The inverting amplifier is the op-amp circuit that flips a signal upside down while making it bigger: a positive input comes out negative, and a negative input comes out positive. It is the partner of the non-inverting amplifier from the morning post — same op-amp, different wiring, and one important difference: its input resistance is set by a resistor, not by the op-amp itself. This post designs one: a 0.5 V signal amplified to −5 V.

Figure 1: The inverting amplifier (a) and its frequency response (b). The input goes through R₁ into the − input, which the feedback holds at 0 V (virtual ground); the gain is −R_F / R₁. (Schematic, not to scale.)

How the gain is set — the virtual ground

With negative feedback the op-amp keeps its two inputs at (almost) the same voltage. Here the + input is tied to ground, so the − input is held at (almost) 0 V — a virtual ground. The op-amp's input draws almost no current, so the current through R₁ has nowhere to go except through R_F:

I = V_in / R₁  ·  V_out = −I × R_F = −(R_F / R₁) × V_in

The minus sign is the inversion: the output must go negative to pull the feedback current through R_F while the − input stays at 0 V. The input resistance of the whole circuit is simply R₁ — the source sees a 1 kΩ load, not the op-amp's 2 MΩ. That is the trade-off against the non-inverting circuit, which keeps the source untouched.

Worked example — 0.5 V in, −5 V out

Take an input of 0.5 V and choose R₁ = 1 kΩ and R_F = 10 kΩ:

Gain A = −10,000 / 1,000 = −10  ·  V_out = −10 × 0.5 = −5 V

Check the current: I = 0.5 V / 1 kΩ = 0.5 mA flows through R₁ and then through R_F, giving V_out = −0.5 mA × 10 kΩ = −5 V — the same answer by the current route. Check the rails: with ±12 V supplies a 741 can swing to about ±10 V, so −5 V is comfortably inside the range and will not clip. The feedback resistor draws only 0.5 mA, which the op-amp supplies easily.

Now check the source. The circuit's input resistance is R₁ = 1 kΩ, so the source's own resistance matters. A sensor with 10 kΩ source resistance forms a divider with R₁: only 0.5 × 1 kΩ / (10 kΩ + 1 kΩ) = 0.0455 V reaches R₁, and the output is only −0.45 V — 9% of what was intended. With a low-impedance source of 100 Ω the loss is small: 0.5 × 1 kΩ / (100 Ω + 1 kΩ) = 0.4545 V in, −4.55 V out (91%). Rule of thumb: drive an inverting amplifier from a low-impedance source, or put a voltage follower (from the earlier post) in front of it.

One small error to check: the op-amp's input bias current (about 80 nA for a 741) flows through R₁ in parallel with R_F (909 Ω), making an input offset of 80 nA × 909 Ω = 0.073 mV, which appears at the output multiplied by the gain: 0.073 mV × 10 = 0.73 mV. Against a −5 V output that is 0.015% — negligible here, but it is why precision circuits use low-bias-current op-amps.

Bandwidth — the gain-bandwidth trade-off

The 741 has a gain-bandwidth product of about 1 MHz: gain × bandwidth stays constant, so the more gain you ask for, the lower the frequency where the gain starts to fall:

f_H = 1 MHz / 10 = 100 kHz

Up to 100 kHz the gain stays at 10 (20 dB). At 100 kHz it has fallen to 10 / √2 = 7.07 (−3 dB), so a 0.5 V input gives only −3.54 V out, and at 1 MHz the gain reaches 1. The output is inverted at every frequency — the minus sign never goes away, only the size shrinks.

Design — gain −5 for a 1 V signal, and a free summing amplifier

Suppose a signal of 1 V must become −5 V. The gain must be −5, so keep R_F = 10 kΩ and choose R₁ = 10 kΩ / 5 = 2 kΩ:

Gain A = −10 / 2 = −5  ·  V_out = −5 × 1 = −5 V  ·  I = 1 V / 2 kΩ = 0.5 mA

2 kΩ is a standard E24 value. The new bandwidth is 1 MHz / 5 = 200 kHz — less gain buys more bandwidth, as always.

The same circuit also adds signals for free. Because the − input is a virtual ground, each input current is independent, so two inputs sum at the output: V_out = −R_F × (V₁ / R₁ + V₂ / R₂). Feed 0.5 V through a 1 kΩ resistor and 0.25 V through a 2 kΩ resistor into the same node with R_F = 10 kΩ:

V_out = −10 kΩ × (0.5 mA + 0.125 mA) = −6.25 V

That is the summing amplifier — the circuit at the heart of audio mixers and digital-to-analogue converters.

Key points to remember

• Inverting gain: A = −R_F / R₁ — the − input is a virtual ground (0 V) because the + input is grounded and feedback holds the two inputs equal.
• Worked example: R₁ = 1 kΩ, R_F = 10 kΩ give a gain of −10, so 0.5 V in becomes −5 V out, with 0.5 mA flowing through both resistors — inside the ±10 V swing of a 741 on ±12 V supplies.
• Input resistance = R₁ = 1 kΩ: a 10 kΩ source loses 91% of its signal (output only −0.45 V), while a 100 Ω source loses 9% (output −4.55 V) — drive it from a low-impedance source or a voltage follower.
• Bias-current offset: 80 nA × 909 Ω = 0.073 mV at the input, 0.73 mV at the output — negligible against −5 V.
• Gain-bandwidth product is constant: with a 1 MHz op-amp, gain 10 gives f_H = 100 kHz (gain 7.07 at the −3 dB point, gain 1 at 1 MHz).
• Design example: for a 1 V signal and gain −5, R₁ = 2 kΩ and R_F = 10 kΩ give −5 V out, with bandwidth 200 kHz.
• Bonus: the same circuit is a summing amplifier — 0.5 V through 1 kΩ plus 0.25 V through 2 kΩ, with R_F = 10 kΩ, gives −6.25 V out.