Solutions Q1

Solution :

Let R₁ = 15 K, R₂ = R, R₃ = 10 k, R₄ = 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 R₁ & R₂ 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

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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, θ₁ = Sin^-1 \left(\frac{2}{4} \right)

θ₁ = π/6

and θ₂ = 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 %

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Ans :

Given,

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

R_A = 1 kΩ

R_B = 3 MΩ

C₁ = 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

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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=\frac{SR}{2\pi f} V_P=\frac{1\times 10^6}{2\pi \times 10^3} V_P =\frac{10^2}{4\pi}

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 –

\frac{V_o}{Gain} = \frac{15.9}{40}

= 0.3975 V

= 397.5 mv

≈ 395 mv

Answer : (ii) 395 mv

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