# Explain to me the analysis of diff amp and op amp

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#### tok47 OP_amp

Dear ALL,

I have confuse by the analysis of diff amp and op amp.

Here, my question. When we try to write a transfer function for a om_amp (please refer to file name op_amp), we will assume the v- is virtually short to v+. Why can we assume that?

Since we know the input part of the op_amp is a diff_amp, when we analysis diff_amp, can we make the same assumtion?

THanks

rdgs
tok

#### VVV Re: OP_amp

Whenever you use and opamp, it usually works in closed loop. Due to the high gain of the opamp (hundreds of thousands to millions) and the closed-loop operation, the voltage difference between the inputs is extremely small and so you can say that they are at the same level.
For example, consider an opamp with a gain of only 100,000. If it is outputting 10V, then the difference between its inputs is only 10V/100,000=0.1mV.
But, I must emphasize, the opamp has to be operating in closed-loop.

In the case of the differential amplifier, since there is no closed loop operation and since the gain is very modest (it is the other stages of the opamp that contribute the most gain, not the input stage), then NO, you cannot apply the same principle to the differential amplifier.

### tok47

Points: 2

#### huojinsi

##### Full Member level 5 Re: OP_amp

For a ideal OAP its gain is very large, and generally the open gain of OAP can reach 90dB. According to formula "Vo=Av*{(V+)-(V-)}",so (V+)-(V-)=Vo/Av. Because Av is very large and equivalent to infinite, (V+)-(V-)=Vo/Av≈0, it is conventional assumption V+=V-.
For ur mentioned diff_amp (as ur diffamp.jpg),it is a single differential input stage, not a whole OAP,so it is not assumed as ideal OAP.

Pls correct me if i am wrong!

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