. The intrinsic concentration of electron-hole pairs is given by,

2in = nenh

⇒ ne = 2i

h

nn

= ( )219

21

1010

= 1017/m3

Target Publications Pvt. Ltd. Chapter 19: Semiconductors

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13. Voltage across RL = 5 V

∴ I = 3L

V 5VR 10

=Ω

= 5 mA

14. For full wave rectifier f0 = 2 fi = 120 Hz. 15. For half wave rectifier f0 = fi = 80 Hz. 16. d. c. output Vo = Id × RL = 30 × 100 = 3000 mV = 3 V

17. RL = o3

d

V 6I 50 10−=

× = 120 Ω

18. VA = o

i

VV

⇒ V0 = 60 × 30 mV

= 1800 mV = 1.8 V

19. VA = β o

i

RR

= 40 800200

⎛ ⎞⎜ ⎟⎝ ⎠

= 160

20. α = 991 100β

=+ β

= 0.99

21. β = 50 / 51 50 51501 51 1151

α= = ×

−α − = 50

22. IC = 90% IE = 0.9 IE and IB = IE − IC = 0.1 IE = 1.2 mA. 24. ∆IB = ∆IE − ∆IC = 6.86 − 6.20 = 0.66 mA = 660 µA

25. β = 1α−α

∴ 1 1−

α β= 1 1−α

−α α

= 1 1− + αα

= 1

126. VA = o

i

RR

β

∴ β = 50 100600× = 50/6

PA = o

i

RR

β2 = 2600 50

100 6⎛ ⎞⎜ ⎟⎝ ⎠

= 416.67

Nearest answer is 420.

27. f = 1 LC2π

1

2

ff

= 2 21 1

1 L CL C

= 1/2 1/2

2 2

1 1

L C 2L 4CL C L C

⎛ ⎞ ×⎛ ⎞=⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠= (8)1/2 = 2 2

⇒ f2 = 1f2 2

⇒ f2 = f2 2

[∵ f1 = f]

[Note: Using shortcut (4)] 28. Voltage gain = Resistance gain × Current gain

o o

i i

V RV R

= × 50

Vo = 4000 1050500 1000

× × = 4 V 29. Af = A

1 A+ β

10 = A1 0.09A+

10(1 + 0.09 A) = A 10 = 0.1 A A = 100 30. As per Berkhausen criterion, Aβ = 1