31 Year NEET Previous Questions: Semiconductor Electronics - 2 Free MCQ

In reverse biasing, the conduction across the
p-n junction does not take place due to
majority carriers but takes place due to
minority carriers if the voltage of external
battery is large. The size of the depletion
region increases thereby increasing the
potential barrier.

In a semiconductor, at room temperature, few of the electrons in the valence band gain enough energy to jump into the conduction band whereas the others remain in the valence band. Hence, the valence band is partially empty and the conduction band is partially filled.

Output will be one if A or B or both are one.
Y= A + B

Electromotive force depends upon intensity
of light falling, it does not depend on
frequency of barrier voltage

Positive terminal is at lower potential (0V)
and negative terminal is at higher potential
+5 V

(a) is true as in case of conductors either the
conduction & valence band overlap or
conduction band is partially filled.
(b) is true as insulators have energy gap of
the order of 5 to 10 eV.
(c) is false as resistivity (opposite of
conductivity) decreases with increase in
temperature.
(d) is true as with increase in temperature
more and more electrons jump to the
conduction band. So, conductivity increases.

Due to strong electronegativity of carbon

In forward bias, p-type is connected to positive terminal of battery and n-type is connected to negative terminal of battery. Hence, carriers are accelerated towards opposite terminals for the conduction thereby increases the number of electrons on n-side.

Number of donor s is more because
electrons from –ve terminal of the cell
pushes (enters) the n side and decreases
the number of uncompensated pentavalent
ion due to which potential barrier is
reduced. The neutralised pentavalent atom
are again in position to donate electrons.

At a certain reverse bias voltage, zener diode
allows current to flow through it and hence
maintains the voltage supplied to any load
Hence, it is used for stabilisation.

On the basis of given graph, following table
is possible.

It is the truth table of AND gate.

In forward biasing of a diode, the emitted
should be at a higher potential, Here, only in
option (c) emitter is at higher potential with
bias voltage 0V compared to the collector,
which is at –2V. So, it is reverse biased.

We use the formula,
R = R₀ A^1/3
This represents relation between atomic mass
and radius of the nucleus.

Therefore truth table :

For a cubic crystal,
a = b = c and α = β = γ = 90°

For a p-type semiconductor, the acceptor
energy level, as shown in the diagram, is
slightly above the top E_v of the valence band.
With very small supply of energy an electron
from the valence band can jump to the level
E_A and ionise acceptor negatively.

Negative feedback is applied to reduce the
output voltage of an amplifier. If there is no
negative feedback, the value of output
voltage could be very high. In the options
given, the maximum value of voltage gain is
100. Hence it is the correct option.

E_g = 2.0 eV = 2 × 1.6 × 10^–19 J
E_g = h_v

Let A and B be inputs and Y the output

Then Y₁ 

Hence, the given circuit is equivalent to a
NOR gate.

(iv), (ii), (i)

Option (iii) represents AND gate.

The wavelength detected by photodiode
should be less than λ_max. Hence it can
detect a signal of wavelength 4000Å.

Isotopes of an element have the same atomic
number but different mass number. A
radioactive substance when emits one alpha
particles  its mass number reduces
by 4 and charge no. reduces by 2 and after
emission of two β-particles its charge no.
increase by 2 thus the charge no. i.e. atomic
number remains the same.