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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 '16 at 7:53
  • \$\begingroup\$ How are you defining the fall time? 90% to 10% or something else? \$\endgroup\$ – Sam May 19 '16 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 '16 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 '16 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 '16 at 18:53
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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 '16 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 '16 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\$ – NoobPointerException May 19 '16 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\$ – NoobPointerException May 19 '16 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 '16 at 5:46

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