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Q. No.
In an \(\mathrm{n\text-}\)type semiconductor, which of the following statement is true?
1. Electrons are minority carriers and pentavalent atoms are dopants.
2. Holes are minority carriers and pentavalent atoms are dopants.
3. Holes are the majority carriers and trivalent atoms are dopants.
4. Electrons are the majority carriers and trivalent atoms are dopants.
Subtopic:  Types of Semiconductors |
 81%
Level 1: 80%+
AIPMT - 2013
Hints

The output \((X)\) of the logic circuit shown in the figure will be: 
  
1. \(X= \overline{A\cdot B}\)
2. \(X = A\cdot B\)
3. \(X= \overline{A+ B}\)
4. None of the above

Subtopic:  Logic gates |
 79%
Level 2: 60%+
AIPMT - 2013
Hints
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Transfer characteristics [output voltage (Vo) vs input voltage (Vi)] for a base biased transistor in CE configurations are as shown in the figure. For using the transistor as a switch, it is used:

1. In region III

2. Both in the region (I) and (III)

3. In region II

4. In region I

 78%
Level 2: 60%+
AIPMT - 2012
Hints
In a forward biasing of the p-n junction:
 
1. the positive terminal of the battery is connected to the p-side and the depletion region becomes thick.
2. the negative terminal of the battery is connected to the n-side and the depletion region becomes thin.
3. the positive terminal of the battery is connected to the n-side and the depletion region becomes thin.
4. the negative terminal of the battery is connected to the p-side and the depletion region becomes thick.

Subtopic:  PN junction |
 76%
Level 2: 60%+
AIPMT - 2011
Hints
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To get an output Y = 1 from the circuit shown below, the input must be:

 

1. A=0 B=1 C=0

2. A=0 B=0 C=1

3. A=1 B=0 C=1

4. A=1 B=0 C=0

Subtopic:  Logic gates |
 91%
Level 1: 80%+
AIPMT - 2010
Hints
A transistor is operated in a common-emitter configuration at Vc = 2 volt such that a change in the base current from 100 μA to 200 μA produces a change in the collector current from 5 mA to 10 mA. The current gain is:
1. 75
2. 100
3. 150
4. 50
 78%
Level 2: 60%+
AIPMT - 2009
Hints

A p-n photodiode is made of a material with a bandgap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly:

1. 10×1014 Hz

2. 5×1014 Hz

3. 1×1014 Hz

4. 20×1014 Hz

Subtopic:  Energy Band theory | Applications of PN junction |
 82%
Level 1: 80%+
AIPMT - 2008
Hints
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The circuit is equivalent to: 
     

1. AND gate
2. NAND gate
3. NOR gate
4. OR gate

Subtopic:  Logic gates |
 77%
Level 2: 60%+
AIPMT - 2008
Hints
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