Charge capacitor to a small voltage

[coolrox86]

Is it possible to charge a capacitor to a small voltage of 3mV with just current 5mA in 3 seconds? The capacitor is 0.667F. I want to fully charge back the capacitor every time after it has used 3mV for sending signal via a small radio module.

Any idea and help is well appreciated. Thanks!

 

[diegobb]

According to the ecuation of the capacitor: \[i(t) = C . \frac{dV}{dt}\], wich can be expressed also as (replacing derivatives by differences) \[i = C . \frac{V}{t}\], and with: C=0.667, V=3mV and t=3sec, you get that the current you need is 0.67mA. So I think that your current (5mA) is enought to charge the capacitor. This is for an ideal capacitor, so I think that with the (almost) x10 factor you have, you will be ok.

Also you can calculate the resistance to charge the capacitor in the needed time, with the source you have. If you remember the charge capacitor equation, \[v(t) = V . (1-e^{\frac{t}{R.C}})\], you can express the resistance as: \[R = - \frac{t}{C . Ln(1-\frac{v(t)}{V})}\] suposse that you have a source of 5V (you need a source with a voltaje much greater of 3mV to charge the capacitor quickly), from the previous equation with: v(t)=3mV, t=3sec, C=0.667 and V=5V, the needed resistance is about R=7.5k ohm.

Regards.

Added after 40 minutes:

I forgot to calculate the max current with R=7.5k, an important parameter of your circuit because your maximum current must be less than 5mA. :!:
Due to the fact that a discharged capacitor is like a closed circuit (a cable), the initial current will be i=V/R, so with V=5V and R=7.5k, the initial current (maximum) will be 0,67mA.

Regards.
PS: you can do the calculations with your voltage source, changin V with the voltage of your source (I mean in the calculation of the needed resistor). I think the source can be as small as 3V, and you can still charge the capacitor properly in 3sec.

 

[coolrox86]

I get it. Thanks for your advice!