Question about Tutorial on Negative Feedback Op amps

[quantumtangles]

I was reading your excellent tutorial on negative feedback operational amplifiers but do not understand a sentence in paragraph 4.

Let's say that our op-amp has a differential voltage gain of 200,000. If Vin equals 6 volts, the output voltage will be 5.999970000149999 volts.

Is the value for differential voltage gain realistic (200,000), and if so, how does that figure relate to the output voltage in terms of the negative feedback loop? More directly, why is it such a big number. Define differential voltage gain. Is this a unit-less ratio?

Such a marvellous site in terms of useful information...really grateful for the effort you have all put into it.

 

[john monks]

When we say that an operational amplifier has a gain of 200,000 we are saying that for every unit of voltage change between the input pins we get a change of 200,000 units of output change.
So to that end if we place 6 volts on a voltage follower that the example uses we must have something less that 6 volts on the output pin.
So whatever it is the output voltage is 200,000 times the differential input pins.
This is a unit less ratio. It is merely the ratio of the output voltage to the differential input voltage.
In this case the output voltage is 6 volts X 200,000 / 200,001 or about 5.99997 volts.

 

[quantumtangles]

Many thanks for such a clear explanation.

I do not need to know all of the technical details for my little Op amp circuits (...really only need reasonable resistor values to set the gain in the feedback loop) but its great to know these things all the same.

Still a lot of stuff that looks really fascinating in the tutorials that I know little about. Sincere thanks

 

[(*steve*)]

The point is, at such high gains, you can generally assume that negative feedback tries to keep both inputs at the same voltage.

 

[quantumtangles]

I get it now. Its a big number for a good reason. Thanks Steve.