1
MCQ (Single Correct Answer)

AIPMT 2015

The input signal given to a CE amplifier having a voltage gain of 150 is Vi = 2cos(15t + $${\pi \over 3}$$). The corresponding output signal will be
A
2cos$$\left( {15t + {{5\pi } \over 6}} \right)$$
B
300cos$$\left( {15t + {{4\pi } \over 3}} \right)$$
C
300cos$$\left( {15t + {{\pi } \over 3}} \right)$$
D
75cos$$\left( {15t + {{2\pi } \over 3}} \right)$$

Explanation

Given, Vi = 2cos(15t + $${\pi \over 3}$$)

and Voltage gain AV = 150

For CE transistor phase difference between input and output signal is $$\pi $$ = 180o

Using formula, AV = $${{{V_0}} \over {{V_i}}}$$

$$ \Rightarrow $$ V0 = AV $$ \times $$ Vi

= 150$$ \times $$2cos(15t + $${\pi \over 3}$$ + $$\pi $$)

= 300cos$$\left( {15t + {{4\pi } \over 3}} \right)$$
2
MCQ (Single Correct Answer)

AIPMT 2014

The barrier potential of a p-n junction depends on
(1)   type of semiconductor material
(2)   amount of doping

(3)   temperature

Which one of the following is correct ?
A
(1) and (2) only
B
(2) only
C
(2) and (3) only
D
(1), (2) and (3)

Explanation

The barrier potential depends on type of semiconductor (For Si, Vb = 0.7 V and for Ge, Vb = 0.3 V), amount of doping and also on the temperature.
3
MCQ (Single Correct Answer)

AIPMT 2014

The given graph represents V-I characteristic for a semiconductor device.



Which of the following statement is correct ?
A
It is V-I characteristic for solar cell where, point A represents open circuit voltage and point B short circuit current.
B
It is for a solar cell and popints A and B represent open circuit voltage and current, respectively.
C
It is for a photodiode and points A and B represent open circuit voltage and current, respectively.
D
It is for a LED and points A and B represent open circuit voltage and short circuit current, respectively.

Explanation

The V-I characteristic for a solar cell is as shown the figure.
4
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

One way in which the operation of a n-p-n transistor differs from that of a p-n-p
A
The emitter junction injects minority carries into the base region of the p-n-p
B
The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n
C
The emitter injects holes into the base of n-p-n
D
The emitter junction is reversed biased in n-p-n

Explanation

In p-n-p transistor holes are injected into the base while electrons are injected into the base of n-p-n transistor. Emitter-base junction is forward biased.

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