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I have some ceramic capacitors and would like to know for sure what capacitance they have. I don't have a multimeter (or any other instrument) that allows me to measure capacitance directly, so I build a simple RC filter that filters a square wave input, I know that the voltage at the output at the time constant R*C is 63% of Vinput ... I reverse the formula and find C.

The question (which is probably stupid) is whether I can use as input the square wave produced by the oscilloscope (is usually used to check before each measurement if the probe is correctly calibrated) .. otherwise I would need a square wave generator :/ enter image description here

enter image description here

After the responses I received, I will upload the photo of the calculations made and ask if they are correct: (565mV is the V measured with the CH1 probe at the ends of 220 ohms) EDIT: missing a multiplication by 2 in the equation! With that multiplication, the result is about 560 ohms which is correct because I used a 500ohm + 100ohm series finding about 1Vpp at the output .. which is half of 2Vpp enter image description here

I tried measuring a 100nF capacitor but obtained 36nF: enter image description here

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    \$\begingroup\$ Not sure if I'm reading your last diagram and interpreting its quantities, correctly. I think you are saying that the 2 Vp-p signal is showing as 0.565 Vp-p with the divider. If so, then I would find: solve(Eq(.565,2*220/(220+Ro)),Ro)[0]= 558.761061946903. \$\endgroup\$ Commented Jan 10 at 10:46
  • \$\begingroup\$ @periblepsis you are right, I forgot to multiply by 2 in the expression ... now it is okay \$\endgroup\$
    – KaleM
    Commented Jan 10 at 11:02
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    \$\begingroup\$ If you've got a PC with a sound card, you might have a look at this GNU Radio program I made a while back. \$\endgroup\$
    – JRE
    Commented Jan 10 at 11:03

5 Answers 5

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That in-scope square wave generator will be fine.

Measure its output resistance, as that will add to the effective value of any series R you are using with your C. Do that by loading it with a resistor. Either do the standard voltage-divider sums, or without calculation, when its output voltage drops to half, the external resistor is equal to the square-wave generator's output impedance. If the output voltage doesn't drop appreciably even with lowish values of loading resistor, then the output impedance is near zero.

IIRC, most oscilloscope calibrators are designed to also deliver a squarewave current into a short circuit, so have a finite output resistance, often 1 kΩ.

If your resistor in your RC is well above 1 kΩ, then you don't need to worry too much about the effect of the calibrator output resistance. Bear in mind though that now the loading resistance of your scope probe now becomes significant, reducing the effective value of the R as it's now in parallel to it. They are typically 1 MΩ on a x1 probe, and 10 MΩ on a x10 probe.

Be aware that many ceramic capacitors, especially ones that have an unusually high value of capacitance for their physical size, have a very large and negative voltage coefficient of capacitance. That means the actual value of C when biassed to 5 V might be less than half of what you measure with a 2 V squarewave.

This voltage coefficient also means that you might not find a consistent time constant for the RC combination versus voltage, as the capacitance is changing as the voltage changes.

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  • \$\begingroup\$ Thank you for your reply, I tried to proceed as you said. To avoid calculations, you suggested that I find the resistor that halved the output voltage so that the voltage divider formed between the oscillator's internal resistor and the external resistor I added would be equal. Did I understand this correctly? Having few resistors available, I had to make do with the ones I had and I did the calculation you can see in the main post I updated (I have 220 ohms available). Is that okay? Is that correct? \$\endgroup\$
    – KaleM
    Commented Jan 10 at 10:27
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    \$\begingroup\$ @KaleM Yes, 560 looks right from the figures. It's an odd value. Maybe they were aiming at 600, which is a sort of 'standard' value for audio work, maybe it just 'enough to protect the output from short circuits'. if your R used in your RC is way above this, then you don't have to worry too much about taking it into account. Then you have to worry about the shunting effect of the scope probe, usually 1 M on a x1 probe, and 10 M on a x10 probe. \$\endgroup\$
    – Neil_UK
    Commented Jan 10 at 11:33
  • \$\begingroup\$ I've updated the main post and I take this opportunity to ask you if I am measuring the capacitance of the capacitor correctly because although it should be 100nF, I measured a couple of the capacitors and they are 36nF . I would like to point out that I had bought them a while ago for $1 on aliexpress so I don't know how reliable they are! \$\endgroup\$
    – KaleM
    Commented Jan 10 at 12:59
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    \$\begingroup\$ @KaleM Your green trace isn't quite square, adjust your probe compensation, that's what the calibrator squarewave is for. But that small error is not going to contribute a factor of 3 in your answer. I don't see anything wrong with your measurements, just your choice of AliExxess as a supplier. Do you have any film/plastic caps to hand, they usually have a much better tolerance than ceramics. A factor of three sounds too big to be voltage coefficient alone, what's the specified voltage rating of the caps? \$\endgroup\$
    – Neil_UK
    Commented Jan 10 at 13:51
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    \$\begingroup\$ @KaleM '100 nF' ceramics caps are often sprinkled around boards like Smarties (tm) (small cheap chocolate sweet in the UK, US equivalent probably M&Ms) one per power pin. There will be so many on the board, and as 100 nF is usually over-kill (indeed I often used 10 nF rather than 100 nF in that role) that almost all users would be unaware of even a factor of 3 shortfall in their capacitance. \$\endgroup\$
    – Neil_UK
    Commented Jan 10 at 13:57
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There's another way to do this which is using the rise time. You can use the rise time approximation for a single pole RC low-pass filter to calculate the bandwidth of a circuit. $$ BW = {0.35 \over t_r} $$ The bandwidth of a simple RC low-pass is: $$ BW = {1 \over 2 \pi RC} $$ Where \$BW\$ is the bandwidth and \$t_r\$ is the rise time between the 10% and 90% points.

Playing some games with algebra: $$ BW = {0.35 \over t_r} = {1 \over 2 \pi RC}$$ $$ C = {t_r \over 2\pi R \;0.35} = {t_r \over 0.7 \; \pi R} $$

enter image description here

If we use 219.24 us for the rise time as measured by the cursors, we end up with C = 9.97 nF which is about a -0.3% error.

As others have stated, know the output impedance of the scope calibration output and make sure the rise time of the calibration output is at least 10 times faster than the rise time of the RC circuit. For smaller capacitors, the probe capacitance (around 13 pF) and resistance may become significant.

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  • \$\begingroup\$ Thank you so much for your response! I had not thought of this possibility \$\endgroup\$
    – KaleM
    Commented Jan 13 at 20:39
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In theory yes, but you likely don't know the specs of the square wave output, so you need to cancel out the effect of the output impedance, to better consider only the effect of the R and C you add.

You can easily generate a square waves good enough for the purpose with MCUs or other digital or analog ICs.

If you have MCUs available, it should be also easy to write a small program to charge and discharge the cap and give a time measurement how long it took or give a frequency output that is related to the capacitance.

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  • \$\begingroup\$ Right! I had not thought about using a microcontroller. I have updated the main post with the measurements I made (then maybe I will proceed using a microcontroller). I also await the response of @Neil_UK \$\endgroup\$
    – KaleM
    Commented Jan 10 at 10:29
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Using the test output is fine. Others have already mentioned the possible effect of the output impedance so I'm not going to repeat.

To decrease the effect of loading the test output or its output impedance, I'd strongly recommend buffering it first (or even amplifying if needed) and use afterwards. The output impedance of an op-amp-based amplifier (diff amp or non-inv amp) would be quite possibly way lower than that of the test output, so it'd be, hopefully, less problematic.

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Measuring capacitance via RC exponential rise using this hack can work. OP's method is slightly incorrect. Time constant is a bit longer, since OP's method doesn't start at the zero-volt reference:
modified scope trace


Many digital 'scopes now allow peak-to-peak measurements. One might use this feature to calculate capacitance. The three known quantities are:

  • Square wave duty cycle is 50% high, 50% low
  • Square wave frequency is 1000 Hz.
  • Known resistance is 560 ohms plus external added resistance.

The easiest way to use this alternate method applies a large-value external resistor, so that the waveform across it appears to be a triangle wave having straight sides and sharp peaks. This will require choosing \$R_{EXTERNAL}\$ so that \$V_C \lt\lt 2V\$.

Then \$ C= {{\Delta t}\over{\Delta V \times R}}\$

\$\Delta V\$ is simply the 'scope's internal peak-to-peak measurement.
Since \$ \Delta t = 500us\$ with a 1kHz square wave, calculation of C is simple. It is also simple because the voltage square wave source is 2V p-p, which makes average voltage 1V. With this method, you can measure \$\Delta V\$ with your 'scope AC-coupled, and avoid the 1V DC offset.
This method fails when \$\Delta V\$ approaches the 2V applied, since triangle waves have paths between peaks that are no longer straight lines.


One must be cautious when measuring small capacitance, since 'scope probe adds significant parallel capacitance to what you're measuring. A 1X 'scope probe adds all the cable capacitance, plus internal 'scope input capacitance. This can add over 100pf to what you're measuring.

A 10X attenuator probe adds up to 9X less capacitance.

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  • \$\begingroup\$ You are right! Unfortunately, you can't see it in the picture because (in live) the waves oscillate on the display .. so I can't really tell where exactly start to measure. This is because the oscilloscope is very cheap \$\endgroup\$
    – KaleM
    Commented Jan 10 at 14:40

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