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My multimeter does not have pF resolution and I am not able to find any tools which can measure a pF capacitance.

My goal is to identify the capacitance of a 0603 capacitor that I have removed from a PCB. I suspect the capacitor is in the order of pF as I am not getting any reading with my Fluke 117 multimeter.

The part that I removed looks similar to this one.

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    \$\begingroup\$ What other equipment do you have? Oscilloscope? Function generator? \$\endgroup\$
    – Mattman944
    Feb 14, 2023 at 3:38
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    \$\begingroup\$ I do have a digital oscilloscope and function generator \$\endgroup\$
    – dinoswords
    Feb 14, 2023 at 3:39
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    \$\begingroup\$ Are you sure it’s a cap and not a resistor or inductor? \$\endgroup\$
    – Michael
    Feb 14, 2023 at 12:10
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    \$\begingroup\$ I'm using a Nano-VNA for measuring the pF range with reasonable accuracy \$\endgroup\$
    – Bart
    Feb 14, 2023 at 13:11
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    \$\begingroup\$ If it's one of the capacitors for a crystal oscillator circuit attached to some IC, the app note for the IC might say. \$\endgroup\$ Feb 14, 2023 at 19:23

9 Answers 9

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If it is really a few pF, then you will need special equipment. If it is 30-100 pF, you can use an RC filter and measure the cutoff frequency. Adjust the frequency until the voltage out is 70%, or the phase shift is 45 degrees.

For more accuracy, measure the capacitance of your scope by measuring without the cap, then subtract the scope capacitance.

$$ C = \frac{1}{2\pi Rfc} $$

schematic

simulate this circuit – Schematic created using CircuitLab

Edit: some quick measurements:

100 pF: fc = 14.4 kHz -> C = 111 pF

50 pF: fc = 26.1 kHz -> C = 61 pF

No cap (scope only): fc = 129 kHz -> C = 12 pF

Subtracting the scope capacitance:

100 pF cap: 111 - 12 = 99 pF

50 pF cap: 61 - 12 = 49 pF

Both well within the tolerance of the components.

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    \$\begingroup\$ or for accuracy, buy a 1% leaded 100pF cap and use it to "cal" your setup ... you should then get 2 or 3% accuracy \$\endgroup\$
    – danmcb
    Feb 14, 2023 at 11:12
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How? Tweezers multimeter.

enter image description here

I think I paid $25 for this one. There's a bit of messing around to get readings at low pF because the meter will read your hand on the probes as capacitance, so you need to mess around a bit with elastic bands and try to tare out the ambient capacitance.

The $300 models may be more convenient to use.

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    \$\begingroup\$ Do you think it's very likely that a $25 tool can measure down to pF with any form of accuracy? I have no good experience of using these at all. \$\endgroup\$
    – Lundin
    Feb 14, 2023 at 13:00
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    \$\begingroup\$ when it I turn it on after about a minute it settles on a reading of 0.063 nF. I press the "rel" button and it reads 0.000nF - looks good enough for low precision (+/- 1 count) picofarad readings to me. you can probably spend more money for a better experience. \$\endgroup\$ Feb 14, 2023 at 20:44
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You need to make a differential measurement. Measure 100pF, or 1 nF, with and without your small cap in parallel. If the meter still offers poor resolution when measuring higher capacitance, build a small RC oscillator round an HC14, and measure the frequency with your oscilliscope.

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The parasitic capacitance in the leads of your Fluke 117 is already on the order of pF and is not compensated for so it can't measure tens of pF with it.

A LCR meter in the form of tweezers is most convenient for measuring SMD components.

But only a good (aka very expensive) bench LCR meter can measure tens of pF. Even then, it's tricky. Parasitic capacitances of boards and wire leads are in the single or double digit pF and can easily swamp your reading if you are not diligent.

You should just get a pair of LCR tweezers. They are fairly affordable and useful. Then you will have a better idea if your capacitor is in the tens of nF and requires a bench meter or not.


You could build a capacitive divider using a second known capacitance and measure the AC or DC voltage and calculate the value of the unknown capacitor. But if you use DC be aware that ceramic capacitors have reduced capacitance under DC bias. And if your capacitor truly is in the tens of pF, the parasitics of the surrounding board are going to throw off your measurement.

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The easiest and cheapest (AFAIK) is to use a capacitive divider.

The Capacitive Divider Schematic Test Fixture

schematic

simulate this circuit – Schematic created using CircuitLab

The principle is to use a known (and hopefully precise) series capacitor C1 to perform voltage division with (See schematic below).

Feed a signal that will make the impedance \$Z_{C_1} + Z_{C_{test}}||Z_{C_{probe}}\$ much larger compared to the 50 ohm termination "Rterm" from your signal generator, such that the signal amplitude you see set in your generator is what you're actually feeding as an input (and not something attenuated).

For this particular test, 100kHz should be good enough.

Before doing anything, you need to know your probe's input capacitance. You can probably find it in the datasheet.

But in case you want to measure it, you can simply use the same capacitive divider test fixture; remove Ctest from the schematic shown below and measure with your probe.

The measurement you see in your oscilloscope or spectrum analyzer will already be considering the probe's capacitance, so you can know its value from the following equation

$$ \frac{V_{measpoint}}{V_1} = \frac{C_1}{C_1 + C_{probe}} $$

Solve for \$C_{probe}\$ in this equation (since all other variables are known) and you're done.

Now, you're ready to measure your unknown capacitance.

$$ \frac{V_{measpoint}}{V_1} = \frac{C_1}{C_1 + C_{probe} + C_{test}} $$

Simply solve for \$C_{test}\$ and you're done with your measurement.

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I've done it with a NanoVNA. I made a test jig, calibrated it with the jig attached but no cap, and then added the cap.

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Here is an active circuit that eliminates several parasitic influences, like the scope probe and doesn't require another small and known capacitor (only a known resistor).

Its precision is probably better than 0.1 pF, the accuracy is limited by the capacitance of the button... I guess also better than 1 pF.

schematic

simulate this circuit – Schematic created using CircuitLab

The trigger button is normally-closed, which will make the output voltage 0.

When the button is pressed, the output voltage will ramp linearly at a rate of:

$$\frac{\text dV}{\text d t}=-\frac{V_\text{in}}{R\cdot C_\text{test}}$$

When the button is released, the output voltage will go back to 0.

Both, R and Vin (e.g. a battery) can be easily measured using a hand-held multimeter.

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Another method which I have used successfully is to put together a rig which has a known inductance - including its lead lengths etc.. This inductance would be then placed in parallel to your unknown cap to form an LC.

Now - instead of trying a frequency sweep, you simply feed a sharp pulse to this via whatever means you have available. I used a uC to generate a square wave, and a schottky hi-current NOT gate. Only the positive edge of the square wave matters - to give a pulse to start it ringing. Couple the output to the tank with a smallesh cap. Exact Value not important.

The LC tank will "ring" at each edge. Note the frequency of ringing. This will be a decayng ring, but the freq will be steady. If there are too few cycles (less than ~3-4), then you need to increase the L somewhat.

Back calculate the value of unknown C, while subtracting for probe capacitance. In fact, calculate probe capacitance & its ground clip impedance without the DUT connected. OR You can even pick up the ringing by simply placing the osc probe tip close to the LC. This would load it much less and give better results.

Care in setting up the test jig will result in some excellent answer

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It is more expensive to obtain 0.1pF measurement accuracy.

If you have to improvise, you need an FM or a 100MHz LC oscillator with some known low pF resonator and measure with an FM tuner the shift in carrier peak amplitude with the signal level meter or Vagc.

e.g. L=250nH, C=10pF f=100MHz Z = 158 ohms.

Meanwhile 0.1 pF is the about the amount added by one fingertip near the coil within a few mm or the equivalent Z of 100*158 ohms = 16k from 10pF/0.1pF=100.

The assumption is you understand how to generate a stable LC signal, have a known L,C values and then compute the relative change when stacked in parallel with tweezers over a similar size SMT cap.

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