How does an op-Amp make the inputs the same?

[Laplace]

so....any ideas how the Vdiff is sustained despite the Vout increasing above 1.9996V ???

Wrong question. It is Vdiff that sustains Vout. Vout cannot increase above 1.9996V with the given parameters. That is the error from the ideal case caused by having a less than infinite open-loop gain.

 

[LvW]

I have trouble understanding how the op amp settles on the Vdiff = 0.0002V
In order the produce this Vout must be 1.9996V and therefore V- will be the 0.9998V like you said

However if the output is now 2V as required, surely this will in turn affect the V- voltage and make it 1V thus producing 0V output!

What means "as required"?
To answer your question, we must avoid any simplifications, which means:
If the feedback factor is exactly 0.5 and the open-loop gain 1E4 the output cannot reach Vout=2V.

The exact gain formula gives us Vout=Vin*[1E4/(1+0.5E4)]=1.9996001 V.
Therefore: V-=0.99980004V and Vdiff= Vin- V- =0.00019996.
This is a stable equilibrium because: V-*1E4=1.9996001.

As a result: The output has a value which exactly produces a differential voltage that perfectly fits to this output voltage when multiplied by 1E4.
Any larger/smaller output voltage causes a small reduction/increase for Vdiff thereby correcting this deviation from the equilibrium. OK?

 

[george2525]

yes OK

Thanks thats done the job. its much easier for me when everything is exact. that was great

 

[LvW]

yes OK

Thanks thats done the job. its much easier for me when everything is exact. that was great

You`re welcome..
I know - from personal experience - about the importance to know (a) why negative feedback is necessary for dc stabilizing purposes and (b) how negative feedback works in detail.
So - I think, for you it was beneficial having asked this question.