Q 1) Attempt the questions.

a) In the circuit given in Fig 1 (a) if the voltage V+ and V- are to be amplified by the same factor, the value of R should be _____ **[01 M]**

(i) 3.3 k

(ii) 33 k

(iii) 330 Ω

(iv) None of these

Justify **[04 M]**

**Solution :**

Let R_{1} = 15 K, R_{2} = R, R_{3} = 10 k, R_{4} = 22 K

Let V_{a} be the voltage at the non-inverting terminal and V_{b} be the voltage at the inverting terminal of the differential amplifier shown in fig. 1 (a).

Resistors R_{1} & R_{2} from a voltage divider network with V_{+} as the input voltage and V_{a }as the output voltage which is applied to the non-inverting terminal. So,

V_{a} = V_{+} \left( \frac{R_2}{R_1+R_2} \right)

Now, if A_{+} is the gain of the non-inverting amplifier and V_{OUT+} is its output then –

V_{OUT+} = A_{+} V_{a}

From the above circuit, we can calculate the non-inverting Gain A_{+} as :

A_{+} = 1 + \frac{R_4}{R_3}

∴ V_{OUT+} = V_{+} \left( \frac{R_2}{R_1+R_2} \right) \left( 1+\frac{R_4}{R_3} \right)

Now, for the inverting output V_{OUT-}, we have gain A_{–} and output V_{OUT-}

∴ V_{OUT-} = A_{–} V_{–}

∴ V_{OUT-} = – \frac{R_4}{R_3} V_{–}

V_{OUT} = V_{+} \left( \frac{R_2}{R_1+R_2} \right) \left( 1+\frac{R_4}{R_3} \right) – V_{–} \frac{R_4}{R_3}

Now, V_{+} & V_{–} are amplified by same factor :

∴ V_{OUT} = 0 & V_{+} = V_{–}

\left( \frac{R_2}{R_1+R_2} \right) \left( 1+\frac{R_4}{R_3} \right) = \frac{R_4}{R_3}

\left( \frac{R_2}{R_1+15} \right) \left( 1+\frac{22}{10} \right) = \frac{22}{10}

\left( \frac{R_2}{R_1+15} \right) \left( \frac{32}{10} \right) = \frac{22}{10}

32R = 22 R + 330

10R = 330

R = 33 k

**Answer :** (ii) 33 k

Q 1 b) If the input to the ideal comparator shown in fig. i (b) is a sinusoidal signal of 8 volt peak to peak without any DC component, then the duty cycle of the output comparator is _____ **[01 M]**

(i) 33.33 %

(ii) 25 %

(iii) 20 %

(iv) None of these

Justify **[04 M]**

**Solution :**

Given,

V_{P-P} = 8V = 2 V_{P}

v(t) = 4 sin ωt

V^{–} = 2 V

∴ Duty cycle = \frac{T_{ON}}{T} = \frac{\theta_2}{\theta_1}

Now, θ_{1} = Sin^{-1} \left(\frac{2}{4} \right)

θ_{1} = π/6

and θ_{2} = π – π/6 = 5π/6

∴ Duty cycle = \frac{\frac{5\pi}{6}-\frac{\pi}{6}}{2\pi} \times 100

= \frac{2\pi}{6\pi} \times 100

= \frac{100}{3}

= 33.33 %

**Answer :** (i) 33.33 %

Q 1 c) What is the frequency of IC 555 astable multivibrator shown in Fig 1(c) ?** [01 M]**

(i) 241 Hz

(ii) 178 Hz

(iii) 78 Hz

(iv) 8 Hz

Justify **[04 M]**

**Ans :**

**Given,**

For the 555 timer operating in Astable mode as shown in fig 1(c) :

R_{A} = 1 kΩ

R_{B }= 3 MΩ

C_{1} = 1 nF

∴ The frequency of oscillation is given by :

f_{o} = \frac{1.45}{\left( R_A+2R_B\right)C}

= \frac{1.45}{\left( 1+2\times 3\times 10^3 \right)\times 10^3 \times 10^{-9}} Hz

= \frac{1.45}{6001 \times 10^{-6}}

= \frac{1.45 \times 10^{6}}{6001} Hz

= 241.62 Hz

≈ 241 Hz

**Answer :** (i) 241 Hz

Q 1 d) An amplifier using OP-AMP with slew rate SR = 1 V/μs has a gain of 40 dB. If this amplifier has to amplify sinusoidal signal of 20 kHz faithfully without any slew rate induced distortion, then the input signal must not exceed _____**[01 M]**

(i) 795 mV

(ii) 395 mV

(iii) 79.5 mV

(iv) 39.5 mV

Justify **[04 M]**

**Ans :**

**Given,**

Slew rate, SR = 1 V/μs

Gain = 40 dB

Input frequency, f = 20 kHz

Let V_{P} = Peak value of the output sine wave (volts)

V_{o} = Peak-to-peak sine wave

Now,

Slew Rate, SR = 2πf V_{P} volts/sec

V_{P} = 7.95 V Peak

or V_{o} = 2 V_{P}

V_{o} = 2 x 7.95 V

V_{o} = 15.9 V

Hence, for the output to be a sine wave, the maximum signal V_{in} max should be less than –

= 0.3975 V

= 397.5 mv

≈ 395 mv

**Answer : ** (ii) 395 mv

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