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Question

Consider the following circuit, the switch S1 allows the output to switch between two ranges of amplitudes from 0-0.1 V and 0-1 V. Arrange these values of R1, R2 and R3 in increasing order.

(A) Value of R1
(B) Value of R2
(C) Value of R3

Choose the most appropriate answer from the options given below :

This question was previously asked in
UGC NET 2023 Electronic Science Question Paper (13-Dec-2023) (Shift 1)
The correct answer is

(C), (A), (B)

The two ranges differ by a factor of ten, and that factor is what fixes the relative sizes of the three resistors: R3 smallest, then R1, then R2 — option 1.

What the switch does. R1 and R2 are in the series path and R3 shunts the input node to ground, so the network is a potential divider whose attenuation the switch changes:

Switch closed, shorting out R2:

\(\dfrac{v_{+}}{v_{i}}=\dfrac{R_{3}}{R_{1}+R_{3}}\)

Switch open, R2 in circuit:

\(\dfrac{v_{+}}{v_{i}}=\dfrac{R_{3}}{R_{1}+R_{2}+R_{3}}\)

Why R3 must be the smallest. Both ranges attenuate the input, so the shunt arm must be small compared with the series arms — otherwise the divider would pass nearly the whole input and no range change would be possible. A shunt resistor comparable with the series resistors would give an attenuation near \(1/2\) rather than the small fractions the two ranges require.

Why R2 must be the largest. Inserting R2 must change the attenuation by the full factor of ten between the ranges, so it has to dominate the series path when it is switched in. Taking the ratio of the two divider expressions:

\(\dfrac{R_{1}+R_{2}+R_{3}}{R_{1}+R_{3}}=10\quad\Rightarrow\quad R_{2}=9\left(R_{1}+R_{3}\right)\)

so R2 is nine times the sum of the other two — comfortably the largest of the three. R1 is therefore left in the middle, and the order is R3 < R1 < R2.

Why the divider feeds the non-inverting input. That terminal draws essentially no current, so it does not load the divider and the ratio above holds exactly. The op-amp then buffers the divided signal, presenting a low output impedance to whatever follows — which is precisely why an attenuator of this kind is built round an amplifier rather than used bare.

The design principle in one line : in a switched-range attenuator the shunt element sets the basic division and the switched series element sets the range ratio, so the shunt is always the smallest and the switched series element the largest.

Hence, the increasing order is (C), (A), (B).

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Important Questions from Op-Amp and Its Applications

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  2. For IC 741 circuit shown below, the maximum input frequency above which the output will be distorted is ____________. 

  3. Assume that an op-amp 741 connected as a unit gain inverting amplifier is applied with an input change of 10 V. Determine the time taken for the output to change by 10 V. [Slew rate for op-amp 741 = 0.5 V/µs]

  4. What is the value of output voltage $V_o$ in the given circuit? Assume op-amp to be ideal and MOSFET operating in saturation region.

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