InterviewSolution

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1.	The forbidden energy gap in germanium is 0.7 eV. The wavelength whose absorption is done by germanium is :(a) 35000 Å(b) 17700 Å(c) 25000 Å(d) 51600 Å
Answer» The answer is (b) 17700 Å

2.	In insulators the forbidden energy gap between valence band and conduction band is of ………. value :(a) 1 eV(b) 0.6 eV(c) 0.1 eV(d) 0.01 eV
Answer» The answer is (b) 0.6 eV

3.	What will be the value of a for that transistor for which β = 19?
Answer» \(\alpha=\frac{\beta}{\beta+1}=\frac{19}{19+1}=\frac{19}{20}=0.95\)

4.	What is the direction of diffusion current in junction diode?
Answer» It flows from P region to N region.

5.	The diode shown in the figure is in which biasing?
Answer» The diode shown in the figure is reversed-biased (∵ V_N > V_P).

6.	Write down the relation between a and p (current amplification factor) of the transistor.
Answer» β = \(\frac{\alpha}{1-\alpha}\) or α = \(\frac{\beta}{\beta+1}\)

7.	The current amplification is 0.99 for any transistor in common-base configuration. Calculate the change in collector current when there is 5.0 milliampere change in emitter current. What will be the change in base current?
Answer» Current amplification α = 0.99 Change in emitter current ΔI = 5 mA ∵ α = \(\frac{\Delta I_{C}}{\Delta I_{E}}\) ∴ ΔI_C = α • ΔI_E = 0.99 × 5 = 4.95 mA ∵ ΔI_E = ΔI_E + ΔI_C => ΔI_B – ΔI_E – ΔI_C = 5 mA – 4.95 mA = 0.05 mA ∴ ΔI_B = 5 × 10^-5 A= 50 × 10^-6 A = 50μA

8.	For any ‘AND GATE’ the three inputs are A, B and C then the output Y will be;(a )Y = A . B + C(b) Y = A + B + C(c ) Y = A + B . C(d) Y = A . B . C
Answer» (d) Y = A . B . C

9.	If a NPN transistor is used as an amplifier then;(a) Electrons move from base to collector(b) Holes move from emitter to base(c) Holes move from base to emitter(d) Electrons move from emitter to base
Answer» (d) Electrons move from emitter to base

10.	In a transistor why is the base made thin in comparison to the emitter and collector?
Answer» The base of a transistor is kept very thin compared to the collector and emitter so that most of the carriers (carriers means electron in case of npn transistor and holes in case of pnp transistor) pass to the collector, injected from the emitter i.e, less number of electron-hole recombination take place and whole current emitted from emitter with a slight difference is passed to the collector.

11.	In a common emitter amplifier an increase of 50 A in base current causes a 1.0 mA increase in collector current. Calculate current gain β. What will be the change in emitter current ? Calculate α with the help of β.
Answer» Change in base current ΔI_E = 50 µA Change in collector current ΔI_C = 1.0 mA \(\therefore \beta=\frac{\Delta I_{C}}{\Delta I_{B}}=\frac{1.0 {mA}}{50 \mu {A}}=\frac{1 \times 10^{-3}}{50 \times 10^{-6}}=\frac{1000}{50}=20\) So, the change in emitter current ΔI_E = ΔI_B + ΔI_C = 50 µA + 1.0 mA = 50 µA + 1000 µA = 1050 µA = 1.050 mA Now, α = \(\frac{\beta}{1+\beta}=\frac{20}{20+1}=\frac{20}{21}\) = 0.9523 α = 0.95

12.	The Boolean equation for given circuit will be :
Answer» The answer is (c)

13.	Write down the name of that logic gate in which the output is 1 when all the inputs are 1.
Answer» The answer is AND Gate.

14.

Make truth table for OR gate.

Answer»

Truth table for OR gate:

A	B	Y = A + B
0	0	0
1	0	1
0	1	1
1	1	1