Op Amp Voltage follower (Op Amp voltage Buffer)
To better understand the operation of a Op Amp Voltage follower (Op Amp Voltage Buffer), we must remember the operation of a Op Amp non-inverting amplifier. See the diagram below.
Op Amp Non-inverting amplifier
The Op-Amp non-inverting amplifier gain is given by the formula: 1 + (R2/R1).
If we make R2 resistor value equal to zero and make the R1 resistor value very large (infinity), we have an amplifier with gain G = 1. This means that the output signal has the same value as the input signal. See the following diagram.
Op Amp Voltage Follower (Op Amp Voltage Buffer)
A buffer has an output that is exactly as the input. This behavior may initially seem useless, but has features that help solve impedance coupling problems.
- The input impedance of a buffer with an operational amplifier is very high, close to infinity
- The output impedance is very low, just a few ohms.
If the input impedance is very high, it does not charge the circuit that is sending the signal, and if its output impedance is low, it can deliver a sufficient amount of current to the circuit receiving the signal.
In other words a Buffer requests very little current from the circuit that gives the signal and greatly increases the capacity to delivery current to the circuit that receives the signal.
Voltage follower advantages
- Provides power gain and current gain. (voltage gain = 1).
- Low output impedance. Loading effects can be avoided.
- High input impedance. Op-amp takes no current from the input.
Voltage follower applications
- Sample and hold circuits.
- Buffers for logic circuits.
- Active filters. Voltage followers can be used to isolate filter stages from each other, when building multistage filters.
Op Amp voltage follower example. Op Amp voltage buffer example
We need to get 6 volts from a 12 volt source to power a 100 ohm load resistor (RL).
We use two 100K resistors in series as a voltage divider (R1, R2). Since the resistors have the same value, the voltage between them is exactly 6 volts (A).
If we connect the 100 ohm load to point A, the voltage will no longer be 6 volts, because the lower 100K resistor is in parallel with the load. The equivalent resistance of parallel resistors is: 100,000 x 100 / (100,000 + 100) = 99.9 ohms. We can safely assume the approximate value of 100 ohms.
Using the voltage divider again, the voltage at point A is: VA = 12 / (100K + 100) x 100 = 0.012 volts. Very different from the expected 6 volts.
To avoid this problem we use a op amp voltage follower. The non-inverting input of op amp is connected to A and the output is connected to the 100 ohm resistor (load). The load will have exactly 6 volts between its terminals.