Semiconductor Electronics: Materials, Devices and Simple Circuits NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
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Every Semiconductor Electronics: Materials, Devices and Simple Circuits question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Semiconductor Electronics: Materials, Devices and Simple Circuits previous year questions
24 questions. Tap an option to check your answer.
Q1NEET 2021
The electron concentration in an n-type semiconductor is the same as hole concentration in a p-type semiconductor. An external field (electric) is applied across each of them. Compare the currents in them.
Not quite — the answer is C.
Current density J = neμE; with equal carrier concentration and the same field, the current depends on mobility. Electrons have higher mobility than holes, so the n-type sample carries more current. Holes do conduct, so p-type current is not zero.
Q2NEET 2020 · Q153
The increase in the width of the depletion region in a p-n junction diode is due to :
Not quite — the answer is B.
A reverse bias adds to the barrier potential and pulls majority carriers away from the junction, so the depletion region widens. Forward bias opposes the barrier and narrows the depletion layer, which is also why forward current rises.
Q3NEET 2019 · Q18
For a p-type semiconductor, which of the following statements is true?
Not quite — the answer is B.
A p-type semiconductor is made by doping with trivalent (acceptor) atoms, which create holes; holes are therefore the majority carriers. Pentavalent dopants produce n-type material with electrons as majority carriers.
Q4NEET 2018 · Q16
In a p-n junction diode, change in temperature due to heating
Not quite — the answer is D.
Heating generates more electron–hole pairs, which changes the reverse saturation current, the barrier potential and the forward current. Hence both forward and reverse behaviour, i.e. the whole V–I characteristic, change; it is not limited to one bias direction.
Q5NEET 2017 · Q174
In a common emitter transistor amplifier the audio signal voltage across the collector is 3 V. The resistance of collector is 3 kΩ. If current gain is 100 and the base resistance is 2 kΩ, the voltage and power gain of the amplifier is:
Not quite — the answer is C.
Voltage gain A_V = β × R_C/R_B = 100 × 3/2 = 150, and power gain = β × A_V = 100 × 150 = 15000. The 3 V signal value is not needed.
Q6AIPMT 2014
The barrier potential of a p-n junction depends on: (a) type of semi conductor material (b) amount of doping (c) temperature. Which one of the following is correct?
Not quite — the answer is C.
The barrier potential depends on the semiconductor (about 0.7 V for Si, 0.3 V for Ge), increases with doping, and falls as temperature rises. So all three factors matter.
Q7NEET 2013
In an n-type semiconductor, which of the following statements is true?
Not quite — the answer is C.
An n-type semiconductor is made by doping with pentavalent atoms (e.g. P, As), which donate electrons. Electrons become the majority carriers and holes the minority carriers. Trivalent dopants would give p-type material.
Q8AIPMT 2012 Mains
The input resistance of a silicon transistor is 100 Ω. Base current is changed by 40 µA which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 kΩ. The voltage gain of the amplifier is
Not quite — the answer is A.
Current gain β = ΔIc/ΔIb = 2 mA/40 µA = 50. Voltage gain = β × RL/Rin = 50 × 4000/100 = 2000.
Q9AIPMT 2012 Prelims
In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ is 2 V. If the base resistance is 1 kΩ and the current amplification of the transistor is 100, the input signal voltage is
Not quite — the answer is A.
Voltage gain of a CE amplifier = β(R_C/R_B) = 100 × (2 kΩ/1 kΩ) = 200. Input voltage = output/gain = 2 V/200 = 0.01 V = 10 mV. Taking 0.1 V comes from using only the resistance ratio or only β instead of their product.
Q10AIPMT 2012 Prelims
C and Si both have same lattice structure, having 4 bonding electrons in each. However, C is insulator where as Si is intrinsic semiconductor. This is because
Not quite — the answer is D.
Carbon's four bonding electrons are in the second shell (n = 2) and silicon's in the third (n = 3). The more tightly bound carbon electrons give a much larger band gap (about 5.5 eV vs 1.1 eV), so diamond is an insulator. A partly filled band at absolute zero would make carbon a conductor, not an insulator.
Q11AIPMT 2011 Mains
Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 × 10¹⁶ m⁻³. Doping by indium increases nh to 4.5 × 10²² m⁻³. The doped semiconductor is of
Not quite — the answer is B.
Indium is trivalent, so the doped crystal is p-type. By the mass-action law ne = ni²/nh = (1.5 × 10¹⁶)²/(4.5 × 10²²) = 2.25 × 10³²/4.5 × 10²² = 5 × 10⁹ m⁻³. The n-type choices are wrong because an acceptor dopant was added.
Q12AIPMT 2011 Prelims
If a small amount of antimony is added to germanium crystal
Not quite — the answer is D.
Antimony is pentavalent; in germanium it donates an extra electron, making an n-type semiconductor with free electrons as majority carriers. It is a donor, not an acceptor, and doping lowers the resistance rather than increasing it.
Q13AIPMT 2011 Prelims
In forward biasing of the p-n junction
Not quite — the answer is A.
In forward bias, the p-side is connected to the positive terminal; this opposes the built-in field and narrows the depletion region, letting current flow. Connecting the positive terminal to the n-side is reverse bias, which widens the depletion layer.
Q14AIPMT 2010 Mains · Q70
For transistor action (a) Base, emitter and collector regions should have similar size and doping concentrations. (b) The base region must be very thin and lightly doped. (c) The emitter-base junction is forward biased and base-collector junction is reverse biased. (d) Both the emitter-base junction as well as the base collector junction are forward biased. Which one of the following pairs of statements is correct?
Not quite — the answer is C.
A transistor works because the base is very thin and lightly doped, and in active mode the emitter-base junction is forward biased while the collector-base junction is reverse biased. The regions are deliberately unequal in size and doping, and forward-biasing both junctions puts the transistor in saturation, not normal transistor action.
Q15AIPMT 2010 Prelims · Q46
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω and an output impedance of 200 Ω. The power gain of the amplifier is
Not quite — the answer is D.
Voltage gain A_V = β × (R_out/R_in), so β = 50 × 100/200 = 25. Power gain = A_V × β = 50 × 25 = 1250. Taking the power gain equal to the voltage gain (50) ignores the current gain.
Q16AIPMT 2010 Prelims · Q47
Which one of the following bonds produces a solid that reflects light in the visible region and whose electrical conductivity decreases with temperature and has high melting point?
Not quite — the answer is B.
Metallic solids reflect visible light (lustre), have high melting points and their conductivity falls as temperature rises because lattice vibrations scatter the free electrons. Covalent and ionic solids are poor conductors whose conductivity, if any, increases with temperature.
Q17AIPMT 2010 Prelims · Q48
The device that can act as a complete electronic circuit is
Not quite — the answer is C.
An integrated circuit contains many transistors, diodes, resistors and capacitors on one chip, so it can work as a complete circuit. A diode or a single transistor is just one component.
Q18AIPMT 2010 Prelims · Q49
Which of the following statements is false?
Not quite — the answer is C.
In n-type semiconductors the majority carriers are electrons, not holes, so that statement is false. Intrinsic resistance does fall with temperature, trivalent doping does give p-type, and electrons are minority carriers in p-type.
Q19AIPMT 2009 Prelims · Q11
A p-n photodiode is fabricated from a semiconductor with a band gap of 2.5 eV. It can detect a signal of wavelength:
Not quite — the answer is C.
The longest detectable wavelength is λ = hc/Eg = 1240 eV nm/2.5 eV ≈ 496 nm ≈ 5000 Å. Only photons of shorter wavelength have enough energy, so 4000 Å (400 nm) is detected. 6000 Å and the nm values of 4000 and 6000 are all longer than the cut-off.
Q20AIPMT 2009 Prelims · Q13
A transistor is operated in common-emitter configuration at Vc = 2 V 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:
Not quite — the answer is C.
Current gain β = ΔIc/ΔIb = (10 − 5) mA/(200 − 100) μA = 5 × 10⁻³ A/10⁻⁴ A = 50. A value of 100 or more comes from mishandling the milli and micro prefixes.
Q21AIPMT 2008 Prelims · Q50
A p-n photodiode is made of a material with a band gap of 2.0 eV. The minimum frequency of the radiation that can be absorbed by the material is nearly
Not quite — the answer is C.
The photon must at least bridge the band gap: ν = Eg/h = (2.0 × 1.6 × 10⁻¹⁹)/(6.63 × 10⁻³⁴) ≈ 4.8 × 10¹⁴ Hz ≈ 5 × 10¹⁴ Hz. 10 × 10¹⁴ Hz would correspond to a 4 eV gap.
Q22AIPMT 2008 Prelims · Q79
With which one of the following elements silicon should be doped so as to give p-type of semiconductor?
Not quite — the answer is A.
A trivalent dopant such as boron has one electron too few to complete the four bonds with silicon, which creates holes, the majority carriers in a p-type semiconductor. Arsenic is pentavalent and gives n-type, while germanium is tetravalent like silicon and creates no extra carriers.
Q23AIPMT 2005 Prelims · Q38
Carbon, silicon and germanium atoms have four valence electrons each. Their valence and conduction bands are separated by energy band gaps represented by (Eg)C, (Eg)Si and (Eg)Ge respectively. Which one of the following relationships is true in their case ?
Not quite — the answer is A.
Band gaps fall down group 14: carbon (diamond) ≈ 5.5 eV, Si ≈ 1.1 eV, Ge ≈ 0.7 eV. So (Eg)C > (Eg)Si; diamond is an insulator, not a material with a smaller gap.
Q24AIPMT 2005 Prelims · Q44
Choose the only false statement from the following :
Not quite — the answer is D.
In a semiconductor more electrons cross the band gap as temperature rises, so conductivity increases and resistivity decreases. The statement that resistivity increases with temperature is therefore the false one; the other three are true.
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