Alright, so I got a circuit that charges a capacitor and I want to charge it to a fixed voltage in a longer time than than the current provided would have (as the current provided would have charged it several times faster).

I can drop the provided current by putting a resistor, but I don't want to create feedback to the circuit that provides the current.

Well, I like minimalistic solutions, so I decided I would simply let the capacitor leak. Basically, I'll be putting a resistor to redirect a good portion of the current.


simulate this circuit – Schematic created using CircuitLab

-- where voltage potential 1 (V1), in respect to ground, may be less than or equal or greater than voltage potential 2 (2), in respect to ground.

I say "leaking" the capacitor because a lot of time, the capacitor wold be directly grounded and the redirected current would not be used anywhere, so it might as well be flowed back to ground. Hence, it's like placing a resistor in parallel with the capacitor, leaking it:


simulate this circuit

I have no idea if this is already being used, but my initial Googling didn't turn up a lot of hits. So I guess even if this is already being used, I don't know of any good sources that will guide me theoretically.

So my question is, given parameters

  • voltage source

  • resistance (R)

  • capacitance (C)

  • time

What are the equations describing this configuration, in ranges

  • V1 < V2

  • V1 = V2

  • V1 > V2

Or maybe just V1 = V2, as I'm using the configuration in figure 2.

I would hazard a guess that it's C = (I*t)/V ... But I don't know how to split the current (I) between the leaking resistor and the capacitor. Also (I'm not that sure) I think one has to take into account that the splitting of the current will change over time, as the capacitor gets more charged. Then again, one can simply assume that the charging is linear like we do with charging/discharging in smoothing capacitors.


The basic formula for a capacitor is Q = CV. If this is differentiated you get: -

\$\dfrac{dq}{dt} = C\dfrac{dv}{dt}\$ and this equals current.

So, for a given current and a given capacitance the voltage rises at a rate of I/C.

You appear to be saying that the circuit that generates the current is "fixed" and so the only option I can see is make the capacitance "several" times bigger in order to lengthen the charge time.

Putting a resistor in parallel will not get you what you want because the capacitor will not charge up to the full voltage.

A more complicated solution is to create a constant current sink (and this is not a simple resistor as per your shunt resistor idea). The sink would be in parallel with the capacitor and basically this diverts current away from the capacitor making the net current into the capacitor smaller and hence increase the charge time.

  • \$\begingroup\$ yes, that's right, the source circuit is a current source (outputs a constant pulse, actually) and i don't want to change that. hmm... larger capacitor... uhh, the problem is, it demands a capacitor in the range of tens of millifarads and that's the real problem. yes, i've taken into account that the pulses from source circuit is about 10% of it's whole cycle. \$\endgroup\$
    – Dehbop
    Mar 15 '15 at 11:26
  • \$\begingroup\$ about the voltage, don't worry, the voltage that powers the whole circuit is higher than the voltage to be charged into the capacitor. what i needed was to stop charging when it reaches that level... but have the charging at a longer time. \$\endgroup\$
    – Dehbop
    Mar 15 '15 at 11:39
  • \$\begingroup\$ Adding a resistor in parallel will lengthen the charge time to a certain voltage level but you have to choose a value that still attains the voltage value you need to reach. The charge will be exponential (non-linear with time) but maybe this would work for you. See this: play-hookey.com/dc_theory/combinations/rc_circuits.html \$\endgroup\$
    – Andy aka
    Mar 15 '15 at 11:40
  • \$\begingroup\$ @Badadeeboop : I'm confused. Is your source a voltage source or a current source? A voltage source has a constant voltage and a current source is a circuit which varies its voltage such that the same current always flows. The latter is optimal for capacitor charging: it needs to be a few engineering tolerances above leakage current for maximal charge time. \$\endgroup\$ Mar 15 '15 at 11:48
  • \$\begingroup\$ @Dan: might as well be a current source... it's an output from a pulse generator at 10% duty. so if you were to simply use it to charge a capacitor, then in regards to the whole cycle, it's a current source... \$\endgroup\$
    – Dehbop
    Mar 15 '15 at 11:58

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