Subtractor circuit impervious to components tolerance?

First question: are you utterly opposed to using trimmers? Because I can't think of any way to get to "100% and absolutely unaffected by components" (if you really, really mean things like "100%" and "absolutely") without some sort of adjustment capability.

Well, depending on how sophisticated you want to get, you CAN pretty much null out any effect of tolerance to within the limits of whatever measuring equipment you're trying to satisfy. But there's nothing truly "perfect" in this here universe.

I've really tempted to point out that the simplest solution is to just abandon all hope of doing this "perfectly" in the analog domain, and simply do it digitally - which can be arbitrarily "perfect," to the limit of your budget.

Bob M.

Well, no, not really. Again, nothing's "perfect" or "exact." A few moments thought should suggest to you several sources of error (even though some are minute beyond belief) that prevent it. But see, you've used "for every practical purpose" and "absolutely exact" in the same sentence. Which do you really mean?

Bob M.

I was hoping that the example/analogy would make it "perfectly clear" (again with the "perfect" attribute!! :-)) what I meant by "absolutely exact".

The conditioned statement (the "for every practical purpose") was to acknowledge that, well, if we get philosophical, then we would have to go and count electrons, and then yes, the signal would not be *absolutely exact* ... Even without going to the extreme of "getting philosophical" --- there's noise; there's the fact that the wire does not have *truly zero* resistance, and the amplifier's input is not a *truly and absolutely open circuit* (it does have a finite, measurable input impedance) ... So the gain is not *truly* one...

But again, no sane practical point of view would really constitute an objection to the fact that the gain in that example is "truly" or "exactly" one... (without having to rely on extra-precise components, and without the extra trouble of having to use trimmers).

Thing is, in that example, the "cleverness" of the design (or the luck of getting away with a gain of 1, if you will) bypasses the complete and direct dependence on the component's tolerance --- it *avoids* having to use a trimmer when there's another solution that gives you an even better result without the trouble of placing a trimmer and adjusting it (and hoping that it will remain adjusted --- not to mention the difference in temperature-related variations in the different components)

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.

Incidentally, I had indeed considered going digital !! But the only part that could arguably justify it is the subtraction of the two signals --- the rest of the system is so trivial and so easy to achieve with analog circuitry that a digital solution seems somewhat overkill.

Thanks,

-Zico

"Zico"

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.

..... Phil

No, and no.

No, there isn't a subtracter circuit that doesn't depend on _some_ component tolerance somewhere in the circuit, and no, there is nothing in our fallen and imperfect world that is 100% and absolutely anything.

Take an op-amp with a 10MHz gain-bandwidth product. Make it into a voltage follower, as you suggest. Gain is pretty close to -A / (s + A), where A = GBW product. Feed a 16-bit ADC with that, and expect the circuit to live up to the op-amp's advertised capability over audio -- because after all, the op-amp's good to 10MHz, right?. You'll be seeing one LSB of error a bit above 150Hz.

Take that same op-amp, and assume it has an open-loop gain of A = 10^6, which is in the ballpark for a lot of devices. Now feed a 24-bit ADC (you can buy them off the shelf if you don't mind that they're not up to audio speeds). Gain is pretty close to -A / (A + 1), for a 1ppm error in your output _at DC_, or a whopping 17 LSB of error.

And I'm not even touching on the op-amp's bias.

Yet what you're demonstrating is a charming naiveté.

Just accept that until the last trump sounds, you are stuck with living in a world where all circuits depend on their components, and even op-amps don't live up to their Platonic ideals.

I have one word: "plastics".

Oops -- wrong world.

I have two words: "capacitive coupling".

You may get good enough, you may get pretty darn good with a specially constructed transformer with careful shielding between primary and secondary, but you won't get "100% perfect".

He only asked that the subtraction arrangement was good for audio band signals and not affected by tolerances in passive components.

I figured he never considered an all passive solution like a transformer.

A good one has a CMRR of over 100dB across the band.

..... Phil

If you need precise ratios use a transformer

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Though at this point I'm convinced that there is no circuit configuration like what I was looking for, I will clarify something (you know, at the risk of continuing to squeeze to death a pointless discussion :-) )

Nicely put, this idea that "even op-amps don't live up to their platonic ideas" :-) (and no, I'm not being sarcastic!) ... I should have expressed better what I intended with the analogy: two parts: (1) the tolerance and imperfections of an op-amp are below my threshold of what would be considered "perfect" and "absolutely precise" (maybe there is a linguistic self-contradiction here, I don't know) ... That is: the errors introduced by the op-amp and the resistance of the wires and traces do not count.

(2) Perhaps more importantly: the op-amp will be part of the circuit anyways, so, by placing resistors, we're *adding* to the imprecisions; not only that, but compensating for those requires trimmers, which at the very least is an extra/house-keeping part of the design and an extra source of trouble (both at manufacture time and at operations time)

So, this was sort of the main intent of the example/analogy ...

Now, perhaps an interesting question that comes to mind now: are the errors introduced by the imperfections of a reasonably good audio-signals op-amp (say, an LF353?) worse than a 0.1% tolerance in the resistors? (doesn't sound like --- 0.1% tolerance in a configuration with linear dependence on the resistor value gives an error 60dB below the signal ... An op-amp does better than that ... right?

Thanks,

-Zico

Not so much a linguistic one as a problem with life. Experienced engineers don't like terms like "perfect" and "absolutely precise", because "perfect" for one person may be "absolute crap" for someone else, or it may be "way too expensive and excessively over-engineered". After you've been around the block a few times you learn that putting in lots of effort to meet vague descriptions just leads to unhappiness. And since no one wants to take responsibility for communications breakdowns, and the grunt engineer is, by definition, not the boss, he automatically loses the subsequent finger-pointing contest.

"Do not count" in this case is just as bad as "perfect", though. Does not count to who? Does not count to you? What doesn't count to you? What are your thresholds? In numbers, please! Or if not in numbers, then in something still vague but a specific application like "consumer audio quality", "pro audio quality", "good enough for my home thermostat", etc.

I suspect you'll find that some folks have objections to that one. IIRC it has visible crossover distortion at audio frequencies on an oscilloscope, which means that it'll definitely be audible. And if it's the one I remember the distortion comes from the slew rate limit so it's getting worse just as the amp's ability to compensate for it gets worse, too.

And my point with the GBW stuff is that at only moderately high frequencies op-amps _don't_ do better than that. My large point is that the whole idealized, infinite gain, negative input equals positive input stuff is a really good crutch to help you start getting your head wrapped around how op-amps behave, but for many many applications you just can't ignore their real characteristics and succeed.

Well, there's a REALLY big part of your problem so far, in terms of getting an answer here - all you've ever said is "perfect," or words to that effect, and still have let anyone in on just what would be considered your "threshold." In short, tell us what you're really trying to achieve, and you'll be a whole lot more likely to get some helpful responses.

Bob M.

A flying capacitor circuit comes very close.

John

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

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.

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

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.

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,

Vout =3D 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 =3D 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

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!

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

No, that's generally referred to as a "wire"...;-)

Bob M.

In my native language (by now you surely noticed that English is not my native language), this would have made an excellent joke/pun --- we have an expression/proverb that involves the phrase "eating wire" ... So, you know, free *lunch* ... Eating wire ... I would have almost thought that you intended the pun! :-)

-Zico