Test: Electronic Devices - 4 | 10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series Electronics and Communication Engineering (ECE)

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Test: Electronic Devices - 4 - Question 1

For a single stage BJT common base amplifier.

A.
Current gain, as well as voltage gain, can be greater than unity
B.
Current gain can be greater than unity but voltage gain is always less than unity
C.
Voltage gain can be greater than unity but current gain is always less than unity
D.
Current gain, as well as voltage gain, is always less than unity

Detailed Solution for Test: Electronic Devices - 4 - Question 1

∴ Voltage gain can be greater than unity but current gain is always less than unity.

*Answer can only contain numeric values

Test: Electronic Devices - 4 - Question 2

A bipolar transistor has an emitter current of 1 mA. The emitter injection efficiency is 0.99, base transit factor is 0.995 and depletion region recombination factor of 0.998. The base current flowing through the transistor is _____μA.

Detailed Solution for Test: Electronic Devices - 4 - Question 2

The common emitter current gain is given by:

α = γ ∗ β ∗ δ

= 0.99 × 0.995 × 0.998 = 0.983

Collector current

I_c = αI_E = 0.983 mA

Base current

I_B = I_E – I_C

= 17 μA

*Answer can only contain numeric values

Test: Electronic Devices - 4 - Question 3

In a Bi-polar junction transistor the base width is 0.54 μm and base diffusion constant is D_B = 25 cm²/sec. The Base transit time time is ________ × 10^-10 sec

Detailed Solution for Test: Electronic Devices - 4 - Question 3

The Base transit time

⇒ 0.5832 × 10^-10

*Answer can only contain numeric values

Test: Electronic Devices - 4 - Question 4

Find the value of bias resistor (in kΩ) if quiescent collector current and voltage value are 4.6 mA and 2.2 V. The transistor has DC gain 110, V_BE = 0.7 V and V_CC = 4.5 V.

Detailed Solution for Test: Electronic Devices - 4 - Question 4

Applying KVL in collector to emitter loop

V_CC = R_LI_C + V_CE

Now applying KVL in collector to base loop

I_BR_B + V_BE = V_CC

Test: Electronic Devices - 4 - Question 5

Consider two pnp bipolar junction transistors. For the first transistor when emitter to collector voltage is 5 V, V_EB is 0.85 V and emitter current is 10 A. (The β for this transistor is 15). Second BJT conducts with a collector current of 1 mA and V_EB = 0.70. The ratio of emitter-base junction area of the first transistor to the second transistor is ______.

(Assume KT = 26 mV)

A.
3.2
B.
15
C.
6.3
D.
29.3

Detailed Solution for Test: Electronic Devices - 4 - Question 5

Given: V_EC = 5V^-, this means pnp transistor is operating in the active node

*Answer can only contain numeric values

Test: Electronic Devices - 4 - Question 6

The leakage current of a transistor with usual notation are I_CEO = 410 μA, I_CBO = 5 μA, and I_B = 30μA. Calculate the I_C ________mA

Detailed Solution for Test: Electronic Devices - 4 - Question 6

I_CEO = 410 μA, I_CBO = 5 μA, I_B = 30 μA

I_C= βI_B + (1 + β) I_CBO

I_CEO = (1 + β) I_CBO

410 = (1 + β)5

β = 81

I_C = 81 × 30 + 82 × 5

I_C = 2.84 mA

Test: Electronic Devices - 4 - Question 7

For what value of current gain β , the given transistor will be in saturation

(Assume V_in = 5V, V_BE(SAT) = 0.8 V , V_CE(SAT) = 0.2V)

A.
40
B.
50
C.
35
D.
38

Detailed Solution for Test: Electronic Devices - 4 - Question 7

Apply kVL in base emitter loop

at saturation V_BE(sat) = 0.8

substituting the value of V_BE

I_B = 0.0525 mA

Apply Kvl , Form the collector to emitter

5 × 10³ × I_C+ V_CE = 12 V

= 2.36 mA

At the edge of saturation

Hence for β value greater than 45 the transistor will be in saturation

*Answer can only contain numeric values

Test: Electronic Devices - 4 - Question 8

An npn bipolar transistor having uniform doping of N_E = 10¹⁸ cm^-3 N_B = 10¹⁶ cm^-3 and N_C = 6 × 10¹⁵ cm^-3 is operating in the inverse-active mode with V_BE = -2V and V_BC = 0.6 V. The geometry of transistor is shown

The minority carrier concentration at x = x_B is _____ × 10¹⁴ cm^-3

(Assume n_i = 1.5 × 10¹⁰/cm³, V_t= 25 mV)

Detailed Solution for Test: Electronic Devices - 4 - Question 8

= 5.96 × 10¹⁴ cm^-3

Test: Electronic Devices - 4 - Question 9

The common emitter forward current gain of the transistor shown is β = 100
The transistor is operating in

A.
Saturation region
B.
Cut-off region
C.
Reverse active region
D.
Forward active region

Detailed Solution for Test: Electronic Devices - 4 - Question 9

Assume Transistor in Active region

20 – 2 (I_B + I_C) – 0.7 – 250 I_B – 10 = 0

20 – 2 × 101 I_B – 0.7 – 250 I_B – 10 = 0

10 – 452 I_B – 0.7 = 0

20 – 2 (I_B + I_C) - V_EC – 2I_C = 0

20 – 202 I_B - V_EC = 0

V_EC = 11.68 V

V_CB = V_CE + V_EB= -V_EC+ V_EB= - 11.68 + .7

V_CB = -10.98

∴ Collector base junction is Reverse Biased

∴ Transistor is operating in forward active region

Second Method

Assume transistor in saturation region

I_B min ≤ I_B

i) V_{EB (sat)} = 0.8 V

ii) V_{EC (sat)}= 0.2 V

I_C ≠ βI_B ----- Because it is valid only active region

20 – 2 (I_B + I_C) – 0.8 - 250 I_B – 10 = 0

20 – 2 I_B – 2I_C – 0.8 – 250 I_B – 10 = 0

10 – 252 I_B – 2I_C – 0.8 = 0

252 I_B + 2I_C= 9.2 -------- (1)

20 – 2 (I_B + I_C) – V_EC(sat)– 2I_C = 0

20 – 2I_B – 2I_C – 0.2 – 2I_C = 19.8 -------- (2)

By solving equation (1) and (2)

I_B = 0.00278 mA

I_C = 4.95 mA

∴ I_B(min)> I_B

∴ Transistor is operating in forward active region

Test: Electronic Devices - 4 - Question 10

In a silicon PNP transistor the mobility of charge carries is μ₄ = 1300 cm²/V-s and μ_p = 450 cm²/V-s and carrier life time τ_p = 0.10 μs. The most appropriate base width for effective transistor function is (Take T = 300° k)

A.
1.08 mm
B.
0.9 mm
C.
0.536 mm
D.
None of the above

Detailed Solution for Test: Electronic Devices - 4 - Question 10

Given μ_n = 1300 cm²/V-s and μ_p = 450 cm²/V-s

D_p = 450 × 0.0259

= 11.655 cm²/s

Base diffusion length = 1.08 mm

For injected holes from emitter to reach the collector they should not be recombined in base. Hence the base width should be very less compared to base diffusion length.

W ≪ L_P

In all options the base width is significant compared to base diffusion length.

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Test: Electronic Devices - 4

10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Electronic Devices - 4

For a single stage BJT common base amplifier.

A bipolar transistor has an emitter current of 1 mA. The emitter injection efficiency is 0.99, base transit factor is 0.995 and depletion region recombination factor of 0.998. The base current flowing through the transistor is _____μA.

In a Bi-polar junction transistor the base width is 0.54 μm and base diffusion constant is D_B = 25 cm²/sec. The Base transit time time is ________ × 10^-10 sec

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10 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | Test: Electronic Devices - 4

For a single stage BJT common base amplifier.

A bipolar transistor has an emitter current of 1 mA. The emitter injection efficiency is 0.99, base transit factor is 0.995 and depletion region recombination factor of 0.998. The base current flowing through the transistor is _____μA.

In a Bi-polar junction transistor the base width is 0.54 μm and base diffusion constant is DB = 25 cm2/sec. The Base transit time time is ________ × 10-10 sec

GATE ECE (Electronics) 2023 Mock Test Series

Up next

Top Courses for Electronics and Communication Engineering (ECE)

GATE ECE (Electronics) 2023 Mock Test Series

Top Courses for Electronics and Communication Engineering (ECE)

Up next

Download free EduRev App

In a Bi-polar junction transistor the base width is 0.54 μm and base diffusion constant is D_B = 25 cm²/sec. The Base transit time time is ________ × 10^-10 sec