Introduction to Unity Gain Buffers

A unity gain buffer (aka voltage follower) is an op amp circuit whose output voltage follows the input voltage. This may not seem to be very useful, but there are all sorts of reasons that you might use one in a circuit. This video describes some of it's uses and then derives the equation for the relationship between input voltage and output voltage (Yes the gain is 1, but that is only when the open loop gain of the op amp is infinite).
This video also shows a simple simulation example using LTSpice
To see other common operational amplifier configurations check out:
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[@boonedockjourneyman7979]

Really good. I’m using your videos with students who are struggling. They help.

[@ElectronXLab]

I'm so glad to hear that. Happy they help!

[@j0mell0]

I have a simple question, and it may depend on its application. But what if we supply(Vin) to the negative(-) input instead while the positive(+) terminal is tied to Vout. I believe this configuration is also Vout = Vin ? It's more common to supply the positive input... why is it ?

[@tusharjamwal]

Hi, just for clarification, "Heavy load" is a load of low impedance which would naturally cause more current to flow through it. Did I get that right?

[@zulvan3880]

I think heavy load means that the impedance is high.Therefore, it causes voltage drop

[@sevildogan8279]

I am also wondering the same thing. The high current it draws is the main problem here from my understanding. The load requires high current which is causing a high voltage drop from the Zout as a result the load is not provided with sufficient voltage. Thats what I have understood from the video.

[@chocolatelasagna3328]

great resource, thanks for sharing

[@dalenassar9152]

GREAT VIDEO! What if you put a feedback resistor in the feedback loop and input a signal in through an input resistor to the "-" input WHERE THE RESISTORS ARE EQUAL. How would the output change as you varied the voltage directly at the "+" (follower) input?

[@RAHUL-xy1ds]

1:05 "The voltage at the output will match the voltage at the input" But a good amount of voltage has already dropped across the high impedance before the unity gain buffer right?

[@ElectronXLab]

No voltage will drop across the high impedance - don't think of the connection between input and output as a continuous connection that current must flow. Think of it more as -a voltage sensor at the input -a driver at the output that will - determine what voltage the output should be based on the "sensors" and the gain of the amplifier - provide power to the load (from the voltage sources) at the voltage determined fro the step above.

[@RAHUL-xy1ds]

@@ElectronXLab Is it right to say that little to no current flows through the source's output impedance BECAUSE of the infinite impedance at the input of the op amp?

[@ElectronXLab]

@@RAHUL-xy1ds For this unity gain buffer circuit, that is correct.

[@RAHUL-xy1ds]

@@ElectronXLab okay Thank you David

[@joyceyu2038]

What kind of op amp was used in the LTspice simulation?

[@thomasrosebrough9062]

I am just not understanding at all. I don't understand how "Vout = Vin" can be true if the op amp is actually doing anything. On the diagram, if R2 caused a voltage drop, why would the op amp not exactly mirror that voltage drop? What causes Vb to match V1, where is the extra difference to make up the drop in voltage coming from?

[@ElectronXLab]

The voltage drop across R2 occurs because the current through R2 is the same as the current through R4. In the op amp circuit, the same voltage drop doesn't occur across R3, because there is almost no current (ideally 0) going through R3. Then the feedback of the op amp causes Vb to be the same as the voltage at the non-inverting input of the op amp. The op amp makes sure that Vb is effectively the exact same as the source voltage...I hope that makes sense

[@grzegorzbrzeczyszczykiewic199]

2:27 how the hell do you add 1 to some voltage? 1 of what?

[@moradtamer]

You add 1 to the voltage gain which is a dimensionless quantity, not the actual voltage

[@user-cq4zk8mw6r]

Hi welcome to you.

[@21thTek]

didnt like it at all, just university very theoric aproach which is excelent but no protoboard real life cicrcuit sample or theoric aplications, good to sleep .....

[@corruptofficial7794]

Hi there! It is not all the ways to use the buffer