Q. No. 1 - 10 Carry One Mark Each
1. Find the maximum output voltage if Vin = 35V, IADJ = 100μA
(A) 30.16V (B) 16.8V (C) 32.2V (D) 1.25V
Correct Answer : A
(A) 30.16V (B) 16.8V (C) 32.2V (D) 1.25V
Correct Answer : A
Solve-1. Vout=VREF(1+R2R1)+IADJR2 = 30.16V
2. What is the transfer gain of the circuit shown below?
(A) RCS2 (B) −1R2S2C2 (C) −R2C2S (D) −R2C2S2
Correct Answer : B
(A) RCS2 (B) −1R2S2C2 (C) −R2C2S (D) −R2C2S2
Correct Answer : B
Solve-2. VoVi=−Y21AY12B = −2(SCR+1)2R(SCR+1)RS2C2=−1R2S2C2
3. The voltage gain verses the frequency curve of an op-amp is shown below, the gain-bandwidth product of the op-amp is
(A) 200Hz (B) 200MHz (C) 200KHz (D) 2MHz
Correct Answer : C
(A) 200Hz (B) 200MHz (C) 200KHz (D) 2MHz
Correct Answer : C
Solve-3. The gain bandwidth product = 20×Av; (Av)dB=20log10(Av)
⟹80=20log10(Av) ⟹Av=104
Gain bandwidth product = 20×104 = 200 KHz
⟹80=20log10(Av) ⟹Av=104
Gain bandwidth product = 20×104 = 200 KHz
4. The current through the resistor R in the circuit shown below is
(A) 1mA (B) 4mA (C) 8mA (D) 10mA
Correct Answer : D
(A) 1mA (B) 4mA (C) 8mA (D) 10mA
Correct Answer : D
Solve-4. We know that the voltage across '+' terminal is equal to '-' terminal in op-amp
So, I_c = \frac {12 - 2}{1KΩ} = 10mA. Since op-amp draws no current I_B = 0
\therefore I_E = I_C \implies The current through resistor R is 10mA.
So, I_c = \frac {12 - 2}{1KΩ} = 10mA. Since op-amp draws no current I_B = 0
\therefore I_E = I_C \implies The current through resistor R is 10mA.
5. Calculate the output of an op-amp circuit shown below.
(A) (1 + \frac {R_2}{R_1})V_i (B) (1 + \frac {2R_2}{R_1})V_i (C) (1 + \frac {R_2}{R_1})V_i (D) 0
Correct Answer : D
(A) (1 + \frac {R_2}{R_1})V_i (B) (1 + \frac {2R_2}{R_1})V_i (C) (1 + \frac {R_2}{R_1})V_i (D) 0
Correct Answer : D
Solve-5. Input current to the op-amp is zero, so output is zero
6. The shunt derived series fed feed back in an amplifier
(A) Increases its output impedance (B) Increases its input impedance
(C) Decreases its output impedance (D) Both B and C
Correct Answer : D
(A) Increases its output impedance (B) Increases its input impedance
(C) Decreases its output impedance (D) Both B and C
Correct Answer : D
7. If Barkhausen criterion is not full filled by an oscillator circuit, it will
(A) Stop oscillating
(B) Produce damping waves continuously
(C) Become an amplifier
(D) Produce high frequency whistles
Correct Answer : A
(A) Stop oscillating
(B) Produce damping waves continuously
(C) Become an amplifier
(D) Produce high frequency whistles
Correct Answer : A
8. Find relation between R_1 and R_2.
(A) R_2 = 3R_1 (B) R_2 = 4R_1 (C) R_2 = 5R_1 (D) R_2 = 6R_1
Correct Answer : C
(A) R_2 = 3R_1 (B) R_2 = 4R_1 (C) R_2 = 5R_1 (D) R_2 = 6R_1
Correct Answer : C
Solve-8. The condition for oscillation is Aβ = 1
\therefore \frac {V_o}{V_f} \frac {V_f}{V_o} = 1 \implies (1 + \frac {R_2}{R_1})\frac {1}{6} = 1 \implies (R_1 + R_2) = 6R_1 \implies R_2 = 5R_1
\therefore \frac {V_o}{V_f} \frac {V_f}{V_o} = 1 \implies (1 + \frac {R_2}{R_1})\frac {1}{6} = 1 \implies (R_1 + R_2) = 6R_1 \implies R_2 = 5R_1
9. The resonant circuit of a tuned collector transistor oscillator has a resonant frequency of 4 MHz, if the value of capacitance is increased by 50% then the new value of resonant frequency is ______ (MHz)
Correct Answer : From: 3.293 To: 3.227
Correct Answer : From: 3.293 To: 3.227
Solve-9. f_o = \frac {1}{2π\sqrt {LC}}
1st case 4 \times 10^6 = \frac {1}{2π\sqrt {LC}}, 2nd cases f_o^1 = \frac {1}{2π\sqrt {L \times 1.5C}}, f_o^1 = \frac {4 \times 10^6}{\sqrt {1.5}} = 3.26MHz
1st case 4 \times 10^6 = \frac {1}{2π\sqrt {LC}}, 2nd cases f_o^1 = \frac {1}{2π\sqrt {L \times 1.5C}}, f_o^1 = \frac {4 \times 10^6}{\sqrt {1.5}} = 3.26MHz
10. An amplifier with an initial open loop gain of 400 is used as negative feedback amplifier. The feedback factor is 0.05, If the gain of the amplifier changes 10% due to temperature, then the closed loop gain will change approximately by _______(%)
Correct Answer : From: 0.505 To: 0.495
Correct Answer : From: 0.505 To: 0.495
Solve-10. We know that \frac {ΔA_f}{A_f} = \frac {dA}{A} \frac {1}{1 + Aβ}; So, \frac {dA_f}{A_f} = \frac {(\frac {10}{100})}{1 + 400 \times 0.05} = 0.5%
Q. No. 11 - 20 Carry Two Mark Each
11. What will be the output waveform for the circuit shown below?
Correct Answer : C
Correct Answer : C
Solve-11. During the negative half cycle of the input voltage diode D basically conducts when V_i is less than -10V. So then the circuit looks as shown
So then V_o = -10V and considering V_{i(max)}
it can be written that
-30 + V_c +10 = 0 \implies V_c = 20 volt
During positive half cycle, the circuit will look like
Here it can be written that
30 + 20 - V_o = 0
\implies V_o = 50 volt
The waveform shown in option (C) will be the proper answer, since in this waveform V_o will not start from zero due to the existence of some non zero capacitive voltage.
So then V_o = -10V and considering V_{i(max)}
it can be written that
-30 + V_c +10 = 0 \implies V_c = 20 volt
During positive half cycle, the circuit will look like
Here it can be written that
30 + 20 - V_o = 0
\implies V_o = 50 volt
The waveform shown in option (C) will be the proper answer, since in this waveform V_o will not start from zero due to the existence of some non zero capacitive voltage.
12. For the circuit drawn below I_{DSS} = 10mA, V_p = -8volt, V_{GSQ} = -2V, Y_{OS} = 20μs . Determine the voltage gain A_v.
(A) -3.61 (B) -3.76 (C) +3.61 (D) +3.76
Correct Answer : A
(A) -3.61 (B) -3.76 (C) +3.61 (D) +3.76
Correct Answer : A
Solve-12. g_{mo} = \frac {2I_{DSS}}{|V_p|} = \frac {2 \times 10}{8} = 2.5ms \therefore g_m = g_{mo}(1 - \frac {V_{GSQ}}{V_p}) = 2.5(\frac {(-2)}{(-8)}) = 1.88ms
r_d = \frac {1}{Y_{os}} = \frac {1}{20μs} = 50kΩ, Z_i = R_G = 1MΩ
Z_o = R_D || r_d = 2kΩ || 50kΩ = 1.92kΩ \therefore A_v = -g_m(R_D || r_d) = -1.88 \times 1.92 = -3.61
r_d = \frac {1}{Y_{os}} = \frac {1}{20μs} = 50kΩ, Z_i = R_G = 1MΩ
Z_o = R_D || r_d = 2kΩ || 50kΩ = 1.92kΩ \therefore A_v = -g_m(R_D || r_d) = -1.88 \times 1.92 = -3.61
13. Calculate A_v for the following circuit which has I_{DS(ON)} = 6mA; V_{GS(ON)} = 8V; V_{GS(Th)} = 3V; Y_{DS} = 20μs; K = 0.24 \times 10^{-3} A/V^2; V_{GSQ} = 6.4V; I_{DQ} = 2.75mA
(A) -4.90 (B) -4.61 (C) -4.75 (D) +4.90
Correct Answer : B
(A) -4.90 (B) -4.61 (C) -4.75 (D) +4.90
Correct Answer : B
Solve-13. g_m = 2k(V_{GSQ} - V_{GS(Th)}) = 2(0.24 \times 10^{-3} A/V^2)(6.4V - 3V) = 1.63ms
r_d = \frac {1}{Y_{DS}} = \frac {1}{20μs} = 50kΩ
\therefore Z_o = R_F || r_d || R_D
= 10MΩ || 50kΩ || 3kΩ = 49.75kΩ || 3kΩ = 2.83kΩ
A_v = -g_m(R_F || r_d || R_D)
= -g_mZ_o = -1.63ms \times 2.8kΩ = -4.6
r_d = \frac {1}{Y_{DS}} = \frac {1}{20μs} = 50kΩ
\therefore Z_o = R_F || r_d || R_D
= 10MΩ || 50kΩ || 3kΩ = 49.75kΩ || 3kΩ = 2.83kΩ
A_v = -g_m(R_F || r_d || R_D)
= -g_mZ_o = -1.63ms \times 2.8kΩ = -4.6
14. A bridge rectifier circuit has a load resistance of 3000 Ω and it is connected across a 200 V (rms) supply. Each diode of the rectifier has a forward resistance of 60Ω. Determine the ripple voltage at the Output.
(A) 43.37 V (B) 6.71 V (C) 73.64 V (D)83.69 V
Correct Answer : D
(A) 43.37 V (B) 6.71 V (C) 73.64 V (D)83.69 V
Correct Answer : D
Solve-14. I_m = \frac {V_m}{2R_f + R_L} = \frac {200\sqrt {2}}{120 + 3000}
I_{dc} = \frac {2I_m}{π} = \frac {2 \times 0.09096}{π} = 0.0576 Amp
Now ripple voltage, I_{rms}'R_L = (I_{rms}^2 - I_{dc}^2)^{\frac {1}{2}} R_L = (0.064^2 - 0.0576^2)^{\frac {1}{2}} \times 3000
I_{rms} = \frac {I_m}{\sqrt {2}} = \frac {0.0906}{\sqrt {2}} = 0.064
\therefore Ripple voltage at the output = 83.69V
I_{dc} = \frac {2I_m}{π} = \frac {2 \times 0.09096}{π} = 0.0576 Amp
Now ripple voltage, I_{rms}'R_L = (I_{rms}^2 - I_{dc}^2)^{\frac {1}{2}} R_L = (0.064^2 - 0.0576^2)^{\frac {1}{2}} \times 3000
I_{rms} = \frac {I_m}{\sqrt {2}} = \frac {0.0906}{\sqrt {2}} = 0.064
\therefore Ripple voltage at the output = 83.69V
15. For the circuit shown below, if R_1 = 2K Ohm, R_2 = 4K Ohm, R_L = 3K Ohm. The value of I_o is ________ (mA), if I_i = 1mA
Correct Answer : From: -1.515 To: -1.485
Correct Answer : From: -1.515 To: -1.485
Solve-15. The given circuit is
Applying Nodal analysis
-1mA + 0 + \frac {0 - V_o}{R_1} = 0 \implies V_o = -1mA \times R_1 = -1mA \times 2kΩ = -2V
I_2 = \frac {-2}{4K} = -0.5mA.
By KCL we can get I_o + 1mA + 0.5mA = 0 \implies I_o = -1.5mA
Applying Nodal analysis
-1mA + 0 + \frac {0 - V_o}{R_1} = 0 \implies V_o = -1mA \times R_1 = -1mA \times 2kΩ = -2V
I_2 = \frac {-2}{4K} = -0.5mA.
By KCL we can get I_o + 1mA + 0.5mA = 0 \implies I_o = -1.5mA
16. For the op-amp circuit shown below, the open loop differential gain is 10^3. The output voltage V_o, for V_i = 2V is _____ (V).
Correct Answer : From: 2.016 To: 1.976
Correct Answer : From: 2.016 To: 1.976
Solve-16. In inverting op-amp the overall voltage gain is given by A_v = -\frac {A_oK}{1 + A_oβ}
Here K = attenuation constant and is given by
K = \frac {R_f}{R_1 + R_f} = \frac {1}{2};
β = Feedback factor, β = \frac {R_1}{R_f + R_1} = \frac {100}{200} = \frac {1}{2}
A_v = \frac {-1000 \times \frac{1}{2}}{1 + 1000 \times \frac {1}{2}} = \frac {-500}{+501} = -0.998
But from op-amp voltage gain is \frac {V_o}{V_i} = -0.998; V_o = -0.998 \times -2 = 1.996V
Here K = attenuation constant and is given by
K = \frac {R_f}{R_1 + R_f} = \frac {1}{2};
β = Feedback factor, β = \frac {R_1}{R_f + R_1} = \frac {100}{200} = \frac {1}{2}
A_v = \frac {-1000 \times \frac{1}{2}}{1 + 1000 \times \frac {1}{2}} = \frac {-500}{+501} = -0.998
But from op-amp voltage gain is \frac {V_o}{V_i} = -0.998; V_o = -0.998 \times -2 = 1.996V
Common Data Questions: 17 and 18
17. Voltage gain for the amplifier is
(A) \frac {-R_2}{R_1R_3} (B) \frac {R_2}{R_1R_3} (C) \frac {-R_2}{R_1} (D) None
Correct Answer : C
Solve-17.
From scale changer
Voltage gain = \frac {V_o}{V_1} = \frac {-R_2}{R_1}
From scale changer
Voltage gain = \frac {V_o}{V_1} = \frac {-R_2}{R_1}
18. Input resistance for the amplifier is
(A) \frac {1}{R_1} + \frac {2R_2}{R_1R_3} + \frac {1}{R_3} (B) \frac {1}{\frac {1}{R_1} - \frac {2R_2}{R_1R_3} + \frac {1}{R_3}}
(C) \frac {1}{\frac {1}{R_1} - \frac {2R_2}{R_1R_3} - \frac {1}{R_3}} (D) None
Correct Answer : B
Solve-18. I = I_2 - I_1 = \frac {V_i}{R_1} - \frac {(-2V_o - V_i)}{R_3}
I = \frac {V_i}{R_1} - \frac {2R_2V_i}{R_1R_3} + \frac {V_i}{R_3}
R_i = \frac {V_i}{I} = \frac {1}{\frac {1}{R_1} - \frac {2R_2}{R_1R_3} + \frac {1}{R_3}}
I = \frac {V_i}{R_1} - \frac {2R_2V_i}{R_1R_3} + \frac {V_i}{R_3}
R_i = \frac {V_i}{I} = \frac {1}{\frac {1}{R_1} - \frac {2R_2}{R_1R_3} + \frac {1}{R_3}}
Statement for Linked Answer Questions: 19 and 20
A differential amplifier has V_{s1} = 10mV; V_{s2} = 9mV; It has a differential mode gain of 60db and CMRR of 80db.
19. Output voltage is
(A) 1.00095 V (B) 1.0095 V (C) 2.5 V (D) None
Correct Answer : A
20. Error voltage is
(A) 0.00025 V (B) 0.00055 V (C) 0.00065 V (D) 0.00095 V
Correct Answer : D
19. Output voltage is
(A) 1.00095 V (B) 1.0095 V (C) 2.5 V (D) None
Correct Answer : A
Solve-19. For Ideal case:
V_o = A_dV_d; V_d = V_{s1} - V_{s2} = 10 - 9 = 1mV
20 log_{10}A_d = 60; A_d = 10^3
V_o = 10^3 \times 1 \times 10^{-3} = 1V
For General Case:
20 log_{10}CMRR = 80
CMRR = 10^4
A_c = \frac {A_d}{CMRR} = \frac {10^3}{10^4} = 1
V_c = \frac {V_{s1} + V_{s2}}{2} = \frac {10 + 9}{2} = 9.5mV
V_o = A_dV_d + A_cV_c = 10^3 \times 1 \times 10^{-3} + 0.1 \times 9.5 \times 10^{-3}
V_o = 1.00095V
V_o = A_dV_d; V_d = V_{s1} - V_{s2} = 10 - 9 = 1mV
20 log_{10}A_d = 60; A_d = 10^3
V_o = 10^3 \times 1 \times 10^{-3} = 1V
For General Case:
20 log_{10}CMRR = 80
CMRR = 10^4
A_c = \frac {A_d}{CMRR} = \frac {10^3}{10^4} = 1
V_c = \frac {V_{s1} + V_{s2}}{2} = \frac {10 + 9}{2} = 9.5mV
V_o = A_dV_d + A_cV_c = 10^3 \times 1 \times 10^{-3} + 0.1 \times 9.5 \times 10^{-3}
V_o = 1.00095V
20. Error voltage is
(A) 0.00025 V (B) 0.00055 V (C) 0.00065 V (D) 0.00095 V
Correct Answer : D
Solve-20. V_{error} = V_{o(General)} - V_{o(Ideal)} = 1.00095 - 1 = 0.00095V