Electronics and Communication Engineering - Analog Electronics
Exam Duration: 45 Mins Total Questions : 30
Consider the given circuit and a waveform for the input voltage. The diode in circuit has cut-in voltage V1 =0.
The waveform of output voltage V0 is
- (a)
- (b)
- (c)
- (d)
The circuit shown in figure with output voltage across AB is that of
- (a)
voltage quadrupler
- (b)
voltage tripler
- (c)
voltage doubler
- (d)
None of these
A diode whose internal resistance is \(10\Omega \), is to power supply to \(1000\Omega \) load from 110 V (rms) source of supply. Calculate the DC and AC load currents.
- (a)
49.02mA,77mA
- (b)
77mA,49.02mA
- (c)
77A, 49.02A
- (d)
49.02A,77A
A diode whose internal resistance is \(10\Omega \), is to power supply to \(1000\Omega \) load from 110 V ( rms ) source of supply. Calculate the DC and AC load currents.
For the above question, find the ripple factor.
- (a)
12.1
- (b)
0.121
- (c)
1.21
- (d)
1.01
While choosing operating point Q, which of the following factors of BJTare considered?
- (a)
Power supply
- (b)
AC and DC load
- (c)
Maximum transistor ratings
- (d)
All of the above
The voltage VCB and the current 18 for the C8 configuration of the figure shown below will be
- (a)
3.4 V and 45.8μA
- (b)
34 V and 4.5 mA
- (c)
6 V and 2.75 mA
- (d)
None of these
Which class of amplifiers operates with least distortion?
- (a)
Class A
- (b)
Class B
- (c)
Class C
- (d)
Class D
For the circuit shown below, the value of V0 is
- (a)
\({4\over 3}\)
- (b)
\(-{2\over3}V\)
- (c)
\({2\over3}V\)
- (d)
\(-{4\over 3}V\)
The output voltage produced by cascaded integrators at t =0.5 s
- (a)
6V
- (b)
4V
- (c)
3V
- (d)
10V
Consider the following circuit:
Input impedance seen by the voltage source is
- (a)
5kΩ
- (b)
6/5kΩ
- (c)
6kΩ
- (d)
\(\infty\)
The input voltage in shown below circuit is Vi = 5 sin cotmY.The current I0 is
- (a)
- 2sin cot ωA
- (b)
- 2.2sin ωA μA
- (c)
-5 sin ωt μA
- (d)
Zero
Assuming the op-amp to be ideal, the voltage gain of the amplifier shown below, is
- (a)
\(-{R_2\over R1}\)
- (b)
\(-{R_3\over R_1}\)
- (c)
\(-{R_2||R_3\over R_1}\)
- (d)
\(-\left(R_2+R_3\over R_1\right)\)
If for a silicon n-p-n transistor, the base to emitter voltage (VBE) is 0.7 V and the collector to base voltage (VCB) is 0.2 V, then the transistor is operating in the
- (a)
normal active mode
- (b)
saturation mode
- (c)
inverse active mode
- (d)
cut-off mode
MOSFET can be used as
- (a)
current controlled capacitor
- (b)
voltage controlled capacitor
- (c)
current controlled inductor
- (d)
voltage controlled inductor
Consider the following two statements:
Statement 1 A stable multi-vibrator can be used for generating square wave.
Statement 2 Bi stable multi-vibrator can be used for storing binary information.
- (a)
only statement 1 is correct
- (b)
only statement 2 is correct
- (c)
both the statements 1 and 2 are correct
- (d)
both the statements 1 and 2 are incorrect
In figure below, the input Vi is a 100 Hz triangular wave having a peak amplitude of 2 V and an average value of zero volt. Given that the diode is ideal,1he average value of the output V0 is
- (a)
0.7
- (b)
0.6
- (c)
0.5
- (d)
0.4
For the circuit given below, the base voltage VB is
- (a)
0.8V
- (b)
1.697V
- (c)
1.632V
- (d)
0.721V
In the op-amp circuit given in figure, the load current IL is
- (a)
\(-\frac { { V }_{ s } }{ { R }_{ 2 } } \)
- (b)
\(\frac { { V }_{ s } }{ { R }_{ 2 } } \)
- (c)
\(-\frac { { V }_{ s } }{ { R }_{ L } } \)
- (d)
\(\frac { { V }_{ s } }{ { R }_{ L } } \)
How can we minimize the errors due to input bias current, input offset current?
- (a)
Use the resistance Rcomp=RF || R1
- (b)
Use of op-amp with small ratings of lios
- (c)
Keep the resistance value as small as possible
- (d)
All of the above
The input to full - wave rectifier shown below is Vi =120 sin 2\(\pi\)60t V. The diode cut - in voltage is 0.7 V. If the output voltage cannot drop below 100 V, the required value of the capacitor is
- (a)
61.2 \(\mu\)F
- (b)
41.2 \(\mu\)F
- (c)
20.6 \(\mu\)F
- (d)
30.6 \(\mu\)F
For the BJT circuit shown in figure below, assume that the \(\beta \) of the transistor is very large and VBE=0.7 V. The mode of operation of the BJT is
- (a)
Cut-off
- (b)
saturation
- (c)
normal active
- (d)
reverse active
An n-p-n transistor has a beta cut - off frequency f\(\beta\) of 1 MHz, and common - emitter short circuit low frequency current gain \(\beta_0\) of 200. If unity gain frequency f\(\gamma\) and the alpha cut-off frequency f\(\alpha\) respectively, are
- (a)
200 MHz, 201 MHz
- (b)
200 MHz, 199 MHz
- (c)
199 MHz, 200 MHz
- (d)
201 MHz, 200 MHz
The phase-shift oscillator in figure below operates at f=80 kHz. The value of resistance RF is
- (a)
\(814\Omega\)
- (b)
\(236\Omega\)
- (c)
\(148\Omega\)
- (d)
\(438\Omega\)
The common-emitter current gain of the transistor is \(\beta=75.\) The voltage VBE in On state is 0.7 V. The value of IE and RC are
- (a)
1.46 mA, 6.74 k\(\Omega\)
- (b)
0.978 mA, 3.04 k\(\Omega\)
- (c)
1.13 mA, 5.98 k\(\Omega\)
- (d)
None of these
Consider the following circuit using an ideal op - amp. The I - V, characteristics of the diode is described by the relation I = I0 (e V / VT - 1), where VT = 25 mV, I0 = 1\(\mu\)A and V0 is the voltage across the diode (taken as positive for forward bias).
For an input voltage Vi = -1 V, the output voltage V0 is
- (a)
Zero
- (b)
0.1 V
- (c)
0.7 V
- (d)
1.1 V
A class-A transformer coupled, transistor power amplifier is required to deliver a power output of 10 W. The maximum power rating of the transistor should not be less than
- (a)
5W
- (b)
10 W
- (c)
20 W
- (d)
40 W
For the 2 N 338 transistor, the manufacturer specifies Pmax =100 mW at 25°C free air temperature and the maximum junction temperature, Tmax =125°C. It's thermal resistance is
- (a)
10°C/W
- (b)
1000°C/W
- (c)
100°C/W
- (d)
None of these
For the Zener diode shown in the figure below the Zener voltage at knee 7, the knee current is negligible and the dynamic resistance is 10 \(\Omega\). if the input voltage (Vi) range is from 10 to 16 V, the output voltage (Vo) ranges from
- (a)
7.00 V to 7.29 V
- (b)
7.14 V to 7.29 V
- (c)
7.14 V to 7.43 V
- (d)
7.29 V to 7.43 V
The circuit shown below is an op-amp based. The ratio Vout / Vin is equal to
- (a)
9
- (b)
10
- (c)
21
- (d)
11
An amplifier has an open loop gain of 100 and its lower and upper cut-off frequency of 100 Hz and 100 kHz respectively. A feedback network with a feedback factor of 0.99 is connected to the amplifier. The new frequencies are
- (a)
10 MHz, 1 Hz
- (b)
1 Hz, 10 MHz
- (c)
10 Hz, 10 Hz
- (d)
None of these