Design of Electronic Devices & Circuits - MCQ from AMIE exams (Winter 2017)

 
Choose the correct answer for the following (10 x 2)
 
1. An op-amp has a voltage gain of 1,00,000 operates with a supply voltage of ± 15 V. If a difference voltage of 1 V is applied at its input, then the output voltage should be
(a) 0
(b) 1,00,000 V
(c) Cannot be more than ± 15 V
(d) 1 V
 
2. A DC power supply has open circuit voltage of 100 V. When the full load current is drawn, the output drops to 80 V. The percentage voltage regulation is
(a) 84%
(b) 25%
(c) 20%
(d) 40 %

3. The minimum number of comparators required to build an 8-bit flash ADC is
(a) 8
(b) 63
(c) 255
(d) 256

4. An AM transmitter has carrier power output of 50 W. Total Power produced with 80 % modulation is
(a) 66 W
(b) 60 W
(d) 40W
(d) 36 W

5. If the differential voltage gain and common mode voltage gain of a differential amplifier arc 48 dB and 2 dB respectively; then its common mode rejection ratio is
(a) 23 dB
(b) 25 dB
(c) 46 dB
(d) 50dB

6. A 741 type op-amp has a gain-bandwidth product of 1 MHz. A non inverting amplifier using this op-amp and having voltage gain of 20 dB will exhibit a 3 dB bandwidth of
(a) 50 kHz
(b) 100 kHz
(c) 1000/17 kHz
(d) 1000/7.07 kHz

(7) The cascade amplifier is a multistage configuration of
(a) CC-CB
(b) CE-CB
(c) CB-CC
(d) CE-CC

(8) In a full wave rectifier, the sinusoidal input voltage has a peak value of 170 V. The average or DC value of the rectified voltage is
(a) 170/πV
(b) (2 x 170)/πV
(c) 170/2πV
(d) 0

9. The frequency (f) produced by a crystal oscillator related with the thickness (t) of the crystal as :
(a) f = t
(b) f ∝ t
(c) f ∝ 1/t
(c) none of the three above

10. Barkhausen criterion for sustained oscillation as:
(a) -Aβ
(b) Aβ = 0
(c) Aβ = l
(d) A = 1/√β

Answers

1. (b)
Vo = Ad x Vid
= 105 x (1)
= 105 V
 
2. (b) Voltage regulation = (No load voltage – Full load voltage)/Full load voltage
VR = (100 - 80)/80 = 25%

3. An 8-bit ADC, for example, will require 256 comparators (2⁸ = 256).

4. (a)
Pt = Pc[1 + (μ2/2)]
Pc = Carrier Power
μ = Modulation index
 
5. (c) 
CMRR = 20log(Ad/Ac)
= 20log(Ad) - 20log(Ac)
= 48 - 2 = 46 dB
where  
Ad = differential voltage gain
Ac = common mode voltage gain
 
6.  (c) 
Non-inverting amplifier voltage gain = 20 dB
20 dB = 20 log10(Av)
Av = 10
gain × bandwidth = 1 MHz
∴ Bandwidth = 1 MHz/10
= 100 kHz
 
7.  (d) 
CE – CB → Cascode connection

CE– CC → Cascade connection

CC – CC → Darlington pair

Where, 
CE = Common Emitter, 
CC = Common Collector, 
CB = Common Base
  • The voltage gain of the amplifier can be increased by using cascading connection of individual stages
  • The Cascode connection is used to obtain large output impedance
  • Darlington Connection is used to obtain high current gain.
  • In a cascade amplifier, the overall gain is equal to the product of gains of individual amplifiers.
  • In cascaded connection, overall gain increases. Hence, bandwidth decreases.
  • Decreasing in bandwidth results increase in lower cut off-frequency and decrease in upper cut-off frequency.
8. (a)

9. (c)

10. (c)

---

The study material for AMIE/Junior Engineer exams is available at https://amiestudycircle.com

Popular posts from this blog

Meaning of AeSI withdrawing case from Delhi High Court and its implications for IEI