GATE 2020 IN Q.21: solution and answer

GATE 2020 · INQ.21 · 1 mark · NAT

Assume that the opamp in the circuit shown is ideal.

The value of $\frac{V _{x}}{I _{x}}$ (in $k Ω$ ) is ____

GATE 2020 IN question 21: An op-amp with its non-inverting (+) input on top and its inverting (−) input below; — Figure from the GATE 2020 IN paper. It is part of the question.

Answer and solution

ANSWER

$\frac{V _{x}}{I _{x}} = - 4 k Ω$ . The output is $3 V_{x}$ , so the current through the $8 k Ω$ resistor flows from the output back into the source.

Official answer: -4Source: Official answer key, GATE 2020 IN (IIT Delhi)

Take the op-amp as ideal and operating in its linear region, as the question intends: no current flows into either input, and the two inputs are at the same voltage (virtual short), $V_{+} = V_{-}$ . $I_{x}$ enters the node wired to the non-inverting input, so $V_{+} = V_{x}$ .
Voltage divider on the inverting side. No current enters the $-$ input, so the $2 k Ω$ and $1 k Ω$ resistors carry the same current and divide $V_{o}$ : $V_{-} = V_{o} \frac{1 k Ω}{1 k Ω + 2 k Ω} = \frac{V _{o}}{3}$
Virtual short: $\frac{V _{o}}{3} = V_{x} \Rightarrow V_{o} = 3 V_{x}$ . Seen from the $+$ input, this is a non-inverting amplifier with gain $1 + \frac{2}{1} = 3$ .
KCL at the input node. The $+$ input draws no current, so all of $I_{x}$ flows through the $8 k Ω$ resistor to the output: $I_{x} = \frac{V _{x} - V _{o}}{8 k Ω} = \frac{V _{x} - 3 V _{x}}{8 k Ω} = - \frac{2 V _{x}}{8 k Ω} = - \frac{V _{x}}{4 k Ω}$
Input resistance: $\frac{V _{x}}{I _{x}} = - 4 k Ω$ The circuit is a negative-impedance converter. In general $R_{in} = - \frac{R _{f}}{A - 1}$ , and here $R_{f} = 8 k Ω$ and $A = 3$ give $- \frac{8}{2} = - 4 k Ω$ . Answer: $- 4$ .

KEY IDEA

With an ideal op-amp in linear operation, the input node sits at $V_{x}$ and the non-inverting gain fixes $V_{o} = (1 + 2/1) V_{x} = 3 V_{x}$ . The $8 k Ω$ feedback resistor is the only path for $I_{x}$ . The far end of that resistor is driven to three times the input voltage, so current flows out of the source terminal instead of in, and the resistance seen at the input is negative: $R_{in} = - R_{f} / (A - 1)$ .

WATCH OUT FOR

Seeing an 8 kΩ resistor from the input to the output and treating it as a load to ground, which gives $+ 8 k Ω$ . Another slip is to use $V_{o} = - 2 V_{x}$ , the inverting gain, because the divider looks like an inverting amplifier. The source drives the $+$ input, so the gain is $+ 3$ , the resistor current is $(V_{x} - 3 V_{x}) /8 k Ω$ , and the sign of the answer must be negative.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2020 IN Q.21: solution and answer

Answer and solution

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