Q. No. 1 - 10 Carry One Mark Each
1. Find the maximum output voltage if \(V_{in}\) = 35V, \(I_{ADJ}\) = 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. \(V_{out} = V_{REF}(1 + \frac {R_2}{R_1}) + I_{ADJ}R_2\) = 30.16V
2. What is the transfer gain of the circuit shown below?
(A) \(\frac {RC}{S^2}\) (B) \(-\frac {1}{R^2S^2C^2}\) (C) \(-\frac {R^2C^2}{S}\) (D) \(-R^2C^2S^2\)
Correct Answer : B
(A) \(\frac {RC}{S^2}\) (B) \(-\frac {1}{R^2S^2C^2}\) (C) \(-\frac {R^2C^2}{S}\) (D) \(-R^2C^2S^2\)
Correct Answer : B
Solve-2. \(\frac {V_o}{V_i} = - \frac {Y_{21}A}{Y_{12}B}\) = \(-\frac {2(SCR + 1)}{2R(SCR + 1)RS^2C^2} = \frac {-1}{R^2S^2C^2}\)
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 = \(20 \times A_v\); \((A_v)dB = 20log_{10}(A_v)\)
\(\implies 80 = 20 log_{10}(A_v)\) \(\implies A_v = 10^4\)
Gain bandwidth product = \(20 \times 10^4\) = 200 KHz
\(\implies 80 = 20 log_{10}(A_v)\) \(\implies A_v = 10^4\)
Gain bandwidth product = \(20 \times 10^4\) = 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