Measure capacitors in parallel and series to increase precision?

Question

I measure a capacitor \$C_1\$ with an instrument that has a tolerance of measurement \$\pm 5\%.\$ I do the same with another cap \$C_2\$. Then I put \$C_1\$ and \$C_2\$ in parallel (values add up) and measure both of them with the same instrument. Did I just dropped the tolerance to \$\pm 2.5\%\$? Can I compute (manually calculate) this way a new value of both \$C_1\$ and \$C_2\$ with a higher precision? (\$2.5\%\$)

I've just done some math and it seems unlikely. It's counterintuitive! (sad face). So here's the thing. When I ADD on paper the values of \$C_1\$ and \$C_2\$, their tolerances don't add up, contrary to what I thought. It still remains \$5\%\$. Then when I MEASURE the same capacitors in parallel, I get the MEASURED value of the sum, with the same tolerance \$5\%\$ as the one CALCULATED by hand on paper. So I win nothing. I get nothing.

However, my intuition tells me that all this is wrong. My intuition tells me that I CAN INCREASE THE PRECISION OF MEASUREMENT by multiple configurations of same two capacitors.

Hear me out. All the theory books tells us, that in order to increase the precision of a measured thing, you measure it repeatedly and average out the results. The more measurements, the higher the precision, the closer you get to the REAL VALUE OF THAT THING. That is in theory.

However, in practice, as usual, things are slightly different. I want to increase the precision of measurement of 2 capacitors with a digital \$LC\$ meter model LC100-S which has a tolerance of every measurement within \$5\%\$ of the REAL VALUE (\$+\$ or \$-5\%\$).

If I measure the same capacitor 10 times, I get the same value each and every time. So you can throw the theory books out the window on this one. Contrary to what the math tells us, my intuition tells me I CAN INCREASE THE PRECISION OF MEASUREMENT and decrease the tolerance from \$5\%\$ to a much lower value. If I measure the two capacitors \$C_1\$ and \$C_2\$ in parallel and in series, I can determine their values and average those values with the values measured individually. Thus I get a much closer value to the TRUE REAL value. WILL IT WORK? How do I do it?

Answer 1

Further edit: in order to answer the observations of Antonio51 and the one of the Asker, I've added some notes on the real structure of the reading precision of instruments (which is independent from the instrument being analog or digital) and on the LC100-S LC meter.

The answer
The calculations are formally impeccable: however the conclusions are flawed by a (quite common) misconception related to the meaning of the (so called) percentage error. Percentage error is only a means to express the absolute error in a more compact way: precisely, when I say that an instrument as an error of \$\pm 5\%\$ I mean it has an absolute error of \$\pm 5\%\$ of its full scale value(s). Thus, if your \$LC\$ meter is set to measure a full scale maximum value of say \$10\text{nF}\$ and you measure a capacity \$C_1=7\text{nF}\$, it is not correct to say that the value of \$C_2\$ lies in the interval \$(7\pm 0.35)\text{nF}\$. The \$\pm 5\%\$ error really means $$ C_1 \in (7\pm5\% \mathbf{fs})\text{nF}=(7\pm0.5)\text{nF}\label{1}\tag{1} $$ where \$\pm5\%=0.5\text{nF}\$ is the absolute error of the given instrument for that full scale value.

In sum, the increase of precision of the measured value in your reasoning is due to the fact that a constant \$\pm 5\%\$ error on the measured value(s) is assumed, which is not true.

Notes

• The complete structure of reading precision, for an analog or digital mesuring instrument. The above formula comes from the following, exact one $$ v_\text{true} \in \big(v_\text{read}\pm e_\% \mathbf{fs}\pm 1\text{lsf}\big), \label{2}\tag{1'} $$ where

  • \$v_\text{true}\$ is the "true" value of the quantity to be measured.
  • \$v_\text{read}\$ is the value given by the instrument,
  • \$e_\% \mathbf{fs}\$ is the absolute error, or the precision of the instrument,
  • \$1\text{lsf}\equiv 1\$ unit on the least significant figure is the reading error i.e. the one done when you read the value on the display of the instrument. It can be the single graduation in the graded index of an analog instrument, or a "\$1\$" one on the least significant digit of your digital instrument.

  In any case, the reading error is never a constant percent of the read value: values near to the low side of the range are measured with a significantly worser precision than values near to the high side.

• The applicability of formulas \eqref{1} and \eqref{2} to a digital instrument. Answering to Niculae's comment, let's see why these formulas are respectively nearly exact and exact by analyzing their application to the LC100-S, which thus is no exception to the rule defined by formula \eqref{2}. This nice instrument offers a reading with four significant digits and an autorange function. This means that, for the case of a capacitance measurement, you have for example the following ranges automatically switched $$ \begin{array}{|c|c|} \style{font-family:inherit}{\text{Reading}} & \style{font-family:inherit}{\text{Full scale }(\mathbf{fs})\text{ range}} \\\hline 5.00\mu\text{F} & 10.00\mu\text{F} \\\hline 7.00\text{pF}. & 10.00\text{pF} \end{array} $$ In the first case we have $$ C_\text{true} \in \big(5.00\pm 0.1\pm 0.01\big)\mu\text{F}= \big(5.00\pm 0.11\big)\mu\text{F}\simeq \big(5.00\pm 0.1)\mu\text{F}, $$ while in the second case we have $$ C_\text{true} \in \big(7.00\pm 0.1\pm 0.01\big)\text{pF}= \big(7.00\pm 0.11\big)\text{pF}\simeq \big(7.00\pm 0.1)\text{pF}, $$ Thus, equation \eqref{1} and \eqref{2} are respectively nearly exact and exact also for digital instruments, in particular for the LC100-S: therefore the objection contained in my answer remains valid.

Answer 2

Did I just dropped the tolerance to ±2.5%?

Nope. Let's say the caps are 1 uF, just to have a number to deal with.

5% of 1 uF is .05 uF, and you know that that is the maximum error for each cap.

If you measure in parallel, your maximum error is 5% of 2 uF, or 0.1 uF. If each cap were high by .05 uF, the parallel combo will be high by 0.1 uF, and the meter won't help.

Answer 3

Assuming you have perfect caps and the error is only on the measurement side.

You measure (C1 + C1err) on C1 and (C2 + C2err) on C2. When you add those, you add errors. So the worst it gets is 1.05*C1 + 1.05*C2 = 1.05*(C1+C2). Obviously, errors can be of the opposite sign, so you result will get more accurate since the errors partially cancel each other out. But it may just as well be in one direction.

The point is, the error is on the measurement side. For your measuring tool, there are no 2 caps of 5nF. There is one cap of 10nF. There are no 10 caps of 5nF. There is one cap of 50nF. Measuring those will give exactly identical results, since, again, it's your tool that sees 50nF wrong (for example, shows 48nF for the real value of 50nF). Your tool doesn't know how many caps there are, it only sees total capacitance and nothing else. It doesn't know what it's measuring.

Now, if you have 10 tools each with 5% error and you measure with them, then the distribution of the error will indeed be tighter, but the maxima and minima will still be 5%.

Things change if you talk about part tolerance - measuring multiple caps is more likely to be close to true value, especially if you use more and more caps, but we are still talking about probabilities. If you imagine error has normal distribution around true value, then more and more caps will make the middle more and more likely (at least as long as caps are of similar order of magnitude, bigger cap obvious has greater individual influence on error, I'm simplifying for the sake of making it more intuitive). But minima and maxima add and will still be +-5%, there is still a chance that all your measurements will be all 5% off to one direction and therefore so will the total error as well.

If you want to get true cap value, it is actually a good idea to measure many caps of the same value in parallel, the chance that you get some weird batch with all of them off in one direction will already be next to zero (assuming your tool is much more precise and the error comes predominantly from parts' tolerance). It doesn't take too much to get probabilities of that comparable to piano falling on you in the street. It's like tossing a coin 1,000,000 times. You can get 900,000 to 100,000 in theory, but you will probably get something like 50.1 to 49.9.

Answer 4

I got somewhere with this. Here's my progress so far. Let's start with this example problem:

Two capacitors \$C_1\$ and \$C_2\$ are given. We are asked to measure their values with maximum precision possible. We are given an LC Meter which has a tolerance of 5% for every measured component. We are given the TRUE REAL values of our caps as \$C_1\$=7nF and \$C_2\$=5nF, but we don't know these values, and we are asked to show on paper how through math, we have calculated(based on measurements) new values for \$C_1\$ and \$C_2\$ which are much closer to their REAL TRUE values.

Problem appears simple, we just measure the caps and we are done, right? Not so fast. This is surprisingly much more complex than that. We are asked to increase the precision of our measurements.

At first we can determine the 5% interval based on their real values. For \$C_1\$ the measurement MUST BE within 6.65-7.35 and for \$C_2\$ the measurement MUST FALL within 4.75-5.25

If we calculate the sum for the minimum values is 6.65+4.75=11.4 which is 5% lower than the TRUE SUM(12) and if we calculate the sum for the maximum values we get 7.35+5.25=12.6 which is 5% higher than the TRUE SUM(which is 12). We thus have proven that if we measure a parallel connection (with 5%) we are no closer or farther from what we can just calculate from their individual measured values. We haven't increased any precision nor decreased any tolerance. We aren't going anywhere with this, or are we?

We start by measuring \$C_1\$. The LC meter reads 6.83nF and for \$C_2\$ it reads 5.15nF. We then put them in parallel and measure 12.1nF Interesting...

We have to think this in reverse. Instead of thinking of our measured value to fall within 5% of the TRUE REAL VALUE (left and right 5%) we think of the REAL VALUE to be within 5% of the measured value (which is true if you think about it). This reversed-thinking is very helpful to proceed further with our calculations. We now can determine that OUR REAL VALUES for \$C_1\$ must be within 6.4885-7.1715 and for \$C_2\$ must be within 4.8925-5.4075.

We then calculate the sum of \$C_1\$ and \$C_2\$=11.98 and we average it with the one we measured 12.1 and we get 12.04 This new sum immediately reduces our intervals, cutting from the extreme ends of them, since not all values added respect this new sum of 12.04

Please feel free to correct me if I'm wrong, but the second interval for \$C_2\$ (4.8925-5.4075) "restrains" the first \$C_1\$ interval from (6.4885-7.1715) to (6.6325-7.1475) which does not seem a lot, but trust me IT IS!!! Put in another way, this new restrained, reduced or cropped interval means our NEW VALUE FOR \$C_1\$ (the middle of the new interval) is 6.89nF \$\pm 3.7\%\$ !!! Amazing! The value for \$C_2\$ remains (in this example) unchanged.

This is extraordinary jaw dropping(at least for me) improvement in measurement accuracy !!! And we can do this just by 3 measurements \$C_1\$, \$C_2\$ and then measure them in parallel. All this calculations can be done in the software for an eventual future build digital LC meter! Amazing!

I have yet to complete similar calculations for a series connection of the same two capacitors.

What am I doing wrong?