No missing half volt? :) Op-amp analysis error?

Lostgallifreyan wrote in news:Xns97A88C0FC6D7Flostgallifreyangmail@140.99.99.130:

Never mind, I guess the 0.5V rise on the non-inverting side means that effectively, there is 0.5V on the inverting input, so 0.5V and -0.5V cancel to make the correct 0V output.

Still looks misleading, like they should have stated that to finish their explanation...

No, it is not that.

You may have started too far into the article.

explains that the voltage gain at the non-inverting input of the op-amp is (1+R1/R2) so V2 has to be potted down to achieve the same output voltage through the circuit. Geddit?

I don't think the article is very well written nor properly (thoroughly) proofread.

HTH

-- Graham W

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"Graham W" wrote in news:4443de54$0$9274$ snipped-for-privacy@ptn-nntp-reader01.plus.net:

Indeed. I got better mileage out of The Art Of Electronics. :) Non- inverting amplifiers were new to me before today.

I'm still a VERY lost Gallifreyan right now though. >:) That offset of gain by 1 would be corrected for a 1V input, but it's fixed, so it won't correct the output for any other value. Also, the feedback is on the inverting input of the differential amplifier in their circuit, so usual rules apply. I'm likely missing something, but I can't see what it is, from what you've told me.

Yes it does! If the left hand R's are both the same value and the right hand R's are also the same value as each other (but at some value that offers the correct gain from the cct) then the assertion is always true.

Lets say that the left ones are 10k while the RHS ones are 100k. That will yeild a gain of ten (100k/10k). If you have a voltage i/p on the -ve i/p at V1 of (say) 2.5vdc and (say) 2.0vdc on the +ve at V2 the output will be -5vdc.

Vo = -V1( rightR/leftR) = - 25vdc

but most of this is balanced by the +ve i/p from V2 which is

Vo= +V2' ( 1 + (rR/lR)) but V2' needs to be found first. This is a simple voltage divider which results in rR/(rR+lR) of the V2 value at the input to the op-amp +ve i/p terminal this will be 2.0 * 10/110 vdc . The cct will amplify this by the factor above ( 1 + rR/lR ) = +20v and subtract the -25v from it yeilding the -5.0vdc o/p predicted.

You haven't appreciated that the gain at the +ve is always 1 more than the gain at the V1 i/p.

Well, try and work through the above - where rR is the RHand R and lR is the Lefthand R.

HTH

-- Graham W

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"Graham W" wrote in news:44443c47$0$9273$ snipped-for-privacy@ptn-nntp-reader01.plus.net:

Ok, thankyou, that's clear, given a correction: "..a simple voltage divider which results in rR/(rR+lR) of the V2 value.." I guess that should have been lR/(rR+lR). As it was it confused me for a while, as did the notation you used, the +V2' bit... A diagram would have helped, but I know that ASCII isn't much of a drawing tool.

Yes, you are correct - a transcription error - it should be lR/(rR+lR). What else do you expect at 3:00 in the morning ! ! !.

The notation was more difficult because of the original article calling them all 'R' !

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"Graham W" wrote in news:44444e98$0$9230$ snipped-for-privacy@ptn-nntp-reader01.plus.net:

Well, if we can both hack this stuff at this time, we'll get by. :) It's the same time here, too.. I won't be forgetting this op-amp rule now.

they are referring to a inverting op-amp using dual voltages, the + would go to ground/common thus giving you 0 out with 0 in. and what they meant by the differential being a the same on the input or a fraction is correct because, both inputs - and + must be allowed to input voltage from a source and thus produce the offset of the difference at the output. and saying that both inputs are the same or just slighly off a bit , they are talking about the offset of the OP-amp being used to produce a differential amp. the ideal Op-amp does not have any offset factors to worry about how ever, there is no such thing as an ideal op-amp.

you can have gain with differential amps if you wish but the term differential is what it means, the difference between one input and the other input. so having a gain of 1 on each input will give you a true real world output of the difference between input 1 and 2.. or (- and +)//

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On Mon, 17 Apr 2006 12:46:09 GMT, Lostgallifreyan put finger to keyboard and composed:

This is how I would analyse it.

V+ = V2 /2 V- = V+ I- = (V2 /2 - V1)/R

where I- is the current flowing on the inverting side

Vout = V- + (R x I-) = V2 /2 + R x [(V2 /2 - V1)/R] = V2 /2 + V2 /2 - V1 = V2 - V1

- Franc Zabkar

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