# Trying to understand op-amps...

#### bob800

Joined Dec 10, 2011
50
Hello,

Here's the diagram that's used: Let's assume that V(in) = 5 volts, and the gain of the op-amp is 4. My understanding is that the op-amp will find the difference between the non-inverting input (5V) and the inverting input (0V... right?)

Assuming that's correct, then the op-amp would find a voltage difference of 5 volts, multiply it by 4, and output 20 volts. The 20 volts would then be fed back into the inverting output, where it would be compared once again to 5 volts (a difference of 15 volts). Anyway, the op-amp would multiply 15 volts by 4 and output 60 volts!! The output voltage gets higher and higher!

Obviously, this is not correct, since the op-amp should eventually output 5 volts... Can anyone please explain where my error lies?

#### bertus

Joined Apr 5, 2008
20,102
Hello,

The schematic shown is a voltage follower (also called buffer).
It has ALWAYS a gain of 1 in that configuration.

Bertus

#### JMac3108

Joined Aug 16, 2010
349
The first problem is that the diagram you show is for a unity gain amplifier, also called a voltage follower. It has a gain of 1.

As a beginner, here is a good way to think about op-amp circuits...

If the circuit has negative feedback, meaning it has a connection from its output back to its inverting input, then the op-amp will drive its output to whatever level is required to make its inputs equal.

So in the case you show where the gain is 1, the op-amp drives its output equal to its input so that both of its inputs are equal.

Does this help clarify it for you?

#### SgtWookie

Joined Jul 17, 2007
22,210
Well, in the case of the way that particular opamp is wired, the gain would be 1; it is wired as a voltage follower.

In the case of negative divided feedback, an opamp tries to adjust its' output so that the voltage on the inverting (-) input is the same as the voltage on the non-inverting (+) input.

In your example, the opamp has a pretty easy time of accomplishing that. There will be a small error, caused by what is called "input offset", which usually amounts to a few millivolts. If the gain is higher than 1, the error will be multiplied by the gain.

You need to go to the next page before you start reading about divided negative feedback.

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#### SgtWookie

Joined Jul 17, 2007
22,210
Dogpile! #### JMac3108

Joined Aug 16, 2010
349 Sometimes it takes more than once to get it!

#### bob800

Joined Dec 10, 2011
50

I understand that, once the op-amp has reached a sort of "equilibrium", that it will assume a gain of 1.

However, what I don't understand is how it gets to that point. I've attached a image of my (obviously flawed) thought process. You'll notice that I've assumed the op-amp has a gain of 4. Please be aware that this is NOT the gain of the whole circuit including the feedback path, but only of the op-amp itself. I know the actual gain would be more like 200,000, but I've just cut it down to 4 to keep my diagram simple (the concept stays the same).

Please let me know if I'm not making any sense #### Attachments

Joined Dec 26, 2010
2,148
No, this is not correct. I think your mistake comes from not realising that because the op-amp is a differential amplifier, the value and sign of its output depends on the difference of the voltages at the non-inverting and inverting inputs. Your example has the sign of the output the wrong way round.

Edit: Another issue is that amplifiers do not work in the kind of step-by-step iterative process you are imagining. The output varies continuously in response to the input, and if the circuit is correctly designed an equilibrium point will indeed be reached.

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#### crutschow

Joined Mar 14, 2008
23,514
If the input node was the negative input and the feedback went to the positive input then what you describe would happen. The voltage would keep getting higher until it reached the maximum output voltage the amp can generate. That mode is sometimes used to make a comparator with a digital type output.

But the feedback goes to the negative input, so it works as Adjuster noted. If the output voltage tries to go more positive then the positive input, then this voltage on the negative input will reduce the output voltage to make it equal to the positive input. This also works if the output tries to go more negative than the positive input. Thus the end result is that the output voltage is (ideally) always the same as the input voltage, giving a voltage follower with a gain of +1.

#### PaulEE

Joined Dec 23, 2011
423
bob800,
Take a look at this derivation. This is the theoretical reason why that circuit physically has a gain of one. Hope this helps you understand in a more intuitive/theoretical fashion.
-Paul Happy new year.

#### rsashwinkumar

Joined Feb 15, 2011
23
First of all, you need to remember that in any analog circuit, the max output voltage is limited by the DC POWER SUPPLY, Op-amps are designed to work within a dc supply range of 3-15V. So if u use a 15V supply, even though the output may tend to go to 20V, the actual voltage u would observe would only be 15V. This is saturation of output voltage.

Now that doesnt actually explain your argument..
Going by your way, op-amp amplifies the difference between its input terminals and provides at the output.

i.e., Vout = A(V+ - V-) ;

Here V- = Vout
Put it in the above equation, and you will derive at the output expression as

Vout = A / (A + 1) * V+;

So this is obviously less than 1. So output doesnt saturate this case.

#### Ron H

Joined Apr 14, 2005
7,014
First of all, you need to remember that in any analog circuit, the max output voltage is limited by the DC POWER SUPPLY, Op-amps are designed to work within a dc supply range of 3-15V. So if u use a 15V supply, even though the output may tend to go to 20V, the actual voltage u would observe would only be 15V. This is saturation of output voltage.

Now that doesnt actually explain your argument..
Going by your way, op-amp amplifies the difference between its input terminals and provides at the output.

i.e., Vout = A(V+ - V-) ;

Here V- = Vout
Put it in the above equation, and you will derive at the output expression as

Vout = A / (A + 1) * V+;

So this is obviously less than 1. So output doesnt saturate this case.
Using our OP's example of V+=5V and A=4,

Vout=(4/5)*5V
Vout=4V

#### thatoneguy

Joined Feb 19, 2009
6,359
In the case of negative divided feedback, an opamp tries to adjust its' output so that the voltage on the inverting (-) input is the same as the voltage on the non-inverting (+) input.
I think if everybody fully understood this, there would be less confusion when it comes to op amps as buffers or amplifiers.

An op amp will do whatever it can to make the voltages at the inverting and non-inverting terminals equal. This is sometimes called "Servo action" in the electrical sense.