op amp basics

[lerameur]

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

 

[Chris]

On May 13, 3:20 am, lerameur <leram...@yahoo.com> wrote:

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

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 www.tech-chat.de)


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 = 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

 

[Tim Wescott]

lerameur wrote:

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

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.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html