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NEET • PHYSICS • SEMI CONDUCTOR DEVICES

Semi conductor devices practice · Mixed difficulty

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91 Medium
91 of 108
Application +4 | -1
A metallic surface with work function of 2  eV2\;eV, on heating to a temperature of 800 KK gives an emission current of 1  mA1\;mA. If another metallic surface having the same surface area, same emission constant but work function 4  eV4\;eV is heated to a temperature of 1600  K1600\;K, then the emission current will be
View answer and explanation
Correct option: C
4 mAmA
Option D (4 mAmA) is correct because the emission current is proportional to T2e−ϕ/kTT^2 e^{-\phi/kT}, and with the given changes in work function and temperature, the current quadruples.
92 Easy
92 of 108
Understanding +4 | -1
If we add impurity to a metal those atoms also deflect electrons. Therefore
View answer and explanation
Correct option: B
The electrical and thermal conductivities both decrease
The electrical and thermal conductivities both decrease because impurities scatter electrons, hindering their movement.
93 Medium
93 of 108
Application +4 | -1
A photodetector used to detect the wavelength of 1700 nm, has energy gap of about
View answer and explanation
Correct option: D
0.73 eV
Option D (0.73 eV) is correct because it is the approximate energy gap calculated using E=1.24eV⋅μmλE = \frac{{1.24 eV \cdot \mu m}}{\lambda}.
94 Medium
94 of 108
Analyzing +4 | -1
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Photodiodes are preferably operated in reverse bias condition for light intensity measurement.

Reason R: The current in the forward bias is more than the current in the reverse bias for a p−np-n junction diode.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: D
Both A and R are true but R is NOT the correct explanation of A
Option D (Both A and R are true but R is NOT the correct explanation of A) is correct because both the assertion and the reason are true, but the reason does not explain why photodiodes are operated in reverse bias.
95 Medium
95 of 108
Application +4 | -1
For the transistor circuit as shown. If β=100\beta \mathrm{ } = \mathrm{ }100, voltage drop between emitter and base is 0.7 V then value of VCE{V_{CE}} will be
View answer and explanation
Correct option: B
15 V
Option B (15 V) is correct because the voltage drop across the collector resistor is calculated using the base current, beta, and collector resistance, and then subtracted from the supply voltage.
96 Medium
96 of 108
Remembering +4 | -1
In NN-type semiconductors, majority charge carriers are
View answer and explanation
Correct option: D
Electrons
Option D (Electrons) is correct because in N-type semiconductors, the majority charge carriers are electrons due to doping with pentavalent impurities.
97 Medium
97 of 108
Remembering +4 | -1
Boolean expression for OROR gate is
View answer and explanation
Correct option: C
Y=A+BY = A + B
Option C (Y=A+BY = A + B) is correct because it represents the Boolean expression for an OR gate, where the output Y is true if either A or B or both are true.
98 Medium
98 of 108
Analyzing +4 | -1
Statement I: In a common-emitter transistor amplifier the input current is much less than output current.

Statement II: The common-emitter transistor amplifier has very high input impedance.
View answer and explanation
Correct option: C
Statement I is true; Statement II is false
Option C (Statement I is true; Statement II is false) is correct because in a common-emitter amplifier, the output current is much larger than the input current, but the input impedance is not very high.
99 Medium
99 of 108
Understanding +4 | -1
The relation between dynamic plate resistance (rp)\left( {{r_p}} \right) of a vacuum diode and plate current in the space charge limited region, is
View answer and explanation
Correct option: D
rp∝1(Ip)1/3{r_p} \propto \frac{1}{{{{\left( {{I_p}} \right)}^{1/3}}}}
Option D (rp∝1(Ip)1/3{r_p} \propto \frac{1}{{{{\left( {{I_p}} \right)}^{1/3}}}}) is correct because in the space charge limited region, the plate current is proportional to the 3/2 power of the plate voltage, which implies that the dynamic plate resistance is inversely proportional to the 1/3 power of the plate current.
100 Medium
100 of 108
Application +4 | -1
The slope of plate characteristic of a vacuum tube diode for certain operating point on the curve is 10−3mAV{10^{ - 3}}\frac{{mA}}{V}. The plate resistance of the diode and its nature respectively
View answer and explanation
Correct option: B
1000 kilokilo-ohmsohms static
Option A (100 kilokilo-ohmsohms static) is correct because the plate resistance is the inverse of the slope of the plate characteristic, and it's a static parameter.
101 Medium
101 of 108
Application +4 | -1
Pure SiSi at 500  K500\;K has equal number of electron (ne)\left( {{n_e}} \right) and hole (nh)\left( {{n_h}} \right) concentrations of 1.5×1016m−31.5 \times {10^{16}}{m^{ - 3}}. Doping by indium increases nh{n_h} to 4.5×1022m−34.5 \times {10^{22}}{m^{ - 3}}. The doped semiconductor is of
View answer and explanation
Correct option: B
p−\mathrm{p} -type having electron concentration ne=5×109m−3{n_e} = 5 \times {10^9}{m^{ - 3}}
Option B is correct because doping with Indium creates a p-type semiconductor, and the electron concentration is calculated using the mass action law.
102 Easy
102 of 108
Application +4 | -1
To write the decimal number 37 in binary, how many binary digits are required
View answer and explanation
Correct option: B
6
6 is correct because 37 in decimal is 100101 in binary, which requires 6 digits.
103 Medium
103 of 108
Remembering +4 | -1
In a triode amplifier, the value of maximum gain is equal to
View answer and explanation
Correct option: B
Amplification factor
Option B (Amplification factor) is correct because the maximum gain achievable in a triode amplifier is ideally equal to its amplification factor (μ\mu).
104 Easy
104 of 108
Application +4 | -1
Conductivity of a photodiode starts changing only if the wavelength of incident light is less than 660 nm660 \mathrm{~nm}. The band gap of photodiode is found to be (X8)eV\left(\frac{X}{8}\right) \mathrm{eV}. The value of XX is (Given, h=6.6×10−34 Js,e=1.6×10−19Ch=6.6 \times 10^{-34} \mathrm{~Js}, e=1.6 \times 10^{-19} \mathrm{C})
View answer and explanation
Correct option: D
15
Option D (15) is correct because the band gap energy can be calculated from the wavelength using E=hcλE = \frac{{hc}}{{\lambda}}, and then solving for X.
105 Medium
105 of 108
Application +4 | -1
A PP-type semiconductor has acceptor levels 57  meV57\;meV above the valence band. The maximum wavelength of light required to create a hole is (Planck’s constant h=6.6×10−34Jh = 6.6 \times {10^{ - 34}}J-s)
View answer and explanation
Correct option: C
217100 A∘\mathop A\limits^ \circ
Option C (217100 A∘\mathop A\limits^ \circ) is correct because it is calculated using E=hcλE = \frac{{hc}}{\lambda} and converting the energy from meV to Joules.
106 Medium
106 of 108
Remembering +4 | -1
Which of the following logic gate is an universal gate
View answer and explanation
Correct option: D
NOR
Option D (NOR) is correct because a NOR gate can be used to implement all other logic gates, making it a universal gate.
107 Medium
107 of 108
Application +4 | -1
The equivalent decimal number of binary number (11001.001)2{\left( {11001.001} \right)_2} is
View answer and explanation
Correct option: D
25.12525.125
Option D (25.125) is correct because it is the decimal equivalent of the given binary number, calculated by summing the weighted values of each bit.
108 Medium
108 of 108
Application +4 | -1
If the forward voltage in a semiconductor diode is changed from 0.5 V to 0.7 V, then the forward current changes by 1.0 mA. The forward resistance of diode junction will be
View answer and explanation
Correct option: C
200  Ω200\;\mathrm{\Omega }
Option C (200Ω200\mathrm{\Omega }) is correct because the forward resistance is calculated as the change in voltage divided by the change in current, R=ΔVΔIR = \frac{{\Delta V}}{{\Delta I}}.
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