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The time $t$ taken by a capacitor of capacitance $C$ in a charging circuit with a resistance $R$ in series with it to accumulate charge $q$ is given by the equation

$$t = \tau \ln\left(\frac{q}{Q-q}\right), $$

where $\tau$ is the time constant given by $\tau = RC$ and $Q$ is the maximum charge the capacitor can have when fully charged in that circuit.

In order to find the time taken by the capacitor to get fully charged we have to put $q = Q$ in the right side of the above equation that gives

\begin{align}t &= τ \ln (Q/0) \\ \implies t &= τ \ln \infty \\ \implies \infty &=\lim_{q\to Q} t\end{align}

This gives me a feeling that a capacitor never gets charged fully. Am I right? Why not?

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    $\begingroup$ Please don't write things like $Q/0 = \infty$. In this case, the reasoning isn't really faulty, but that just not how one should deal with limits and infinities, or dividing by zero. $\endgroup$ – ACuriousMind Apr 4 '15 at 15:44
  • $\begingroup$ I know. You are correct. Q/0 does not have any mathematical meaning. I wrote that just to cut the words short. $\endgroup$ – BibThePhysicist Apr 4 '15 at 16:31
  • $\begingroup$ However when q approaches Q, how would you reduce the equation t=τln(q/(Q−q)) further down to the final line? $\endgroup$ – BibThePhysicist Apr 4 '15 at 16:36
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As others have mentioned, for all intents and purposes, yes it reaches %99 charge after 5 tau.

However, as the current gets smaller and smaller as we reach full charge, technically it will never become 'fully' charged, even in practice. The current will continue to get smaller and smaller, until it is unmeasurable and therefore negligible.

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This gives me a feeling that a capacitor never gets charged fully. Am I right? Why not?

In the context of ideal circuit theory, it is true that the current through the capacitor asymptotically approaches zero and thus, the capacitor asymptotically approaches full charge.

But this is of no practical interest since this is just an elementary mathematical model that cannot be applied outside the context in which its assumptions hold.

For example, this model assumes that $Q(t)$ is continuous and that there is no noise. However,

  • electron current is not continuous; the charge changes in discrete units, the charge of an electron.

  • there is ever-present and random noise and, after some number of time constants, the 'charge current' predicted by the simple model is below the noise floor.

Since the capacitor goes from zero charge to better than 99% charged in $5\tau$, we typically use this as the time required to 'fully' charge the capacitor.

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  • $\begingroup$ So there is no 100% charging of the capacitor, huh? $\endgroup$ – BibThePhysicist Apr 4 '15 at 16:31
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    $\begingroup$ @ThePhysicist, that's not what I wrote or implied. There's no 100% charging according to the simple model in your question but that simple model is not applicable when the charging current is very small such that, e.g., noise currents are much larger than than the charge current. $\endgroup$ – Alfred Centauri Apr 4 '15 at 16:52
  • $\begingroup$ This makes me ask the root question. Went through Johnson–Nyquist noise calculations. If the surrounding temperature and the charging current is kept under such control that the noise current and thermal disturbance is negligible, how do you find the time t for the complete charging of a capacitor of capacitance C in an RC circuit of resistance R with a voltage source of voltage V? Assuming the capacitor has no charge in the beginning, do we still use that simple mathematical model I mentioned or is there a better way to do that? $\endgroup$ – BibThePhysicist Apr 4 '15 at 18:03
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    $\begingroup$ @ThePhysicist, I suppose you could stipulate, for argument's sake, that the capacitor is fully charged when the total charge on the capacitor is within one electron charge of $CV$. I'm not sure what practical value this calculation is though. $\endgroup$ – Alfred Centauri Apr 4 '15 at 19:08
  • $\begingroup$ In this negligible-noise model, it's fully charged when the current is negligible. How to define "negligible"? Well, if you can say the noise is negligible, then use the same criteria :-) The point is if you're trying to talk about arbitrarily small charges and currents then the noise cannot be negligible, it will affect things when the signal gets small enough. Same goes for the quantization. $\endgroup$ – Steve Jessop Apr 5 '15 at 2:47
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To fully charge a capacitor to 5 Volts, say, you could connect it to a 10 Volts source until it is half charged, then connect it to your 5 V source. This is of courcse a ridiculous method, since you could hardly hit the moment of correct charge so precisely; any micorvolt error would start an exponential curve as in your original setup. That being said, there is a reason why we need only wait until the difference is negligible: Soon enough the error will be less than noise and errors introduced by thermal effects, by quantisaion effects, or simply by the fact that the targe tvoltage and the capacitor's capacity are not so constant in the first place (e.g., temperature changes might change the geometry and hence capacity). The (more or less) exact meaning of "soon enogh" of course depends on your requirements, measuring instruments, etc.

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In computer science, there's a semi-joking saying that $\log n < 50$ for all $n$. Of course this isn't true – as you say the logarithm is unbounded. But what is true is that it grows so slowly that, if you can only put in quantities like memory size, mass of some material, time etc., then the logarithm is “effectively bounded”, because it is (literally) exponentially expensive to grow it any further.

At first glance, this doesn't seem to apply to your question, because you're only shrinking something to zero. In this case, the cost is not of supplying an anfeasible amount of something, but of making the circuitry behave so nearly ideal that the tiny residual voltage can still be measured. In particular, you need to reduce the noise; if there's more noise than signal voltage you can't measure the latter. Experimentally, this pretty much comes down to cooling down the measurement amplifiers (in nuclear physics, often detectors need liquid nitrogen cooling simply for the preamplifiers to work more accurately!), and reducing interference by electromagnetic shielding.

Either that or, relevant here: you can also reduce noise by introducing large capacitors into the circuit (in this case, there is still noise, but it's “filtered out” again). That's often avoided because, obviously, it makes the circuit's response slow; in this particular experiment it would mean that we continue to measure a voltage difference, but we can't really tell whether it is from the capacitor we're actually interested in, or from the one we just used in order to make the measurement more “accurate”.

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