# Energy subtracted from a capacitor for a given voltage drop

I am trying to estimate, from a total energy of a capacitor (which is the output of a charge pump circuitry), how much energy do I have to power up an LED before the voltage goes below a certain threshold.

At first, I guessed is the $$E = \frac{1}{2} C (dV)^2$$ where dV is the voltage drop I can allow to happen before I turn off the LED and keep the remaining energy in the capacitors (and allowing them to charge again).

But I have the feeling is not that simple. For example, I was forgetting what was analyzed here in this question, and made me suspicious on my simplistic view. Or can you confirm mine is correct?

• (A-B)^2 is not the same as A^2-B^2, to be short the energy formula comes without d, so yours is wrong. Mar 20 '20 at 13:59
• The other question does not apply as long as you dissipate energy from the capacitor in the load (LED) directly and do not "store" it in another capacitor at lower voltage. Mar 20 '20 at 22:02
• @MarkoBuršič does that mean doing the A^2-B^2 (so applying superposition) will also make sense? Mar 21 '20 at 11:16

$$Q = CV \rightarrow \dfrac{dQ}{dt} = C\dfrac{dV}{dt} = \text{current}$$
So, if you know the LED current and you know what the initial capacitor voltage is you can estimate dV/dt as a slope taking you from $$\V_{INITIAL}\$$ to $$\V_{FINAL}\$$ in so-many seconds (dt). You could make this a bit more sexy by factoring in the dwindling LED current as V falls.