Is there a way to average resistors together to get a tighter overall resistance tolerance?

Question

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?

Answer 1

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).

Answer 2

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:

^{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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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:

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

Answer 7

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.

Answer 8

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.

Answer 9

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.

Answer 10

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.

Answer 11

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.