---
title: "Input impedance of inverting op-amp"
subject: "Advanced Electronics"
chapter: 2
key: "semester_4/advanced_electronics:inverting-input-impedance"
source: https://www.bctengineer.com/q/semester_4/advanced_electronics/inverting-input-impedance
site: BCT Engineering
---

# Input impedance of inverting op-amp

Advanced Electronics · Chapter 2: Operational Amplifier Characterization

- Asked in: 2072 Kartik, 5 marks
- Video: [L2 1 3 Input Resistance of OpAmp Configurations](https://www.youtube.com/watch?v=rUiAK-qrMDc) — Lee Brinton
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/advanced_electronics/inverting-input-impedance
- Whole chapter: https://www.bctengineer.com/md/semester_4/advanced_electronics/chapter-2

## Question

Draw a circuit diagram of inverting op-amp and derive the expression of input impedance.

## Short answer

Input impedance of an inverting amplifier is Rin = R1 + Rf/(1+A), where A is open-loop gain. Because the inverting terminal acts as a virtual ground, this simplifies to Rin ≈ R1.

## Answer

The input impedance of the inverting amplifier is the resistance seen by the signal source, Rin = vin/iin. Take a practical op-amp with open-loop gain A and very large input resistance Ri.

```text
                Rf
       +------/\/\/-----+
       |                 |
 vin --R1--+--(-)        |
  -> i1    |        >----+---- vo
       GND-+--(+)
```

The input current i1 flows through R1 into node v− (the op-amp input current is negligible since Ri is very large).

```text
vo = -A·v-          (v+ = 0)
Current through Rf:
  iF = (v- - vo)/Rf = v-(1 + A)/Rf
So, seen from node v-, Rf looks like
  Rf/(1 + A)        (Miller effect)
Node v- to ground: Ri ‖ Rf/(1 + A) ≈ Rf/(1 + A)
Input resistance:
  Rin = vin/i1 = R1 + Rf/(1 + A)
Since A is very large:
  Rin ≈ R1
```

So the input resistance of the inverting amplifier is set by R1, because the inverting input is a **virtual ground**. Example: R1 = 10 kΩ, Rf = 100 kΩ, A = 2 × 10⁵: Rin = 10 kΩ + 0.5 Ω ≈ 10 kΩ.

**Answer: Rin = R1 + Rf/(1 + A) ≈ R1.**

This is low compared with the non-inverting amplifier (Rif = Ri(1 + Aβ)), so the source must be able to drive R1; R1 is chosen large enough (e.g. 10 kΩ or more) not to load the source.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is the input resistance of an inverting amplifier about R1?
   A. The inverting input is a virtual ground, so the source sees R1 to ground
   B. The op-amp input resistance Ri is small
   C. Rf is in series with the source
   D. The non-inverting input draws large current
   Answer: A. With v- near 0, i1 = vin/R1, so Rin is about R1. (Key point: Virtual ground)

2. Which relation gives the current through Rf in the derivation?
   A. vo = +A*v-, so iF = 0
   B. vo = -A*v-, so iF = v-(1 + A)/Rf
   C. iF = vin/Rf
   D. vo = -v-/A
   Answer: B. iF = (v- - vo)/Rf with vo = -A v- gives v-(1 + A)/Rf. (Key point: Output relation used)

3. The derived expression for input resistance is:
   A. Rin = R1 || Rf
   B. Rin = R1(1 + A)
   C. Rin = R1 + Rf/(1 + A), which is about R1
   D. Rin = Ri(1 + A*beta)
   Answer: C. Rf/(1 + A) is tiny for large A, so Rin is about R1. (Key point: Result)

4. For R1 = 10 kohm, Rf = 100 kohm, A = 2x10^5, Rin is about:
   A. 110 kohm
   B. 100 kohm
   C. 10 Mohm
   D. 10 kohm (R1 + 0.5 ohm)
   Answer: D. Rf/(1+A) = 100k/200001 = 0.5 ohm. (Key point: Numerical check)

## Related in Advanced Electronics

- [Find coefficients of op-amp summing circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Asumming-amplifier-coefficients) — Advanced Electronics, chapter 1
- [Short note: Noise in op-amp circuit](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Ashort-note-opamp-noise) — Advanced Electronics, chapter 2
- [Output voltage of two op-amp IA](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Atwo-opamp-ia-output) — Advanced Electronics, chapter 4
- [Single op-amp IA with mismatched resistances](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Asingle-opamp-ia-mismatched) — Advanced Electronics, chapter 4
- [IA features; 3 op-amp IA output voltage](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Aia-features-3opamp-output) — Advanced Electronics, chapter 4
- [3 op-amp IA output and common-mode rejection](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Athree-opamp-ia-cm-rejection) — Advanced Electronics, chapter 4

## Related in other subjects

- [Ideal op-amp properties; non-inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 3×
- [Ideal op-amp properties; inverting gain](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-props-inverting-gain) — Basic Electronics Engineering, chapter 4, asked 3×
- [Voltage gain of non-inverting op-amp](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Aopamp-noninverting-gain) — Basic Electronics Engineering, chapter 4, asked 2×
- [Non-inverting op-amp voltage gain](https://www.bctengineer.com/question.md?key=semester_1%2Ffundamental_of_electrical_and_electronics_engineering%3Anon-inverting-amplifier-gain) — Fundamental of Electrical and Electronics Engineering, chapter 6