Combining tolerances of resistors

Discussion in 'Math' started by davebee, Nov 4, 2013.

  1. davebee

    Thread Starter Well-Known Member

    Oct 22, 2008
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    How do tolerances combine when equal valued resistors are paralleled?

    I'm making a project, a NiMH battery charger, in which I'm trying to measure the battery current by measuring the voltage across a low value shunt resistor of about 0.5 ohms.

    The shunt resistor is just a junkbox 10% wirewound, and I want to estimate its value to quite a bit better than that.

    (This is a hobby, not a commercial project, so I'm not interested in buying a precision shunt; the point of this project is learning how to do this sort of design.)

    I have a digital voltmeter that measures using a precision reference of about 0.2% tolerance, and I have a box of several dozen 250 Ohm precision resistors rated at 0.01% tolerance.

    My thought was to parallel a bunch of the 250 Ohm resistors to make a quite low valued precision resistor and connect that in series with the shunt resistor, then apply a voltage to the series pair to send the same current through both resistors. Then by comparing the voltages across the resistors, the resistance of the shunt can be expressed to much better than its 10% stated tolerance.

    So if several of the 0.01% resistors are paralleled, how do their tolerances combine?

    Paralleling two of the 250 Ohm resistors gives a nominal 125 Ohms. The parallel of two maximum toleranced resistors gives 125.0125, and two having the minimum at their stated tolerance gives 124.9875, so with a combined resistance of 125 Ohms +/- 0.0125 Ohms, it seems like the tolerance of the combined pair is unchanged at 0.01%.

    Is that right?

    It would be great if it were really that simple, because with that, I could parallel 10 or even 25 of the 250 Ohm resistors and still have a 0.01% combination.

    Is combining tolerances (of equal valued, equal toleranced resistors) really that easy?
     
  2. adam555

    Active Member

    Aug 17, 2013
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    I guess you got it right: tolerance would remain the same whether you have a single resistor, 2 resistors in series, or 2 resistors in parallel.

    For example: if you have 2 x 100 ohms resistors in series with 5% tolerance, and both are at the lower limit, you'll have 190 ohms total resistance, which is still 5%.

    And the same goes for parallel: if you have the same 2 resistors in parallel you will have 47.5 ohm, which is again 5%.

    Same goes for resistors of different values; a percentage always remains the same.
     
  3. Wendy

    Moderator

    Mar 24, 2008
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    When paralleling resistors the highest values have the least effect. Program it in a BASIC program and play with it, it is bread and butter to simple calculators.

    The extremes are if everything is at a low 5% or a high 5%. This will give you the final numbers. Everything in between is a bell curve.

    At Rockwell Int. Collins Div. they had a problem with the then new laser trimmers stopping at -5%, always. This affected performance of some circuits. The answer was to program the software (on really primitive computers) for -0.1%/+5%. Problem solved.
     
  4. DerStrom8

    Well-Known Member

    Feb 20, 2011
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    When paralleling resistors of the same resistance value, the combination is only as tolerant as the least tolerant resistor. So if you have 4 resistors in parallel--all 100 ohm--with tolerances 0.1%, 1%, 5%, and 20%, the overall tolerance rating will be 20%.

    At least I'm pretty sure that's how it works....

    Matt

    EDIT: Keep in mind that if you have a higher tolerance and the resistance is significantly higher than the others, it will force more current through the other resistors.
     
  5. Wendy

    Moderator

    Mar 24, 2008
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    Like I said, it is the smallest value that rules. Take 4 conditions, 1KΩ ± 5% and 100KΩ ± 5%.

    Perfect World: 1KΩ ║ 100KΩ ≈ 990Ω
    1KΩ - 5%, 100KΩ - 5%: 950Ω ║ 95KΩ ≈ 940.6Ω
    1KΩ - 5%, 100KΩ + 5%: 950Ω ║ 105KΩ ≈ 941.5Ω
    1KΩ + 5%, 100KΩ - 5%: 1050Ω ║ 95KΩ ≈ 1.038KΩ
    1KΩ + 5%, 100KΩ + 5%: 1050Ω ║ 105KΩ ≈1.039KΩ

    See what I mean?

    I use something very similar to tweak a 5% value to something much more precision (if it doesn't drift).

    BASIC programming is your friend, as is any math type programming language like FORTRAN.
     
    DerStrom8 likes this.
  6. WBahn

    Moderator

    Mar 31, 2012
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    One thing to keep in mind is that putting resistors in parallel introduces a systematic bias toward lower values.

    To see what I mean, consider two resistors in which one resistor is above the mean and the other resistor is below the average value. So

    R1 = R + ΔR
    R2 = R - ΔR

    If you these are in series, the total resistance is

    Req = R1 + R2 = 2R

    The errors in the resistors that are high cancel out the errors in the resistors that are low in such a way that, if you put a whole lot of resistors in series you expect to get total resistances that that converge on the nominal value. So say you had 100 10Ω/10% resistors and you put them all in series. Assuming (and it is very likely not a good assumption given binning and systematic errors in lots) that the values are normally distributed over the range, you would actually expect to end up with a 1000Ω total resistance that is within 1% of the target value. This is basically a case of averaging a signal in order to average out the noise and the noise generally falls off as the square-root of the number of samples.

    But no consider the case of those same two resistors put in parallel. The total resistance in this case is

    Req = R1||R2 = (R^2 - (ΔR)^2)/(2R)
    Req = (R/2)*(1 - (ΔR/R)^2)

    So if you put two 10% resistors in parallel, you expect see an average total resistance that is about 1% lower than it should be. As you put more and more resistors in parallel, the amount of the bias grows (but the rate at which it grows slows).

    If you are looking at putting enough 250Ω resistors in parallel to get down to something comparable to the 0.5Ω you are wanting to characterize, you would need 500 resistors and even with 0.01% resistors this bias might well end up being significant.
     
  7. Tesla23

    Active Member

    May 10, 2009
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    For reference, assuming your resistors are samples of a random variable uniformly distributed in the range (R(1-δ), R(1+δ))

    then if you connect N in parallel, the expected value of the resistance of the combination is

    \frac{R}{N}\left(1-\left(1-\frac{1}{N}\right)\frac{\delta^2}{3}\right)

    I think you are a bit high in your estimate WBahn, for two 10% resistors (δ=0.1) then the expected value is only 0.167% low. This increases to 0.25% for N=4 and to 0.33% as N gets very large.

    For 5% resistors the effect is only 1/4 as big.
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    I realize I was a bit too loose with my phrasing. What I meant was that if the two resistors you use are up/down by 10%, then you expect to be about 1% low. That was what I meant, but I was sloppy and then that led me to draw a conclusion that was unwarranted because clearly this is the extreme case and you have to average it over all the other cases.

    Thanks for the catch and the correction.
     
  9. Tesla23

    Active Member

    May 10, 2009
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    It was the high value that made me take notice and work it out.

    It's interesting that for N>3/\delta that the bias exceeds the standard deviation of the resultant resistance.
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    It's very interesting. This seems like a good problem for a prob/stats course because it is a very real-world situation in which you can't just go assuming unbiased results.

    It would be interesting to look at the behavior for real distributions, particularly distributions typical "back in the day". It's hard to really even guess what it was like. Ont he one hand, I can easily see that you just made batches of resistors not caring at all about process variation. In fact, you might want a lot of process variation so that in any given set up you got resistors that varied over a wide range that you then binned into several different standard values (and tolerances) before having to change the setup. If that's the case, then the distribution might end up being fairly uniform between the tolerance limit and the next tighter tolerance limit. On the other hand, if you are nominally set up to make a particular resistor you probably wouldn't care too much about controlling the mean tightly and so, after binning, the distributions might be very asymmetrical about the nominal mean.
     
  11. Wendy

    Moderator

    Mar 24, 2008
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    Collins Radio had many circuits that were very precision on the alarms, but due to variations in components they were not predictable. So we used 1% parts to control drift, then paralleled them. CRG (Customer Returned Goods) were a lot worse, they had nothing but 5% parts, and soldering them caused lots of value drift. It was very aggravating, but we had to make do. Modern 5% parts are much more stable, indeed, many times they use the same exact process to make 1% parts, it is easier to just label them at a lower tolerance and then sell them.

    Lets take a real world example though. You need 1150Ω 1%. Your meter is .2% accurate. Since I am talking hobbyist use we'll treat the meter as completely accurate.

    A 1100Ω 5% might hit this value, but maybe none of your parts are there. So we look at a 1.2KΩ 5% part. It variance is 1140-1260. It may still be unusable, but we measured 1182Ω. This we can use.

    So the parallel resistor we would need is

    R2 = 1 / (( 1 / Rt) - (1 / R1))
    R2 = 1 / (( 1 / 1150Ω) - ( 1 / 1182Ω)) = 42478Ω ≈ 43KΩ 5%

    So the 43KΩ can be 40.85KΩ to 45.15KΩ

    Lets see what the tolerance will be with this unmeasured resistor (43KΩ).

    Rt = 1 / (( 1 / R1 ) + ( 1 / R2 ))
    Rt = 1 / (( 1 / 1182Ω ) + ( 1 / 40.85KΩ)) = 1148.5Ω which is -0.4%
    Rt = 1 / (( 1 / 1182Ω ) + ( 1 / 45.15KΩ)) = 1151.8Ω which is +0.2%

    Works like a charm. Sometimes there is no substitute for doing the math.

    Always remember though, whenever you solder a resistor there is a very good chance it will change value.
     
  12. davebee

    Thread Starter Well-Known Member

    Oct 22, 2008
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    Thank you all who responded. The tolerance remaining the same is how it seemed like it would work, but somehow it seemed too simple.

    I connected 25 of the 250 +/= 0.01 resistors to form 10.00 Ohms, so the unknown resistance should be R = 10.00 * V1/V2 where V1 and V2 are the voltages across each resistance.

    Now I'm trying to figure out which would dominate the result - the resistance uncertainty or the voltage uncertainty.

    Wbahn, that's interesting. I see what you mean, after doing the calculation for a parallel pair (250 + 0.01) with (250 - 0.01). Learning that kind of thing is the purpose of what I'm doing. But for this particular case, I don't think it will affect the result, because my suspicion is that the voltage uncertainties will be much greater than that resistance uncertainty.

    Even if I wanted to figure out the parallel resistance to that level, I don't think I could, because I don't know the resistance distribution of all the precision resistors within that tolerance and I don't have any instruments or references better than 0.01% to compare the resistors to.

    The voltages come from a computer-controlled 24 bit ADC with external voltage reference, op-amp buffer and CMOS switch, and so far it seems that the two dominant uncertainties are from random noise (which can be reduced by averaging) and thermal drift of the op-amp offset and the voltage reference chip (which may be able to be minimized if it turns out to be a major error source.

    Anyway, it's proving to be an interesting project.
     
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