22216 Basic Electronics MCQ for Electronics & Telecommunication Engineering (BEL) 2nd Sem I – Scheme MSBTE.
A. lower than
B. higher than
C. same as
D. none of the mentioned
Answer: B
Clarification: C filter improves the efficiency and eliminates ripples by keeping the output voltage constant.
2. An inductor filter connected in series with a resistive load provides a
A. smoothing of the output voltage waveform
B. smoothing of the input voltage waveform
C. smoothing of the output current waveform
D. smoothing of the input current waveform
Answer: C
Clarification: Filter is always connected to the load side (output). Inductor has the property of keeping the current smooth and constant.
3. In the below given configuration, L is connected as a filter across the R load.
The average ammeter current is
A. 2Vm/R
B. 2Vm/πR
C. Vm/πR
D. Vm/R
Answer: B
Clarification: I = Vo/R
Vo = 2Vm/π.
4. The current ripple factor (CRF) is the ratio of
A. Average value/RMS value
B. RMS value/Average value
C. Average value/Maximum value
D. Maximum value/RMS value
Answer: B
Clarification: CRF = Ir/Io.
5. In case of an L filter, the ripple current increases with
A. increase in Load
B. decrease in Load
C. increase in the value of L
D. ripple current never increases
Answer: B
Clarification: When the load is reduced the resistance R is increased & ripple current increases.
6. In case of a C filter, if R (load resistance) is increased
A. ripple factor is reduced
B. ripple factor is increased
C. ripple factor is not affected
D. increases noise in the circuit
Answer: A
The filter connected is most likely a
A. L filter
B. C filter
C. LC filter
D. None of the above mentioned
Answer: A
Clarification: The above waveform is that of the output current. The peak per half cycle is on the right, hence it is more likely to be an L filter.
8. C filters are suitable for ___________ load resistances and L filters are suitable for _____________ load resistances.
A. low, low
B. high, high
C. high, low
D. low, high
Answer: C
Clarification: In case of L filter, if R is lowered, the time constant (L/R) increases, therefore ripple factor reduces. In case of a C filter, the time constant RC is increased, therefore, ripple factor is reduced.
9. An LC filter will have ripple factor value___________ (For the same value of L & C.
A. lower than that obtained by L filter but higher than that obtained by C filter
B. lower than that obtained by C filter but higher than that obtained by L filter
C. lower than that obtained by either L or C filter
D. higher than that obtained by either L or C filter
Answer: C
Clarification: Simply, an LC filter combines both the advantages of L and C filters.
10. In a single-phase full wave rectifier ___________ order harmonics are the most dominant
A. first
B. second
C. third
D. fourth
Answer: B
Clarification: The output voltage is given by,
Vo = (2Vm/π) – (4Vm/3π)cos 2ωt – (4Vm/15π)cos 4ωt . . . .
Hence the other values after the second factor (4Vm/3π) become very very small hence they can be eliminated.
Power Electronics Multiple Choice Questions on “1-Phase-Diode Rectifiers HW-1”.
11. In the process of diode based rectification, the alternating input voltage is converted into
A. an uncontrolled alternating output voltage
B. an uncontrolled direct output voltage
C. a controlled alternating output voltage
D. a controlled direct output voltage
Answer: B
Clarification: Rectification is AC to DC. In DIODE biased rectification, control is not possible.
12. In a half-wave rectifier, the
A. current & voltage both are bi-directional
B. current & voltage both are uni-directional
C. current is always uni-directional but the voltage can be bi-directional or uni-directional
D. current can be bi-directional or uni-directional but the voltage is always uni-directional
Answer: C
Clarification: Current is always in one direction only, but voltage can be bi-directional in case of an L load.
13. For a certain diode based rectifier, the output voltage (average value) is given by the equation
1/2π [ ∫Vm sin ωt d(ωt) ]
Where the integral runs from 0 to π
The rectifier configuration must be that of a
A. single phase full wave with R load
B. single phase full wave with RL load
C. single phase half wave with R load
D. single phase half wave with RL load
Answer: C
Clarification: Integration is 0 to π from base period of 1/2π so it is a half wave R load.
14. For a single phase half wave rectifier, with R load, the diode is reversed biased from ωt =
A. 0 to π, 2π to 2π/3
B. π to 2π, 2π/3 to 3π
C. π to 2π, 2π to 2π/3
D. 0 to π, π to 2π
Answer: B
Clarification: Diode will be reversed biased in the negative half cycles.
15. For the circuit shown below,
The secondary transformer voltage Vs is given by the expression
Vs = Vm sin ωt
Find the PIV of the diode.
A. √2
B. Vs
C. Vm
D. √2 Vm
Answer: C
Clarification: PIV = √2 Vs = Vm.
16. For the circuit shown below,
The peak value of the load current occurs at ωt = ?
A. 0
B. π
C. 2π
D. Data is insufficient
Answer: B
Clarification: Due to the L nature, load current is maximum when the diode will be com-mutated i.e at π.
17. Find the rms value of the output voltage for the circuit shown below.Voltage across the secondary is given by Vm sinωt.
A. Vm
B. 2Vm
C. Vm/2
D. Vm2/2
Answer: C
Clarification: The above is a HW diode rectifier, the RMS o/p voltage equation is given by
Vor = √ [ (1/2π) ∫π Vm2sin2ωt. d(ωt) ]
Solving above equation we get, Vor = Vm/2.
18. In a 1-Phase HW diode rectifier with R load, the average value of load current is given by
Take Input (Vs) = Vm sinωt
A. Vm/R
B. Vm/2R
C. Vm/πR
D. Zero
Answer: C
Clarification: Vo = √ [(1/2π) ∫π Vm sinωt. d(ωt)]
Vo = Vm/π
I = Vo/R = Vm/πR.
19. In the circuit shown below,
The switch (shown in green) is closed at ωt = 0. The load current or capacitor current has the maximum value at ωt =
A. 0
B. π
C. 2π
D. none of the mentioned
Answer: A
Clarification: The instant switch is closed the load current will be zero due to the nature of the capacitor.
20. Find the average value of output current for a 1-phase HW diode rectifier with R load, having RMS output current = 100A.
A. 200R A
B. 100/R√2 A
C. 200/R√2 A
D. 200/Rπ A
Answer: D
Clarification: I(rms) = Vm/2R
Therfore, Vm = 200R
I(avg) = Vm/πR = 200R/πR.
Multiple Choice Questions on “Semiconductor Electronics – Junction Transistor”.
21. How many types of transistors are there?
a) 2
b) 5
c) 1
d) 7
Answer: a
Clarification: A transistor has three doped regions forming two p-n junctions between them. There are two types of transistors, namely n-p-n transistors and p-n-p transistors. In n-p-n transistors, the two segments of the n-type semiconductor are separated by a segment of p-type semiconductor, and in a p-n-p transistor, it’s just the opposite scenario.
22. Which of the following supplies charge carriers in a transistor?
a) Collector
b) Base
c) Emitter
d) Charger
Answer: c
Clarification: Emitter is the section or region on one side of the transistor that supplies charge carriers. It is heavily doped and is always kept forward biased with respect to the base so that it can supply a large number of charge carriers to the base.
23. Which among the following is larger compared to the other regions of a transistor?
a) Emitter
b) Collector
c) Base
d) Charger
Answer: b
Clarification: The collector is the section on the other side of the transistor that collects the charge carriers supplied by the emitter. It is moderately doped but large in size and is always kept in reverse bias with respect to the base.
24. The base forms two p-n junctions with the emitter and the collector.
a) True
b) False
Answer: a
Clarification: Yes, this is a true statement. The base is the middle section of the transistor that forms two p-n junctions with the emitter and the collector. It is very thin and lightly doped so as to pass most of the emitter injected charge carriers to the collector.
25. On which of the following does base current not depend on?
a) The thickness of the base
b) The shape of the transistor
c) Doping levels
d) Number of charge carriers
Answer: d
Clarification: In an n-p-n transistor, as the base is very thin and lightly doped, a very few electrons from the emitter combine with the holes of the base, giving rise to base current and the electrons finally collected by the positive terminal of the battery gives rise to collector current. This base current is a small fraction of collector current depending on the shape of a transistor, thickness of the base, doping levels, and bias voltage.
26. Identify the relationship between base current amplification (α) and emitter current amplification (β).
a) β=(frac {alpha }{1- alpha})
b) β=(frac {1 – alpha }{alpha})
c) β=(frac {alpha }{1 + alpha})
d) β=(frac {1 + alpha }{1 – alpha})
Answer: a
Clarification: IE=IB+IC
(frac {I_E}{I_C} =frac {I_B}{I_C}) + 1
(frac {1}{alpha }=frac {1}{beta }) + 1
α = (frac {beta }{1 + beta})
Therefore, β=(frac {alpha }{1- alpha})
27. The current gain of a transistor in a common emitter configuration is 50. If the emitter current is 5.5 mA, find the base current.
a) 0.203 A
b) 0.107 mA
c) 0.107 A
d) 0.203 mA
Answer: b
Clarification: The expression for current gain is given as:
β=(frac {I_C}{I_B})
IC=50IB
Since, IE=IB+IC ➔ IE=IB+50IB
IE=51IB
IB=(frac {I_E}{51}=frac {5.5}{51})=0.107 mA
28. DC current gain is the ratio of change in collector current to the change in base current.
a) True
b) False
Answer: b
Clarification: No, this is a false statement. DC current is defined as the ratio of the collector current (IC) to the base current (IB). The ratio of change in the collector current to the change in the base current is known as AC current gain.
29. A transistor having α = 0.90 is used in a common base amplifier. If the load resistance is 5.0 kΩ and the dynamic resistance of the emitter junction is 50 Ω, then calculate the voltage gain.
a) 50
b) 100
c) 70
d) 90
Answer: d
Clarification: The expression of voltage gain is as follows:
AV=(frac {αR_L}{R_e})
Given: α = 0.90, RL = 5.0 kΩ = 5000 Ω, Re = 50 Ω
AV=(frac {0.90×5000}{50})
AV = 90
30. What will the power gain of a transistor if it’s α value is 0.80 and the voltage gain is 95?
a) 70
b) 80
c) 76
d) 75
Answer: c
Clarification: Power gain is defined as the ratio of output power to input power. It can also be determined as the product of current gain and voltage gain.
Given: α = 0.80, voltage gain (AV) = 95
The required equation ➔ AP = α × AV
AP=0.80×95
AP=76
Analog Circuits Multiple Choice Questions on “Diode Clipper and Clamper”.
31)What is the circuit in the given diagram called?
A. Clipper
B. Clamper
C. Half wave rectifier
D. Full wave rectifier
Answer: A
Clarification: The circuit given above is a clipper. The diode conducts when it is forward biased, i.e, whenever the input vi is greater than 5V (for ideal diode). For lower voltages, the diode does not conduct and the output is zero.
32. What is the circuit in the given diagram called?
A. Clipper
B. Clamper
C. Half wave rectifier
D. Full wave rectifier
Answer: B
Clarification: During the positive half cycle, the diode is forward biased and no signal appears across the output. The capacitor holds the charge in that state. During negative cycle, diode is reverse biased and diode does not conduct. The charge in capacitor is released and is obtained at the output.
33. For a sinusoidal input of 20 Vpeak to the given circuit, what is the peak value of the output waveform?
A. 20 V
B. 25 V
C. 0 V
D. -25 V
Answer: B
Clarification: In the given circuit, the output becomes zero for vi less than -5 V. Hence, the peak value of the output is 25 V owing to the additive effect of V for vi.
34. For the given circuit for a 20 Vpeak sinusoidal input vi, what is the value of vi at which the clipping begins?
A. 5 V
B. 0 V
C. -5 V
D. Clipping doesn’t occur
Answer: C
Clarification: Considering the connection of diode, it is evident that the diode becomes reverse biased when vi < -5 V. Hence, clipping starts at -5 V.
35. For a sinusoidal input of 20 Vpeak to the given circuit, what is the minimum value of the output waveform?
A. 20 V
B. 25 V
C. -25 V
D. 0 V
Answer: D
Clarification: The given circuit is a clipper that cuts off a part of the negative cycle of the input sinusoid i.e. the output becomes zero for a certain region of the input waveform. Hence, the minimum value is 0 V.
36. For the given input waveform to the given circuit, what is the peak value of the output waveform?
A. 0 V
B. 16 V
C. 12 V
D. 0 V
Answer: B
Clarification: In the given circuit, the diode is in the off stage when vi > 4 V. Hence, when vi > 4 V, vo = vi and hence the peak value of vo = the peak value of vi = 16 V.
37. For the given input waveform to the given circuit, what is the minimum value of the output waveform?
A. 4 V
B. 16 V
C. 12 V
D. 0 V
Answer: A
Clarification: The circuit above is a parallel clipper. When the input is less than 4V, then diode is forward biased and thus output voltage is 4V. When input increases above 4V, the diode is reverse biased and output is equal to the input. Hence, minimum output is 4V.
38. Which of the following is not a necessary component in a clamper circuit?
A. Diode
B. Capacitor
C. Resistor
D. Independent DC Supply
Answer: D
Clarification: Diode, Capacitor and Resistor are necessary to build a clamper circuit. An independent DC supply is required to bring an additional shift.
39. For the given circuit, what is the minimum peak value of the output waveform if the input waveform is 10V square wave with switching time of 1 second?
Assume that the input switches between +10V and -10V DC levels.
A. 0 V
B. -5 V
C. -20 V
D. -10 V
Answer: C
Clarification: For the positive half of the input, the diode is in the on state and hence acts as a short circuit and hence vo = 0 V. For the negative half cycle, the resistor receives voltage input both from the source and the capacitor which is charged during the positive half of the input. Hence, vo = -20 V.
40. For the given circuit and input waveform, the peak value of the output is +30V.
A. True
B. False
Answer: A
Clarification: The given circuit is a clamper with an independent DC supply of +10 V. Keeping in mind the connection of the diode and the DC supply, we see that the output waveform is clamped at +10V i.e. it shifts up by +10V. Hence, the maximum value of vo=+30 V.
Electronic Devices and Circuits Multiple Choice Questions on “Power BJTs”.
41. A power transistor is a ____________
A. three layer, three junction device
B. three layer, two junction device
C. two layer, one junction device
D. four layer, three junction device
Answer: b
Clarification: A power BJT has three layers, p-n-p or n-p-n forming two junctions.
p-n-p: two positive (p) layers and one negative (n) layers in between them.
n-p-n: two negative(n) layers and one positive(p) layers.
42. For a power transistor, if base current IB is increased keeping Vce constant; then______
A. IC increases
B. IC decreases
C. IC remains constant
D. IC changes sinusoidal
Answer: a
Clarification: IB is directly proportional to Ic. The IC curve is linearly distributed when using IB as a parameter, and exponentially distributed using VBE as a parameter. So when VBE is constant, the transistor current IC is almost linear with respect to IB.
43. Which one is the most suitable power device for high frequency (>100 kHz) switching application?
A. BJT
B. Power MOSFET
C. Schottkey diode
D. Microwave transistor
Answer: b
Clarification: Power MOSFET has a low turn off time. So it can be operated in a frequency range of 1 to 10 MHz.
44. Insulated-gate bipolar transistor (IGBT) has combinational advantages of ______
A. BJTs and SITs
B. BJTs and MOSFETs
C. SITs and MOSFETs
D. FETs and BJTs
Answer: b
Clarification: IGBT combines advantages of BJTs and MOSFETs. IGBT process high input impedance like a MOSFET and has low on state power loss as in BJTs.
45. A Gate Turn Off (GTO) can be turned on by applying _______
A. positive gate signal
B. positive drain signal
C. positive source signal
D. negative source signal
Answer: a
Clarification: A Gate Turn Off (GTO) like and SCR is a device four layer, three junction semiconductor device with three external terminal (anode, cathode, and gate). GTO can be turned ON and OFF by positive pulse or signal respectively, to the gate terminal.
46. SITH is also known as ________
A. Field controlled diode
B. Field controlled rectifier
C. Silicon controlled diode
D. Silicon controlled rectifier
Answer: a
Clarification: The static induction thyristor (SITHs) is a thyristor with a buried gate structure in which the gate electrodes are placed in n-base region. They are normally on-state, gate electrodes must be negatively biased to hold-off state. It is a self controlled GTO like device. Hence it is sometimes called the field controlled diode.
47. The turn on time of an SCR with inductive load is 20 µs. The pulse train frequency is 2.5 KHz with a mark/space ratio of 1/10, and then SCR will ______
A. Turn On
B. Not turn on
C.Turn on if inductance is removeC.
D. Turn on if pulse frequency is increased to two times
Answer: A
Clarification: Pulse repetition rate (PRR) = 1/(2.5 – 103) = 0.4 ms = 400 µs.
Mark/space ratio = 1/10.
Pulse width = 400/11 = 36.4 µs.
The SCR will ‘turn on’ as the pulse width is greater than SCR turn on time.
48. What are the three terminals of a power MOSFET called?
A. Collector, emitter, Gate
B. Drain, source, gate
C. Collector, emitter, base
D. Drain, emitter, base
Answer: B
Clarification: A power MOSFET has three terminals called Drain, source and gate in place corresponding to the three terminals Collector, emitter and base for BJT.
49. A thyristor can be termed as ______
A. AC switch
B. DC switch
C. Wave switch
D. Square wave switch
Answer: b
Clarification: Thyristor is a unidirectional device, that is it will only conduct current in one direction only, but unlike a diode, a thyristor can be operate as either ran open circuit switch or as a rectifying diode depending on how the thyristor gate is triggereC. In other words, the thyristor can operate only in switching mode.
50. If anode current is 400 A, then the amount of current required to turn off the GTO is about ________
A. 20A
B. 200A
C. 400A
D. 100A
Answer: D
Clarification: Generally, anode current required of GTO is four times of turn off current. So the amount of current required to turn off GTO is 400/4 = 100A.
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