I am creating a microprocessor design that will have a battery backup in case the mains power goes down.

My plan is to have 2 3.6 V batteries in parallel, so that either may be swapped in/out while maintaining battery power to the CPU.

The batteries are rated 2.4 Ah, and the CPU should only draw about 8 uA when on battery backup; the CPU should survive for a very long time on battery power.

I want to also include a cap in parallel with the batteries, that would keep the CPU powered for minutes / hours. My question is, if I used something like this http://media.digikey.com/pdf/Data%20Sheets/United%20Chemi-Con%20PDFs/KMH%20Series.pdf which has a leakage current of approximately 3 mA, am I going to start burning the batteries through the Cap's leakage current?

Reference schematic below. Assume SW1 only closes once main's power has been disconnected.


simulate this circuit – Schematic created using CircuitLab

So here are my questions:

  1. Is the 3 mA leakage current in the datasheet realistic? Or is this a drastic over-estimate?

  2. Are there alternative cap's which have leakage currents levels of magnitudes lower?

  3. Will the batteries leak through the cap even when "SW1" is open?

  4. Are there any widely-used ways to implement my idea (cap in parallel) that won't drain the batteries?

  • 2
    \$\begingroup\$ You shouldn't put batteries in parallel like this, without diodes to make sure that one doesn't try to charge the other. \$\endgroup\$
    – user1844
    Oct 20, 2015 at 20:59
  • \$\begingroup\$ I agree with Will. But the easiest solution would probably be to just use a single battery (or two in series) for backup and get rid of the capacitor, or use a 22uF ceramic cap with low leakage current if you really feel like you need a cap. The battery will last years if the processor is only drawing 8uA, so you don't need to worry about changing batteries. For longest shelf life, I would suggest you use 2 AA size Li-FeS2 disposable batteries. \$\endgroup\$
    – user57037
    Oct 21, 2015 at 2:05

3 Answers 3

  1. That cap has a leakage quoted as

0.02CV (uA) or 3mA, whichever is smaller, after 5 minutes at20C. Where I = Max. leakage current (uA), C = Nominal capacitance (uF) and V= Rated voltage (V)

So for a 6.3V 4700uF cap, the leakage would be 592uA, which is much smaller than 3mA, but still roughly 100x the circuit's standby power consumption, which does seem a bit sad.

  1. Yes, there are lots of caps specifically made as 'low leakage' - obviously that can be a bit like 'low cost', in that it's relative to someone else's expectations, but you can do much, much better than a generic electrolytic. There are also caps made specifically for what you're trying to do here, called 'supercaps' or 'ultracaps' - they're so big and give such long retention times at your low microamp currents that you could probably get rid of your second battery. Here are some supercaps which have very low leakage: http://www.maxwell.com/images/documents/hcseries_ds_1013793-9.pdf

  2. Yes.

  3. Use a more suitable cap, see above.

A general note about doing low microamp-level microprocessor projects - you have to be really very careful about everything to get down to that sort of level, and even trivial mistakes with a single pull-up turned on in a micro, or a carelessly chosen component can leave you absolutely orders of magnitude away from where you need to be. You need to make sure you have the kit to make trustworthy measurements of low current.


The more important question you need to be asking: What do my batteries do with regards to charge/discharge?

From voltages and their capacity, I am assuming you use Li-Ion.

The comment Will made on your question is especially valid for Li-Ion, since they can cause huge currents to flow when they aren't of equal charge, which happens when you replace one. But certainly still valid for many other types.

But, what I am mainly getting at: Li-Ion batteries will discharge from full to empty in pretty much a fixed time at a reasonably steady rate, which depends on their quality. High end cells from a reputable source will last between 1 and 2 years on a full charge. Generic brand batteries from a decent source will last about a year on a full charge. eBay/Ali batteries anywhere between a few weeks and a year.

And that's with no load added to them.

At 2.4Ah with 1 year of self discharge time, that's 365.25days * 24hours = 8766 hours on average. Which makes for a self discharge current of approximately 0.273mA. That means your MCU will not be contributing to the discharge time at all.

Of course it's still "only" 0.5-ish mA for two cells, so 3mA in a capacitor would impact the time. Effectively making it less than 3 months if it does follow the worst case limit (which the 3mA is). But if you already have a double battery with diodes, what's the point of a capacitor higher than a couple of μF?

What is much more common for your problem is using prime Litium batteries, like the coin cells "CR2032" and "CR2016", which have a self discharge time of 10years or more. At about 240mAh energy content in a larger cell, over three years your 8μA MCU will eat away 8μA * 8766hours * 3 = 210mAh, which means the cell will be about empty. That's $2.50 per 3 years of battery costs, but with a very good reliability of it working for long stretches of time.

It is then usually set up as:


simulate this circuit – Schematic created using CircuitLab

You don't need a switch, because the diodes will take care of that. When the main power is larger than the battery voltage, 3.3V for example, it will supply the MCU, when it falls away, the batteries will share the load through the diodes over time.

I selected BAT54, but they may not be the best. It's best to use a diode with a reverse leakage current of (much) less than half a μA, while also giving a reasonably low forward voltage at 2 times your expected MCU average drain current. If the reverse leakage current at 3V (we don't care about 50V, because we only have 3V batteries) is a few micro amp, that will contribute noticably, because D3 will let the batteries leak that amount back into the power supply.

As long as you always change the batteries one by one, the MCU will always have a battery supplying it and the two caps will easily cover small transients.

  • 1
    \$\begingroup\$ Various regulatory bodies (e.g. UL) will want you to have two devices between a lithium battery and a supply which could charge it, in case one fails. \$\endgroup\$
    – user1844
    Oct 20, 2015 at 23:39
  • 1
    \$\begingroup\$ And for an 8 uA load, one of the devices can be a 1k resistor. \$\endgroup\$
    – user57037
    Oct 21, 2015 at 2:07
  • \$\begingroup\$ @WillDean Fair enough. But I didn't expect this to be UL ready yet, so to speak, from the question content. \$\endgroup\$
    – Asmyldof
    Oct 21, 2015 at 5:55
  1. The datasheet seems to imply this is a worst-case specification. It probably isn't unreasonable, but it might be a little higher than the nominal case. Buy one and measure the leakage current!

  2. Yes - you have not disconnected the capacitor from the batteries.


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