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This mock test of Electronic Devices 4 for Electronics and Communication Engineering (ECE) helps you for every Electronics and Communication Engineering (ECE) entrance exam.
This contains 10 Multiple Choice Questions for Electronics and Communication Engineering (ECE) Electronic Devices 4 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

For a single stage BJT common base amplifier.

Solution:

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

*Answer can only contain numeric values

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.

Solution:

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

QUESTION: 3

In a Bi-polar junction transistor the base width is 0.54 μm and base diffusion constant is D_{B} = 25 cm^{2}/sec. The Base transit time time is ________ × 10^{-10} sec

Solution:

The Base transit time

⇒ 0.5832 × 10^{-10}

*Answer can only contain numeric values

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.

Solution:

Applying KVL in collector to emitter loop

V_{CC} = R_{L}I_{C} + V_{CE
}

Now applying KVL in collector to base loop

I_{B}R_{B} + V_{BE} = V_{CC
}

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)

Solution:

Given: V_{EC} = 5V^{-}, this means pnp transistor is operating in the active node

*Answer can only contain numeric values

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

Solution:

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

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)

Solution:

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 × 10^{3} × 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

QUESTION: 8

An npn bipolar transistor having uniform doping of N_{E} = 10^{18} cm^{-3} N_{B} = 10^{16} cm^{-3} and N_{C} = 6 × 10^{15} 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 _____ × 10^{14} cm^{-3}

(Assume n_{i } = 1.5 × 10^{10}/cm^{3}, V_{t }= 25 mV)

Solution:

= 5.96 × 10^{14} cm^{-3}

QUESTION: 9

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

The transistor is operating in

Solution:

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

QUESTION: 10

In a silicon PNP transistor the mobility of charge carries is μ_{4} = 1300 cm^{2}/V-s and μ_{p} = 450 cm^{2}/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)

Solution:

Given μ_{n} = 1300 cm^{2}/V-s and μ_{p} = 450 cm^{2}/V-s

D_{p} = 450 × 0.0259

= 11.655 cm^{2}/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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