The non-inverting amplifier is the op-amp circuit you reach for when a small sensor signal needs to be made bigger without loading the sensor. The signal goes into the + input, the gain is set by two resistors, and the input resistance is so high that the sensor barely notices the amplifier is there. This post designs one: a 0.5 V sensor signal amplified to 5.5 V.

Figure 1: The non-inverting amplifier (a) and its frequency response (b). The divider R₁–R_F samples the output and feeds a fraction back to the − input; the gain is 1 + R_F / R₁. (Schematic, not to scale.)

How the gain is set

With negative feedback the op-amp keeps its two inputs at (almost) the same voltage — the virtual short. The + input sits at V_in, so the − input sits at V_in too, and R₁ and R_F form a divider between the output and ground with V_in at their junction. That divider gives:

V_out / V_in = 1 + R_F / R₁

Unlike the inverting amplifier, the input signal goes straight into the op-amp's + terminal, which draws almost no current. The input resistance of the whole circuit is the op-amp's own input resistance — about 2 MΩ for the classic 741 — so a sensor with a 10 kΩ source resistance loses only 10 kΩ / (2 MΩ + 10 kΩ) = 0.5% of its signal: 0.5 V arrives at the + input as 0.4975 V. And the gain can never fall below 1 — with R_F = 0 the circuit becomes the voltage follower covered in the earlier post.

Worked example — 0.5 V in, 5.5 V out

Take a sensor that gives 0.5 V and choose R₁ = 1 kΩ and R_F = 10 kΩ:

Gain A = 1 + 10,000 / 1,000 = 11  ·  V_out = 11 × 0.5 = 5.5 V

Check the supply rails: with ±12 V supplies a 741 can swing to about ±10 V, so 5.5 V is comfortably inside the range — the output will not clip. The feedback divider itself draws only V_out / (R₁ + R_F) = 5.5 / 11,000 = 0.5 mA, which the op-amp supplies easily.

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 × 11 = 0.8 mV. Against a 5.5 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

An op-amp cannot hold its gain at all frequencies. 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 / 11 = 90.9 kHz

Up to 91 kHz the gain stays at 11 (20.8 dB). At 91 kHz it has fallen to 11 / √2 = 7.78 (−3 dB), and at 1 MHz it reaches 1. For audio (20 kHz) and slow sensor signals this is plenty; for a 500 kHz signal you would need a faster op-amp or less gain.

Design — gain 21 for a 0.25 V sensor

Suppose a different sensor gives only 0.25 V and you need about 5 V out. The gain must be 5 / 0.25 = 20, so aim a little higher, 21, keeping R₁ = 1 kΩ:

R_F = (21 − 1) × 1 kΩ = 20 kΩ  ·  V_out = 21 × 0.25 = 5.25 V

20 kΩ is a standard E24 value. Check the rails again: 5.25 V is inside the ±10 V swing, so no clipping. The new bandwidth is 1 MHz / 21 = 47.6 kHz — the price of the extra gain. If you needed both the gain of 21 and 100 kHz bandwidth, a single 741 could not do it (21 × 100 kHz = 2.1 MHz > 1 MHz); you would split the gain across two stages or pick a faster op-amp.

Key points to remember

• Non-inverting gain: A = 1 + R_F / R₁ — the input goes to the + terminal, so the gain is never less than 1 and the input resistance is the op-amp's own (about 2 MΩ for a 741).
• Worked example: R₁ = 1 kΩ, R_F = 10 kΩ give a gain of 11, so 0.5 V in becomes 5.5 V out — inside the ±10 V swing of a 741 on ±12 V supplies.
• Bias-current offset: 80 nA × 909 Ω = 0.073 mV at the input, 0.8 mV at the output — negligible against 5.5 V.
• Gain-bandwidth product is constant: with a 1 MHz op-amp, gain 11 gives f_H = 90.9 kHz (gain 7.78 at the −3 dB point, gain 1 at 1 MHz).
• Design example: for a 0.25 V sensor and gain 21, R_F = 20 kΩ gives 5.25 V out, with bandwidth 47.6 kHz.
• A sensor with 10 kΩ source resistance loses only 0.5% of its signal into a 2 MΩ input — that is why the non-inverting circuit is the right choice for weak sensors.