Inverting &non-inverting Op-Amps are at the same potenti

principal of opam

In op-amp circuits analysis it is taken as both inverting and non inverting terminals are at same potetial, how it could be proved?
why there is no potential difference eventhough there is a high input resistance between the two terminals(current is small, but resistance is very high)?

what are assumptions made to take inverting terminal at virtual ground (in some circuit like summing amps etc.)?

Re: Op-Amp

In op-amp circuits analysis it is taken as both inverting and non inverting terminals are at same potetial, how it could be proved?

Click to expand...

This is only the case if there is atleast one feedback loop from the output to one of the inputs (or both).

Think of an opamp as an infinite gain amplifier.

The output is then

vout = A(vp-vn)

If A is infinite, then vout is infinite as long as vp - vn is not zero. The feedback prevents vout from being infinite, so vp - vn must be zero.

Re: Op-Amp

You can assume that the inverter pin is an virtual ground only if no invertin pin is conected to the ground.
The diference betewen the 2 is minimal

you can analyse the opamp using an simple diferential circuit

As you know the current in the emitter is always the same and in given by Ie1 +Ie2 .

If the transistor are equal the BE junction is equal the the emmiter potencial is always Ve = Vin - 0.7V
If one of the entrace is not conected and you insert tension in the other, the base tension is equal to the other entrance.
The input tension produce that the current on both transistor are diferent but the add of the to curren in the emitter resistance is always the same.
thats why most opamp have an offset adjust that connect an potentiometer bettewen two pins and -VEE. This pot equals the current on both transistors to be equal and 1/2 the Emitter current.
I think that is the principal of the opamp analyse regarding the entrance .
The other stages are currennt amplifier an tension.
Of course an real opamp has current source the stabilize the emitter current and to increse the input impedance.
The other entrace must be always compensated by the impedance that the otehr input has viewd fron the base to match the base current and decrease the offset tension.

Regards

Op-Amp

If you want learn all about op-amp, you can read the book "Desing with op-amp and linear Integrated Circuit" by Franco.

Re: Op-Amp

The difference between the + and - pins of the opamp is ideally 0. This is called the virtual short principle.

==> The virtual short principle is ONLY true when there is MORE negative feedback around the opamp than positive feedback.

For instance, in a basic inverting opamp, the negative feedback from the output pin is returned to the - terminal. There is NO positive feedback in this configuration because the + pin is connected to ground.

In other configurations, it is important to make sure that there is more negative feedback than positive feedback. If there is more positive feedback than negative feedback, then ALL bets are off in terms of the virtual short principle, and thus you can not be guaranteed that the + and - terminals are at the same potential.

Re: Op-Amp

I'm agree with my colleges in their answears.

why there is no potential difference eventhough there is a high input resistance between the two terminals(current is small, but resistance is very high)?

Click to expand...

The negative feedback poduces that effect. When there are no negative feedback, there are no V+ = V-.

Re: Op-Amp

Hi,
This is a feature of Op Amp. The gain of the Op Amp is so high that even for a small difference between the voltages at two input ports the output is high.
So you can say that both the inputs are shirted.

Thanks
Shaikh Sarfraz