A voltage follower is the strangest amplifier in electronics: its gain is exactly one — the output is just a copy of the input. So why use an amplifier that does not amplify? Because it fixes a loading problem. A weak sensor that cannot drive a load directly can drive a follower easily, and the follower then drives the load with the full signal. This post shows the numbers: the same sensor and load, with and without the buffer.

Figure 1: The voltage follower (a) and what it does to the load voltage (b). Direct connection loses 99% of the signal in the source resistance; the follower delivers the full 2.5 V. (Schematic, not to scale.)

The problem — a weak source driving a heavy load

Take a sensor that produces 2.5 V but has an internal (source) resistance of R_S = 10 kΩ, and connect it straight to a 100 Ω load. The source resistance and the load form a voltage divider, so the load gets only

V_load = 2.5 × 100 / (10,000 + 100) = 2.5 × 0.0099 = 24.8 mV

— barely 1% of the sensor's voltage. The power in the load is V²/R = 0.0248² / 100 = 6.13 µW. The rest of the signal is wasted across the source resistance. Any high-resistance source — a sensor, a potentiometer wiper, a microphone — has this problem when it meets a low-resistance load.

The follower — gain one, impedance transformed

Connect the sensor to the op-amp's non-inverting (+) input and wire the output straight back to the inverting (−) input, as in Figure 1(a). With negative feedback the op-amp drives its output until the two inputs are equal (V₋ = V₊), so V_out = V_in. This is the non-inverting amplifier formula with no feedback resistor: gain = 1 + R_f/R₁ = 1 + 0 = 1.

What changes is the impedance. An op-amp input draws almost no current, so almost nothing flows through R_S and almost no voltage is lost across it. The output, meanwhile, is a low-impedance source that can push real current into the load. The load current comes from the op-amp's power supply, not from the sensor.

Worked example — the same sensor and load, buffered

With the follower in place, the load sees the full 2.5 V:

V_load = 2.50 V  ·  I_out = 2.5 / 100 = 25.0 mA  ·  P_load = 2.5² / 100 = 62.5 mW

That is 101 times the voltage and roughly 10,000 times the power of the direct connection (62.5 mW / 6.13 µW ≈ 10,200). Nothing was amplified — the follower simply stopped the source resistance from eating the signal.

How exact is it? A real op-amp draws a tiny input bias current — for a common general-purpose op-amp of the 741 class, typically around 80 nA. That current flows through R_S and drops 80 nA × 10 kΩ = 0.8 mV, an error of just 0.03% of 2.5 V. With a modern FET-input op-amp the bias current is thousands of times smaller still, so the error vanishes. (The follower's input impedance is, in theory, the open-loop gain — about 200,000 for a 741 — times the op-amp's own input resistance: effectively hundreds of gigaohms.)

Two limits to respect

First, the output cannot go beyond the supply rails: a follower on a single 5 V supply cannot deliver a clean 5 V output, only something a volt or two below it — use a rail-to-rail op-amp if the output must swing close to the rails. Second, the output current is limited: a classic 741-class op-amp can source only about 25 mA, so the 25 mA in this example is right at its limit. For a heavier load, pick an op-amp rated for more output current, or add a transistor after the follower.

Key points to remember

• A 2.5 V sensor with 10 kΩ source resistance delivers only 24.8 mV (6.13 µW) into a 100 Ω load directly — 99% of the signal is lost in the source resistance.
• The voltage follower (output wired to the − input) has gain 1 + R_f/R₁ = 1, so V_out = V_in — but it presents a very high impedance to the source and a very low impedance to the load.
• Buffered, the same load gets the full 2.50 V, 25.0 mA and 62.5 mW — 101× the voltage and about 10,000× the power; the current comes from the op-amp's supply, not the sensor.
• Error is tiny: a typical 80 nA input bias current drops only 0.8 mV across the 10 kΩ source resistance (0.03% of 2.5 V).
• Respect the limits: the output cannot exceed the supply rails, and a 741-class op-amp can source only about 25 mA — check the datasheet before driving a heavy load.