How long does it take to charge a capacitor given constant power?

I know how long it takes to charge a capacitor given constant voltage (that's the first thing everyone learns about capacitors). In my search to answer this question for constant power, I discovered the answer instead for constant current here.

My first thought was $$\Q=CV\$$ and $$\I=C\frac{dv}{dt}\$$ and $$\P=IV\$$. You can combine all these formulae and cancel the two $$\V\$$'s to get $$\Q=\frac P{\frac{dv}{dt}}\$$

It was at this point that I realized I had no idea what the heck this equation was supposed to mean. I get what $$\P\$$ is, but where do I get $$\\frac{dv}{dt}\$$? So I thought "maybe I should have rearranged it differently". Perhaps like this: $$\Q=\frac{CP}{I}\$$

That at least makes sense, because it now depends on the capacitance of the capacitor. But where do I get the current to plug into this equation? I want power to be constant, so voltage and current are going to change.

I understand I could get $$\I\$$ or $$\V\$$if I chose a specific configuration of battery cells in parallel or series. But will that make a difference in the total time it takes to charge the capacitor? (I could use boosters to up the voltage and lower current or vice versa, so it seems the relevant thing is the power, not the specific battery configuration)

motivation:

This question came up because I want to charge a capacitor in a very short period of time, so I need to know how large of a battery I need. I can get the power density of LiPO battery's, multiply by the power will give me the weight of the battery.

This question also seems apt because in "the real world", where ever that is, neither voltage nor current are constant. If a battery supplies "a voltage", but you short it, the voltage between the terminals will drop, because there is some internal resistance, so it's not really supplying that voltage. Similarly, if something supplies "a current" to something like a capacitor, well... the capacitor will act more and more like an open circuit as it charges, until it explodes and literally becomes and open circuit. So while constant current and constant voltage equations are "good enough" for situations when things are almost constant, it will never be accurate when things vary.

• You can arrive at this by letting $P = v(t) i(t)$ (power equation) and $dv(t) /dt = C i(t)$ (capacitor equation), then solving the resulting differential equation -- but you end up doing a lot of math to get to the same solution as you do just using the equation for energy stored in a capacitor. Commented Mar 8 at 21:36
• Unless you have an interesting/less usual constant power charging circuit you will not have much use for this information. To charge at maximum rate you would use the largest current your charge circuit can sustain until your set voltage is achieved. Limiting to a certain power will reduce the charge speed unless you have a power limited source (generator/photovoltaic system). To remain efficient you would probably use a DC/DC switching circuit. Commented Mar 9 at 13:02
• "How long it takes to charge a capacitor given constant voltage" is awfully complex to estimate (it depends on imperfections of the capacitor and constant voltage source), and definitely not "the first thing everyone learns about capacitors", which is more t = ΔU C / I corresponding to constant current I. Commented Mar 10 at 10:03
• @KalleMP, no, I don't have fancy circuitry to maintain max power, but I still found this info incredibly useful as it sets an upper bound, even if I can never actually reach it. I now know that to recharge my capacitor in less than a second, I'd need a battery greater than 50 pounds. That is extremely useful info, even if I can never optimize it to be exactly 50 pounds. Commented Mar 12 at 16:20

This question came up because I want to charge a capacitor in a very short period of time, so I need to know how large of a battery I need.

To charge a capacitor to a certain voltage you need energy and, that energy (work) is: -

$$W = \dfrac{1}{2}\cdot C\cdot V^2$$

So, if you calculated that the energy required by your capacitor is 100 joules then, you could regard this as 100 watts for a duration of 1 second or, 1 watt for a duration of 100 seconds.

But, most importantly, this is the energy taken from the battery. You can assume an energy transfer efficiency of usually more than 90% for small situations and, up to around 99% for very large capacitor charging power supplies (an actual thing).

The time T to charge capacitance C to voltage V (starting from 0V) with constant power P is just

$$\T = \frac{C V^2}{2P}\$$

so the capacitor voltage as a function of time would be:

$$\v(t) = \sqrt{\frac{2Pt}{C}}\$$

So, for example, if we are charging 1000uF with 0.5W, the voltage will rise like this (Wolfram Alpha):

• Mathematica finds also the solution with negative current/voltage ;-) Commented Mar 9 at 16:52

Let the voltage and current of capacitor be $$\v(t)\$$ and $$\i(t)\$$ respectively. The main equations are $$p(t) = v(t)i(t) \ \ \text{and} \ \ i(t)=C\frac{dv(t)}{dt}$$ We are interested in the case that $$\p(t) = K \gt 0\$$. By substitution we have $$K = Cv(t)\frac{dv(t)}{dt} \tag{1}$$Multiplying both sides by $$\dt\$$ and integrating yields $$Kt =C\frac{v^2(t)}{2} + C_0 \tag{2}$$where $$\C_0\$$ is a constant which depends on the initial condition. If we assume $$\v(0)=0\$$ then $$\C_0 = 0\$$. So in this case, the solution for $$\v(t)\$$ is $$v(t) = \pm\sqrt{\frac{2K}{C}t}$$ If you are uncomfortable with multiplying by $$\dt\$$, recall that $$\\frac{d}{dt}(v^2(t)) = 2v(t)\frac{dv(t)}{dt}\$$ and use it in the equation $$\(1)\$$. The result should be same as $$\(2)\$$.