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enter image description here

I have a very simple circuit as above, there is a 1 uF bypass capacitor connected parallel to the VCC & GND of the IC.

I need to detect the existence of this capacitor through the testpad, but my tester not able to measure capacitance, only voltage/current.

I tried to measure the voltage when capacitor is still charging. Apply current 100 uA for 3 ms on VCC, can observed the voltage difference through the testpad (~250 mV when capacitor exist, ~500 mV when capacitor not exist), as shown below:

enter image description here

now my problem is the old tester can only supply current for more than 20 ms. when 20 ms, the capacitor is already fully charged and act as open, so I can't observe the differences from voltage measurement.

Is there any way to slow down the capacitor charging without modify the circuit in order to detect the the capacitance existence through voltage measurement? If not, any other method to measure the existence of this capacitor?

Some suggestion to add a series resistor to decrease the rate of energy delivered to the capacitor. Tried simulate with low/high resistor in series/parallel as below, only observed the capacitance charging curve changing, but still can't observe the difference to detect the existence of bypass capacitor.

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2 Answers 2

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is there any way to slow down the capacitor charging without modify the circuit ?

Send the current elsewhere. Attach any load, and solve the result system of equations. Since you know your total current, adding a load doesn't add new unknowns, so it's essentially the same solvable problem, mathematically.

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  • \$\begingroup\$ Not sure if I understand correctly, I tried to add a resistor(5.2k ohm) parallel to the bypass capacitor to divert the current. the voltage measurement I got is 250mV with/without 1uF capacitor, no much differences to detect whether the capacitor exist or not. \$\endgroup\$
    – ahbear
    Commented Jul 13, 2021 at 8:53
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    \$\begingroup\$ well, this is not hard to see: your resistor is too large in value. I mean, instead of measuring, have you tried sitting down and writing the equations for voltage over the capacitor? It's not hard! \$\endgroup\$ Commented Jul 13, 2021 at 8:58
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    \$\begingroup\$ (I say "it's not that hard", but it might be for you. In that case "How do I calculate the voltage over a capacitor with a parallel resistor attached to a current source" would be a good question to ask. You'd have to show all the calculations you managed to do: then we could help you on. All we can do with the things you've told us so far is say that you do seem to prefer experimenting over calculating, but this will not give you the answer you're looking for.) \$\endgroup\$ Commented Jul 13, 2021 at 9:18
  • \$\begingroup\$ Statically, you can not detect the capacitor. \$\endgroup\$
    – Antonio51
    Commented Jul 13, 2021 at 9:25
  • \$\begingroup\$ But if you add a little board ... and you can measure ac signals with your tester, you can. \$\endgroup\$
    – Antonio51
    Commented Jul 13, 2021 at 10:44
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Maybe you can remove the supply and measure the voltage (say) 1 ms later. The time constant of 1 uF and 5200 ohms is about 5ms so if you can measure with reasonable time resolution you can detect the capacitor. By measuring the current with the supply on and the approximate time constant you can even get a rough idea of the value of the capacitance (eg. did someone populate 100 nF or 100 pF rather than 1 uF- they often almost identical and are rarely marked).

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