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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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