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Obtaining an accurate resistor
I need two accurate resistors.
The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? Thanks, Ian |
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On Fri, 07 Oct 2005 18:44:17 GMT, Ian Iveson wrote:
I need two accurate resistors. The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? Thanks, Ian If you want a really accurate single resistor (I'm presuming you have the means to measure accurately), but the next lower value, measure it and then pad it up to the correct value with a smaller resistor or resistors. For the equal value resistor pair, just pad up the smaller of the two with a suitable small resistor. Some resistance wire will allow you to get as close as you need to an accurate match. d |
#3
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"Don Pearce" wrote
If you want a really accurate single resistor (I'm presuming you have the means to measure accurately), but the next lower value, measure it and then pad it up to the correct value with a smaller resistor or resistors. For the equal value resistor pair, just pad up the smaller of the two with a suitable small resistor. Some resistance wire will allow you to get as close as you need to an accurate match. Thanks, Don. The possibility of using strings of resistors of different values adds another dimension to the plot! Although it would clearly provide a simpler solution, "Trim on test" is not allowed here. Home-made resistors are definitely not an option in this case. Only selection from like-value resistors is allowed. If I were stuck with the options I have outlined, which would be best? cheers, Ian On Fri, 07 Oct 2005 18:44:17 GMT, Ian Iveson wrote: I need two accurate resistors. The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? |
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"Ian Iveson" wrote in message k... : I need two accurate resistors. : : The first, (R1) is for a tuned analogue circuit in which all the : other components are already fixed. Correct tuning of the circuit : requires an accurate resistor as close as possible to its nominal : value. : : The second (R2) is for a matched pair. Neither needs to be close to : the nominal value, but they must be matched as accurately as : possible. : : The kind of resistors I want are only available at 1% tolerance. I : am only building one example of the circuit. I don't care about : anything except accuracy in both cases. : : For each resistor, R1 and R2, am I better off: : : a) Buying one resistor of the right nominal value. : : b) Buying X resistors each of 1/X th of the nominal value and : connecting them all in series. : : c) Buying Y resistors of the correct nominal value and selecting the : best one. : : d) Buying X * Y resistors, each of 1/X th of the nominal value, and : selecting the best series combination of X resistors. : : If I decide to produce the circuit in quantity, should that make a : difference to which I choose? : : Thanks, Ian See my measurements post - if you're interested, could send you the data, comma delimited or so, for this particular batch and brand (BC 0.6W 50 ppm 1% mf) distribution is rapidly falling off towards the extremes, so series connecting does sharpen the resulting peak, so increases the change that the compound value is closer to the nominal series value. For tube circuits, the voltage rating is much improved with series connecting, eg. a 0.6 W resistor can handle some 300 V without adverse effects, smaller resistors are usually 200 V tops. 50 ppm/K, 2ppm/Vr resistors only cost about 1 eurocent, in bulk, so easy to just find out by gettin' some . Cheers, Rudy nb increasing the change of tighter tolerance is not the same as equals tighter tolerance, so multiple series connecting seems overkill to no avail. |
#5
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You state that you need them in volume which changes the methods available.
What kind of volumes are you looking at and what values? For high volumes, wirewound could give tight accuracy--and you could get them made and sorted in China. Thus would require tooling and set up costs. For really low volumes--you could wind or sort by hand. THe problem is for medium volumes (1,000 to 100,000). Tecktronics has solved that problem somehow because I see surplus components from them with ..01% tolerance. Maybe give them a call and find out where their stock comes from? "Ian Iveson" wrote in message k... I need two accurate resistors. The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? Thanks, Ian |
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On Fri, 07 Oct 2005 18:44:17 GMT, "Ian Iveson"
wrote: I need two accurate resistors. The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? Without any cost constraints, all answers are equal. Just buy a billion and sort in-house. You're obviously after something more significant, so maybe the question should be framed more tightly. Good fortune, Chris Hornbeck |
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"Chris Hornbeck" wrote
Without any cost constraints, all answers are equal. Just buy a billion and sort in-house. You're obviously after something more significant, so maybe the question should be framed more tightly. You have chosen to select rather than connect all billion in series, which says something...hopefully more than a recognition of the problems of finding billionth-ohm resistors, soldering them all, and fitting them in the chassis. I reckon getting the question right is always the hardest part. I can't put a price on quality, but where strategies give the same quality, I want to choose the cheapest. So, if I use series strings rather than individual selected resistors, would I be likely to get an equally accurate result from less than a billion resistors? And is the answer the same for the prototype as it would be for a production run? cheers, Ian in message ... On Fri, 07 Oct 2005 18:44:17 GMT, "Ian Iveson" wrote: I need two accurate resistors. The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? Good fortune, Chris Hornbeck |
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On Sat, 08 Oct 2005 01:50:03 GMT, "Ian Iveson"
wrote: I reckon getting the question right is always the hardest part. Yeah, and it seems that it's the part gotten wrongest in our modern world-a-go-go. I can't put a price on quality, but where strategies give the same quality, I want to choose the cheapest. Then two new factors enter: cost of sorting (in-house) and absolute quantity. The second is because the number of rejects falls for your second case (matching) with larger absolute values. Without defining those values, though, I don't think that there's any real answer to your question. So, if I use series strings rather than individual selected resistors, would I be likely to get an equally accurate result from less than a billion resistors? And is the answer the same for the prototype as it would be for a production run? A million years ago, when we were young, pretty accurate military accessments of reliability for electronic equipment were taken as the reciprocal of the number of solder joints. With modern flow soldering, etc, this is just a passing fancy, but.... FWIW. Thanks, as always, Chris Hornbeck |
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So, if I use series strings rather than individual selected
resistors, would I be likely to get an equally accurate result from less than a billion resistors? Hi Ian, Suppose you bought a bag of 100 1k resistors and 200 500R resistors. And, without measuring them, soldered together pairs of 500R resistors. You would then have two bags of 1k resistors both of which will have the same variation. Neither set will will be of tighter tolerance than the other. However, there are 19,900 different pairs that can be created from the set of 200 500Rs. This sample space is equivalent to a bag of 19,900 1k resistors. Given enough resistors you are likely to find one that is very very close to 1k as well as one that is very very close to 1k +/- 1%. If both the 500R and 1k resistors were rated at 1/2W, then, in service, the resistors in the series combination will be stressed less and so should be more accurate in service. Lastly, if you string together resistors you will add reactance to your circuit which, depending on what you are doing, can affect performance. Joe |
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On Fri, 07 Oct 2005 22:04:59 -0500, flipper wrote:
He's not really seeking advice or suggestions, he's trying to demonstrate, using a 'real world example', his statement in the "Where to get 1watt 1% resistors" thread. Namely that "Using resistors in series or parallel IMPROVES the % accuracy of the final value." Problem is, it's the selecting of values that achieves it and not the series or parallel connection itself although allowing multiples increases the odds he'll find more suitable pairs than seeking singles only. But the resultant value of any two random 1% resistors in series or parallel can still be 1% off nominal and that's what one must design to, unless you do selection. Beautifully framed; thanks very much. The gods designed us by incorporating (nudge-nudge) the selection process into our universe. Should we do less? Arf! Chris Hornbeck |
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"flipper" wrote
He's not really seeking advice or suggestions, he's trying to demonstrate, using a 'real world example', his statement in the "Where to get 1watt 1% resistors" thread. Namely that "Using resistors in series or parallel IMPROVES the % accuracy of the final value." Sneer all you like, I'm just an amateur trying to learn. This is a different thread, specifically about accuracy. I have tried to frame it in practical terms, and narrow its scope, precisely to focus on what I am wondering about. Is that supposed to be a quote from me, BTW? Problem is, it's the selecting of values that achieves it and not the series or parallel connection itself although allowing multiples increases the odds he'll find more suitable pairs than seeking singles only. Thank you. That's a very complicated sentence though. Could you explain please? But the resultant value of any two random 1% resistors in series or parallel can still be 1% off nominal and that's what one must design to, unless you do selection. That is a different issue, but thanks for the thought anyway. in message ... On Fri, 07 Oct 2005 23:22:36 GMT, Chris Hornbeck wrote: On Fri, 07 Oct 2005 18:44:17 GMT, "Ian Iveson" wrote: I need two accurate resistors. The first, (R1) is for a tuned analogue circuit in which all the other components are already fixed. Correct tuning of the circuit requires an accurate resistor as close as possible to its nominal value. The second (R2) is for a matched pair. Neither needs to be close to the nominal value, but they must be matched as accurately as possible. The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. c) Buying Y resistors of the correct nominal value and selecting the best one. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. If I decide to produce the circuit in quantity, should that make a difference to which I choose? Without any cost constraints, all answers are equal. Just buy a billion and sort in-house. You're obviously after something more significant, so maybe the question should be framed more tightly. Good fortune, Chris Hornbeck He's not really seeking advice or suggestions, he's trying to demonstrate, using a 'real world example', his statement in the "Where to get 1watt 1% resistors" thread. Namely that "Using resistors in series or parallel IMPROVES the % accuracy of the final value." Problem is, it's the selecting of values that achieves it and not the series or parallel connection itself although allowing multiples increases the odds he'll find more suitable pairs than seeking singles only. But the resultant value of any two random 1% resistors in series or parallel can still be 1% off nominal and that's what one must design to, unless you do selection. |
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"Joseph Meditz" wrote
Suppose you bought a bag of 100 1k resistors and 200 500R resistors. And, without measuring them, soldered together pairs of 500R resistors. You would then have two bags of 1k resistors both of which will have the same variation. Neither set will will be of tighter tolerance than the other. Thanks Joe It may be true that tolerance limits are not improved. That would depend on whether they are all measured, and whether the quoted tolerance is valid only at the time of measuring. Unless both of these are true, the quoted tolerance is not absolute, but is rather a statistical measure with some confidence limit. I am quite happy to accept that tolerance is absolute, but haven't found definitive evidence of this. In any event, I am not looking for better tolerance limits, I am looking for more accurate resistors. It is certainly not true that the 500R pairs would have the same variation, unless you mean that the 1k resistors would have more variation than the 500R to begin with, or that by "variation" you mean "tolerance". My contention is that averaging always reduces variance. This certainly improves the chance of accurate matching, but it is a bit less certain that it reduces the deviation from the nominal value. However, there are 19,900 different pairs that can be created from the set of 200 500Rs. This sample space is equivalent to a bag of 19,900 1k resistors. Given enough resistors you are likely to find one that is very very close to 1k as well as one that is very very close to 1k +/- 1%. Right. I wonder whether, if I decide that no resistors must be discarded, the final bag of 500R pairs could be more accurate than the bag of 1k as a result of the selective pairing? Again, I am trying to focus on accuracy rather than tolerance. Working this out needs more statistics than I have in the top of my head. I feel sure they would be probably more accurate as a result of the series connection, however. If both the 500R and 1k resistors were rated at 1/2W, then, in service, the resistors in the series combination will be stressed less and so should be more accurate in service. That is a good point. If the 500R were 1/4W, would the probability of drift in service be the same for the pairs as for the 1/2W 1k? Lastly, if you string together resistors you will add reactance to your circuit which, depending on what you are doing, can affect performance. Reactance is another good point. I am a bit finnicky, so I tend to match for both inductance and resistance. Perhaps that's more than just finnicky, on reflection. Maybe it's a fettish, considering the insignificance of MF resistor reactance. Anyway, with resistors from the same batch, I have found that inductance varies in keeping with the variation in resistance. This is not true of resistors from different batches, and certainly not of those of different manufacture. Different resistances of the same manufacture and series do not have similar inductance. Based on a small sample of less than a hundred measurements, the 500R resistors would have less inductance, but not half. So the series connection would be worse than the 1k singles. Considering inductance is always dominant, that would be a case for parallel connection. cheers, Ian |
#13
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Hello!
I've read through the threads, and thought I'd respond :-) I'll state now, up-front, that I'm taking 'tolerance' to mean 'certainly no more than [whatever the tolerance is] away from the nominal value' (and that any resistors which fall outside their stated tolerances are therefore duff, and to be regarded as defective). Ian Iveson wrote: I need two accurate resistors. [...] The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. For R1, it could be as much as 1% above or below the nominal value (obviously). For R2, you could end up with one of the pair being as much as 2.020202...% above the other (if one's 1% up, and the other's 1% down). b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. That'll tend to 'home in' on the mean value of those resistors. The bigger X is, the better (but you'll still occasionally be unlucky). However, the mean value of a bunch of resistors is not the same as the nominal value. Indeed, the mean doesn't even have to be particularly close to the nominal for the resistors to meet the tolerance requirements (the smaller the range of actual values, the further from nominal the mean can be). So, for R1, this isn't much good. But, for the pair of R2s, this will tend to be good, as long as each R2 in a pair is taken from the same, well-mixed bunch of resistors. But it's still possible to be unlucky, and get R2s in a pair that differ by up to about 2% (from each other). c) Buying Y resistors of the correct nominal value and selecting the best one. That's much better than (b) for R1, but there's still no guarantee of a close match. All Y resistors might be, say, 0.3% to 0.7% below nominal. d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. That won't work for R1, as you'll just end up with Y networks tending to be of the mean, rather than nominal, value. You might, just possibly, be (very) lucky, but you're much more likely to find a better match with (c). For the R2s, assuming the X*Y resistors are in a well-mixed batch, this is the best method, yet. It's like (b), but with lots of networks to choose from. With Y such networks, you'd have Y*(Y - 1)/2 possible pairs :-) If I decide to produce the circuit in quantity, should that make a difference to which I choose? Well, for a single R1, (c) is the best of the options, even though it's not guaranteed to work very well in practice (but would still work better than the other options). Options (b) and (d) are just lousy. So, for producing the circuit in quantity, you could either settle for 1% plain and simple with option (a), or you could, perhaps, automate the measuring and selecting process for option (c). For pairs of R2s, it's either option (b) or (d), depending on whether or not you want to do the measuring. With (d) you have to measure and select, but with (b) you can just get on with it (as long as the batches of resistors are well-mixed, of course). With (b), though, there might be quality control issues, as it's always possible for bad matches to occasionally occur by chance. But, for both options, have a large X. Personally, though, I don't much like these kinds of approaches. I'd much rather make the need for calibration a feature and a selling point, complete with exciting panel meters and the like :-) Simon |
#14
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"Ruud Broens" wrote
See my measurements post - if you're interested, could send you the data, comma delimited or so, Yes please Rudy! Even though I record all the values of resistors I use, I have used less than 100 in my life, and I have no more than 9 of any value in stock. Even so, my experience selecting from sets of 10 is that they seem to be in tight clusters centred off the nominal value. That would be consistent with a production process using blanks with a constant or slowly changing variation from the one used to set up the trimming machine. Or just a hysterical trimming machine. for this particular batch and brand (BC 0.6W 50 ppm 1% mf) distribution is rapidly falling off towards the extremes, so series connecting does sharpen the resulting peak, My contention is that this is true regardless of the original distribution, simply because averaging always reduces variance. so increases the chance that the compound value is closer to the nominal series value. Doesn't quite necessarily follow. Certainly increases the chance that it is closer to the mean of the original distribution, but our measurements suggest that is not the same as the nominal value. Good for matching, but there is still an outside chance in my mind that it may not improve the probability of getting closer to nominal. It is possible to contrive a distribution from which more series pairs would be further from nominal than the singles, and contrive an argument to suggest this would worsen the chance of getting closer to the nominal value. Consider a distribution that is symmetrical and bell-shaped, but shifted to one side of the nominal value, with tails still fitting entirely within the tolerance limits. Averaging by series connection will reduce variance, and so emphasise the bell shape. The mean variation from nominal will not change, so on average the series connection will not result in being closer to nominal. Fewer will be very close to nominal, but fewer will be very far. In other words, they will be more consistently inaccurate. Now consider the same distribution, skewed in the opposite direction to the shift. Averaging will in this case move the *mode* (distribution peak) away from the nominal value. You could then say that the most common series value is less accurate than the most common single value. Variance would still be reduced, and the mean variation from nominal would still be the same I think (not sure about this though). For tube circuits, the voltage rating is much improved with series connecting, eg. a 0.6 W resistor can handle some 300 V without adverse effects, smaller resistors are usually 200 V tops. 50 ppm/K, 2ppm/Vr resistors only cost about 1 eurocent, in bulk, Right. I must admit that I don't understand resistor voltage ratings. The only way to get 1W through 1Mohm for example, is to put 1kV across it. So what does a 1W, 1Mohm, 500V spec mean? A resistor that never gets very hot, I suppose. so easy to just find out by gettin' some . Next opportunity, I'll pick up a whole roll of surplus resistors, just for the hell of it. nb increasing the change of tighter tolerance is not the same as equals tighter tolerance, so multiple series connecting seems overkill to no avail. To most of the world, overkill is what we seem to be here for! thanks cheers, Ian |
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On Sun, 09 Oct 2005 08:48:52 GMT, "Ian Iveson"
wrote: Right. I wonder whether, if I decide that no resistors must be discarded, the final bag of 500R pairs could be more accurate than the bag of 1k as a result of the selective pairing? Only if you are able to measure them accurately, and select complementary pairs. Otherwise, you'll gain no improvement over the bag of 1k resistors. Again, I am trying to focus on accuracy rather than tolerance. Working this out needs more statistics than I have in the top of my head. I feel sure they would be probably more accurate as a result of the series connection, however. Well, you're wrong. The statitstics are very simple, and given a Gaussian distribution of the values within that 2% tolerance band, the vaiation will be identical for both bags, hence a random pairing will produce 1k resistors of statistically the same variation as the bag of 1k resistors. If both the 500R and 1k resistors were rated at 1/2W, then, in service, the resistors in the series combination will be stressed less and so should be more accurate in service. Only if stress is an issue in the application for which they are required. That is a good point. If the 500R were 1/4W, would the probability of drift in service be the same for the pairs as for the 1/2W 1k? Yes, as the thermal cycling would be identical in each case. Lastly, if you string together resistors you will add reactance to your circuit which, depending on what you are doing, can affect performance. Eh? If (and only if) the original resistors have identical L and C parasitics, what you'll do is double the inductance but halve the capacitance for a serial pair connection. The term 'add reactance' is much too vague, and not necessarily accurate. Reactance is another good point. I am a bit finnicky, so I tend to match for both inductance and resistance. Perhaps that's more than just finnicky, on reflection. Maybe it's a fettish, considering the insignificance of MF resistor reactance. Depends on the bandwidth. Get into a few Megs, and these things become significant. For audio use though, I'd agree that MF parasitics are negligible for any reasonable resistance value. Anyway, with resistors from the same batch, I have found that inductance varies in keeping with the variation in resistance. This is not true of resistors from different batches, and certainly not of those of different manufacture. Different resistances of the same manufacture and series do not have similar inductance. Based on a small sample of less than a hundred measurements, the 500R resistors would have less inductance, but not half. So the series connection would be worse than the 1k singles. Considering inductance is always dominant, that would be a case for parallel connection. Indeed it would, but you'd have to work out if this has any real significance at 100kHz. Another advantage of parallel connection is that if one resistor should fail open-circuit, the other one will maintain some kind of connection. Resistors hardly ever fail short-circuit, that would need an unusual construction such as a bifilar wirewound. If your resistors are physically small, you should also consider that a parallel connection doubles the capacitance. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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"Stewart Pinkerton" wrote in message ... On Sun, 09 Oct 2005 08:48:52 GMT, "Ian Iveson" wrote: Right. I wonder whether, if I decide that no resistors must be discarded, the final bag of 500R pairs could be more accurate than the bag of 1k as a result of the selective pairing? Only if you are able to measure them accurately, and select complementary pairs. Otherwise, you'll gain no improvement over the bag of 1k resistors. Again, I am trying to focus on accuracy rather than tolerance. Working this out needs more statistics than I have in the top of my head. I feel sure they would be probably more accurate as a result of the series connection, however. Well, you're wrong. The statitstics are very simple, and given a Gaussian distribution of the values within that 2% tolerance band, the vaiation will be identical for both bags, hence a random pairing will produce 1k resistors of statistically the same variation as the bag of 1k resistors. No I don't think I am wrong, and I think you are confusing mean with variance. The mean would stay the same, but the variance would reduce. For a normal distribution, doesn't variance decrease in proportion to root n, where n is the number of resistors in the chain? Stewart, you have inspired the Ultimate Solution (cue phil...), thanks. If I get lots of resistors of different manufacture, I should get close to a normal distribution centred on the nominal value. If I then reduce variance by series connection, I get more accurate resistors. thanks, Ian If both the 500R and 1k resistors were rated at 1/2W, then, in service, the resistors in the series combination will be stressed less and so should be more accurate in service. Only if stress is an issue in the application for which they are required. That is a good point. If the 500R were 1/4W, would the probability of drift in service be the same for the pairs as for the 1/2W 1k? Yes, as the thermal cycling would be identical in each case. Lastly, if you string together resistors you will add reactance to your circuit which, depending on what you are doing, can affect performance. Eh? If (and only if) the original resistors have identical L and C parasitics, what you'll do is double the inductance but halve the capacitance for a serial pair connection. The term 'add reactance' is much too vague, and not necessarily accurate. Reactance is another good point. I am a bit finnicky, so I tend to match for both inductance and resistance. Perhaps that's more than just finnicky, on reflection. Maybe it's a fettish, considering the insignificance of MF resistor reactance. Depends on the bandwidth. Get into a few Megs, and these things become significant. For audio use though, I'd agree that MF parasitics are negligible for any reasonable resistance value. Anyway, with resistors from the same batch, I have found that inductance varies in keeping with the variation in resistance. This is not true of resistors from different batches, and certainly not of those of different manufacture. Different resistances of the same manufacture and series do not have similar inductance. Based on a small sample of less than a hundred measurements, the 500R resistors would have less inductance, but not half. So the series connection would be worse than the 1k singles. Considering inductance is always dominant, that would be a case for parallel connection. Indeed it would, but you'd have to work out if this has any real significance at 100kHz. Another advantage of parallel connection is that if one resistor should fail open-circuit, the other one will maintain some kind of connection. Resistors hardly ever fail short-circuit, that would need an unusual construction such as a bifilar wirewound. If your resistors are physically small, you should also consider that a parallel connection doubles the capacitance. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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ps. this might help
http://syque.com/quality_tools/toolb..._variation.htm I tried to explain the central limit theorem in the "how to find 1W 1% resistors" thread. Not very well it seems. Above has pictures, which might be easier for some. Ian |
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pps
If I get lots of resistors of different manufacture, I should get close to a normal distribution centred on the nominal value. If I then reduce variance by series connection, I get more accurate resistors. ....probably. |
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On Sun, 09 Oct 2005 14:13:46 GMT, "Ian Iveson"
wrote: "Stewart Pinkerton" wrote in message .. . On Sun, 09 Oct 2005 08:48:52 GMT, "Ian Iveson" wrote: Right. I wonder whether, if I decide that no resistors must be discarded, the final bag of 500R pairs could be more accurate than the bag of 1k as a result of the selective pairing? Only if you are able to measure them accurately, and select complementary pairs. Otherwise, you'll gain no improvement over the bag of 1k resistors. Again, I am trying to focus on accuracy rather than tolerance. Working this out needs more statistics than I have in the top of my head. I feel sure they would be probably more accurate as a result of the series connection, however. Well, you're wrong. The statitstics are very simple, and given a Gaussian distribution of the values within that 2% tolerance band, the vaiation will be identical for both bags, hence a random pairing will produce 1k resistors of statistically the same variation as the bag of 1k resistors. No I don't think I am wrong, and I think you are confusing mean with variance. The mean would stay the same, but the variance would reduce. No, the mean *and* the variance would stay the same. The mean is simply the central value, and is independent of the number of items. For a normal distribution, doesn't variance decrease in proportion to root n, where n is the number of resistors in the chain? No, since they still lie within a 2% tolerance band. Stewart, you have inspired the Ultimate Solution (cue phil...), thanks. If I get lots of resistors of different manufacture, I should get close to a normal distribution centred on the nominal value. If I then reduce variance by series connection, I get more accurate resistors. Alternatively, get some 910 ohm resistors, and put them in series with some others between 82 and 110 ohms to get a value within 0.1% of 1kohm without much difficulty. -- Stewart Pinkerton | Music is Art - Audio is Engineering |
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"Ian Iveson" wrote in message . uk... : "Ruud Broens" wrote : : See my measurements post - if you're interested, could send you : the data, comma delimited or so, : : Yes please Rudy! consider it done :-) : Even though I record all the values of resistors I use, I have used : less than 100 in my life, and I have no more than 9 of any value in : stock. shame on you ;-) : : Even so, my experience selecting from sets of 10 is that they seem : to be in tight clusters centred off the nominal value. That would be : consistent with a production process using blanks with a constant or : slowly changing variation from the one used to set up the trimming : machine. Or just a hysterical trimming machine. heh. emotions in power tools - no thanks, dangerous enuf already :-) : : for this particular batch and brand (BC 0.6W 50 ppm 1% mf) : distribution is rapidly falling off towards the extremes, so : series connecting does sharpen the resulting peak, : : My contention is that this is true regardless of the original : distribution, simply because averaging always reduces variance. : yep, the watering down effect : so increases the : chance that the compound value is closer to the nominal series : value. : : Doesn't quite necessarily follow. Certainly increases the chance : that it is closer to the mean of the original distribution, but our : measurements suggest that is not the same as the nominal value. Good : for matching, but there is still an outside chance in my mind that : it may not improve the probability of getting closer to nominal. It : is possible to contrive a distribution from which more series pairs : would be further from nominal than the singles, and contrive an : argument to suggest this would worsen the chance of getting closer : to the nominal value. well, i'd expect a feedback process to target the nominal value probably set up more accurately than my DVM - it has a basic 0.15 % accuracy, that i assume is a systematic error, repeatability is much better, so could be all measurements were 0.15 % too low : : Consider a distribution that is symmetrical and bell-shaped, but : shifted to one side of the nominal value, with tails still fitting : entirely within the tolerance limits. Averaging by series connection : will reduce variance, and so emphasise the bell shape. The mean : variation from nominal will not change, so on average the series : connection will not result in being closer to nominal. Fewer will be : very close to nominal, but fewer will be very far. In other words, : they will be more consistently inaccurate. yes, i had that inprecisely formulated, thanks : Now consider the same distribution, skewed in the opposite direction : to the shift. Averaging will in this case move the *mode* : (distribution peak) away from the nominal value. You could then say : that the most common series value is less accurate than the most : common single value. Variance would still be reduced, and the mean : variation from nominal would still be the same I think (not sure : about this though). ok, needin' some coffee ... report back l8er maybe ;-) : : For tube circuits, the voltage rating is much improved with : series connecting, eg. a 0.6 W resistor can handle some 300 V : without adverse effects, smaller resistors are usually 200 V tops. : 50 ppm/K, 2ppm/Vr resistors only cost about 1 eurocent, in bulk, : : Right. I must admit that I don't understand resistor voltage : ratings. The only way to get 1W through 1Mohm for example, is to put : 1kV across it. So what does a 1W, 1Mohm, 500V spec mean? A resistor : that never gets very hot, I suppose. : : so easy to just find out by gettin' some . : : Next opportunity, I'll pick up a whole roll of surplus resistors, : just for the hell of it. : : nb increasing the change of tighter tolerance is not the same : as equals tighter tolerance, so multiple series connecting : seems overkill to no avail. : : To most of the world, overkill is what we seem to be here for! : : thanks : : cheers, Ian : : |
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Suppose you bought a bag of 100 1k resistors and 200 500R
resistors. And, without measuring them, soldered together pairs of 500R resistors. You would then have two bags of 1k resistors both of which will have the same variation. Neither set will will be of tighter tolerance than the other. Thanks Joe It may be true that tolerance limits are not improved. That would depend on whether they are all measured, and whether the quoted tolerance is valid only at the time of measuring. Unless both of these are true, the quoted tolerance is not absolute, but is rather a statistical measure with some confidence limit. I am quite happy to accept that tolerance is absolute, but haven't found definitive evidence of this. In any event, I am not looking for better tolerance limits, I am looking for more accurate resistors. It is certainly not true that the 500R pairs would have the same variation, unless you mean that the 1k resistors would have more variation than the 500R to begin with, or that by "variation" you mean "tolerance". To be precise, I mean the variance of the distribution. If you have two independent distributions and add them together, the mean of the sum is the sum of the means and the variance of the sum is the sum of the variances. The variance is a measure of the spread of values. The spread for the 1k resistors is +/- 10R, and the spread of the 500R resistors is +/- 5R. So, when you connect two 500Rs together, i.e., sum their resistances, the spread of the sum is now +/- 10R. My contention is that averaging always reduces variance. This certainly improves the chance of accurate matching, but it is a bit less certain that it reduces the deviation from the nominal value. If I understand you, the you are actually referring to the variance of the estimate of the mean. A set of a very large number of randomly distributed resistors has a mean value. Let's call this the true mean. Because of the large numbers involved, you can't ever know the true mean, you can only estimate it from a bag of resistors. The larger the bag of resistors, the closer your estimate of the mean will get to the true mean. Joe |
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whoa there joe
apologies for no capitals or anything that requires a shift key. knackered keyboard. To be precise, I mean the variance of the distribution. If you have two independent distributions and add them together, the mean of the sum is the sum of the means and the variance of the sum is the sum of the variances. The variance is a measure of the spread of values. The spread for the 1k resistors is +/- 10R, and the spread of the 500R resistors is +/- 5R. So, when you connect two 500Rs together, i.e., sum their resistances, the spread of the sum is now +/- 10R. you say you mean variance, but then you confuse variance with tolerance limits. they are very different measures. variance is the sum of squares of differences from the mean of all members of the population. i got the formula wrong and suggested the ratio is 1/rootN. that is for the standard deviation. the variance of the sample mean is actually 1/N times the variance of the population. check last para of http://www.gps.caltech.edu/~tapio/ac...estimation.pdf it is from this relationship that you get... If I understand you, the you are actually referring to the variance of the estimate of the mean. A set of a very large number of randomly distributed resistors has a mean value. Let's call this the true mean. Because of the large numbers involved, you can't ever know the true mean, you can only estimate it from a bag of resistors. The larger the bag of resistors, the closer your estimate of the mean will get to the true mean. that last sentence is very tricky. on average, the mean of a sample will be the same as the population mean regardless of sample size. perhaps you mean that the variance of your estimates will reduce. then i am right with you. if you take many bags of n resistors and find the mean of each bag, then the variance of the bag means would be 1nth of the variance of the population. Hence, if you connect the resistors in each bag in series, which effectively averages them, the resulting resistances would show a variance of 1nth of the variance of the population. if i string n resistors together to make a number of compound resistors, then the standard deviation of the compound values would be 1/rootn that of the singles. standard deviation is the most widely accepted measure of variation within a population. note this is variation from the mean, not from the nominal resistor value. if i take resistors from various manufactures, however, the mean of the population would be very likely to be normal *and* the mean would be very probably the same as the nominal value. in other words, accuracy would, on average, improve by a factor of root n, where n is the number of resistors in series. thanks, that's a precise enough theory to partly test with rudy's data. it will take me a while though...and its all from one manufacture so i can only illustrate the effect of reducing variation from the mean, rather than from the nominal value. woo hoo. thanks cheers, ian |
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"Simon G Best" wrote in message ... Hello! I've read through the threads, and thought I'd respond :-) I'll state now, up-front, that I'm taking 'tolerance' to mean 'certainly no more than [whatever the tolerance is] away from the nominal value' (and that any resistors which fall outside their stated tolerances are therefore duff, and to be regarded as defective). Fair enough. Still some cloud around Phil's contention that service life is included in the quoted tolerance, which of course would mean there would be tails on the final distribution regardless of rejection at the factory. Does seem unlikely though so I will assume as you have done. Ian Iveson wrote: I need two accurate resistors. [...] The kind of resistors I want are only available at 1% tolerance. I am only building one example of the circuit. I don't care about anything except accuracy in both cases. For each resistor, R1 and R2, am I better off: a) Buying one resistor of the right nominal value. For R1, it could be as much as 1% above or below the nominal value (obviously). For R2, you could end up with one of the pair being as much as 2.020202...% above the other (if one's 1% up, and the other's 1% down). OK. same as for singles. b) Buying X resistors each of 1/X th of the nominal value and connecting them all in series. That'll tend to 'home in' on the mean value of those resistors. The bigger X is, the better (but you'll still occasionally be unlucky). However, the mean value of a bunch of resistors is not the same as the nominal value. Indeed, the mean doesn't even have to be particularly close to the nominal for the resistors to meet the tolerance requirements (the smaller the range of actual values, the further from nominal the mean can be). Yes. From the few we have measured. For example, Rudy's set of 159 56k 1% can be seen here http://www.ivesonaudio.pwp.blueyonde.../rudysdata.gif It just about conforms to a worst case scenario: it is offset in one direction, and skewed in the other. Hence the mode will move away from the nominal value. The mean will of course stay the same, and the distribution variance will decrease, resulting in a set of more accurately inaccurate strings. So, for R1, this isn't much good. But, for the pair of R2s, this will tend to be good, as long as each R2 in a pair is taken from the same, well-mixed bunch of resistors. But it's still possible to be unlucky, and get R2s in a pair that differ by up to about 2% (from each other). Yes. So accurate matching may be more likely, but if for a one-off circuit that needs to be right, there is no substitute for selection and trimming. c) Buying Y resistors of the correct nominal value and selecting the best one. That's much better than (b) for R1, but there's still no guarantee of a close match. All Y resistors might be, say, 0.3% to 0.7% below nominal. OK d) Buying X * Y resistors, each of 1/X th of the nominal value, and selecting the best series combination of X resistors. That won't work for R1, as you'll just end up with Y networks tending to be of the mean, rather than nominal, value. You might, just possibly, be (very) lucky, but you're much more likely to find a better match with (c). Yes For the R2s, assuming the X*Y resistors are in a well-mixed batch, this is the best method, yet. It's like (b), but with lots of networks to choose from. With Y such networks, you'd have Y*(Y - 1)/2 possible pairs :-) Woohoo! If I decide to produce the circuit in quantity, should that make a difference to which I choose? Well, for a single R1, (c) is the best of the options, even though it's not guaranteed to work very well in practice (but would still work better than the other options). Options (b) and (d) are just lousy. So, for producing the circuit in quantity, you could either settle for 1% plain and simple with option (a), or you could, perhaps, automate the measuring and selecting process for option (c). Yes. I'm getting worried about this agreeing thing. For pairs of R2s, it's either option (b) or (d), depending on whether or not you want to do the measuring. With (d) you have to measure and select, but with (b) you can just get on with it (as long as the batches of resistors are well-mixed, of course). With (b), though, there might be quality control issues, as it's always possible for bad matches to occasionally occur by chance. But, for both options, have a large X. Yes. sigh... But what about if I get many smallish batches from different makers and mix them. Then they should be gaussian and with a mean of the nominal value, considering the nominal value is the only target value. I have a bridge with a parallel port. Always wondered how to use it. Might be interesting to find out. But my mind is already wandering to valve matching, sepp, pse, and randomly selected massively parallel mixed pentode and triode. RSMPMPT. Catchy eh? Personally, though, I don't much like these kinds of approaches. I'd much rather make the need for calibration a feature and a selling point, complete with exciting panel meters and the like :-) And a bluetooth-enabled microprocessor-controlled "total precision management" calibration system with optional manual override and a "reset defaults" option. Thanks Simon cheers, Ian |
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Ian Iveson wrote: whoa there joe apologies for no capitals or anything that requires a shift key. knackered keyboard. To be precise, I mean the variance of the distribution. If you have two independent distributions and add them together, the mean of the sum is the sum of the means and the variance of the sum is the sum of the variances. The variance is a measure of the spread of values. The spread for the 1k resistors is +/- 10R, and the spread of the 500R resistors is +/- 5R. So, when you connect two 500Rs together, i.e., sum their resistances, the spread of the sum is now +/- 10R. you say you mean variance, but then you confuse variance with tolerance limits. they are very different measures. variance is the sum of squares of differences from the mean of all members of the population. i got the formula wrong and suggested the ratio is 1/rootN. that is for the standard deviation. the variance of the sample mean is actually 1/N times the variance of the population. check last para of http://www.gps.caltech.edu/~tapio/ac...estimation.pdf it is from this relationship that you get... If I understand you, the you are actually referring to the variance of the estimate of the mean. A set of a very large number of randomly distributed resistors has a mean value. Let's call this the true mean. Because of the large numbers involved, you can't ever know the true mean, you can only estimate it from a bag of resistors. The larger the bag of resistors, the closer your estimate of the mean will get to the true mean. that last sentence is very tricky. on average, the mean of a sample will be the same as the population mean regardless of sample size. perhaps you mean that the variance of your estimates will reduce. then i am right with you. if you take many bags of n resistors and find the mean of each bag, then the variance of the bag means would be 1nth of the variance of the population. Hence, if you connect the resistors in each bag in series, which effectively averages them, the resulting resistances would show a variance of 1nth of the variance of the population. if i string n resistors together to make a number of compound resistors, then the standard deviation of the compound values would be 1/rootn that of the singles. standard deviation is the most widely accepted measure of variation within a population. note this is variation from the mean, not from the nominal resistor value. if i take resistors from various manufactures, however, the mean of the population would be very likely to be normal *and* the mean would be very probably the same as the nominal value. in other words, accuracy would, on average, improve by a factor of root n, where n is the number of resistors in series. thanks, that's a precise enough theory to partly test with rudy's data. it will take me a while though...and its all from one manufacture so i can only illustrate the effect of reducing variation from the mean, rather than from the nominal value. woo hoo. thanks cheers, ian |
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