Subtractor circuit impervious to components tolerance?
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From: Zico on 25 May 2010 19:14 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 voilà --- 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
From: Bob Myers on 25 May 2010 19:54 On 5/25/2010 5:14 PM, Zico 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. > 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. > 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. > 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. > 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; 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.
From: Zico on 25 May 2010 20:17 > > 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 voilà --- this will give, from > > any conceivable point of view (at least for every practical > > purposes), an *absolutely exact* gain of 1; > > 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? 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
From: Phil Allison on 25 May 2010 21:21 "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. ** A transformer can be used to subtract one signal from another IF the ends of the primary are connected to each signal source - due to the fact that windings respond only to the difference between the ends. ...... Phil
From: Tim Wescott on 26 May 2010 01:30
On 05/25/2010 04:14 PM, Zico 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. 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. > 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; 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. > 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. Yet what you're demonstrating is a charming naivet�. > 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) 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. -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com |