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I have a very sensitive, kHz frequency level, application where I need two matched resistors of the same resistance better than 0.05%. Like maybe by one magnitude (0.005%).

I am basically needing to match two resistors together so as to form a differential bridge in order to measure a specific resistor. 50 kHz 180 degrees signal → 75 ohm → Resistor ← 75 ohm ← 50 kHz 0 degrees signal

Here is the schematic:

I need the 75 ohms to match so the signals across the 10 ohm resistor are symmetric in magnitude.

I can buy two 75-ohm 0.05% resistors, but, to get a lower tolerance than that, is there a way to make a circuit that uses a whole slew of 0.05% resistors and then averages them together to get better performance?

Peter Mortensen
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Jirhska
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    You can see mathematically that neither serial nor parallel connection of resistors is changing their overall precision. – Eugene Sh. Nov 03 '20 at 20:49
  • Yes series and parallel won't amount to any tighter tolerance. I meant more of an op amp circuit to do the averaging. – Jirhska Nov 03 '20 at 20:50
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    Then you probably need to describe more details of the application. – Eugene Sh. Nov 03 '20 at 20:52
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    What are you making? How will you control the resistors' temperature? How would an op-amp make a resistor better? – Transistor Nov 03 '20 at 20:52
  • @EugeneSh. I can't go in to too much detail because it is rather complicated but it is like a voltage divider/ bridge to measure resistance. It requires two known resistors. Signal -->75--> measurement resistor <--75<--Other signal. In this way I need matching 75 ohm resistors. – Jirhska Nov 03 '20 at 20:57
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    You can buy 0.005% resistors, although that level of absolute accuracy is not going to be cheap. Usually you use a trim pot for this and calibrate the device. – user1850479 Nov 03 '20 at 21:00
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    There may be a way for you to rearrange your design so that you are only using a single resistor for both signals. E.g., a relay or perhaps some type of chopping or self-calibration method. You should consider asking a question with more detail about your design so you can get some alternatives that don't need this very high precision. – Justin Nov 03 '20 at 21:04
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    You can use really crappy resistors and a controlled switching circuit to get what you appear to be saying what you want but, it's not entirely clear what you want so I'll say no more. Ditto what @Justin says. Therefore to answer your question (as it stands) the answer is an emphatic yes. However, that makes it a poor question and, to improve it, details are needed. – Andy aka Nov 03 '20 at 21:05
  • @Justin good idea but I really think it is a critical part of the design to have 2 resistors :/ . I believe if I gave too many design details it would overcomplicate things – Jirhska Nov 03 '20 at 21:06
  • @Andyaka Yeah things are going on simultaneously with this circuit Signal -->75--> measurement resistor <--75<--Other signal. So it cannot be switched unfortunately. – Jirhska Nov 03 '20 at 21:07
  • What signal bandwidth? – Andy aka Nov 03 '20 at 21:08
  • @Andyaka Signals are both at 1.2Vpp 50kHz. one at 0 degrees and the other 180 degrees. Tight bandwidth variation of .1Hz on those signals – Jirhska Nov 03 '20 at 21:09
  • I don't see switching would be a problem. – Andy aka Nov 03 '20 at 21:10
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    You can pay for laser trimmed $$$ – crasic Nov 03 '20 at 21:11
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    It's a differential signal needing 150 ohm differential termination? Are you trying to measure the common mode signal? – Justin Nov 03 '20 at 21:12
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    @Justin I am trying to measure resistance like a kelvin four wire method but as an AC method instead, and differentially. The measurement resistor does not have symmetric I versus V curves so it would be tricky to implement a switching method – Jirhska Nov 03 '20 at 21:14
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    I am voting to close because you are leaving out way to many details to allow us to give you a useful answer. – The Photon Nov 03 '20 at 21:32
  • Digital trimming by external means? Laser trimmed ceramic hybrid? – winny Nov 03 '20 at 21:54
  • @winny I wonder if Digital trimming a resistor in parallel with a 75R 0.05% ohm resistor would work. It would be on the order of 50k to 75k – Jirhska Nov 03 '20 at 22:10
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    You seem to be residing telling us what we need to know to really help. You have some world class experts here. Making best use of them is often a good idea. – Russell McMahon Nov 04 '20 at 00:13
  • @RussellMcMahon I've said most everything besides discrete ICs the best I can do without overcomplicating everything is this simplified schematic: https://imgur.com/a/CiTvj00 I need the 75 ohms to match so the signals across the 10 ohm resistor is symmetric in magnitude. Edited to the original post – Jirhska Nov 04 '20 at 00:50
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    Does this answer your question? [Reducing the tolerance of resistors manually](https://electronics.stackexchange.com/questions/77325/reducing-the-tolerance-of-resistors-manually) – Dmitry Grigoryev Nov 04 '20 at 09:39
  • Do you mean DC resistance or AC impedance? – Andrew Morton Nov 04 '20 at 19:53
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    Without reading too deep, this sounds like a poor design approach – Scott Seidman Nov 05 '20 at 00:05
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    The probability distribution will definitely change. But without knowing the probability distribution to begin with it is hard to say HOW it will change. Trimming seems more promising, although you will need to use fairly precise instrumentation to trim to that precision. – user57037 Nov 05 '20 at 00:27
  • Adding more resistors will only ever decrease the accuracy, not increase it. – Mast Nov 05 '20 at 12:29
  • How many of these do you need? Can you buy a pack and test each one with a high quality ohmmeter? – Joshua Nov 05 '20 at 19:53
  • @Joshua Yes I think that may be the next step. Buy once Cry once for the high quality ohmmeter that is – Jirhska Nov 05 '20 at 20:47

11 Answers11

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No.

Because the notion of averaging a bunch of resistor values only works if you can be sure that the error in their values is random, and has a zero-mean distribution.

Typically, neither of these is the case. First, because the resistors have already been selected for their value at the factory. Second, because there's no guarantee that what's coming out of the factory that day isn't biased in one direction or another. So you may get a lot of resistors one day that are right on value, the next they may all be \$75\Omega + 0.025\%\$, the day after that they may all be \$75\Omega - 0.025\%\$

You need to either design your circuit to be trimmed, or you need to pay for the super-precise resistors. (And note -- even .05% is getting absurdly precise, and you're going to start seeing all sorts of confounding effects from thermal and mechanical issues. Getting down to 0.005% is going to make things all the worse).

TimWescott
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    You can get all the way to 0.01% for not too much money (maybe $1.00 each). See the Stackpole RNCF series. – user4574 Nov 04 '20 at 05:43
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    0.01% of 75 ohm is well below the usual PCB etching tolerances. Is there a better PCBs? – fraxinus Nov 04 '20 at 09:05
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    @fraxinus I think using more copper would make process variation less meaningful in this context . – crasic Nov 05 '20 at 18:13
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    Sure. All factors considered, one may as well check for parasitic thermocouples between solder and copper. And probably other things as well. – fraxinus Nov 05 '20 at 20:12
27

You can buy resistor networks that have superior matching characteristic relative to tolerance. However 75 ohms is rather low for that.

Vishay has their excellent bulk metal foil resistors in 75 ohms with 0.01% tolerance and (just as important) +/-2ppm/°C drift.

If you are looking for a half bridge you can easily trim the midpoint in with a few additional resistor that can be a cheap 0.1% part and a cermet trimpot. For example:

schematic

simulate this circuit – Schematic created using CircuitLab

However the 2ppm/°C drift will remain. Two identical resistors should track fairly well if kept in close proximity.

You will also have to take care with inductance at such high frequencies and precision, 0.005% of 75\$\Omega\$ is only 3.75m\$\Omega\$. At 50kHz that represents around 12nH, about 10 or 15mm of straight wire. Ordinary wirewound resistors are about useless in that situation. Penetration in copper is ~0.3mm at 50kHz so wire will also have more resistance than at DC.

Frankly, the preferable solution for dividing an AC voltage when precision is required is to use a ratio transformer. You can get stability in the ppb, at least a couple orders of magnitude better than resistors.

Spehro Pefhany
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  • Thanks I suspect I will have to do something similar to this. Forgive my lack of knowledge but are there discrete options for ratio transformers that are more affordable than the one linked in the picture? I really am trying to do things discretely. – Jirhska Nov 04 '20 at 00:59
  • You can wind fixed ratio transformers on a ferrite core. Such things tend to be custom. – Spehro Pefhany Nov 04 '20 at 10:28
13

A 0.05% resistor is guaranteed to be no more than that tolerance from its nominal value.

When you connect for instance four resistors in a series and parallel arrangement to get the same value, there is nothing to stop them all being high, or low, so this will not reduce the worst case deviation.

However, if you took many sets of four resistors, then the spread of total deviation would be expected to be tighter than for the individual resistors. The problem is you're not interested in the statistics of many sets, you're only interested in the one set you have in front of you.

The benefit to buying tight tolerance resistors is that you often get a much tighter tempco (temperature coefficient of resistance) and better long term stability than is offered with 1% resistors. The manufacturers realise there's no point in selling a close tolerance resistor, if it doesn't stay that way with time and temperature changes. This allows you to trim them manually to a very close balance, and have a reasonable hope that they will stay balanced.

With 75 Ω resistors, you can get 0.01% adjustment by paralleling them with resistors of the order of 1 MΩ. These trimming resistors do not have to be high tolerance or tempco. There are ways to measure the balance of resistors with a DMM which itself has low resolution, for instance substitution in a Wheatstone bridge.

Neil_UK
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  • What is a "tempco"? Googling for it just gives me various brands which use that as their name. Is it an abbreviation for "temperature coating"? Why would that be correlated to a tolerance specification? Can you please [edit] your answer to clarify the meaning of this term? Thanks! – Cody Gray - on strike Nov 04 '20 at 08:00
  • @CodyGray Yes, take your point. When search for 'tempco resistor', the first proper explanation is 6 places down the results. – Neil_UK Nov 04 '20 at 08:50
  • @CodyGray "tempco" = temperature coefficient. All resistors will change their resistance if you change the temperature. – Simon B Nov 04 '20 at 23:19
  • Everything changes one way or another with temperature. Everything can be a thermometer if you have a way to measure the change. – user57037 Nov 05 '20 at 00:12
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    Actually you suggest a solution. If he uses only resistors from the same lot, even though their overall precision isn't going to be any better, they'll match each other better. – Joshua Nov 05 '20 at 19:55
8

You can get all the way down to 0.01% for about $1~$2 per resistor with the Stackpole RNCF series of resistors.

If you need to achieve an absolute resistance value then averaging lots of resistors won't work. If you just need to achieve a ratio, then yes, averaging a lot of resistors in a grid can in theory improve tolerance.

But 0.005% of 75 ohms is 3.75 milli ohms.

The problem is that even if you could get a 0.005% resistor the traces on your circuit board are going to contribute much more than 3.75 mOhms of error. Standard 1oz copper on a circuit board has a resistance of 0.5mOhms per square. So a 10 mil wide by 100 mil long trace will contribute 5 mOhms of copper resistance to your circuit. All the traces, wiring, solder joints, and vias together might contribute much more.

Even if you accounted for the trace resistances by careful layout; copper has a very high temperature coefficient (like 0.393% per degree C) so the resistance value is going to drift a lot with temperature. Also, once you run current through your 75 ohm resistor, its value is going to drift also with temperature (but not as much as the copper).

Additionally the thickness of copper on a PWB also has a lot of tolerance in it.

Basically your only option is to use a resistor close to 75 ohms (like 73.2) and then use a small trimmer potentiometer (like 5 ohms). Make sure to use a multi-turn trim pot so you can make fine adjustments.

To actually make the adjustments with the potentiometer you may need to add some test points on the board for you to make a precision resistance measurement. The location of the test points is critical because the value will change based on how many traces the measurement goes through.

Secondly since the board copper, the trim pot, and the actual resistor will all drift with temperature; you may need to keep the board in a controlled temperature environment. Usually this is done by enclosing the circuit in a box with a heating element to regulate the temperature. The trimming measurement and pot adjustment would need to be done while the board is at the correct regulated temperature.

user4574
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7

So, you are thinking that if you put ten 750 ohm resistors in parallel, the resulting tolerance would be better. This would only be true if you had a batch of resistors that had a gaussian distribution around the target resistance. This is most likely not the case. Manufacturing tolerances will often cause the average to be higher or lower than the target.

You also must consider temperature variances. The tighter tolerance resistor will have less temperature variation.

Mattman944
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  • I feel like there should be a way to match two resistors together. For resistive divider applications temperature coefficients would be ratiometric so that would be ok – Jirhska Nov 03 '20 at 21:00
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    @Joshua: Maybe but we don't know if you're doing ratiometric so it's not OK. You could save a lot of time wasting if you edit your question to explain your application. – Transistor Nov 03 '20 at 21:10
  • @Transistor My mistake, Edited. – Jirhska Nov 03 '20 at 21:17
  • Actually, the distribution does not need to be Gaussian. To be predictable, the mean would need to match the target as stated, but the distributions merely need to be i.i.d. in order to follow the inverse square root behavior for reducing the standard deviation. Typically, an untrimmed resistor will have a Gaussian distribution but a trimmed resistor will have something approaching a uniform distribution. – FrontRanger Nov 04 '20 at 17:08
  • Even with a Gaussian distribution with a non-deviating average, how much better would it be (on average) for 10 resistors? Sqrt(10) = 3.2 times better? – Peter Mortensen Nov 05 '20 at 03:58
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You would need to know how the manufacturing process takes place, at first. The resistors are produced in the same batch and then measured and sorted. In each bin they are thown according to the their tolerance. For example +/1%, +/-0.5%, +/-0.1%,... In the bin of +/-1% you would never find a resitor that is close to nominal value, it could be +0.5% to +1% or -1% to -0.5%.

As for your question, there is a Gaussian distribution:

$$E_{total}= \dfrac{1}{N}\sqrt{E_1^2+E_2^2+E_3^2+ ... E_N^2}$$

This would be valid only if all resistors would fall in the same bin, but also if you have in hand the resistors from the most accurate bin. For all others, this is not valid since the tolerances are already missing.

EDIT:

For example: Putting 10 resistors in parallel, from a most accurate bin, within that batch, you get:

$$E_{total}= \dfrac{1}{10}\sqrt{10}\cdot E= 0.32\cdot E$$

Marko Buršič
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    +1 for this. It is an old engineering adage that resistors marked in a given tolerance will *not* be within a better tolerance range, because it was tested as being outside that range, or else it would have been marketed as such. – CCTO Nov 04 '20 at 14:13
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    @CCTO: I used to think this was always true, and for the same reason, namely that more money could be made by marketing the component with a tighter tolerance. But a few weeks ago, I came across a batch of 5% 0603 resistors that defied this belief. I tested a few dozen with the same DMM, and not one of them was worse than 0.2% error. Not only that, but all the errors were the same polarity, so it was predominantly a systematic error. I was quite surprised. – FrontRanger Nov 04 '20 at 17:15
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    @FrontRanger They have all rights to pack all in the same bin. It could be that the process had this 0.2% drift and they just discarded this batch and sold it as 5% tolerance. – Marko Buršič Nov 04 '20 at 17:38
  • @MarkoBuršič Just to clarify your example: if you have ten 5% resistors in parallel, the achieved tolerance is 0.32(5) = 1.6% ? – S.s. Nov 05 '20 at 14:52
  • @S.s. No. Read my answer: the 5% bin is already missing those accurate resistors, so the deviation probability of those 5% doesn't follow a Gaussian distribution, the formula is valid only for Gaussian distribution. – Marko Buršič Nov 05 '20 at 16:55
  • @CCTO: Me looks at the 10% resistors we had in physics lab that were dead on at the start of the semester, then somebody cooked one and that one just wasn't. – Joshua Nov 05 '20 at 21:27
2

No.

You would need to start with precision resistors and then measure them on a bridge to ensure you have a matched pair.

Note that resistors are measured before the tolerance is assigned to them. A 20% resistor is guaranteed to be at least +/- 10% off. A 10% resistor is guaranteed to be at least +/- 5% off. Ditto for 5% and 1% likewise being off, and all the way to the finest resolution that company sells.

user267965
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Yes!

... but it might become impractical to do it many times. As stated by @timwescott the notion that resistor actual values are zero-mean distributed within the tolerance range is false. However you can buy 100 resistors and measure them all, select 2 (or more) with opposing variances and use those to achieve your desired accuracy.

  1. You might achieve a high enough tolerance on many circuits by buying a large number of resistors with a single order, and measuring them all (binning).

  2. Alternatively to avoid measuring you could try to achieve a distribution with a mean closer to the rated resistor value by purchasing many resistors from many batches or factories and mixing them together before random sampling.

Adam Ratcliff
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No.

But you can buy a bunch of reasonably good resistors, and measure them precisely at a couple of temperaturs. In the bunch, you'll quite surely have two matching ones. That's especially the case if you don't care about them being both at a precise value (then the expected number of required resistors is quadratic in precision), but you only care about being them close to each other (that's only linear). Resistors from the same manufacturing series will also likely have similar thermal characteristics, so your chances aren't too bad.

Then you only need to make sure that they are at the same temperature, so probably placed very close to each other physically and possibly also glued (with thermopaste) to a single heatsink.

yo'
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  • Why would similar thermal characteristics help? Wouldn't the best be opposite sign (but equal magnitude)? – Peter Mortensen Nov 05 '20 at 04:10
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    @PeterMortensen Well, if I read the question correctly, the OP needs two resistors that are both ~75 ohms, but they are very close to each other. So having two resistors that are both 75.1 at 20 degrees and 74.9 at 80 degrees is fine with him. And in that case, he needs (1) identical thermal characteristics and (2) ensure the temperature is equal. Or do I read it wrong? – yo' Nov 05 '20 at 15:20
0

However this is AC and to control the 'amplitude' of a sin wave you must use an inductance and not a resistance ( that work only in DC based on Ohm's law). Calculations will differ accordingly so ideally one would use a coil-inductance and take the midpoint of it to obtain two equal amplitude sin waves. You can then be as accurate as you wish and build a specific inductance calculation wire type/size/etc.

WaveForm01
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  • No. Many circuits, that do work, wouldn't work if resistor (networks) didn't change the amplitude of AC signals. – Peter Mortensen Nov 05 '20 at 04:13
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    @Waveform01, the potentiometers on my electric guitars, radios, signal generator and oscilloscope all "control the amplitude" of alternating signals. Ohm's law applies to resistance no matter whether it's DC or AC. Some correction to your thinking is required. You can [edit] your post to correct it. Welcome to EE.SE. – Transistor Nov 05 '20 at 18:22
0

It's a lottery. With two resistors in parallel, at worse they won't make any difference because their bias will be identical. At best they will compensate perfectly, their bias being opposite. Most of the time you will have something in between.

To increase your chances of having complementary resistances, you should buy them form two different manufacturers.

Fredled
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  • Well first of all, using just one resistor, at worst it won't make any difference because it will have the worst variation. At best it will be perfect on its own. Most of the time you will have something in between. - so that part about two resistors in parallel is somewhat useless – quetzalcoatl Jan 12 '21 at 15:08
  • Second - two Rs in parallel may both have deviation of +MAX or -MAX. Even buying from different sources, you have no idea what deviation will be.. how large and in which direction. Even by buying from multiple sources you still may get all +x or all -y and you can't buy 2 from 2 suppliers, connect them in parallel, and live happily ever after. The only reason for connecting more than needed, and/or buying deliberately from different sources is when you're going to measure and match them manually before use and you count on getting them from various batches to get more spread. – quetzalcoatl Jan 12 '21 at 15:13
  • @quetzalcoatl I agree. I'm not advocating this method. – Fredled Jan 12 '21 at 18:40