MCQ on Semiconductor Devices

The following simple question appeared in Kerala Medical Entrance 2006 Test paper:

A PN junction diode is connected to a battery of emf 5.5 V and external resistance 5.1 kΩ. The barrier potential in the diode is 0.4 V. The current in the circuit is
(a) 1.08 mA (b) 0.08 mA (c) 1 mA (d) 1 A (e) 2 mA

Out of 5.5 volts, 0.4 volt (barrier potential) is dropped across the forward biased diode and the remaining 5.1 volts appears across the 5.1 kΩ resistance. So, the current through the resistance is (5.1V/5.1kΩ), which is equal to 1mA [Option (c)].
The following MCQ on half wave rectifier also is simple:
A sinusoidal voltage of peak to peak value 310 V is connected in series with a diode and a load resistance R so that half wave rectification occurs. If the diode has negligible forward resistance and very high reverse resistance, the r.m.s. voltage across the load resistance is
(a) 310 V (b) 155 V (c) 219 V (d)109.5 V (e) 77.5 V
If V_m is the peak value (maximum value) of an alternating voltage, its r.m.s. value is Vm/√2. The r.m.s. value of full wave rectifier output also is V_m/√2. Therefore, the mean square value of the full wave rectifier output is V_m²/2. The mean square value of the half wave rectifier output is half of this, which is equal to V_m²/4. Therefore, r.m.s. value of half wave rectifier output voltage is V_m/2.
Note that the peak to peak value (310 V) is given in the question. The peak value V_m is 155 volts and hence the r.m.s. value is 155/2 = 77.5 V.
Here is a question which may confuse you and you may be tempted to pick out the wrong answer:
A common emitter low frequency amplifier has a collector supply voltage (V_CC) of 9 V. The amplifier has input resistance 1kΩ and collector load resistance 5 kΩ. The transistor used has a common emitter current gain (β) of 100. What will be the peak to peak output signal voltage if an input peak to peak signal voltage of 50 mV from a low impedence source is applied to this amplifier?
(a) 250 V (b) 25 V (c) 18 V (d) 12.5 V (e) 9 V
The voltage gain (amplification) of the amplifier is, A_v = β (R_L/R_i) where β is the common emitter current gain, R_L is the load resistance and Ri is the input resistance.Therefore, A_v = 100×5/1 = 500.
( Note that questions on calculation of voltage gain are often seen in Medical and Engineering Entrance test papers. They are simple to solve at Higher Secondary and Plus Two levels, as we hve done above).
The important thing to remember (especially in the context of the present question) is that you will get the full gain only if the input voltage is small. If the input voltage were 5 mV peak to peak for instance, you would have obtained a peak to peak output voltage of 500 times the input voltage, which is 500×5 mV = 2500 mV = 2.5 V.
But since the input is 50 mV, you cannot obtain 500×50 mV (= 25000 mV = 25 V) for the simple reason that the collector supply voltage is 9 volts only. The maximum possible peak to peak signal voltage will be 9 volts only given in option (e).
[The transistor will swing between saturation and cut off as the signal voltage swings between the positive and negative peaks and you will get a clipped output signal which has a peak to peak value of approximately 9 volts in the present case].

You will find more multiple choice questions in this section at physicsplus: Multiple Choice Questions from Electronics

Questions on Communication Systems

Let us discuss the following multiple choice question which appeared in the Kerala Engineering entrance 2005 test paper:
If a radio receiver is tuned to 855kHz radio wave, the frequency of local oscillator in kHz is
(a) 1510 (b) 455 (c) 1310 (d) 1500 (e) 855

Since the radio frequency mentioned in the question is 855 kHz, it must an AM (amplitude modulation) receiver. AM receivers are super heterodyne receivers with intermediate frequency (IF) equal to 455 kHz, by convention.. So the local oscillator in the receiver has a frequency higher by 455 kHz compared to the frequency of the received radio wave. Its frequency is therefore (855+455) kHz = 1310 kHz.
Consider the following MCQ which high lights the component frequencies in an AM wave:
A carrier of frequency 1.2 MHz is amplitude modulated by a microphone output of frequency 800 Hz. The frequencies present in the amplitude modulated carrier are (in kHz)
(a)1.2 MHz and 800 Hz (b) 400 kHz, 1.2 MHz and 2 MHz (c)1.04 MHz,1.2 MHz and 2 MHz (d) 1.1992 MHz, 1.2 MHz and 1.2008 MHz (e) 1.1992 MHz and 1.2008 MHz
The correct option is (d) since the AM wave contains the carrier frequency, the upper side frequency (which is the sum of the carrier frequency and the modulating signal frequency) and the lower side frequency (which is the difference between the carrier frequency and the modulating signal frequency). While finding the upper and lower side frequencies, it will be convenient if you convert the frequencies in to kilohertz or megahertz. Thus, the lower side frequency = (1200 – 0.8)kHz = 1199.2 kHz = 1.1992 MHz. Similarly, the upper side frequency = (1200 + 0.8) kHz = 1200.8 kHz = 1.2008 MHz.
Now consider the following question:
A semiconductor is to be selected for use as a photo detector to detect radiation of wavelength 620 nm. The appropriate semiconductor will have a band gap of
(a) 2eV (b) 2.2eV (c) 2.4eV (d) 2.8eV (e) both (a) and (b)
The band gap of the semiconductor should not exceed the energy of the photon. The wavelength of the photon being 620 nm which is equal to 6200 Angstrom Units (A.U), its energy is 12400/6200 = 2.0 eV (Note that the product of the wave length in A. U. and the energy in eV for a photon is 12400). The option (a) alone is therefore suitable.
The following MCQ appeared in Kerala Engineering Entrance 2006 test paper:
If α and β are the current gains in the CB and CE configurations respectively of the transistor circuit, then (β – α)/αβ =
(a) ∞ (b) 1 (c) 2 (d) 0.5 (e) zero
Since β = α/(1- α), the given ratio, (β – α)/αβ =[α/(1- α) – α]/αβ = [1/(1- α) -1]/β =[1-(1- α)]/(1-α)β = α/(1- α)β = β/β =1. So, the correct option is (b).

Intrinsic and Extrinsic Semiconductor

Intrinsic semiconductor is pure (un-doped) semiconductor whereas extrinsic semiconductor is impure (doped) semiconductor. The intrinsic carrier number density Ni and extrinsic carrier number densities densities Ne and Nh are related by the law of mass action as
Ni^2 = NeNh
Note that Ni is the number density of mobile electrons as well as holes in the pure semiconductor where as Ne and Nh are their number densities in the doped semiconductor. Simple problems based on this simple law often find place in Medical and Engineering entrance test papers. Consider the following M.C.Q. which appeared in Kerala Engineering entrance test paper of 2006:
The number densities of electrons and holes in pure germanium at room temperature are equal and its value is 3×10^16 per m^3. On doping with aluminium the hole density increases to 4.5×10^22 per m^3. Then the electron density in doped germanium is

(a) 2×10^10/m^3 (b) 5×10^9/m^3 (c) 4.5×10^9/m^3 (d) 3×10^9/m^3 (e) 4×10^10/m^3

From the law of mass action we have, Ne = Ni^2 /Nh = (9×10^32)/4.5×10^22 = 2×10^10. The correct option therefore is (a).