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Semiconductor Electronics · Question 55

AIPMT 2012 (Mains)MCQEasy📈 Topic in 75% papers+4−1
1

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

Answer

(2) A 1, B 0, C 1

Solution

A and B are fed to an OR gate; its output and C are fed to an AND gate.

⇒ Y = (A + B) C

∵ The AND gate gives 1 only when both its inputs are 1

⇒ C = 1 and (A + B) = 1

Option (2): Y = (1 + 0) 1 = 1

Options (1), (3), (4) have C = 0 ⇒ Y = 0

Answer: (2)

AIPMT 2012 (Mains)

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2

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)
3

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)
4

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)
5

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)
6

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)
7

Transfer characteristics [output voltage (V0) vs input voltage (Vi)] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used [figure]

  1. in region III
  2. both in region (I) and (III)
  3. in region II
  4. in region I
Out of syllabusAIPMT 2012 (Prelims)
8

A Zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the diode is [figure]

  1. 5 mA
  2. 10 mA
  3. 15 mA
  4. 20 mA
AIPMT 2011 (Mains)
9

In the following figure, the diodes which are forward biased, are [figure]

  1. (A), (B) and (D)
  2. (C) only
  3. (C) and (A)
  4. (B) and (D)
AIPMT 2011 (Mains)
10

Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 1016 m-3. Doping by indium increases nh to 4.5 1022 m-3. The doped semiconductor is of

  1. p-type having electron concentration ne = 5 109 m-3
  2. n-type with electron concentration ne = 5 1022 m-3
  3. p-type with electron concentration ne = 2.5 1010 m-3
  4. n-type with electron concentration ne = 2.5 1023 m-3
AIPMT 2011 (Mains)
11

In forward biasing of the p-n junction

  1. the positive terminal of the battery is connected to p-side and the depletion region becomes thick.
  2. the positive terminal of the battery is connected to n-side and the depletion region becomes thin.
  3. the positive terminal of the battery is connected to n-side and the depletion region becomes thick.
  4. the positive terminal of the battery is connected to p-side and the depletion region becomes thin.
AIPMT 2011 (Prelims)

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