Q. No. 1 - 10 Carry One Mark Each
1. A triangular to square wave generator uses
(A) A sine wave oscillation and a comparator
(B) An integrator and a comparator
(C) A differentiator and a comparator
(D) A sine wave oscillator and a clipper
Correct Answer : C
(A) A sine wave oscillation and a comparator
(B) An integrator and a comparator
(C) A differentiator and a comparator
(D) A sine wave oscillator and a clipper
Correct Answer : C
Solve-1. A triangular to square wave generator uses a diflerentiator and a comparator.
2. Which of the following can be used to generate a square wave from a sinusoidal input signal?
1. Schmitt Trigger circuit
2. Clippers and amplifiers
3. Mono stable multi vibrators
(A) 1 alone (B) 1 and 2 (C) 2 and 3 (D) 1, 2 and 3
Correct Answer : A
1. Schmitt Trigger circuit
2. Clippers and amplifiers
3. Mono stable multi vibrators
(A) 1 alone (B) 1 and 2 (C) 2 and 3 (D) 1, 2 and 3
Correct Answer : A
Solve-2. Among the given a generators only Schmitt trigger circuit produces square wave as output by given any input.
3. An amplifier has a mid frequency gain of \(A_o\) = 100 and upper cut off frequency of \(f_H\) = 40kHz. For a feedback transfer ratio β = 0.1, find upper cut off frequency with negative feedback.
(A) 44 kHz (B) 440 kHz (C) 50 kHz (D)30 kHz
Correct Answer : B
(A) 44 kHz (B) 440 kHz (C) 50 kHz (D)30 kHz
Correct Answer : B
Solve-3. \(f_{Hf} = f_H(1 + βA_o)\) = 440kHz
4. In the circuit shown below, the Zener diode has a knee current of 5 mA and a maximum allowed power dissipation of 240 mW. What are the minimum and maximum load currents that can be drawn safely from the circuit, keeping the output voltage \(V_o\) at 6 V?
(A) 0 mA, 180 mA (B) 5 mA, 110 mA
(C) 20 mA, 55 mA (D) 60 mA, 180 mA
Correct Answer : C
(A) 0 mA, 180 mA (B) 5 mA, 110 mA
(C) 20 mA, 55 mA (D) 60 mA, 180 mA
Correct Answer : C
Solve-4. \(I_{Z(min)}\) = 5mA \(V_o\) = 6V = \(V_z\)
\(P_{max}\) = 240mW \(\therefore I_{z(max)} = \frac {240}{V} = \frac {240}{6}\) = 40mA
Now I = \(\frac {9 - 6}{50}\) = 60mA. \(\therefore I_{L(min)} = I - I_{z(max)}\) = 60 - 40 = 20mA
\(I_{L(max)} = I - I_{z(min)}\) = 60 - 5 = 55mA
\(P_{max}\) = 240mW \(\therefore I_{z(max)} = \frac {240}{V} = \frac {240}{6}\) = 40mA
Now I = \(\frac {9 - 6}{50}\) = 60mA. \(\therefore I_{L(min)} = I - I_{z(max)}\) = 60 - 40 = 20mA
\(I_{L(max)} = I - I_{z(min)}\) = 60 - 5 = 55mA
5. In the circuit given below \(V_z\) = 5V and β = 80. The value of \(I_{CQ}\) and \(V_{CEB}\) are
(A) 15.75 mA, 5.7V (B) 15.75 mA, 4.3V
(C) 12.47 mA, 4.3V (D) 12.47 mA, 5.7V
Correct Answer : A
(A) 15.75 mA, 5.7V (B) 15.75 mA, 4.3V
(C) 12.47 mA, 4.3V (D) 12.47 mA, 5.7V
Correct Answer : A
Solve-5. Applying KVL, it can be written that
12 - 5 - 0.7 - 400\(I_C\) = 0 \(\implies\) 12 - 5.7 - 400\(I_C\) = 0 \(\implies I_{CQ}\) = 15.75mA
Now 12 - \(400 \times (15.75 \times 10^{-3}) - V_{CEQ}\) = 0 \(\implies V_{CEQ}\) = 5.7V
12 - 5 - 0.7 - 400\(I_C\) = 0 \(\implies\) 12 - 5.7 - 400\(I_C\) = 0 \(\implies I_{CQ}\) = 15.75mA
Now 12 - \(400 \times (15.75 \times 10^{-3}) - V_{CEQ}\) = 0 \(\implies V_{CEQ}\) = 5.7V
6. Find the time period of multivibrator shown below (\(R_2 = 0.86R_1\)).
(A) 0.5ms (B) 2ms (C) 0.86ms (D) 20ms
Correct Answer : D
(A) 0.5ms (B) 2ms (C) 0.86ms (D) 20ms
Correct Answer : D
Solve-6. T = 2\(R_f\)C = 20ms
7. Find the capacitor voltage \(V_c\), in an astable multi-vibrator using 555 timer with \(V_{cc}\) = 9V.
(A) 3 to 6V (B) 3 to 4V (C) 5 to 6V (D)3 to 5V
Correct Answer : A
(A) 3 to 6V (B) 3 to 4V (C) 5 to 6V (D)3 to 5V
Correct Answer : A
Solve-7. From pin-diagram of 555 timer the threshold voltage = 2/3VCC = 2/3 \(\times\) 9 = 6 V
The trigger voltage = \(\frac {1}{3} V_{cc}\) = 3V
The trigger voltage = \(\frac {1}{3} V_{cc}\) = 3V
8. The output waveform for the circuit shown below is
Correct Answer : D
Correct Answer : D
Solve-8. When \(V_i\) = 10V, diode is forward biased.
\(V_o\) = 2V
-10 + \(V_c\) + 2 = 0, \(V_c\) = +8V
When \(V_i\) = -10V, diode is reverse biased.
\(V_i + 8 + V_o\) = 0
10 + 8 +\(V_o\) = 0
\(V_o\) = -18V
\(V_o\) = 2V
-10 + \(V_c\) + 2 = 0, \(V_c\) = +8V
When \(V_i\) = -10V, diode is reverse biased.
\(V_i + 8 + V_o\) = 0
10 + 8 +\(V_o\) = 0
\(V_o\) = -18V
9. The percentage change in the open loop gain is 2O% and desensitivity factor is equal to 10 then the percentage change in the closed loop gain is ______ (%)
Correct Answer : From: 2.02 To: 1.98
Correct Answer : From: 2.02 To: 1.98
Solve-9. Desensitivity factor D = 1 + Aβ = 10
\(\frac {dA_f}{A_f} = \frac {dA/A}{1 + Aβ}\) = \(\frac {20%}{10}\) = 2%
\(\frac {dA_f}{A_f} = \frac {dA/A}{1 + Aβ}\) = \(\frac {20%}{10}\) = 2%
10. For the circuit shown \(R_i\) = 1kΩ, \(R_f\) = 99k, A = 500. The exact closed loop gain is ___.
Correct Answer : From: 84.133 To: 82.467
Correct Answer : From: 84.133 To: 82.467
Solve-10. β = \(\frac {R_i}{R_i + R_f}\) = 0.01, A = \(\frac {A}{1 + βA}\) = 83.3
Q. No. 11 - 20 Carry Two Mark Each
11. If R = 10kΩ and C = 0.001μF then what is the cutoff frequency?
(A) 15.9kHz (B) 1MHz (C) 10GHz (D) 159MHz
Correct Answer : A
(A) 15.9kHz (B) 1MHz (C) 10GHz (D) 159MHz
Correct Answer : A
Solve-11. R = \(\frac {1}{ω_cC} = \frac {1}{2πf_cC}\); \(f_c = \frac {1}{2πRC}\) = 15.9kHz
12. A 50 Hz symmetric square wave is applied to the RC-circuit shown in the diagram given below. Which one of the following is the correct output waveform for given input?
Correct Answer : A
Correct Answer : A
Solve-12. Figure shown in this given problem is low pass filter. Low pass filter can act as an integrator so, by integrating step signal (or) square wave we can get, triangle wave form.
13. Consider an RC phase oscillator with R = 1 KΩ, and C = 500 PF. The frequency at which the phase angle produced is 45° is given by
(A) \(1.5 \times 10^5\)Hz (B) \(3.2 \times 10^5\)Hz (C) \(2.5 \times 10^5\)Hz (D) \(2.8 \times 10^5\)Hz
Correct Answer : B
(A) \(1.5 \times 10^5\)Hz (B) \(3.2 \times 10^5\)Hz (C) \(2.5 \times 10^5\)Hz (D) \(2.8 \times 10^5\)Hz
Correct Answer : B
Solve-13. In RC-phase shift oscillator, the inverting op-amp provides 180° phase, and the combination of three RC combinations provides 180°.
So, \(θ = tan^{-1}(\frac {1}{ωRC})\)
Here we must choose R, C values to get the θ as 180°.
Here given θ = 45°
ωRC = tanθ \(\implies\) ωRC = 1 \(\implies ω = \frac {1}{RC}\)
f = \(\frac {1}{2πRC}\) = \(\frac {1}{2π \times 1 \times 10^3 \times 500 \times 10^{-12}}\) \(\implies f = 3.2 \times 10^5\)Hz
So, \(θ = tan^{-1}(\frac {1}{ωRC})\)
Here we must choose R, C values to get the θ as 180°.
Here given θ = 45°
ωRC = tanθ \(\implies\) ωRC = 1 \(\implies ω = \frac {1}{RC}\)
f = \(\frac {1}{2πRC}\) = \(\frac {1}{2π \times 1 \times 10^3 \times 500 \times 10^{-12}}\) \(\implies f = 3.2 \times 10^5\)Hz
14. What is the gain of the inverting amplifier if the op-Amp and diodes are ideal?
(A) \(\frac {-R_1}{R_2}, V_i > 0; \frac {-R_3}{R_1}, V_i ≤ 0\) (B) \(\frac {-R_2}{R_1}, V_i > 0; \frac {-R_1}{R_3}, V_i ≤ 0\)
(C) \(\frac {-R_2}{R_1}, V_i > 0; \frac {-R_3}{R_1}, V_i ≤ 0\) (D) \(\frac {-R_3}{R_1}, V_i > 0; \frac {-R_2}{R_1}, V_i ≤ 0\)
Correct Answer : C
(A) \(\frac {-R_1}{R_2}, V_i > 0; \frac {-R_3}{R_1}, V_i ≤ 0\) (B) \(\frac {-R_2}{R_1}, V_i > 0; \frac {-R_1}{R_3}, V_i ≤ 0\)
(C) \(\frac {-R_2}{R_1}, V_i > 0; \frac {-R_3}{R_1}, V_i ≤ 0\) (D) \(\frac {-R_3}{R_1}, V_i > 0; \frac {-R_2}{R_1}, V_i ≤ 0\)
Correct Answer : C
15. The value of \(V_o\) for the circuit shown below is ____ (V).
Correct Answer : From: 7.07 To: 6.93
Correct Answer : From: 7.07 To: 6.93
Solve-15. \(V_1 = \frac {2 \times 10}{10}\) = 2V, \(V_o = (1 + \frac {50}{20})\)2V = 7V
16. For a BJT phase shift amplifier shown below, f = 5MHz then the value of 'C' is ___ (pF).
Correct Answer : From: 0.869 To: 0.851
Correct Answer : From: 0.869 To: 0.851
Solve-16. f = \(\frac {1}{2π\sqrt {6} RC}\) \(\implies\) 5MHz = \(\frac {1}{2π\sqrt {6} 15K C} \implies\) C = 0.86pF
Common Data Questions: 17 and 18
Consider the Astable multivibrator shown below with \(R_A\) = 6.8kΩ, \(R_B\) = 3.3kΩ
17. The value of free running frequency is ___ (KHz).
Correct Answer : From: 1.081 To: 1.059
18. The value of duty cycle is ___.
Correct Answer : From: 0.758 To: 0.743
17. The value of free running frequency is ___ (KHz).
Correct Answer : From: 1.081 To: 1.059
Solve-17. f = \(\frac {1}{T} = \frac {1.44}{(R_A + 2R_B)C}\) = 1.07kHz
18. The value of duty cycle is ___.
Correct Answer : From: 0.758 To: 0.743
Solve-18. D = \(\frac {t_{on}}{T} = \frac {R_A + R_B}{R_A + 2R_B}\) = 0.75
Statement for Linked Answer Questions: 19 and 20
Circuit diagram for an Astable multivibrator is shown below where \(R_A\) = 5K, \(R_B\) = 5K, C = 1μF, (ln2 = 0.693)
19. What is the value of ON-time of the output waveform in the multivibrator shown?
(A) \(1.7325 \times 10^{-4}\) sec (B) \(6.93 \times 10^{-3}\) sec
(C) \(6.93 \times 10^{-7}\) sec (D) \(3.465 \times 10^{-3}\) sec
Correct Answer : D
20. Duty cycle of the output waveform of the above system is
(A) 1 (B) 0.5 (C) 0.33 (D) 0.67
Correct Answer : B
19. What is the value of ON-time of the output waveform in the multivibrator shown?
(A) \(1.7325 \times 10^{-4}\) sec (B) \(6.93 \times 10^{-3}\) sec
(C) \(6.93 \times 10^{-7}\) sec (D) \(3.465 \times 10^{-3}\) sec
Correct Answer : D
Solve-19. \(T_{ON} = 0.693 R_AC\) = \(3.465 \times 10^{-3}\) sec.; Because of the diode, \(R_B\) is bypassed.
20. Duty cycle of the output waveform of the above system is
(A) 1 (B) 0.5 (C) 0.33 (D) 0.67
Correct Answer : B
Solve-20. Duty cycle = \(\frac {T_{ON}}{T_{ON} + T_{OFF}}\) = \(\frac {R_A}{R_A + R_B}\) = 0.5