GATE 2023 IN Q.26: solution and answer

GATE 2023 · INQ.26 · 1 mark · NAT

The opamp in the circuit shown is ideal, except that it has an input bias current of 1 nA and an input offset voltage of 10 $μ$ V. The resulting worst-case output voltage will be $\pm$ ______ $μ$ V (rounded off to the nearest integer).

GATE 2023 IN question 26: Inverting-amplifier configuration with no input signal. — Figure from the GATE 2023 IN paper. It is part of the question.

Answer and solution

ANSWER

The worst-case output voltage is ±1110 µV: the offset voltage contributes 10 µV × 101 = 1010 µV, and the bias current contributes 1 nA × 100 kΩ = 100 µV.

Official answer: 1110Source: Official answer key, GATE 2023 IN (IIT Kanpur)

Model. Treat the non-ideal op-amp as an ideal op-amp plus (i) an offset source $V_{O S} = 10 μ V$ in series with the non-inverting input and (ii) a bias current $I_{B} = 1 nA$ flowing into each input. No signal is applied, so everything at $V_{o}$ is error. The circuit is linear, so find each error term by superposition. Here $R_{1} = 1 k Ω$ (inverting node to ground) and $R_{f} = 100 k Ω$ (inverting node to output).
Offset-voltage term (bias currents set to zero). $V_{O S}$ sits at the + input, and $R_{1}$ to ground with $R_{f}$ as feedback make a non-inverting amplifier. The noise gain applies: $V_{o 1} = V_{O S} (1 + \frac{R _{f}}{R _{1}}) = 10 μ V \times (1 + \frac{100 k Ω}{1 k Ω}) = 10 \times 101 = 1010 μ V$
Bias current into the + input. The + input is tied straight to ground. Its bias current flows through zero resistance, produces no voltage, and adds nothing to $V_{o}$ .
Bias current into the − input ( $V_{O S}$ set to zero). The virtual short holds $v_{-} = v_{+} = 0$ , so no current flows in the 1 kΩ resistor, because both its ends are at 0 V. All of $I_{B}$ must therefore reach the − input through the 100 kΩ feedback resistor, from the output: $V_{o 2} = I_{B} R_{f} = 1 nA \times 100 k Ω = 100 μ V$
Worst case. The polarity of $V_{O S}$ is not specified, so the worst case is when the two terms add: $∣ V_{o} ∣_{m a x} = 1010 μ V + 100 μ V = 1110 μ V$
So $V_{o} = \pm 1110 μ V$ .

KEY IDEA

The input offset voltage is amplified by the noise gain $1 + R_{f} / R_{1}$ , whatever configuration the circuit is drawn in. When the + input is grounded directly, the bias current into the − input is forced through $R_{f}$ alone, because the virtual short leaves no voltage across $R_{1}$ . That gives $I_{B} R_{f}$ . Superpose the two and add magnitudes for the worst case.

WATCH OUT FOR

Multiplying $V_{O S}$ by the inverting gain $R_{f} / R_{1} = 100$ gives 1000 µV instead of 1010 µV, which would be the correct figure only for a signal applied through $R_{1}$ . Other errors are dropping the bias-current term (answer 1010), or letting part of $I_{B}$ flow through the 1 kΩ resistor, when that resistor has 0 V across it.

From the GATE 2023 IN paper. Every question of this paper · Download the full paper · All 36 Op-amps questions, with notes

GATE 2023 IN Q.26: solution and answer

Answer and solution

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