Buffer amplifier

A buffer amplifier has roughly unity voltage gain, very high input resistance, and very low output resistance. It doesn’t make a signal bigger, it isolates a weak source from a heavy load so the connection itself doesn’t destroy the signal.

The loading problem it solves

Any real source can be drawn as its Thévenin equivalent: an ideal voltage $v_S$ behind a source resistance $R_S$. Connect it to a load $R_L$ and the load only receives the divided voltage

$v_L=v_S\frac{R_L}{R_S+R_L}$

(a Voltage divider). If $R_S$ is comparable to or larger than $R_L$ (a high-impedance sensor driving a low-impedance stage, a long cable, a divider tap feeding the next block), most of the signal is lost across $R_S$ before it reaches the load. The fix isn’t more gain, it’s impedance transformation. Insert a buffer:

• high input resistance ($R_{in}\to \infty$) means it draws almost no current from the source, so $v_S$ appears at the buffer input undivided;
• low output resistance ($R_{out}\to 0$) means it presents a stiff voltage to $R_L$, so the load gets the full voltage with no further division.

The voltage is unchanged (gain $\approx 1$) but the impedance chain is broken: the source now sees infinite load, the load now sees zero source. It’s the Input and output resistance (amplifier) idea applied deliberately.

Where it shows up

The cleanest buffer is the op-amp Voltage follower (op-amp): infinite $R_{in}$, zero $R_{out}$, gain exactly $1$ from Negative feedback. The single-transistor versions are the MOSFET Source follower (common-drain) and the BJT Emitter follower (common-collector). Both have voltage gain just under unity, high input resistance, and low output resistance, which is why they sit at the output of a multi-stage amplifier driving the final load, or between a high-gain stage and a low-impedance line. A buffer is a real circuit’s best attempt at an ideal voltage source: it copies a voltage and supplies whatever current the load demands without letting the load disturb the original.