LinkedIn Insight Solutions for Exercises of Semiconductor Devices - Grad Plus

Solutions for Exercises of Semiconductor Devices

1. Choose the correct option.

i) In a BJT, largest current flow occurs

(A) in the emitter

(B) in the collector

(C) in the base

(D) through CB junction

Answer: (A) in the emitter


ii) A series resistance is connected in the Zener diode circuit to

(A) Properly reverse bias the Zener

(B) Protect the Zener

(C) roperly forward bias the Zener

(D) Protect the load resistance

Answer: (B) Protect the Zener


iii) A LED emits visible light when its

(A) junction is reverse biased

(B) depletion region widens

(C) holes and electrons recombine

(D) junction becomes hot

Answer: (C) holes and electrons recombine


iv) Solar cell operates on the principle of:

(A) diffusion

(B) recombination

(C) photo voltaic action

(D) carrier flow

Answer: (D) carrier flow


v) A logic gate is an electronic circuit which:

(A) makes logical decisions

(B) allows electron flow only in one direction

(C) works using binary algebra

(D) alternates between 0 and 1 value

Answer: (C) works using binary algebra


2 Answer in brief.

i) Why is the base of a transistor made thin and is lightly doped?

 

11) The common-base DC current gain of a transistor is 0.967. If the emitter current is 10mA. What is the value of base current?

α = 0.967, Ig = 10mA = 10-3 × 10-2A = 10A,

IB = ?

$latex \alpha\;=\;\frac{I_C}{I_E}$

$latex \therefore\;I_C\;=\;\alpha I_E$

= 0.967× 10-2A

Now we have,

IE = IB + IC

∴ IB = (10-2) − (0.967 × 10-2)

∴ IB = (1 − 0.967) × 10-2

∴ IB = 0.033 × 10-2

∴ IB = 0.33 × 10-3 = 0.33mA


12) In a comman-base connection, a certain transistor has an emitter current of 10mA and collector current of 9.8 mA. Calculate the value of the base current.

IE = 10mA, IC = 9.8mA, IB =?

For transistor, we have,

IE = IB + IC

= 10 − 9.8

IE = IB + IC = 0.2mA


13) In a common-base connection, the emitter current is 6.28mA and collector current is 6.20 mA. Determine the common base DC current gain.

IE = 6.28mA, IC = 6.20mA, α =?

$latex \therefore\;\alpha\;=\;\frac{I_C}{I_E}$

$latex \therefore\;\alpha\;=\;\frac{6.20}{6.28}$

$latex \therefore\;\alpha\;=\;0.9872$


 

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