Analog Circuits Sectional Test One

Q. No. 1 - 10 Carry One Mark Each

1. In the circuit shown below,

$D_1$ and

$D_2$ are ideal diodes. The current

$i_1$ and

$i_2$ respectively is

(A) 0, 4 mA (B) 4 mA, 0 (C) 0, 8 mA (D) 8 mA, 0

Correct Answer : A

Solve-1. The voltage across the anode terminal of diode

$D_1$ is less than the voltage across cathode terminal. Therefore,

$V_C$ >

$V_A$ . Diode

$D_1$ reverse biased

$i_1$ =0. Therefore

$D_2$ is forward biased, and

$i_2 = \frac{5-2}{500} = 4 mA$

2. What is the status of the ideal diodes of the circuit shown below?

(A) D1 ON, D2 OFF (B) D1 OFF, D2 ON (C) D1 ON, D2 ON (D) D1 OFF, D2 OFF

Correct Answer : A

Solve-2. Analyzing the circuit, we can see that

$D_1$ forward bias and

$D_2$ is in reverse bias. But no current flow through

$D_2$ , because current choose shortest path through

$D_1$ . (Current choose always low resistance path only)

3. Class AB operation is often used in power (large signal) amplifier in order to

(A) Get maximum efficiency (B) Remove even harmonics (C) Overcome cross over distortion (D) Reduce collector dissipation

Correct Answer : C

Solve-3. Class AB operation is used in power amplifier in order to overcome cross-over distortion.

4. Compared to full wave rectifier using 2 diodes, 4 diode bridge rectifiers has dominant advantage of

(A) Higher current carrying capacity (B) Lower peak inverse voltage requirement (C) Lower ripple voltage (D) Higher efficiency

Correct Answer : B

5. A relaxation oscillator is one which

(A) Has two stable states (B) Oscillates continuously (C) Relaxes indefinitely (D) produces non-sinusoidal output

Correct Answer : D

Solve-5. A relaxation oscillator is one which produces non-sinusoidal output.

6. The circuit shown in the figure represents a

(A) Low pass filter (B) High pass filter (C) Band pass filter (D) Band stop filter

Correct Answer : B

Solve-6. The figure shown in the problem is High Pass Filter.

7. When two identical stages, with upper cut off frequency

$\omega_n$ are cascaded the overall upper cutoff point will be at

(A)

$\omega_n$ (B) 2

$\omega_n$ (C) 0.5

$\omega_n$ (D) 0.64

$\omega_n$

Correct Answer : D

Solve-7.

$f_H^* = f_H \sqrt {2^\frac {1}{2} -1}$ ; Where n = number of stages;

$f_H$ = individual amplifier's upper cut off frequency;

$f_H^*$ = overall upper cut off frequency. Here n = 2,

$f_H =\omega_n$ ,

$f_H^* = f_H \sqrt {2^\frac {1}{2} -1} = 0.64\omega_n$ .

8. A two stage amplifier is required to have an upper cutoff frequency of 2 MHz and a lower cutoff frequency of 30 MHz. The upper and lower cutoff frequency of individual stage are respectively are

(A) 4 MHZ, 60 MHz (5) 3 MHz, 20 MHz (C) 3 MHZ, 60 MHz (D) 4 MHz, 20 MHz

Correct Answer : B

Solve-8. We know that

$f_L^* = \frac {f_L}{\sqrt {2^\frac {1}{n} - 1}}$ and

$f_H^* = f_H \sqrt {2^\frac {1}{n} - 1}$ , n = 2 (given)
30 =

$\frac {f_L}{0.64} \implies$

$f_L = 30 \times 0.64$ = 19.2

$\approx$ 20 MHz
2MHz =

$f_H \sqrt {2^\frac {1}{2} - 1}$

$\implies f_H = \frac {2MHz}{0.64}$ = 3.125MHz

$\approx$ 3MHz

9. If the power input to the amplifier is 2 μW and the power gain of the amplifier is 400 dB, then the output power of the amplifier is _____ (mW).

Correct Answer : From: 20.2 To: 19.8

Solve-9.

$10\log_{10} A = 40;$ A =

$10^4$ ,

$P_o = P_i A = 20mW$

10. The power output to an amplifier is 2 watts at 5 KHz, and 0.5 watts at 50 Hz. If input power is constant at 10 mW, the variation of power gain in dB at the given frequencies is _________ (dB)

Correct Answer : From: 6.06 To: 5.94

Solve-10. At 5 kHz

$P_{01}$ = 2W, At 50 kHz

$P_{02}$ = 0.5W
Variation of power gain =

$10\log_{10} \frac {P_{01}}{P_{02}}$ =

$10\log_{10} \frac {2}{0.5}$ =

$10\log_{10} 4$ = 6dB

Q. No. 11 - 20 Carry Two Mark Each

11. The value of

$V_o$ , for op-amp circuit shown below is

(A) -2V (B) -1V (C) -0.5V (D) 0.5V

Correct Answer : C

Solve-11. The giver Op-amp is a combination of both inverting and non inverting OP-amp's
In inverting

$V_o = -\frac {R_F}{R_1}V_i$ , in non inverting

$V_o = (1 + \frac {R_F}{R_1})V_i$
Here

$V_i$ = 1V
The Thevenin's voltage accross '+' terminal is

$V_{th} = \frac {1 \times 1KΩ}{2KΩ} = \frac {1}{2}$ = 0.5V
In non inverting OP-amp,

$V_o = 0.5(1 + \frac {2}{1})$

$\implies V_o = 3 \times 0.5$ = 1.5V
In inverting Op-amp

$V_o = -\frac {2}{1} \times 1$ = -2V

12. The given figure shows a silicon transistor connected as a common emitter amplifier, the quiescent collector voltage of the circuit approximately as

(A)

$\frac {20}{3}V$ (B) 11V (C) 14V (D) 20V

Correct Answer : C

Solve-12. Since transistor is in saturation

$I_C = \frac {20 - 0.2}{R_C + R_E + \frac {R_E}{β}}$ = 1.3mA

$I_B = \frac {20 - 0.7}{100K + 1 + 100 \times 10K}$ = 17.38μA

$\therefore I_B > \frac {I_C}{β} \implies$ Transistor is in saturation

$\therefore V_C = 20 - 1.3mA \times 5K$ =

$13.5 \simeq 14V$

13. An amplifier with an initial open loop gain of 400 is used as a 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 changes approximately by

(A) 0.05 % (B) 0.1 % (C) 0.5 % (D)1 %

Correct Answer : C

Solve-13. We know that

$\frac {\Delta A_f}{A_f} = \frac {dA}{A} \frac {1}{1 + Aβ}$ ;

$\frac {dA}{A}$ , β, A values are given
So,

$\frac {dA_f}{A_f}$ =

$( \frac {10}{100} ) \times \frac {1}{1 + 400 \times 0.05} \times 100$ = 0.5%

14. An op-amp has a slew rate of 5V/μs. The largest sine wave output voltage possible at a frequency of 1 MHz is

(A) 10π Volts (B) 5 Volts (C)

$\frac {5}{π}$ Volts (D)

$\frac {5}{2π}$ Volts

Correct Answer : D

Solve-14. We know the relationship between signal voltage and S.R is 2πKf ≤ SR
Here, K = Signal voltage and f = frequency

$K ≤ \frac {S.R}{2πf}$

$\implies K ≤ \frac {5 \times 10^6}{2π \times 10^6}$ ,

$k = \frac {5}{2π}$ Volts

15. An op-amp with input voltage of

$V_{i1}$ = 150μV,

$V_{i2}$ = 140μV and if the differential amplifier has a differential gain of

$A_d$ = 4000 and the value of CMRR is 100. Then the output voltage of op-amp is ____(mV).

Correct Answer : From: 46.258 To: 45.342

Solve-15.

$V_o = A_dV_d (1 + \frac {1}{CMRR} \frac {V_C}{V_d})$
Here

$V_d = V_{i1} - V_{i2}$ = 150 -140 = 10μV,

$V_c = \frac {V_{i1} + V_{i2}}{2} = \frac {150 + 140}{2}$ = 145μV

$V_o$ =

$4000 \times 10μV(1 + \frac {1}{100} \times \frac {145}{10})$ =

$0.04 + \frac {0.04 \times 145}{1000}$ =

$0.04 + 5.8 \times 10^{-3}$ = 0.0458

$V_o$ = 45.8mV

16. Total offset voltage for the circuit shown below with given input offset voltage

$V_{IO}$ = 4mV and the input offset current

$I_{IO}$ = 150nA will be ____ (mV).

Correct Answer : From: 483.79 To: 474.21

Solve-16. Offset voltage due to

$V_{IO} = V_{IO}(\frac {R_1 - R_f}{R_1})$ =

$4mV(1 + \frac {500}{5}) = 404mV$
Output voltage due to offset current is =

$I_{IO}R_f = 150 \times 10^{-9} \times 500KΩ$ = 75mV
Total offset = 404 + 75 = 479mV

Common Data Questions: 17 and 18

An RC phase shift oscillator designed using FET having

$g_m$ = 5000μs,

$r_d$ = 40KΩ and the feedback circuit value of R = 10K. Gain of the amplifier A = 40.

17. The value of 'C' for oscillator at 1KHz is ____ (nF).

Correct Answer : From: 6.565 To: 6.435

18. The value of

$R_D$ required is ____ (KΩ)

Correct Answer : From: 10.1 To: 9.9

Solve-17 and 18.

$f = \frac {1}{2πRC\sqrt 6}$

$C = \frac {1}{6.28 \times 10K \times 1K \times \sqrt 6}$ = 6.5nF

$\because A = 40 = g_mR_L \implies R_L = \frac {40}{5000μs}$ = 8KΩ
And alse

$R_L = \frac {R_Dr_d}{r_d + R_D} = 8K$

$\implies R_D = 10KΩ$

Statement for Linked Answer Questions: 19 and 20

In the amplifier circuit shown in the figure, the values of

$R_1$ and

$R_2$ are such that the transistor is operating at

$V_{CE}$ = 3V and

$I_C$ = 1.5mA when it's β is 150.

19. If the β of the transistor is 200, the operating collector current

$I_C$ is

(A) 2mA (B) 3mA (C) 1mA (D) 4mA

Correct Answer : A

20. If the β of the transistor is 200, then the operating voltage

$V_CE$ (in volts) is

(A) 2V (B) 3V (C) 4V (D) 5V

Correct Answer : A

Solve-19 and 20. The given circuit is in fixed bias stabilization method. Apply KVL to outer loop then

$V_{CC} - I_CR_C - V_{CE} = 0$ ,

$R_C = \frac {V_{CC} - V_{CE}}{I_C}$ =

$\frac {6 -3}{1.5mA}$ =

$\frac {3}{1.5} \times 1000$ = 2kΩ
And we know that

$I_C = βI_B$

$I_B = \frac {I_C}{β} = \frac {1.5mA}{150} = 10μA$
Apply KVL to inner loop

$V_{CC} - I_BR_B - V_{BE} = 0$ . For active region

$V_{BE}$ = 0.7

$\implies R_B = R_1 = \frac {V_{CC} - V_{BE}}{I_B}$ =

$\frac {6 - 0.7}{10μA}$ =

$0.53 \times 10^6 = 530KΩ$
Then, given β is changed from 150 to 200.
So,

$I_C = βI_B$ =

$200 \times 10μA = 2mA$
By applying KVL to outer loop we can get

$V_{CEQ}$

$\implies V_{CC} - I_CR_C - V_{CEQ} = 0$

$\implies V_{CEQ} = 6 - 2 \times 2$ = 6 - 4 = 2V

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