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
(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 %
(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
(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
(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
(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 V2. (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 Powerμ = Modulation index
5. (c)
CMRR = 20log(Ad/Ac)
= 20log(Ad) - 20log(Ac)
= 48 - 2 = 46 dBwhere
Ad = differential voltage gain
Ac = common mode 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
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– 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)
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