op amp basics

hello,

I am trying to understand the fundamental of op-amp. I found this link to be very interesting: courses.ece.uiuc.edu/ece486/labs/lab1/386_op_amp.pdf I would like to know why that when the output of an opamp is connected to the input, like a voltage follower, the voltage is equal to the other input. The pdf I mentionned says that the 2 input tries to be the same voltage. So if you put 5 volt in one input, you will get 5v on the other input. Looking a the basic of opamp, v2-v1, that would have an output of 0, Then one input would be zero, It does not make sense that on the same wire, there are two voltages

thanks

k
Reply to
lerameur
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Hi, K. The tutorial you mention doesn't include non-inverting amplifiers, of which the voltage follower is the simplest:

| Voltage Follower | | Vin |\\| | o-----|+\\ Vout | | >----o---o | .--|-/ | | | |/| | | | | | | | | | | | '----------' | | (created by AACircuit v1.28.6 beta 04/19/05

formatting link

Assume the gain of the op amp is arbitrarily high (a couple hundred thousand is very common, use 100,000 for this example), and you apply

5.000V at the input. The output will be 99,999/100,000 of the input, using your equation

Vout =3D Av (v2 - v1)

Which means the output will be 4.99995V, essentially the same voltage as the input.

The main source of error with an op amp configured as a voltage follower for low frequency or DC inputs is voltage offset (typically several millivolts or so).

Hope this has been of help.

Chris

Reply to
Chris

This sort of first-order op-amp analysis assumes that the op-amp has infinite gain. Practically that's meaningless, but mathematically it means that V2 - V1 _can_ be zero, yet the output can be anything.

In practice the gain is very large (DC gains of over 10^5 are just about assumed, 10^6 isn't uncommon). So the offset between V2 and V1 due to the gain not being infinite may well be less than the offset due to the amplifier input stage not being ideal.

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Tim Wescott
Wescott Design Services
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Tim Wescott

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