op amp basics
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Posted by lerameur on May 13, 2008, 4:20 am
Please log in for more thread options 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 | ||||||||||||||||
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Posted by Chris on May 13, 2008, 8:17 am
Please log in for more thread options 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 =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 | ||||||||||||||||
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Posted by Tim Wescott on May 13, 2008, 11:50 am
Please log in for more thread options 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 | ||||||||||||||||
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>
> 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 =A0output 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, =A0v2-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