Q. No. 1 - 10 Carry One Mark Each
1. In an L section filter, a resistance is connected across load
(A) Provides good regulation for all values to load
(B) Ensures lower PIV of diodes
(C) Ensures lower values of capacitance in filter
(D)Reduces ripple current
Correct Answer : A
(A) Provides good regulation for all values to load
(B) Ensures lower PIV of diodes
(C) Ensures lower values of capacitance in filter
(D)Reduces ripple current
Correct Answer : A
2. If T is the duration of linear sweep and τ is the time constant, then slop error \(e_s\) is given by
(A) \(e_s = \frac {τ}{T}\) (B) \(e_s = \frac {T}{τ}\) (C) \(e_s = \frac {8τ}{T}\) (D) \(e_s = \frac {8T}{τ}\)
Correct Answer : B
(A) \(e_s = \frac {τ}{T}\) (B) \(e_s = \frac {T}{τ}\) (C) \(e_s = \frac {8τ}{T}\) (D) \(e_s = \frac {8T}{τ}\)
Correct Answer : B
3. To obtain very high input and output impedance in a feedback amplifier, the topology, mostly used is
(A) Voltage series (B) Current series (C) Voltage shunt (D)Current shunt
Correct Answer : B
(A) Voltage series (B) Current series (C) Voltage shunt (D)Current shunt
Correct Answer : B
4. The Barkhausen criterion gives
(A) Condition for stability
(B) Maximum gain for which there are no oscillation
(C) Phase shift needed for oscillation
(D) Maximum feedback for a stable amplifier
Correct Answer : C
(A) Condition for stability
(B) Maximum gain for which there are no oscillation
(C) Phase shift needed for oscillation
(D) Maximum feedback for a stable amplifier
Correct Answer : C
5. The lowest input impedance is obtained by
(A) CE (B) CC (C) CB (D) Both CB and CC
Correct Answer : C
(A) CE (B) CC (C) CB (D) Both CB and CC
Correct Answer : C
6. The emitter resistance in a differential amplifier is used to
(A) Provide high output impedance (B) Provide a high gain (C) Bias the transistor in active region (D) Provide a constant current source
Correct Answer : D
(A) Provide high output impedance (B) Provide a high gain (C) Bias the transistor in active region (D) Provide a constant current source
Correct Answer : D
7. An ideal op-amp is an ideal
(A) Voltage controlled current source (B) Voltage controlled voltage source (C) Current controlled current source (D)Current controlled voltage source
Correct Answer : B
(A) Voltage controlled current source (B) Voltage controlled voltage source (C) Current controlled current source (D)Current controlled voltage source
Correct Answer : B
Solve-7. An ideal Op-amp is an ideal voltage controlled voltage source because in inverting and non- Inverting Op-amp by giving input voltage and the output also in terms of voltage.
8. Match the List-I (Devices) with List II (Associated terms) and select the correct answer using the code given below
(A) A=3 B=2 C=4 D=1 (B) A=2 B=3 C=4 D=1
(C) A=3 B=2 C=1 D=4 (D) A=3 B=1 C=2 D=4
Correct Answer : A
| List - I (Devices) | List - II (Associated term) |
|---|---|
| (A) Diode (B) FET (C) SCR (D) BJT | 1. Saturation region 2. Pinch off voltage 3. Shockley's equation 4. Firing angle |
(A) A=3 B=2 C=4 D=1 (B) A=2 B=3 C=4 D=1
(C) A=3 B=2 C=1 D=4 (D) A=3 B=1 C=2 D=4
Correct Answer : A
9. In class A series amplifier using a transistor under ideal conditions, the maximum a.c. power is 2 W. The maximum transistor dissipation capability has to be ____ (Watt).
Correct Answer : From: 4.04 To: 3.96
Correct Answer : From: 4.04 To: 3.96
Solve-9. The class A power amplifier has only 25% efficiency and given that the maximum a.c power is 2 watts. So, the maximum transistor dissipation capability has to be 4 watts.
\(\frac {P_{ac}}{P_{dc}} = 25% \implies P_{dc} = 8W\)
Transistor power dissipation = 4W
\(\frac {P_{ac}}{P_{dc}} = 25% \implies P_{dc} = 8W\)
Transistor power dissipation = 4W
10. The unity gain frequency of an op-amp Is 1 MHz for a non-inverting amplifier. The value of rise time when a negative feedback is used with \(A_f = 1000\) is ____ (msec).
Correct Answer : From: 0.354 To: 0.347
Correct Answer : From: 0.354 To: 0.347
Solve-10. We know that the relationship between \(t_r\) and B.W. is given by
\(t_r = \frac {0.35}{B.W.} = \frac {0.35}{1K} = 0.35\) m sec
\(t_r = \frac {0.35}{B.W.} = \frac {0.35}{1K} = 0.35\) m sec
Q. No. 11 - 20 Carry Two Mark Each
11. Match the List-I (Devices) with List - II (Applications) and select the correct answer using the code given below
(A) P=3 Q=4 R=2 S=1 (B) P=4 Q=3 R=2 S=1
(C) P=4 Q=3 R=1 S=2 (D) P=3 Q=4 R=2 S=1
Correct Answer : C
| List - I (Devices) | List - II (Applications) |
|---|---|
| (P) Varactor diode (B) Schottky barrier diode (C) Laser diode (D) Tunnel diode | 1. CD players 2. High frequency oscillators 3. Microwave mixer 4. Tuning element in tank circuit |
(A) P=3 Q=4 R=2 S=1 (B) P=4 Q=3 R=2 S=1
(C) P=4 Q=3 R=1 S=2 (D) P=3 Q=4 R=2 S=1
Correct Answer : C
12. A Hartley oscillator uses a FET with a \(g_m\) of 3 ms and \(r_d\)=2O kΩ. The total coil inductance is 20 μF with a turn’s ratio of input side to output side is 1:10. It is tuned with a 20 pf capacitor. Find the frequency of oscillation and minimum gain for maintaining oscillation.
(A) 8.37MHz, 7 (B) 7.96MHz, 6 (C) 7.96MHz, 7 (D) 8.37MHz, 6
Correct Answer : B
(A) 8.37MHz, 7 (B) 7.96MHz, 6 (C) 7.96MHz, 7 (D) 8.37MHz, 6
Correct Answer : B
Solve-12. \(f_o = \frac {1}{2π\sqrt {LC}} \)= \(\frac {1}{2π\sqrt {20 \times 10^{-6} \times 20 \times 10^{-12}}} \)= 7.96MHz
\(μ = g_mr_d\) = \(3 \times 10^{-3} \times 20 \times 10^3\) = 60; for oscillation, \(A_{loop} = μ\frac {N_1}{N_2} \)= \(60 \times \frac {1}{10} = 6\)
\(μ = g_mr_d\) = \(3 \times 10^{-3} \times 20 \times 10^3\) = 60; for oscillation, \(A_{loop} = μ\frac {N_1}{N_2} \)= \(60 \times \frac {1}{10} = 6\)
13. The output voltage \(V_o\)(t) for the circuit below is
(A) \(e^{-t/10}u(t) v\) (B) \(-e^{-t/10}u(t) v\) (C) \(e^{-t/1.6}u(t) v\) (D) \(-e^{-t/1.6}u(t) v\)
Correct Answer : A
(A) \(e^{-t/10}u(t) v\) (B) \(-e^{-t/10}u(t) v\) (C) \(e^{-t/1.6}u(t) v\) (D) \(-e^{-t/1.6}u(t) v\)
Correct Answer : A
Solve-13. Voltage follower since \(V_o = V_- = V_+\)
\(V_+(0^+)\) = \(5 \times 10^{-3} \times (250||1000)\) = 1V; \(V_+(∞) = 0\) (\(\because\) capacitor brcomes open)
\(τ = 8 \times 10^{-3}\)(1000 + 250) = 10s
\(\therefore V_+\) = \(V_+(∞) + {V_+(0^+) - V_+(∞)}e^{-t/τ}\) = \(0 + (1 - 0)e^{-t/10}u(t)\) = \(e^{-t/10}u(t)V\)
\(V_+(0^+)\) = \(5 \times 10^{-3} \times (250||1000)\) = 1V; \(V_+(∞) = 0\) (\(\because\) capacitor brcomes open)
\(τ = 8 \times 10^{-3}\)(1000 + 250) = 10s
\(\therefore V_+\) = \(V_+(∞) + {V_+(0^+) - V_+(∞)}e^{-t/τ}\) = \(0 + (1 - 0)e^{-t/10}u(t)\) = \(e^{-t/10}u(t)V\)
14. The circuit below is at steady state before the switch opens at t = 0. The \(V_c(t)\), for t>0 is
(A) 10 - \(5e^{-12.5t}\)V (B) 5 + \(5e^{-12.5t}\)V (C) 5 + \(5e^{-t/12.5}\)V (D) 10 - \(5e^{-t/12.5}\)V
Correct Answer : A
(A) 10 - \(5e^{-12.5t}\)V (B) 5 + \(5e^{-12.5t}\)V (C) 5 + \(5e^{-t/12.5}\)V (D) 10 - \(5e^{-t/12.5}\)V
Correct Answer : A
Solve-14. \(V_c(0^-) = 5V = V_c(0^+) = 5V\)
For t>0 the equivalent circuit is
(As output of op-amp is 5V and \(V_c(0^+)\) = 5V add to give 10V)
\(\therefore\) Time constant \(τ = 20k \times 4μ = 0.08s\); \(V_i = V(0^+) = 5V\); \(V_F = 10V\)
\(V_c = V_F + (V_i - V_F)e^{-t/τ}\) = \(10 + (5 - 10)e^{-t/0.08}\) = \(10 - 5e^{-12.5t}\) for t>0
For t>0 the equivalent circuit is
(As output of op-amp is 5V and \(V_c(0^+)\) = 5V add to give 10V)
\(\therefore\) Time constant \(τ = 20k \times 4μ = 0.08s\); \(V_i = V(0^+) = 5V\); \(V_F = 10V\)
\(V_c = V_F + (V_i - V_F)e^{-t/τ}\) = \(10 + (5 - 10)e^{-t/0.08}\) = \(10 - 5e^{-12.5t}\) for t>0
15. If the voltage \(V_A\) is limited to ±10V, then the frequency of oscillation is ____ (Hz) Where, C = 10μF, \(R_1\) = 10kΩ, \(R_2\) = 200Ω, \(R_3 = R_4\) = 5kΩ
Correct Answer : From: 4.596 To: 4.505
Correct Answer : From: 4.596 To: 4.505
Solve-15. \(T = 2R_1C\ln {(\frac {1 + β}{1 - β})}\); \(β = \frac {R_4}{R_3 + R_4}\) = \(\frac {5k}{10k} = \frac {1}{2}\)
T = \(2 \times 10 \times 10^3 \times 10 \times 10^{-6}\ln {\frac {1 + \frac {1}{2}}{1 - \frac {1}{2}}}\) = \(0.2\ln 3 = 0.22\)
\(f = \frac {1}{T} = 4.55\)Hz
T = \(2 \times 10 \times 10^3 \times 10 \times 10^{-6}\ln {\frac {1 + \frac {1}{2}}{1 - \frac {1}{2}}}\) = \(0.2\ln 3 = 0.22\)
\(f = \frac {1}{T} = 4.55\)Hz
16. For the circuit below, the 3 dB cut-off frequency is ____ (KHz).
Correct Answer : From: 1.606 To: 1.574
Correct Answer : From: 1.606 To: 1.574
Solve-16. Let \(R_1 = 2k, R_2 = 6k, C = 50nF\)
\(\frac {V_i}{R_1||(\frac {1}{SC})} + \frac {V_i - V_o}{R_2} = 0\) \(\implies \frac {V_i}{\frac {R_1}{1 + SR_1C}} + \frac {V_i}{R_2} = \frac {V_o}{R_2}\) \(\implies V_i[ \frac {R_2 + \frac {R_1}{1 + SR_1C}}{R_2(\frac {R_2}{1 + SR_1C} 1)} ] = \frac {V_o}{R_2}\) \(\implies \frac {V_o}{V_i} = 1 + \frac {R_2}{R_1} + SR_2C\)
\(\therefore F_{3dB} = \frac {1}{2π(R_1||R_2)C} \) = \((1 + \frac {R_2}{R_1})(1 + s(R_1||R_2)C) \) = \(\frac {1}{2π(3k||6k)50n} \) = \(\frac {1}{2π(2k)50n} = 1.59KHz\)
\(\frac {V_i}{R_1||(\frac {1}{SC})} + \frac {V_i - V_o}{R_2} = 0\) \(\implies \frac {V_i}{\frac {R_1}{1 + SR_1C}} + \frac {V_i}{R_2} = \frac {V_o}{R_2}\) \(\implies V_i[ \frac {R_2 + \frac {R_1}{1 + SR_1C}}{R_2(\frac {R_2}{1 + SR_1C} 1)} ] = \frac {V_o}{R_2}\) \(\implies \frac {V_o}{V_i} = 1 + \frac {R_2}{R_1} + SR_2C\)
\(\therefore F_{3dB} = \frac {1}{2π(R_1||R_2)C} \) = \((1 + \frac {R_2}{R_1})(1 + s(R_1||R_2)C) \) = \(\frac {1}{2π(3k||6k)50n} \) = \(\frac {1}{2π(2k)50n} = 1.59KHz\)
Common Data Questions: 17 and 18
A common emitter amplifier working from a voltage square of \(R_s = 300\) ohms has a collector load \(R_c = 5\) kohms. The transistor parameters are \(g_m = 0.05, r_π = 2\)kohms, \(C_π = 19.5\)pF and \(C_π = 0.5\)pF.
17. The |\(A_{vo}\)| (mid band gain) is ___ .
Correct Answer : From: 219.574 To: 215.226
17. The |\(A_{vo}\)| (mid band gain) is ___ .
Correct Answer : From: 219.574 To: 215.226
Solve-17. \(β_o = g_mr_π = 0.05 \times 2k = 100\); \(A_{vo} = -\frac {β_oR_c}{R_s + r_π}\) \(\implies |A_{vo}| = \frac {100 \times 5k}{300 + 2000}\) = 217.4
18. The upper cutoff frequency is ___ (MHz).
Correct Answer : From: 4.262 To: 4.178
Correct Answer : From: 4.262 To: 4.178
Solve-18. \(A_v = -\frac {217.4}{1 + jω/ω_H}\); \(ω_H = \frac {1}{RC}\); \(R = R_s||r_π = 300Ω || 2k = 260Ω\)
\(C = C_π + C_μ(1 + g_mR_c)\) = \(19.5 + 0.5{1 + (0.05 \times 5k)}\) = 145pF
\(ω_H = \frac {1}{260 \times 145 \times 10^{-12}}\) = 26.5Mrad/sec; \(f_H = \frac {ω_H}{2π}\) = 4.22MHz
\(C = C_π + C_μ(1 + g_mR_c)\) = \(19.5 + 0.5{1 + (0.05 \times 5k)}\) = 145pF
\(ω_H = \frac {1}{260 \times 145 \times 10^{-12}}\) = 26.5Mrad/sec; \(f_H = \frac {ω_H}{2π}\) = 4.22MHz
Statement for Linked Answer Questions: 19 and 20
A Schmitt Tigger circuit is shown below
19. Find the values of \(V_TH\) and \(V_TL\)
(A) 1.8V and 0V (B) 2V and 0.2V (C) 2V and 0V (D) 1.8V and -0.2V
Correct Answer : D
20. Find the value of switching voltage
(A) 0.9V (B)2V (C) 0.8V (D) 1V
Correct Answer : C
19. Find the values of \(V_TH\) and \(V_TL\)
(A) 1.8V and 0V (B) 2V and 0.2V (C) 2V and 0V (D) 1.8V and -0.2V
Correct Answer : D
Solve-19. \(\frac {R_1}{R_1 + R_2}V_o + V_{ref}\frac {R_2}{R_1 + R_2}\) = \(V_k \frac {R_3}{R_3 + R_4} + V_T\frac {R_4}{R_3 + R_4}\)
\(V_{ref}\) = 1V, \(V_k\) = 2V
\(R_1\) = 50k, \(R_2\) = 500k, \(R_3\) = 10k, \(R_4\) = 100k
Putting \(V_o\) = +10V, we get \(V_{TH}\) = 1.8V
Pitting \(V_o\) = -10V, we get \(V_{TH}\) = -0.2V
\(V_{ref}\) = 1V, \(V_k\) = 2V
\(R_1\) = 50k, \(R_2\) = 500k, \(R_3\) = 10k, \(R_4\) = 100k
Putting \(V_o\) = +10V, we get \(V_{TH}\) = 1.8V
Pitting \(V_o\) = -10V, we get \(V_{TH}\) = -0.2V
20. Find the value of switching voltage
(A) 0.9V (B)2V (C) 0.8V (D) 1V
Correct Answer : C
Solve-20. Switching voltage \(V_s = \frac {V_{TH} + V_{TL}}{2} = 0.8V\)