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While testing the capacitor voltage fall times for my board at 3 different voltage levels I see that for 6V the fall time is higher than for 18V. I checked the same on a breadboard with a capacitor in parallel with a resistor and DC voltage supply across it. If time constant is only dependent on R and C then why is it that the fall time is varying? Or is it because the internal resistance of the voltage source varies with voltage and this value has to be considered with the R in my as well?

Thanks,

Edit 1 - Added the schematic

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    \$\begingroup\$ are you discharging a capacitor accross a resistor only? how are you measuring the voltage? how big is your capacitor? if it is small then maybe the impedance of the measurement instrument is contributing in discharging the capacitor. \$\endgroup\$
    – fhlb
    May 19, 2016 at 7:53
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    \$\begingroup\$ How are you defining the fall time? 90% to 10% or something else? \$\endgroup\$
    – Sam
    May 19, 2016 at 7:59
  • \$\begingroup\$ You should supply a schematic of your circuit. So we can take into account all elements and connections. \$\endgroup\$
    – Ariser
    May 19, 2016 at 8:02
  • \$\begingroup\$ My circuit is a parallel R and C circuit with DC voltage source.across it. Yes, I am discharging the capacitor across resistor only. I am measuring the voltage in an oscilloscope. I am defining the fall time as 90% to 10% of voltage level. \$\endgroup\$
    – Akella
    May 19, 2016 at 9:05
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    \$\begingroup\$ Are you using a ceramic capacitor? The capacitance of a ceramic capacitor can be strongly dependent on the applied voltage. \$\endgroup\$
    – rioraxe
    May 19, 2016 at 18:53

2 Answers 2

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The discharge time of a capacitor in a RC - circuit should be an exponential function with a time constant of RC. This is quite well established model and should be quite accurate under reasonable circumstances.

In your setup it is important to know how you define fall-time. If you use a fixed voltage level, for example 0,5V as the end point for the discharge process it will take different amount time for voltages in range of 6 -18V. However I assume you use a level which corresponds to a percentage of the charging voltage.

Next step is identifying your voltage source. If you use a variable power supply you must look into the specifications, typically you would need a circuit for it or a good data sheet. One thing to consider is that many of these devices have a capacitance on the output, anywhere in the range of 1muF to 470muF which is parallel to the capacitor under test. But that one has same value regardless of voltage.

If you are using batteries you will definitely be in trouble – depending on the type and size their resistance varies, but for a alkaline AA 1,5V you can expect something like 0,15 Ohm at room temperature.

Generally speaking you can calculate the internal resistance of your voltage source by measuring the voltage at no load and at load Rl. However this will be accurate only for unregulated voltage sources.

Rint = (Vnl/Vfl - 1)Rl

However to determine exactly what causes the discharge time to vary full information needs to be provided for your voltage source.

A simple solution would be to simply remove it from the circuit at the start of the discharge process. But that was not what you were asking about.

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  • \$\begingroup\$ I am measuring the and rise and fall times as a percentage of the voltages. Yes, i am using a variable DC power supply with a 0-30V voltage range and 0-2A current range. Like you said, I am now planning to disconnect the DC supply from the circuit using a switch. I will post results soon. \$\endgroup\$
    – Akella
    May 19, 2016 at 9:11
  • \$\begingroup\$ I did perform the test as per the below schematic using a switch. I still get a difference of 0.4ms between the fall times of 6V and 14V. Not sure of this difference is okay. Any ideas? \$\endgroup\$
    – Akella
    May 19, 2016 at 12:49
  • \$\begingroup\$ According to the simple mathematical model this makes no sense, general discharge time is covered in this topic; electronics.stackexchange.com/questions/4951/… Intuitively I would say that 0.4ms difference is okay, but at this point I have no good explanation for it. \$\endgroup\$ May 19, 2016 at 12:50
  • \$\begingroup\$ Even though I can't find anything useful to this case there are some potentially useful information in the search-words "capacitor non-ideal-behaviour". en.wikipedia.org/wiki/Capacitor#Non-ideal_behavior The resistor is unlikely to be the bad guy here, unless it gets really hot. Othervise the resistors are VERY linear in their DC behaviour. \$\endgroup\$ May 19, 2016 at 13:13
  • \$\begingroup\$ Then the question is how much variation is okay and wouldn't quantifying it be difficult? If the variable parameters of a capacitor such as parasitic inductance etc as given in the wiki link are likely to blame, then I guess it leaves it ambiguous. Also acc to @rioraxe answer the value of capacitor varies with voltage, so that could also be a reason. \$\endgroup\$
    – Akella
    May 20, 2016 at 5:46
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Fall-time/rise-time = 2.2 * timeConstant, approximately.

It is necessary to understand to what extent the measurements are dissimilar, so a cause of imperfection can be sought for.

Internal resistance is usually very low compared to 10k, which dominates. The problem is likely to be due to capacitor nonlinearities. What sort of capacitor have you experimented with?

Troubleshooting first step boils down to evaluate how different the numbers are.

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  • \$\begingroup\$ you'll want to put this as a comment under the original question as it isn't an answer. \$\endgroup\$
    – RJR
    Jun 8, 2020 at 10:42

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