Q. No. 1 - 10 Carry One Mark Each
1. Find the condition for LED ON.
(A) \(V_i\) > 0, \(V_i\) < 5V (B) \(V_i\) > 5V (C) \(V_i\) < 0 (D) None
Correct Answer : B
(A) \(V_i\) > 0, \(V_i\) < 5V (B) \(V_i\) > 5V (C) \(V_i\) < 0 (D) None
Correct Answer : B
Solve-1. \(V_1 = \frac {10 \times 10}{20}\) = 5V \(\therefore V_i > V_1\)
2. The signal conditioning amplifier is shown in figure. Find the voltage gain if \(V_s\) = 5V and the diode D is consider to be ideal and R = 1kΩ
(A) 5 (B) 4 (C) 10 (D) 1.42
Correct Answer : D
(A) 5 (B) 4 (C) 10 (D) 1.42
Correct Answer : D
Solve-2. When \(V_s\) > 0 diode ideally short circuit
\(R_{feq} = \frac {(2R)5R}{2R + 5R}\) = \(\frac {10}{7}R\), \(A_v = \frac {-R_{feq}}{R_1}\) = \(\frac {-10R/7}{R} = \frac {-10}{7}\) = -1.42, |\(A_v\)| = 1.42
\(R_{feq} = \frac {(2R)5R}{2R + 5R}\) = \(\frac {10}{7}R\), \(A_v = \frac {-R_{feq}}{R_1}\) = \(\frac {-10R/7}{R} = \frac {-10}{7}\) = -1.42, |\(A_v\)| = 1.42
3. What is the output of the circuit below?
(A) 3\(V_s\) (B) \(\frac {1}{3}V_s\) (C) 4\(V_s\) (D) \(V_s\)
Correct Answer : C
(A) 3\(V_s\) (B) \(\frac {1}{3}V_s\) (C) 4\(V_s\) (D) \(V_s\)
Correct Answer : C
Solve-3. \(\frac {V_o}{4R} = \frac {V_s}{R}\), \(V_o = 4V_s\)
4. High power efficiency of the push pull amplifier is due to the fact that
(A) Each transistor conducts on different cycle of the input
(B) Transistors are placed in CE configuration
(C) There is zero collector current under no signal condition
(D) Low forward biasing voltage is required
Correct Answer : C
(A) Each transistor conducts on different cycle of the input
(B) Transistors are placed in CE configuration
(C) There is zero collector current under no signal condition
(D) Low forward biasing voltage is required
Correct Answer : C
Solve-4. High power efficiency of the push pull amplifier is due ti the fact that there is no quiescent collector current.
5. Consider the following:
1. Distortion 2. Gain 3. Bias stabilization 4. Sensitively 5. Frequency response
Which of these are the properties of a good power amplifier circuit?
(A) 1,2 and 3 (B) 1,3 and 5 (C) 2,3 and 4 (D) 4 and 5
Correct Answer : A
1. Distortion 2. Gain 3. Bias stabilization 4. Sensitively 5. Frequency response
Which of these are the properties of a good power amplifier circuit?
(A) 1,2 and 3 (B) 1,3 and 5 (C) 2,3 and 4 (D) 4 and 5
Correct Answer : A
Solve-5. To design a good power amplifier circuit we must consider the following properties, Distortion, Gain, Bias stabilization.
6. In common collector amplifier, the emitter resistance provides
(A) Voltage shunt feedback (B) Voltage series feedback
(C) Current series feedback (D) Current shunt feedback
Correct Answer : B
(A) Voltage shunt feedback (B) Voltage series feedback
(C) Current series feedback (D) Current shunt feedback
Correct Answer : B
Solve-6.
Clearly, the total output voltage drop \(V_o\) across \(R_E\) is feedback into the input loop it he above figure. It is incorporated as a series voltage drop. So it is voltage series feedback.
Clearly, the total output voltage drop \(V_o\) across \(R_E\) is feedback into the input loop it he above figure. It is incorporated as a series voltage drop. So it is voltage series feedback.
7. In a transistor push pull amplifier
(A) There is no d.c. present in the output
(B) There is no distortion in the output
(C) There are no odd harmonics in the output
(D) None of these
Correct Answer : A
(A) There is no d.c. present in the output
(B) There is no distortion in the output
(C) There are no odd harmonics in the output
(D) None of these
Correct Answer : A
Solve-7. Push-Pull amplifiers are operated in class B.
In class B, there is no quiescent current as biasing is at cut-off and the output current \(i_L\) = 2(\(A_1 cos ωt + A_3 cos ωt + ...\))
Hence there is no dc current and even harmonics are cancelled out.
\(\therefore\) Option (A) is correct.
In class B, there is no quiescent current as biasing is at cut-off and the output current \(i_L\) = 2(\(A_1 cos ωt + A_3 cos ωt + ...\))
Hence there is no dc current and even harmonics are cancelled out.
\(\therefore\) Option (A) is correct.
8. Find out the output voltage for circuit shown below.
(A) \(V_o = i_R + i_DV_i\) (B) \((i_n + i_R)R ln \frac {V_i}{ηT}\) (C) \(-I_oRe^{\frac {V_i}{ηV_T}}\) (D) \(-I_oRe^{\frac {V_T}{ηV_i}}\)
Correct Answer : C
(A) \(V_o = i_R + i_DV_i\) (B) \((i_n + i_R)R ln \frac {V_i}{ηT}\) (C) \(-I_oRe^{\frac {V_i}{ηV_T}}\) (D) \(-I_oRe^{\frac {V_T}{ηV_i}}\)
Correct Answer : C
Solve-8. \(i_D = i_R = I_oe^{V_i/ηV_T}\), \(V_D = -i_RR = -i_oRe^{V_i/ηV_T}\)
9. The output voltage for circuit shown below is _____ (V).
Correct Answer : From: 10.1 To: 9.9
Correct Answer : From: 10.1 To: 9.9
Solve-9. \(V_o = \frac {R_2}{R_1}(V_{S_1} - V_{S_2})\) [\(\because R_2 = 2kΩ, R_1 = 1kΩ\)]
=10V
=10V
10. If \(V_{S_1} = 5V\) and \(V_{S_2} = 12V\) is applied to circuit shown below, the output voltage is ____ (V).
Correct Answer : From: 51.51 To: 50.49
Correct Answer : From: 51.51 To: 50.49
Solve-10. \(V_o = V_{o_1} + V_{o_2}\) = \(\frac {1}{2}[1 + \frac {R_2}{R_1}](V_{S_1} + V_{S_2})\) = 51V
Q. No. 11 - 20 Carry Two Mark Each
11. Pulses of definite width can be obtained from irregular shaped pulses
(A) When it is given as input to a mono stable multi vibrator
(B) When it is given as triggering signal to astable multi vibrator
(C) When it is used as input to a Schmitt trigger
(D)When it is used as input to pulse transformer
Correct Answer : A
(A) When it is given as input to a mono stable multi vibrator
(B) When it is given as triggering signal to astable multi vibrator
(C) When it is used as input to a Schmitt trigger
(D)When it is used as input to pulse transformer
Correct Answer : A
12. What is the output voltage \(V_o\)?
(A) \(V_2 - V_1\) (B) \(3V_2 - 2V_1\) (C) \(2V_2 - 3V_1\) (D) \(\frac {2V_2}{4.2V_1}\)
Correct Answer : C
(A) \(V_2 - V_1\) (B) \(3V_2 - 2V_1\) (C) \(2V_2 - 3V_1\) (D) \(\frac {2V_2}{4.2V_1}\)
Correct Answer : C
Solve-12. \(V_+ = \frac {V_2}{2}\); \(V_- = V_1\frac {3}{4} + V_o\frac {1}{4}\), \(V_o = \frac {0.5}{0.25}V_2 - \frac {0.75}{0.25}V_1\) = \(2V_2 - 3V_1\)
13. The output of Schmitt trigger shown below is limited to 10V and -5V. What are the upper trip and lower trip voltage?
(A) 1V, -2V (B) 3V, -1V (C) 0.5V, -1V (D) 4V, 1V
Correct Answer : C
(A) 1V, -2V (B) 3V, -1V (C) 0.5V, -1V (D) 4V, 1V
Correct Answer : C
Solve-13. \(V_{UT} = -(-V_{SAT})\frac {R}{dR}\) = \(-(-5)\frac {1}{10} = 0.5V\), \(V_{LT} = -V_{SAT}\frac {R}{dR}\) = \(-10\frac {1}{10} = -1V\)
14. What is the ratio of \(\frac {I_o}{I_i}\) in figure shown below?
(A) \(\frac {R_L}{R_L + R}\) (B) \(R_L\) + R (C) \(\frac {R_L + R}{R_L}\) (D) \(\frac {R_L}{R}\)
Correct Answer : A
(A) \(\frac {R_L}{R_L + R}\) (B) \(R_L\) + R (C) \(\frac {R_L + R}{R_L}\) (D) \(\frac {R_L}{R}\)
Correct Answer : A
Solve-14. \(I_o = I_i\frac {R_L}{R_L + R}\), \(\frac {I_o}{I_i} = \frac {R_L}{R_L + R}\)
15. A single stage amplifier with gain = 1000 has negative feedback = \(\frac {1}{10}\), applies \(f_L\) = 20Hz and \(f_H\) = 50kHz for the basic amplifier. Find the value of \(f_L\) and \(f_H\) with feedback.
(A) 0.2Hz and 5.05MHz (B) 0.2Hz and 3.05MHz
(C) 0.02Hz and 3.05MHz (D) 0.02Hz and 5.05MHz
Correct Answer : A
(A) 0.2Hz and 5.05MHz (B) 0.2Hz and 3.05MHz
(C) 0.02Hz and 3.05MHz (D) 0.02Hz and 5.05MHz
Correct Answer : A
Solve-15. \(A_f = \frac {A_m}{1 + A_mβ}\) = \(\frac {1000}{1 + 1000 \times \frac {1}{10}} = \frac {1000}{101}\) = 10, \(ω_H' = ω_H(1 + βA_m)\)
\(\therefore f_H' = 50kHz(1 + \frac {1}{10} \times 1000)\) = \(50 \times 101kHz\) = 5.05MHz, \(f_L' = \frac {f_L}{1 + βA_m} = \frac {20}{101}\) = 0.2Hz
\(\therefore f_H' = 50kHz(1 + \frac {1}{10} \times 1000)\) = \(50 \times 101kHz\) = 5.05MHz, \(f_L' = \frac {f_L}{1 + βA_m} = \frac {20}{101}\) = 0.2Hz
16. In a practical integrator circuit shown below,
(A) \(R_F\) enables low frequency gain to be limited
(B) Larger the value of \(R_s\), larger the gain
(C) Effective integration occurs at w less than \(\frac {1}{R_FC_f}\)
(D) Basic elements for integration are R and \(C_F\)
Correct Answer : A
(A) \(R_F\) enables low frequency gain to be limited
(B) Larger the value of \(R_s\), larger the gain
(C) Effective integration occurs at w less than \(\frac {1}{R_FC_f}\)
(D) Basic elements for integration are R and \(C_F\)
Correct Answer : A
Common Data Questions: 17 and 18
For the circuit shown below, the diode are assumed to have a constant 0.7V drop when conducting and op-amp saturates at ±12V
17. The threshold levels are
(A) ± 0.1V (B) ± 0.2V (C) ± 0.7V (D) ± 0.3V
Correct Answer : A
18. The maximum diode current is
(A) 12 mA (B) 1 mA (C) 1.13 mA (D) 1.12 mA
Correct Answer : D
17. The threshold levels are
(A) ± 0.1V (B) ± 0.2V (C) ± 0.7V (D) ± 0.3V
Correct Answer : A
Solve-17. Output levels = ± 0.7V
Thresholds levels = \(\frac {±10}{10k + 60k} \times 0.7\) = ±0.1V
Thresholds levels = \(\frac {±10}{10k + 60k} \times 0.7\) = ±0.1V
18. The maximum diode current is
(A) 12 mA (B) 1 mA (C) 1.13 mA (D) 1.12 mA
Correct Answer : D
Solve-18. \(I_{D Max} = \frac {12 - 0.7}{10k} - \frac {0.7}{10k + 60k}\) = 1.12mA
Statement for Linked Answer Questions: 19 and 20
The Zener break down voltage \(V_z\) = 12V and \(R_L\) = 1kΩ
19. Current through resistor R is
(A) 3.64 mA (B) 36.4 mA (C) 6.34 mA (D) 63.4 mA
Correct Answer : B
20. Current through Zener diode is
(A) 35.7 mA (B) 3.57 mA (C) 53.7 mA (D) 5.37 mA
Correct Answer : A
19. Current through resistor R is
(A) 3.64 mA (B) 36.4 mA (C) 6.34 mA (D) 63.4 mA
Correct Answer : B
Solve-19. \(I_R = \frac {20V - 12V}{220Ω}\) = 36.4mA
20. Current through Zener diode is
(A) 35.7 mA (B) 3.57 mA (C) 53.7 mA (D) 5.37 mA
Correct Answer : A
Solve-20. For \(R_L\) = 1kΩ,
\(I_L = \frac {V_o}{R_L} = \frac {11.3V}{1kΩ}\) = 11.3mA
\(I_B = \frac {I_c}{β} = \frac {11.3mA}{50}\) = 226μA
\(I_z = I_R - I_B\) = 36.4mA - 226μA = 35.7mA
\(I_L = \frac {V_o}{R_L} = \frac {11.3V}{1kΩ}\) = 11.3mA
\(I_B = \frac {I_c}{β} = \frac {11.3mA}{50}\) = 226μA
\(I_z = I_R - I_B\) = 36.4mA - 226μA = 35.7mA