For the circuit with an ideal OPAMP shown in the figure. V_{REF} is fixed.

If V_{OUT} = 1 for V_{IN} = 0.1 volt and V_{OUT} = 6 volt for V_{IN} = 1 volt, where V_{OUT} is measured across R_{L} connected at the output of this OPAMP, the value of R_{F} / R_{IN} is

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GATE EC 2021 Official Paper

Option 2 : 5.555

CT 1: Ratio and Proportion

3536

10 Questions
16 Marks
30 Mins

__Concept__**:**

When V_{IN} is active and V_{REF} = 0, the circuit is drawn as:

\({V_{OUT1}} = - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}\)

When V_{IN} = 0 and V_{REF} is active, the circuit is drawn as:

\({V_{OU{T_2}}} = \left( {\frac{{{V_{REF}} - {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\)

Total V_{out} = V_{out1} + V_{out2}

\({V_{out}} = \left( {\frac{{{V_{REF}} - {R_2}}}{{{R_1} + {R_2}}}} \right)\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right) - \frac{{{R_E}}}{{{R_{IN}}}}\;{V_{IN}}\)

Given V_{REF} is fixed then V^{+} is fixed, i.e.

\({V^ + } = \frac{{{V_{REF}} - {R_2}}}{{{R_1} + {R_2}}}\)

\({V_{OUT}} = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}}\;{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}{V_{IN}}\)

For V_{IN} = 0.1 V, we have V_{OUT} = 1 V

\(1 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}}\;{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)\) ---(i)

For V_{IN} = 10 V, we have V_{OUT} = 6 V

\(6 = \left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}}\;{R_2}}}{{{R_1} + {R_2}}}} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)\) ---(ii)

From (i)

\(\left( {1 + \frac{{{R_F}}}{{{R_{IN}}}}} \right)\left( {\frac{{{V_{REF}} \cdot {R_2}}}{{{R_1} + {R_2}}}} \right) = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right)\)

Putting above value in eq (ii)

\(6 = 1 + \frac{{{R_F}}}{{{R_{IN}}}}\left( {0.1} \right) - \frac{{{R_F}}}{{{R_{IN}}}}\left( 1 \right)\)

\(5 = \frac{{{R_F}}}{{{R_{IN}}}}\left( { - 0.9} \right)\)

\(\frac{{{R_F}}}{{{R_{IN}}}} = - \frac{5}{{0.9}} = - 5.55\)

Since only positive values are given in the options:

\(\frac{{{R_F}}}{{{R_{IN}}}} = 5.55\)