There are two questions: the number of charging and the time of charge.

The 2450 battery has about 620mAh of energy stored. (dataheet from digikey). I'm assuming that you want to charge the capacitor with the same voltage as the battery: 3v. Therefore to fully charge a 470uF you'd need:

\$ Q_{cap} = C \times U \$

\$ Q_{cap} = 470 \mu F \times 3V = 1.41 mC \$

\$ Q_{bat} = 620mAh \$

\$ Q_{bat} = 620mAh \times \dfrac{3600 s}{1 h} \times \dfrac{1 A}{1000 mA} = 2232 As = 2232 C \$

\$ Cycles = \dfrac{Q_{bat}}{Q_{cap}} ~= 1582978 \$

Therefore the battery can charge the capacitor over 1.5 million times. If you want to "fully charge" the capacitor with 45v this number would drop to 100 thousand times, more than enough.

Remenber that I'm not assuming losses or other circuitry attached to the battery

The second question is if the capacitor can be charged under 6 seconds. We need to know how is the max currency we can get from the battery. Back to the datasheet this number is 0.2mA.

\$ Q_{cap} = 1.41 mC \$ (from earlier calculations)

In order to feed this amount of charges in 6 seconds we would have a currency of:

\$ I_{cap} = \dfrac{Q_{cap}}{\Delta T} = \dfrac{1.41mC}{6 s} = 0.235mA \$

As the required currency is lower than the one needed, there is no way to charge the capacitor under 6 seconds. The number from the datatheet is the "continuous standard load". It may be possible to charge the capacitor in less than 6 seconds, but it may be dangerous.

You may try another manufacturer aside the one I showed, they may have a similar battery with more "continuous standard load".

Another point to mention is that you need to consider the current drawn from the circuit you're building to charge the capacitor and subtract it from the available current from the datasheet.

The calculation is straightforward. The capacitor size is simply a question of how much voltage drop you can tolerate over the duration of the pulse. The average current from the battery is a function of the duty cycle.

ΔV = I × Δt / C

Solving for C gives:

C = I × Δt / ΔV

Let's assume you can allow ΔV = 0.1V. For your first example, this works out to:

C = 25 mA × 25 ms / 0.1 V = 6.25 mF

The average current draw is 25 mA * 25 ms / 2.5 s = 0.25 mA.

For the second example, the numbers work out to:

C = 50 mA × 100 ms / 0.1 V = 50 mF

Average current = 50 mA * 100 ms / 1.0 s = 5 mA.

## Best Answer

A CR2032 holds about 2400 useful Joules of energy. Operating entirely within specification, it can supply \$200\:\mu\textrm{A}\$.

If applied subcutaneously, \$10\:\mu\textrm{A}\$ can be sufficient to cause fibrillation. So if you could apply the battery directly near the heart and underneath the skin, I suppose it could directly kill someone without any additional circuitry.

From an editorial in

, June 2010, Volume 110, Number 6, International Anesthesia Research Society, pp 1517-1518, "Electrical Safety in the Operating Room: Dry Versus Wet," by Steven J. Barker, PhD, MD, and D. John Doyle, MD, PhD, FRCPC, the following quote is found:Anesthesia & Analgesia(The above information was pointed out in a comment here by Russell McMahon and is a substantial improvement over citing a Wiki page.)

I've read that death can be caused with as little as 50 Joules. But I think 100 Joules is a more certain estimate. Lethality (with AC, anyway) is pretty common even at \$200\:\textrm{V}\$. So if I had to make an educated guess, I'd probably guess that using a Sanyo OS-CON Aluminum-Polymer (way too expensive and you'd need lots of them) or aluminum capacitor (such as Vishay BCcomponents' Aluminum Electrolytics), with \$200\:\textrm{V}\$ and 100 Joules would be sufficient. This suggests a value of \$5\:\textrm{mF}\$.

However, it would take a while to achieve. Assume you can design a circuit that is, overall, 50% efficient in charging this capacitor from a single CR2032 while staying fully within specs and drawing just \$200\:\mu\textrm{A}\$ from it. Then on first blush it would take 10000 seconds or about \$2\:\frac{3}{4}\$ hours to charge it up for one such use if you could sustain \$200\:\mu\textrm{A}\$ throughout the process. But the CR2032 is only capable of sustaining about \$600\: \frac{\mu\textrm{J}}{\textrm{s}}\$ of power. So really, I think this would take closer to four days to achieve. (And that doesn't account for capacitor leakage. With the Vishay capacitor mentioned above, leakage power may be below charging power near the end, but it probably will add a fair bit more time to the process.)

So the answer is probably "technically, yes" but rather unlikely as it would be quite odd to open someone up in order to stick a button battery across some tissues inside their body near the heart (read: very low probability) and it is similarly unusual to find a circuit designed to charge up a large, low ESR capacitor from a battery supplying just \$600\:\mu\textrm{W}\$ continuous and requiring almost a week to charge up (read: low probability.)

Of course, now that someone is thinking this way, I am sure such a circuit will be promptly designed and then sold as pet rocks to millions of happy consumers, making this a significant problem in the wild. ;)