GATE 2018 EE Q.15: solution and answer

GATE 2018 · EEQ.15 · 1 mark · MCQ

The op-amp shown in the figure is ideal. The input impedance $\frac{v _{in}}{i _{in}}$ is given by

GATE 2018 EE question 15: An ideal op-amp with its non-inverting (+) input at the top and its inverting (-) input at the bottom; — Figure from the GATE 2018 EE paper. It is part of the question.

(A)
$Z \frac{R _{1}}{R _{2}}$
(B)
$- Z \frac{R _{2}}{R _{1}}$
(C)
$Z$
(D)
$- Z \frac{R _{1}}{R _{1} + R _{2}}$

Answer and solution

ANSWER

The input impedance is $\frac{v _{in}}{i _{in}} = - Z \frac{R _{2}}{R _{1}}$ , which is option (B).

Official answer: BSource: Official answer key, GATE 2018 EE (IIT Guwahati)

Assumptions. The op-amp is ideal (infinite gain, zero input current, as the question states) and works in its linear region. So there is a virtual short, $v_{+} = v_{-}$ , and no current flows into either input terminal.
Inverting input. No current enters the $-$ terminal, so $R_{1}$ and $R_{2}$ form a plain voltage divider from $V_{o}$ to ground: $v_{-} = V_{o} \frac{R _{2}}{R _{1} + R _{2}}$
Non-inverting input. The source is connected straight to the $+$ terminal, so $v_{+} = v_{in}$ . The virtual short gives $v_{in} = V_{o} \frac{R _{2}}{R _{1} + R _{2}} \Rightarrow V_{o} = v_{in} (1 + \frac{R _{1}}{R _{2}})$
KCL at the non-inverting input node. The ideal op-amp draws no input current, so all of $i_{in}$ flows through $Z$ from that node to the output: $i_{in} = \frac{v _{in} - V _{o}}{Z}$
Substitute for $V_{o}$ : $i_{in} = \frac{v _{in} - v _{in} ( 1 + \frac{R _{1}}{R _{2}} )}{Z} = - \frac{v _{in} R _{1}}{R _{2} Z}$
Input impedance: $\frac{v _{in}}{i _{in}} = - Z \frac{R _{2}}{R _{1}}$ This is option (B). The circuit is a negative impedance converter. Because $V_{o} > v_{in}$ , current flows out of the input node, back toward the source.

KEY IDEA

The $R_{1}$ – $R_{2}$ network makes a non-inverting amplifier, so $V_{o} = v_{in} (1 + R_{1} / R_{2})$ is larger than $v_{in}$ . The only path for the input current is through $Z$ to the output. Since the output sits above the input, current is pushed back into the source, so the input impedance comes out negative: $- Z R_{2} / R_{1}$ . This is a negative impedance converter.

WATCH OUT FOR

Answering $Z$ (option C) by treating $Z$ as if it simply loads the source, which ignores the fact that the far end of $Z$ is the amplified output $V_{o}$ , not ground. Another slip is writing the current through $Z$ as $(V_{o} - v_{in}) / Z$ while still calling it $i_{in}$ . That drops the minus sign, the key feature of the result.

From the GATE 2018 EE paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2018 EE Q.15: solution and answer

Answer and solution

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