Semiconductor

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Semiconductor

Test1 Max. Marks = 40

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1. In forward bias, the width of potential barrier in a P-N junction diode

4 points

(a) Increases

(b) Decreases

(d) First increases then decreases

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2. In a PN-junction diode not connected to any circuit

4 points

(a) The potential is the same everywhere

(b) The P-type is a higher potential than the N-type side

(d) There is an electric field at the junction directed from the P-type side to the N-type side

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3. A potential barrier of 0.50 V exists across a P-N junction. If the depletion region is m wide, the intensity of the electric field in this region is

4 points

(a) 1 x 10⁶ V /m

(b) 1 x 10⁵ V /m

(d) 2 x 10⁶ V /m

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4. A common emitter amplifier is designed with NPN transistor (α = 0.99). The input impedance is 1 KΩ and load is 10 KΩ. The voltage gain will be

4 points

(a) 9.9

(b) 99

(d) 9900

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5. In a PNP transistor working as a common-base amplifier, current gain is 0.96 and emitter current is 7.2 mA. The base current is

4 points

(a) 0.4 mA

(b) 0.2 mA

(d) 0.35 mA

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4 points

(a)

(b)

(c)

(d)

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7.The combination of ‘NAND’ gates shown here under (figure) are equivalent to

4 points

(a) An OR gate and an AND gate respectively

(b) An AND gate and a NOT gate respectively

(d) An OR gate and a NOT gate respectively.

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8. The logic behind ‘NOR’ gate is that it gives

4 points

(a) High output when both the inputs are low

(b) Low output when both the inputs are low

(d) None of these

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9. How many NAND gates are used to form an AND gate

4 points

(a) 1

(b) 2

(d) 4

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10. What will be the input of A and B for the Boolean expression

4 points

(a) 0, 0

(b) 0, 1

(d) 1, 1

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