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I was wondering what would happen if you took a 2.7v super-cap and began to charge it with 2.5V then while continuing to charge it the input voltage drops to 1V. Also what would happen the other way around if you began charging with 1V and then raised the voltage to 2.5V.

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  • \$\begingroup\$ I don't see why you'd have any special behaviour besides the capacitor charging and discharging accordingly. \$\endgroup\$ – Dr Coconut May 23 '15 at 6:53
  • \$\begingroup\$ I = C*dv/dt.... \$\endgroup\$ – John D May 23 '15 at 13:13
  • \$\begingroup\$ What is the application? Wha behavior are you trying to get? \$\endgroup\$ – Gregory Kornblum May 23 '15 at 14:45
  • \$\begingroup\$ Voltage differences cause current to flow (more-or-less). If you connect a capacitor with a voltage of 2.5V across it, to a source with a voltage of 1V across it (presumably with some resistance as well), the capacitor will discharge through the source until the voltage across it is 1V. \$\endgroup\$ – user253751 Oct 6 '15 at 10:39
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I was wondering what would happen if you took a 2.7v super-cap and began to charge it with 2.5V then while continuing to charge it the input voltage drops to 1V.

This question carries an assumption that has an error. Once the capacitor is charged to 2.5V, if you reduced the supply voltage, the capacitor would have charge taken away from it i.e. the action of lowering the terminal voltage "discharges" the capacitor.

Q = CV under all circumstances and assumign C is constant (not always a good assumption), if you lower the voltage you lower the charge stored in the device.

If you started with 1V and raised it to 2.5V then you are indeed charging the capacitor. Another formula derived from Q = CV is the formula for current into (or out of) a capacitor. Differentiate both sides with respect to time (assuming C is constant) and you get: -

\$\dfrac{dQ}{dt} = C\dfrac{dV}{dt}\$

And of course dQ/dt = current. This tells you that if you apply a raming voltage (1 volt per second) a 1 farad capacitor, you will cause a constant current into the cap of 1 amp.

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  1. after charging to say 2.5 and if u want to maintain the charge, do'nt derease the voltage to 1 volt, instead add a resistance between supply and the cap.
  2. by charging it by 1 volt and then by 2.5 volt , it may reduce the loading of the voltage source else NO change. vtingole
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