Electronic – Pulse-powering heavy loads with a coin cell

Quick Rules of Thumb

1. 10mA is your maximum current draw.
2. Buy fresh batteries. Always verify this by running lifetime tests on your batteries when you buy from a new supplier, or a new batch.

Our General Conclusions

When testing with coin cell batteries at my last job we found a number of things:

1. Surpassing 10mA per cell greatly reduced battery life.
2. 20-30mA was normally around your "maximum" current you could draw, but this is not dependable, the 10mA line being the highest current we could pull for deterministic function.
3. 1mA was as high as you could go without significantly degrading rated capacity(get nearly the published capacity).
4. Staying below the .1 mA suggested line would normally meet the rating of the capacity.
5. Someone selling you old batteries can do more to harm your tests than almost anything else. The batteries purchased directly from china gave the best results, the nice company in new york had battery cells that performed worse that batteries we had on site in storage for 4 years.

Capacity Rating

These results all seemed relatively consistent across different rated capacities, giving the same results with 50mAh batteries and 400mAh batteries. We chose to use 200mAh batteries on devices that needed to pull 20mA, they would significantly outperform 800mAh coin cells(the big ones you can buy) as passing 10mA hurt the battery life significantly.

When we passed the 10mA line the lifetime of the battery had a very large variance, we attributed this to manufacturing differences, but the same batch could have vastly large differences.

How I know this.

We did testing on batches of 20-50 coin cells to choose our rating and package. After this point we ran tests on more than 500 batteries under different conditions to verify results and predict lifetime. We ran tests where the current was pulsed, tests in different temperatures, and tests were we let devices run for months to try to test what our expectation would be. I am sorry I do not have references, as this testing was done at a company and was not published in any form, I have nothing I can reference for this.

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.