What is a Voltage Follower?
A voltage follower (also called a unity-gain amplifier, a buffer amplifier, and an isolation amplifier) is a op-amp circuit which has a voltage gain of 1.
This means that the op amp does not provide any amplification to the signal. The reason it is called a voltage
follower is because the output voltage directly follows the input voltage, meaning the output voltage is the same as the input
voltage. Thus, for example, if 10V goes into the op amp as input, 10V comes out as output. A voltage follower acts as a buffer, providing
no amplification or attenuation to the signal.
What is the Purpose of a Voltage Follower
One may ask then, what is the purpose of a voltage follower? Since it outputs the same signal it inputs, what is its purpose in a circuit? This will now be explained.
An op amp circuit is a circuit with a very high input impedance. This high input impedance is the reason voltage
followers are used. This will now be explained.
Voltage Followers Draw Very Little Current
When a circuit has a very high input impedance, very little current is drawn from the circuit. If you know ohm's law, you know that current, I=V/R. Thus, the greater the resistance, the less current is drawn from a power source. Thus, the power of the circuit isn't affected when current is feeding a high impedance load.
Let's look at both illustrations below:
The below circuit is a circuit in which a power source feeds a low-impedance load.
In this circuit above, the load demands and draws a huge amount of current, because the load is low impedance. According to ohm's law, again, current, I=V/R. If a load has very low resistance, it draws huge amounts of current. This causes huge amounts of power to be drawn from the power source and, because of this, causes high disturbances and use of the power source powering the load.
Now let's look at the circuit below, connected to an op-amp voltage follower:
This circuit above now draws very little current from the power source above. Because the op amp has such high impedance, it draws very little current. And because an op amp that has no feedback resistors gives the same output, the circuit outputs the same signal that is fed in.
This is one of the reasons voltage followers are used. They draw very little current, not disturbing the
original circuit, and give the same voltage signal as output. They act as isolation buffers, isolating a circuit so that
the power of the circuit is disturbed very little.
Voltage Followers Consume Practically All of the Voltage from a Voltage Divider Circuit
So current, as explained above, is one of the reasons voltage followers are used. They simply don't draw a lot of current, so they do not load down the power source.
Another reason voltage followers are used is because of voltage dividers. This again deals with ohm's law. According to ohm's law, voltage= current x resistance (V=IR).
In a circuit, voltage divides up or is allocated according to the resistance or impedance of components.
Because an op amp has a very high input impedance, the majority of voltage will fall across, since it's so high impedance. So it's very valuable when used in a voltage divider because it guarantees that it will receive the majority of voltage if placed in a voltage divider circuit.
This will now be illustrated so you can see.
So let's say we have a circuit shown below which represents a voltage divider with a load attached to the output.
So the above circuit will not work and it will be explained now why not.
So in the circuit above, we have a voltage divider between the 2 10KΩ resistors. Since the resistors are of equal value and the power source is 10V, the voltage at the output where the load is connected to is 5V.
In this circuit, there's actually 2 voltage dividers. The first is between the 2 10KΩ resistors, which gives an output of 5V. And the second voltage is now between the bottom 10KΩ resistor and the 100Ω load. So there's actually 2 voltage dividers.
Now let's say the load needs about 5V to operate, which is why you have the first voltage divider, to give 5V out of the 10V power supply. However, it will not receive 5V.
We can see why if we look at ohm's law. According to ohm's law, voltage is allocated according to V=IR.
We have a voltage divider between a 10KΩ resistor and a 100Ω load, dividing the 5V output from the first voltage divider. According to ohm's law, the 10KΩ resistor will receive 4.95V out of the 5V. This is formula for voltage allocation in a voltage divider is V= VIN(R2/R1 + R2, which is V= 5V (10KΩ/10.1KΩ =4.95V). The 100Ω load receives about 0.5V (V= 5V(100/10.1KΩ). Therefore, it does not receive close to the 5V it needs to operate and it does not work.
This is because the impedance is too low compared to the 10KΩ resistor.
However, if you take an op amp (with its very high impedance) and input the voltage from the first voltage divider, it has an extremely high impedance (several megohms) compared to the 10KΩ resistor and receives practically all of the 5V from the voltage divider.
This is shown below.
So this circuit above now works.
After the first voltage divider, 5V is output.
We then buffer the circuit using a voltage follower so that practically all of the 5V gets allocated across the op amp (and not the 10KΩ resistor). This ensures that the load receives enough power to operate.
So these are the 2 chief reasons we use voltage followers. We either don't want to load down the current or we want to buffer the output voltage from a circuit so that practically all of the voltage gets allocated across the voltage follower, which can power on a load.
Voltage followers are important to isolate or buffer a low impedance load from a voltage source. This
means that rather than
connect a relatively low value of load resistance across the terminals of the power source, the op amp can be used to
eliminate any loading that might occur. Thus, the power source will not be loaded down. And since the voltage follower has very little output imepdance, the circuit acts as an ideal voltage
source with nearly zero internal impedance, outputs the full voltage to the output.