# CR2032 alternative for higher current [closed]

I'm designing a small battery powered circuit which will draw around 50mA for a millisecond every couple of seconds. I was planning on using a CR2032 coin cell but have recently learned that a coin cell has a rather low peak current of a couple of milliamps at best.

Are there any alternatives in a similar form factor which can be used?

• coin cell in parallel with a big capacitor? – The Photon Jul 12 '16 at 15:55
• How would I calculate the size of the capacitor required? – bitshift Jul 12 '16 at 16:01
• First decide how much you can allow the voltage to ripple, and the maximum current draw x time the load will be on. Then choose the capacitor big enough so the maximum load pulse won't pull the voltage around by more than the maximum ripple. – The Photon Jul 12 '16 at 16:04
• Then add 20 to 50% for tolerancing. – The Photon Jul 12 '16 at 16:05

50 mA is too much for a "coin cell". A CR2032 is already a fairly large coin cell, although there are larger ones.

What you should do is put a large enough capacitor in parallel with the coin cell. The cap provides the short term burst of power, and is then recharged by the coin cell more slowly over the next couple seconds until the next pulse.

Let's say you want the cap voltage not to drop more than 100 mV by the end of the pulse. (50 mA)(1 ms)/(100 mV) = 500 µF. At this low voltage even twice that is still relatively small.

You might also want to put a resistor between the coin cell and the cap. This will spread out the charging current over time, so decrease the max current the cell sees. If the voltage drop is small enough and the internal resistance of the cell large enough, this may not be needed.

However, the way to calculate the resistance is to look at the time to recharge the cap. The cap needs to be full again in 2 seconds. Let's say you want the cap to charge within 99% of full before the next pulse. That's 4.6 time constants, so one time constant is 430 ms. Let's say you chose a 1 mF cap. (430 ms)/(1 mF) = 430 Ω. That's the total effective resistance in series with the cap and cell. That includes the cell's internal resistance, which could be a significant part of that.

I'd probably not put additional resistance there. The initial recharge current will be the voltage drop divided by the cell's internal resistance. With 1 mF, the voltage will only drop 50 mV, so the current should be low enough to not appreciably shorten the life of the cell without any additional resistance. For example, the internal resistance would have to be 50 Ω for the initial recharge current to be 1 mA.

• What sort of capacitor would be best when taking leakage into account? – bitshift Jul 13 '16 at 13:04
• @bit: At 1 mF you won't have any choice. It will need to be electrolytic. Look at datasheets carefully and find one with low leakage. Consider that 50 mA for a ms every two seconds is 25 uA average. – Olin Lathrop Jul 13 '16 at 15:30