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From: John Fields on 26 May 2010 13:17
On Wed, 26 May 2010 08:21:39 -0700, John Larkin
<jjlarkin(a)highNOTlandTHIStechnologyPART.com> wrote:

>On Tue, 25 May 2010 16:14:05 -0700 (PDT), Zico <zico(a)mailinator.com>
>wrote:

>>So, my question: what about for a circuit that subtracts two
>>signals? Or, equivalently (and even better for audio signals),
>>a circuit that adds two signals + an inverting circuit with gain 1 ?
>>(the standard solutions I know for these two rely on components
>>precision/tolerance)
>>
>>Thanks,
>>-Zico
>
>A flying capacitor circuit comes very close.

---
To what?

From: Tim Wescott on 26 May 2010 13:56
On 05/26/2010 08:21 AM, John Larkin wrote:
> On Tue, 25 May 2010 16:14:05 -0700 (PDT), Zico<zico(a)mailinator.com>
> wrote:
>
>> Hi,
>>
>> I'm wondering if there's some circuit configuration (presumably
>> an op-amp based design) that would subtract two signals (say,
>> two audio signals) in a way that is *100% and absolutely
>> unaffected* by components tolerance.
>>
>> I do not mean using 0.1% tolerance resistors, or matched resistor
>> networks, or adding a potentiometer for fine adjustment. Those
>> solutions *minimize* the effect of components tolerance.
>>
>> For example (more like an analogy): if I need a non-inverting
>> buffer/amplifier with absolutely precise gain, say 2, then I could
>> try the standard op-amp circuit, and use two identical resistors,
>> for a gain of 2. Problem is, gain *can not* be exactly two (well,
>> it can not be *expected* to be exactly two); I *can* obtain an
>> absolutely precise gain of *1* ... Connect output terminal
>> *directly* to the inverting input, and voil� --- this will give, from
>> any conceivable point of view (at least for every practical
>> purposes), an *absolutely exact* gain of 1; where I'm trying
>> to get is: the solution in this example goes beyond the highest
>> available precision components; it goes beyond the most
>> expensive and most precise matched pairs of resistors, etc.
>>
>> So, my question: what about for a circuit that subtracts two
>> signals? Or, equivalently (and even better for audio signals),
>> a circuit that adds two signals + an inverting circuit with gain 1 ?
>> (the standard solutions I know for these two rely on components
>> precision/tolerance)
>>
>> Thanks,
>> -Zico
>
> A flying capacitor circuit comes very close.

I've put on a flying capacitor circus or two (they usually involve
electrolytics, and errors in magnitude or sign). Does that count?

--
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com
From: whit3rd on 26 May 2010 15:20
On May 25, 5:17 pm, Zico <z...(a)mailinator.com> wrote:

[given signals A and B, form a third signal A-B]

> So, basically, I was trying to find out whether I might be lucky
> and there might be some "clever" configuration in which a
> subtractor circuit could be obtained that is not affected by
> any component's tolerance.

Still, that's not clear. The transformer idea, for instance,
is good but limited to the 'ideal' high input impedance of
the unloaded transformer (and at DC, that always fails).

It seems to me, your question is really about the difference
equation having no reference to other units than the inputs,
i.e. no 'ohms' value is in the ideal difference equation.
The differencing amplifier built from op amps, however,
DOES have resistances (ratios of 'em, anyhow) all over
the operating equation.

The answer is, therefore, you can connect a floating voltmeter
from signal A to signal B, and it measures A-B without any
component-value elements in the equation (other than the
voltmeter innards). A transformer winding is a very good
approximation to that floating measurement circuit. So is
an op amp difference circuit, but it DOES depend on resistor
matching.

The use of floating (isolated, or battery-powered) meters and
sources is a very powerful technique for high precision work.
From: stan on 26 May 2010 18:15
Zico wrote:
> I'm wondering if there's some circuit configuration (presumably
> an op-amp based design) that would subtract two signals (say,
> two audio signals) in a way that is *100% and absolutely
> unaffected* by components tolerance.

If this "configuration" has components they will certainly have
tolerances and the resulting circuit will likewise have tolerances.

> I do not mean using 0.1% tolerance resistors, or matched resistor
> networks, or adding a potentiometer for fine adjustment. Those
> solutions *minimize* the effect of components tolerance.

Sounds like you are looking for a pair of matching magic components
with a fixed gain over some unstated audio (simple telephony vs CD
quality) frequency range. But you aren't interested in tight tolerance
components. Sounds like a free lunch hunt. Actually that's not quite
right; sounds like a free gourmet lunch hunt. Good luck and if you
find it, please drop some crumbs.
From: Zico on 26 May 2010 21:23
On May 26, 6:15 pm, stan <smo...(a)exis.net> wrote:
> Zico wrote:
> > I'm wondering if there's some circuit configuration (presumably
> > an op-amp based design) that would subtract two signals (say,
> > two audio signals) in a way that is *100% and absolutely
> > unaffected* by components tolerance.
>
> If this "configuration" has components they will certainly have
> tolerances and the resulting circuit will likewise have tolerances.

Well, again, maybe the components could have been "wires" or
traces, which can be reasonably taken as 0 ohms with zero
tolerance (from the point of view of the given application). Like
in the example --- I mean, there *is* the configuration of the
voltage follower that achieves a gain of one with higher accuracy
and lower housekeeping and trouble (compared to using trimmers
for fine-tuning).

Another (completely hypothetical) example: as you point out,
if there are components, there is tolerance; but imagine that
someone had come up with a clever configuration (again,
*hypothetically*) in which the output voltage is given by, say,

Vout = V1 (A + R1/R2) / (A + R2/R1) - V2

Where A is the open-loop gain of the op-amp.

(please don't ask me how I would achieve that --- *hypothetical
example* ... The thing is, just because I can not figure out a
way of doing it does not mean someone else can not)

You see, R1 and R2 could have 20% error, and still, the output
would be V1*1.0000004 - V2 (with A = 10^6)... So, yes, you're
going to tell me that "there, see? it's not *exact*" ... But using
components with 20% tolerance that have an net effect of a
0.00004% tolerance on the output.... well, sue me for calling
that "unaffected by components tolerance" ...

Ok, yes, yes, I did not specify "good enough for what" --- because
yes, if I feed that to a A/D converter with 24 *actual* bits of
accuracy, then yes, that A/D converter will be able to tell the
difference between that and the *exact* V1-V2 ... But then
again, that's what I get for using 20% components!! :-) If in this
same hypothetical example I had used 1% resistors, I would
actually be at 0.000002% of the *exact* V1-V2; three times
more precise than the 24-bit A/D converter. (as opposed to
a circuit configuration that gives me a gain that directly depends
on, say, R2/R1, in which case, using resistors with 1% tolerance,
a 10-bit A/D converter could tell the difference!

> components. Sounds like a free lunch hunt. Actually that's not quite
> right; sounds like a free gourmet lunch hunt.

Well, if I was not aware of the existence of that voltage follower
configuration, then an amplifier with gain "exactly" 1 would have
sounded like a free lunch to me ... So, the thing is, I was posting
here because there may be some clever circuit configuration
that someone invented a while ago that I was not aware of.
You know, not knowing about something does not constitute
proof that that something does not exist!! Right?

Ok, I guess now the subject has been officially squeezed to
waaaaay beyond its death!!! So I will respectfully withdraw from
the discussion --- with thanks to all that replied!

Regards,
-Zico
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