1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

A realistic graph depicting the variation of the reciprocal of input resistance in an input characteristics measurement in a commonemitter transistor configuration is :
A
B
C
D

Explanation



For common emitter configuration, the input characteristic graph is shown above,

ri $$=$$ $${{\Delta {V_{BE}}} \over {\Delta {{\rm I}_B}}}$$

$$ \Rightarrow $$   $${1 \over {{r_i}}} = {{d{{\rm I}_B}} \over {d{V_{BE}}}}$$ $$=$$ shope of this curve.

Upto knee voltage 0.7 V the shope is almost constant. Then it increases sharply.

So, option (c) is the correct choice.
2
MCQ (Single Correct Answer)

JEE Main 2017 (Offline)

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be:
A
180°
B
45°
C
90°
D
135°

Explanation

In common emitter configuration for n-p-n transistor input and output signals are 180° out of phase i.e., phase difference between output and input voltage is 180°.
3
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

What is the conductivity of a semiconductor sale having electron concentration of $$5 \times {10^{18}}\,\,{m^{ - 3}},$$ hole concentration of $$5 \times {10^{19}}\,\,{m^{ - 3}},$$ electron mobility of 2.0 m2 V$$-$$1 s-1 and hole mobility of 0.01 m2 V$$-$$1 s$$-$$1 ?

(Take charge of electronas 1.6 $$ \times $$ 10 $$-$$19 c)
A
1.68 ($$\Omega $$-m)$$-$$1
B
1.83 ($$\Omega $$-m)$$-$$1
C
0.59 ($$\Omega $$-m)$$-$$1
D
1.20 ($$\Omega $$-m)$$-$$1

Explanation

Conductivity of semiconductor,

$$\sigma $$ = e$$\left( {{\eta _e}{\mu _e} + \eta '{\mu _h}} \right)$$

= 1.6 $$ \times $$ 10$$-$$19 (5 $$ \times $$ 1018 $$ \times $$ 2 + 5 $$ \times $$ 1019 $$ \times $$ 0.01)

= 1.6 $$ \times $$ 1.05

= 1.68
4
MCQ (Single Correct Answer)

JEE Main 2017 (Online) 8th April Morning Slot

The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is :

A
10
B
10$$-$$6
C
106
D
100

Explanation

Forward bias resistance

R1 = $${{\Delta V} \over {\Delta I}}$$ = $${{0.8 - 0.7} \over {\left( {20 - 10} \right) \times {{10}^{ - 3}}}}$$ = 10 $$\Omega $$

Reverse bias resistance,

R2 = $${{10} \over {1 \times {{10}^{ - 6}}}}$$ = 107 $$\Omega $$

$$\therefore\,\,\,$$ Ratio of resistance

= $${{{R_1}} \over {{R_2}}}$$ = $${{10} \over {{{10}^7}}}$$ = 10$$-$$6

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