Why do op-amps want their inputs to be the same?

[TimNJ]

Hi everyone,

It's well known that an op-amp wants to have its inverting (V-) and non-inverting (V+) inputs to be the same...but why is this the case? I've heard (and even taught to people) that the op-amp does whatever it needs to internally to keep those two inputs the same, via feedback.

It's easy enough to calculate the gain/behavior of an op-amp configuration using the ideal op-amp rules and KCL.

It’s confusing to me because in an open-loop configuration, keeping the inputs the same doesn’t seem to be a priority…but then once you add feedback, all of the sudden the op-amp wants its inputs to be the same. Does an open-loop configuration also want its inputs to be the same…but it physically can’t control them, so it hopelessly amplifies by a huge number?

Thanks!

[Kremmen]

Well, the opamp doesn't of course "want" anything, that is just the way it works in certain circuit configurations.
The differential input arrangement combined with very high amplifier gain and _negative_ feedback produces this result. It is not a question of "wanting", just a consequence of the feedback circuit topology.
For an ideal (infinite open loop gain) amp the closed loop gain is determined by the feedback loop gain and that only. The closed loop gain is
A = G / (1 + GH) where

A = closed loop gain
G = open loop gain (input to output),
H = feedback gain (output to input)

as G approaches infinity you can approximate A ~ 1/H. Say H = 0.1 then A = 10 and so on.

P.S.
Yes, an open loop amplifier has gain G since there is no feedback. Typically such a circuit works as a comparator.

[Galaxyrise]

Does an open-loop configuration also want its inputs to be the same…but it physically can’t control them, so it hopelessly amplifies by a huge number?

A difference between the inputs causes the output to change proportional to the difference.  With feedback, that change in output can change the difference between the inputs and potentially achieve a steady state, but without feedback, the output will change until it hits its limits.

[XFDDesign]

"Wanting to keep the inputs the same," is assigning a kind of intelligence that isn't really there, to simplify the approach to the problem. It's actually how negative feedback works, and not necessarily a function of the op-amp itself (it just happens to be really good at it).

OpAmps work through having unconditionally massive amounts of small signal gain. Typically, a good DC performance amplifier will have 120dB of "open loop" gain, or about 1,000,000V out per Volt of input. This is a scalar though. If you could get 1uV put into the non-inverting terminal with 0V on the inverting, you will get 1.000V out, in open loop mode (ideally).

What you actually get between the inputs is an error voltage. The output will be the voltage difference between the inputs multiplied by the open loop gain.

We express this mathematically as Vo = A(Vp -Vn) where Vp=non ivnerting, and Vn=inverting.

Suppose Vp were grounded. Vp=0, then Vo=-AVn.

Let us put a 10kOhm resistor from the node Vn to Vo. Then another 10kOhm resistor from an input called Vin to Vn.
The ideal opamp does not sink any current into the inputs, so the node voltage Vn is developed only by the currents through the resistors.

Let's do a Nodal analysis on the node Vn:

(Vn-Vin)/10k + (Vn-Vo)/10k = 0

We can multiply both sides by 10k to make them vanish:

Vn - Vin + Vn - Vo = 0

As we saw before, Vo = -AVn

Vn - Vin + Vn - (-AVn) = 0

Let's add Vin to both sides:

Vn + Vn + AVn = Vin

Simplify:

Vn (2 + A) = Vin

And let's add a test voltage to Vin, Vin=1, and choose an open loop gain of A=100,000

Vn (2 + 100,000) = 1

Vn = 1/100,002 = 9.9998uV

What is Vo when Vn=9.9998uV?

Vo = -100,000*0.0000099998 = -0.99998V

What you see with Vn= ... is that negative feedback is using the large amounts of gain  to reduce the voltage seen by the node to the point that the error is as small as possible. You do not get negative feedback if you tie the same resistors to the non-inverting input, which would be the expected case with the idea that "the opamp tries to make the inputs equal" - which implies _any_ input to be equal.

[Simon]

The op-amp amplifies the difference between it's inputs, an open loop gain is usually 10'000+ so say you have 12V supply just a 0.0012V (1.2mV) difference between inputs will cause full swing either way. As opamps have some internal error between inputs which can be as much as 2mV, without something going in and closed loop gain control it will just as others have said act as a comparator.

[David Hess]

You could examine the difference between a voltage feedback operational amplifier and a current feedback operational amplifier.  The later enforces a condition where the inverting input follows the non-inverting input whether there is feedback or not.

As pointed out in previously in the discussion, feedback around a voltage feedback operational amplifier enforces a condition where the inverting input follows the non-inverting input but if you remove that feedback or make it positive, then you end up with a comparator and the input differential voltage may vary from zero.  Operational amplifier tend to make poor substitutes for comparators which are designed to operate under these conditions and recover from saturation quickly.

For real fun, you can drive the output of a voltage feedback operational amplifier into a low impedance summing node and take the output from its supply pins.  Then you end up with balanced high input impedance inverting and non-inverting inputs and a low impedance inverting input similar to the inverting input of a current feedback operational amplifier.  This is a handy way to get higher gain-bandwidth product and higher slew rate out of a voltage feedback amplifier.

[tggzzz]

It's well known that an op-amp wants to have its inverting (V-) and non-inverting (V+) inputs to be the same...but why is this the case? I've heard (and even taught to people) that the op-amp does whatever it needs to internally to keep those two inputs the same, via feedback.

Only when the circuit ensures that there is negative feedback around the opamp. Specifically not true when there is no feedback or positive feedback.

[TimNJ]

Thank you everyone for your replies. I have been reading them through and I really appreciate it. I have a big physics test  tomorrow and I'm really stressed about it so I'll get back to this later!