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If a ceramic capacitor of 94 uF (2x47 uF in parallel) is charged with 3.2 V, then starts from it a constant current consumption of 1.2 uA, how long will take for it to reach 1.5 V?

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    \$\begingroup\$ Why do you need an online calculator? Just do the math yourself, it's not complicated. \$\endgroup\$
    – Hearth
    Oct 23 at 14:47
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    \$\begingroup\$ Charge (Coulombs) = Volts (Volts) * Capacitance (Farads) = Current (Amps) * Time (seconds). That's all you need to know to compute your answer. Equate the charge = (3.2-1.5) * your_capacitance with the charge = 1.2 uA * t \$\endgroup\$
    – Neil_UK
    Oct 23 at 15:12
  • \$\begingroup\$ @Neil_UK I tried. (3.2-1.5) * 94*10^-6 = 1.2*10^-6 * t. t = ( (3.2-1.5) * 94*10^-6 ) / 1.2*10^-6 = 1.33*10^-10. Probably this is wrong. \$\endgroup\$
    – abomin3v3l
    Oct 23 at 16:27
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    \$\begingroup\$ @abomin3v3l The trivial numbers, 133 are right. The very important numbers, 10^-10, are wrong by 12 orders of magnitude. Dividing by 10^-6 is the same as multiplying by 10^6. That's of course assuming your capacitor is not a high K ceramic, which will have a fraction of its advertised capacitance at its rated voltage. \$\endgroup\$
    – Neil_UK
    Oct 23 at 16:33
  • \$\begingroup\$ @Neil_UK Thanks very much. Then the correct is ( (3.2-1.5) * 94 ) / 1.2 = 133 seconds. The capacitor is "47uF +-20% 10V 0805 X5R PN GRM21BR61A476ME15L (Murata)". And yes, I am aware that the capacitor have leakage current, what will help to self-discharge it, the question was "in theory". Regards. \$\endgroup\$
    – abomin3v3l
    Oct 23 at 16:42
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Simon gave the clean math answer. dv/dt = i/C, so it's a rule of three. If this is a homework question, then that's the good one. If this is an engineering question about real world stuff, then...

If it is a 47µF ceramic capacitor then I'll assume its dielectric will be high-K class-2 like X7R, Y5V, something like that. So, if you bought some 47µF caps, what is the actual capacitance is a good question...

enter image description here (source)

So X7R holds up pretty well with temperature, but please don't use Y5V/Z5U for electronics that have to operate outside of an air conditioned office.

Then of course, capacitance of these dielectrics has a huge dependency on E-field strength.

enter image description here (source)

This means, for the same capacity, a capacitor that is bigger will have more spacing between plates, so it will have lower E-field, so it will lose less capacitance.

For example among the capacitors tested, if you pick the tiniest one, at 3V3 it no longer has 4.7µF capacitance, but only 3µF. It would be quite a headache to calculate the constant current discharge curve when the capacitance value is changing, so I would recommend using a Spice simulation, if you can get a spice model that accurately tracks C versus V, which is quite hard to find...

The curves are easy to find for your particular cap, usually in the datasheet or the manufacturer's website, so using Simon's equation with the actual capacitance value at 3V3 would provide a good worst-case approximation.

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