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Analog10 min read
Voltage Buffers
Learn why a unity-gain circuit is useful even though its output voltage is almost equal to its input voltage.
Mental model
A buffer is an interpreter between a weak source and a demanding load: it repeats the same voltage but supplies the current so the source does not collapse.
Vin enters the + input, and Vout connects directly to the − input. Negative feedback gives approximately unity voltage gain while isolating the signal source from the load. Op-amp supply pins are omitted.Scroll sideways to view the complete circuit.UNDERSTAND THE BEHAVIOR
01
A source with nonzero output resistance loses voltage when a load draws current. A buffer places a very high input impedance at the source, so almost no input current flows and almost no source voltage is lost.
02
Negative feedback connects the op-amp output directly to the inverting input. The op-amp drives its output until V− nearly equals V+, producing a closed-loop voltage gain close to one while greatly reducing output impedance.
03
Unity gain does not mean unlimited performance. Confirm that the op-amp is unity-gain stable, accepts the input common-mode voltage, reaches the required output voltage, supplies the load current, and responds fast enough.
Why it matters
Buffers isolate sensor dividers, references, filters, ADC inputs, and other stages that should not load one another.
Core ideas
An ideal voltage buffer has infinite input impedance, zero output impedance, and gain of one.
The source sees very little loading while the buffer drives the next stage.
Real buffers are limited by common-mode range, output swing, output current, bandwidth, slew rate, and stability.
1A sensor behaves like a 100 kΩ source and produces 2.5 V.
2Connecting a 10 kΩ load directly forms a divider and gives only 2.5 × 10/(100+10) ≈ 0.227 V.
3A suitable buffer presents a much larger input impedance, so it reads nearly 2.5 V.
4Its low output impedance then supplies about 2.5/10 kΩ = 0.25 mA to the load.
Common mistake
Do not assume every op-amp is stable at gain one. Some high-speed devices require a specified minimum closed-loop gain.
RememberAn ideal voltage buffer has infinite input impedance, zero output impedance, and gain of one.
DEEPER EXPLANATION
How can gain one still make a circuit useful?
An ideal op-amp voltage follower connects its output directly to its inverting input and applies the source to its non-inverting input. Negative feedback then drives Vout close to Vin. The voltage gain is near one, but the buffer changes how much current the source must supply to the next stage.
A high input impedance draws very little current from a weak source. A comparatively low output impedance helps the op-amp drive a load without the source voltage collapsing. Imagine a sensor modeled as a 10 kΩ source resistance feeding a 10 kΩ load: without a buffer, the load sees only half the source's open-circuit voltage. With a suitable follower between them, loading of the sensor is much smaller.
A real buffer is limited by its supply rails, input common-mode range, output current and bandwidth. Some op-amps are unstable at unity gain, and a capacitive load can make stability worse. Check the datasheet and output load before selecting a part.
Walk through another example
A 1 V source with 10 kΩ series resistance drives a 10 kΩ load directly: the divider gives 1 × 10/(10 + 10) = 0.5 V at the load.
Insert an ideal voltage follower: its input draws negligible current, so its input remains near 1 V.
Its output drives the 10 kΩ load near 1 V, provided its output swing and current limits allow this voltage.
Check your intuition
The buffer does not create voltage gain above one. Its advantage is isolating a source from loading by the next stage.
Where this model can mislead you
Do not assume a follower can supply unlimited current or work outside its specified common-mode and output range.
TEST YOUR UNDERSTANDING
Three quick checks
Choose one answer per question, then check your work. Each answer includes an explanation.
Practice and check your understanding
Try each question before opening the worked answer.
1. An ideal unity-gain buffer receives 1.2 V and drives 1 kΩ. Find Vout and load current.
Show answer and reasoning
Vout = 1.2 V. Iload = Vout/Rload = 1.2/1000 = 1.2 mA. A real device must support that current and voltage; the energy comes from its supply, not from the ideal signal source.
2. Why use a buffer if it does not increase voltage?
Show answer and reasoning
It isolates a high-impedance source from a load. Its high input impedance reduces source loading, and its low output impedance can supply load current. Check unity-gain stability and capacitive-load behavior.