Voltage follower...help?

[therecordingart]

Can someone explain how taking the output of the amplifier and applying it to the inverting input doesn't cause phase cancellation at the output?

[familyortiz]

I am sure someone can go thru the circuit analysis of the opamp, but the deal is, the inverting input is not out of phase with noninverting input. The opamp will try to equalize the inputs, which is how you get a voltage follower.

[alanlan]

Simple model:
vo = A (v+ - v-) where A is very large so:

as vo = v- and vice-versa

vo = A(v+ - vo)

vo = Av+ - Avo

vo(1+A) = Av+

vo= Av+ / (1+A)

As A is very large we can forget the "1"

vo = v+

Hope this is helpful. 

[therecordingart]

Simple model:
vo = A (v+ - v-) where A is very large so:

as vo = v- and vice-versa

vo = A(v+ - vo)

vo = Av+ - Avo

vo(1+A) = Av+

vo= Av+ / (1+A)

As A is very large we can forget the "1"

vo = v+

Hope this is helpful. 

Doesn't compute, but I really appreciate the reply.

[alanlan]

Well there is a form of cancellation going on.  The op-amp has a huge amount of gain and by feeding back the output to the (-ve) input we are applying negative feedback which cancels out all that huge amount of gain in the overall circuit.  The op-amp still has a huge amount of gain in itself - that can't be changed and by inference, if the output voltage is within a normal range of values i.e. within the output swing then it follows that the signal between the +ve and -ve inputs is tiny.  May be this helps a little more?

[Gurner]

There's no phase inversion becuase the signal goes into the non inverting input - that's the simple answer! (ie if you put a signal into the non-inverting input, it comes out the same polarity)

In this config, the feedback into the inverting input is acting like a brake - with no feedback (ie break that loop) the opamp 'gain' engine would be going full RPM (I'm using car engine analogy here!) ...it'll simply clip (even the tiniest of signals), by taking the output and feeding 100% back into the inverting input essentially ...you're applying the brake fully on - an overall opamp gain of one.

Don't get too hung up on phase here....you're essentially adding two signals together at the opamp inputs one is inverted one isn't ...the output is multiple of the difference.

Billiard table analogy - one snooker ball heading down the table fast ....and another snooker ball coming the other way but slower (ie it has the opposite polarity) ...the end result is that the once they collide (add), the main original ball still travels forward ...albeit slower, but it's phase hasn't reversed.

[therecordingart]

There's no phase inversion becuase the signal goes into the non inverting input - that's the simple answer! (ie if you put a signal into the non -nverting input, it comes out the same polarity)

In this config, the feedback into the inverting input is acting like a brake - with no feedback (ie break that loop) the opamp 'gain' engine would be going full RPM (I'm using car engine analogy here!) ...it'll simply clip (even the tiniest of signals), by taking the output and feeding 100% back into the inverting input essentially ...you're applying the brake fully on - an opamp gain of one.

That makes sense. Thank you!