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Semiconductor Electronics · Question 50

NEET 2013MCQEasy📈 Topic in 75% papers+4−1
1

The output (X) of the logic circuit shown in figure will be [figure]

  1. X = A . B
  2. X = A + B
  3. X = A . B
  4. X = A . B

Answer

(1) X = A . B

Solution

Output of the first NAND gate: Y1 = A . B

The second NAND gate has both inputs joined, so it works as a NOT gate:

X = Y1 . Y1 = Y1

X = A . B = A . B

The circuit works as an AND gate.

Answer: (1)

NEET 2013

Keep practising — Semiconductor Electronics

2

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be

  1. 13G
  2. 54G
  3. 23G
  4. 1.5G
Out of syllabusNEET 2013
3

The output from a NAND gate is divided into two in parallel and fed to another NAND gate. The resulting gate is a [figure]

  1. AND gate
  2. NOR gate
  3. OR gate
  4. NOT gate
NEET 2013 (Karnataka)
4

In an unbiased p-n junction, holes diffuse from the p-region to n-region because of

  1. The attraction of free electrons of n-region.
  2. The higher hole concentration in p-region than that in n-region.
  3. The higher concentration of electrons in the n-region than that in the p-region.
  4. The potential difference across the p-n junction.
NEET 2013 (Karnataka)
5

One way in which the operation of an n-p-n transistor differs from that of a p-n-p is

  1. The emitter junction injects minority carriers into the base region of the p-n-p
  2. The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n
  3. The emitter injects holes into the base of n-p-n
  4. The emitter junction is reverse biased in n-p-n
Out of syllabusNEET 2013 (Karnataka)
6

To get an output Y = 1 in given circuit which of the following input will be correct ? [figure]

  1. A 1, B 0, C 0
  2. A 1, B 0, C 1
  3. A 1, B 1, C 0
  4. A 0, B 1, C 0
AIPMT 2012 (Mains)
7

The input resistance of a silicon transistor is 100 . Base current is changed by 40 which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 k . The voltage gain of the amplifier is

  1. 2000
  2. 3000
  3. 4000
  4. 1000
Out of syllabusAIPMT 2012 (Mains)
8

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is [figure]

  1. 0.75 A
  2. zero
  3. 0.25 A
  4. 0.5 A
AIPMT 2012 (Prelims)
9

C and Si both have same lattice structure; having 4 bonding electrons in each. However, C is insulator whereas Si is intrinsic semiconductor. This is because

  1. In case of C the valence band is not completely filled at absolute zero temperature.
  2. In case of C the conduction band is partly filled even at absolute zero temperature.
  3. The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.
  4. The four bonding electrons in the case of C lie in the third orbit, whereas for Si they lie in the fourth orbit.
AIPMT 2012 (Prelims)
10

The figure shows a logic circuit with two inputs A and B and the output C. The voltage wave forms across A, B and C are as given. The logic circuit gate is [figure]

  1. OR gate
  2. NOR gate
  3. AND gate
  4. NAND gate
AIPMT 2012 (Prelims)
11

In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k is 2 V. If the base resistance is 1 k and the current amplification of the transistor is 100, the input signal voltage is

  1. 0.1 V
  2. 1.0 V
  3. 1 mV
  4. 10 mV
Out of syllabusAIPMT 2012 (Prelims)

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