AN-02 · Chapter 2 · Analog & Op-Amps

Inverting gain of −10, non-ideal

Q: Draw an op-amp with an inverting gain of −10. How does a real op-amp change your design?

Say this first: R1 = 10 kΩ input, Rf = 100 kΩ feedback, + input to ground (or to a bias voltage for single supply). Gain = −Rf/R1 = −10. Real-world effects Input bias current × Rf creates an output offset; add R = R1‖Rf (9.1 kΩ) on the + input to cancel it (bipolar inputs). Input offset voltage × noise gain (1 + Rf/R1 = 11) appears at the output. Bandwidth = GBW / noise gain = GBW/11, not GBW/10. Input impedance is only R1 (10 kΩ) — it loads the source. Resistor values: too low wastes current and loads the output; too high adds Johnson noise and makes bias-current errors worse.

Draw an op-amp with an inverting gain of −10. How does a real op-amp change your design?

Level	● screening
Track	General board
Really testing	Bias current, offset, GBW, and choosing resistor values.

Free full answer — one of 12 questions answered in full on this site.

Think first. Before you read on: which gain sets the bandwidth and the output offset — the signal gain of −10, or something else? And what impedance does the source see?

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Say this first: R1 = 10 kΩ input, Rf = 100 kΩ feedback, + input to ground (or to a bias voltage for single supply). Gain = −Rf/R1 = −10.

Real-world effects

Input bias current × Rf creates an output offset; add R = R1‖Rf (9.1 kΩ) on the + input to cancel it (bipolar inputs).
Input offset voltage × noise gain (1 + Rf/R1 = 11) appears at the output.
Bandwidth = GBW / noise gain = GBW/11, not GBW/10.
Input impedance is only R1 (10 kΩ) — it loads the source.
Resistor values: too low wastes current and loads the output; too high adds Johnson noise and makes bias-current errors worse.

Common mistake. Saying the bandwidth is GBW/10. Bandwidth and offset both scale with the noise gain, 1 + Rf/R1 = 11.

Going deeper

Put numbers on every non-ideal effect (example op-amp: GBW 1 MHz, Vos 2 mV, bipolar Ib 100 nA, Ios 10 nA):

Bandwidth = GBW / noise gain = 1 MHz / 11 ≈ 91 kHz — not 100 kHz.
Offset at the output = 2 mV × 11 = 22 mV.
Bias current: 100 nA × 100 kΩ = 10 mV at the output. Add 9.1 kΩ (R1‖Rf) in the + input and only Ios × Rf = 1 mV remains.

Choosing the resistor scale. 1 kΩ/10 kΩ: the feedback resistor loads the output, and the source sees only 1 kΩ. 100 kΩ/1 MΩ: 1 MΩ adds about 128 nV/√Hz of noise and makes bias-current errors 10× worse. 10 kΩ/100 kΩ is the usual middle.

Single supply. Tie the + input to a mid-rail reference (bypassed), and AC-couple the input — otherwise the output sits at a rail.

Where people go wrong

Saying the bandwidth is GBW/10.
Adding the bias-cancel resistor on a CMOS-input op-amp — it adds noise and fixes nothing (Ib is pA).
Forgetting that R1 is the input impedance.

More like this in the book

This is AN-02 from Hardware Engineer Interview Questions. The book has 16 questions on analog & op-amps and 141 in all, each answered the same way: what to say first, how to explain it, the numbers worth quoting and the follow-ups that come next.

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Inverting gain of −10, non-ideal

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