 Solutions Q1 - Grad Plus

# Solutions Q1

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 R1 = 15 K, R2 = R, R3 = 10 k, R4 = 22 K

Let Va be the voltage at the non-inverting terminal and Vb 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 Va as the output voltage which is applied to the non-inverting terminal. So,

Va = V+ \left( \frac{R_2}{R_1+R_2} \right)

Now, if A+ is the gain of the non-inverting amplifier and VOUT+ is its output then –

VOUT+ = A+ Va

From the above circuit, we can calculate the non-inverting Gain A+ as :

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

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

Now, for the inverting output VOUT-, we have gain A and output VOUT-

∴ VOUT- = A V

∴ VOUT- = – \frac{R_4}{R_3} V

VOUT = 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 :

∴ VOUT = 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

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,

VP-P = 8V = 2 VP

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 %

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) :

RA = 1 kΩ

RB = 3 MΩ

C1 = 1 nF

∴ The frequency of oscillation is given by :

fo = \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

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 VP = Peak value of the output sine wave (volts)

Vo = Peak-to-peak sine wave

Now,

Slew Rate, SR = 2πf VP 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}

VP = 7.95 V Peak

or Vo = 2 VP

Vo = 2 x 7.95 V

Vo = 15.9 V

Hence, for the output to be a sine wave, the maximum signal Vin max should be less than –

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

= 0.3975 V

= 397.5 mv

≈ 395 mv