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

L

Lostgallifreyan

Jan 1, 1970
0
http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

Error in article, surely? Their analysis of the differential amp says that
the output is V2-V1, where V2 goes to the noninverting input. So far, so
good.

Then they say that "the analysis is essentially the same as that of an
inverting amplifier, except that the noninverting input (+) of the op-amp
is at a voltage equal to a fraction of V2, rather than being connected
directly to ground".

Isn't this wrong? Assuming all resistors to be equal, that's unity gain, so
1V on the inverting side gets -1V out, but the 1V into the divider on the
non-inverting side gets 0.5V on the output, so a total of -0.5V out.

This looks like an analysis failure like that old game about three man and
a hotal porter and a missing dollar, but I can't see where it goes wrong.
Their analysis can't account for the missing half volt in the non-
inverting input, yet we know there is really no half volt missing. :)
 
L

Lostgallifreyan

Jan 1, 1970
0
http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

Error in article, surely? Their analysis of the differential amp says
that the output is V2-V1, where V2 goes to the noninverting input. So
far, so good.

Then they say that "the analysis is essentially the same as that of an
inverting amplifier, except that the noninverting input (+) of the
op-amp is at a voltage equal to a fraction of V2, rather than being
connected directly to ground".

Isn't this wrong? Assuming all resistors to be equal, that's unity
gain, so 1V on the inverting side gets -1V out, but the 1V into the
divider on the non-inverting side gets 0.5V on the output, so a total
of -0.5V out.

This looks like an analysis failure like that old game about three man
and a hotal porter and a missing dollar, but I can't see where it goes
wrong. Their analysis can't account for the missing half volt in the
non- inverting input, yet we know there is really no half volt
missing. :)

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...
 
G

Graham W

Jan 1, 1970
0
Lostgallifreyan said:
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.

http://www.allaboutcircuits.com/vol_3/chpt_8/6.html 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
 
L

Lostgallifreyan

Jan 1, 1970
0
No, it is not that.

You may have started too far into the article.

http://www.allaboutcircuits.com/vol_3/chpt_8/6.html 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 http://www.gcw.org.uk/ PGM-FI page updated, Graphics
Tutorial WIMBORNE http://www.wessex-astro-society.freeserve.co.uk/
Wessex Dorset UK Astro Society's Web pages, Info, Meeting Dates,
Sites & Maps Change 'news' to 'sewn' in my Reply address to avoid my
spam filter.

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.
 
G

Graham W

Jan 1, 1970
0
Lostgallifreyan said:
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.

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.
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.

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
 
L

Lostgallifreyan

Jan 1, 1970
0
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 http://www.gcw.org.uk/ PGM-FI page updated, Graphics Tutorial
WIMBORNE http://www.wessex-astro-society.freeserve.co.uk/ Wessex
Dorset UK Astro Society's Web pages, Info, Meeting Dates, Sites & Maps
Change 'news' to 'sewn' in my Reply address to avoid my spam filter.

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.
 
G

Graham W

Jan 1, 1970
0
Lostgallifreyan said:
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' !
 
L

Lostgallifreyan

Jan 1, 1970
0
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' !

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.
 
J

Jamie

Jan 1, 1970
0
Lostgallifreyan said:
http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

Error in article, surely? Their analysis of the differential amp says that
the output is V2-V1, where V2 goes to the noninverting input. So far, so
good.

Then they say that "the analysis is essentially the same as that of an
inverting amplifier, except that the noninverting input (+) of the op-amp
is at a voltage equal to a fraction of V2, rather than being connected
directly to ground".
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.
Isn't this wrong? Assuming all resistors to be equal, that's unity gain, so
1V on the inverting side gets -1V out, but the 1V into the divider on the
non-inverting side gets 0.5V on the output, so a total of -0.5V out.

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 +)//
 
F

Franc Zabkar

Jan 1, 1970
0
http://www.allaboutcircuits.com/vol_3/chpt_8/9.html

Error in article, surely? Their analysis of the differential amp says that
the output is V2-V1, where V2 goes to the noninverting input. So far, so
good.

Then they say that "the analysis is essentially the same as that of an
inverting amplifier, except that the noninverting input (+) of the op-amp
is at a voltage equal to a fraction of V2, rather than being connected
directly to ground".

Isn't this wrong? Assuming all resistors to be equal, that's unity gain, so
1V on the inverting side gets -1V out, but the 1V into the divider on the
non-inverting side gets 0.5V on the output, so a total of -0.5V out.

This looks like an analysis failure like that old game about three man and
a hotal porter and a missing dollar, but I can't see where it goes wrong.
Their analysis can't account for the missing half volt in the non-
inverting input, yet we know there is really no half volt missing. :)

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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