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I have designed a micropump driver powered by 2x AA (Li) batteries, the pump is piezoelectric and requires 27V.

The pump requires a switching waveform, at the battery side this draws (including inefficiencies and driver consumption) 400mA and 900mA at the battery voltage which is nominal 3V (depending on the pressure being created by the pump), the batteries won't really see the switching because I am using a H-bridge and so there are two signals that are orthogonal to eachother and other than the small (ns) deadtime between them the boost converter will only see a DC drain (and even the deadtime will get lost in the feedback loop response time of the SMPS).

Several problems have arisen:

  1. After a failed life-cycle test I measured the voltage across the batteries, one was sitting at 0.2V the other 1.47V, they are used in series, so why would one battery have such drastically different voltage? Does this suggest damage? or simply one is much more drained that the other?

  2. I'm using boost solution that can boost from ~1.5V up to the required 27V. I have noticed that the someway into a test the battery voltage drops off during a pump cycle, very quickly dropping below 1V and after the cycle ends quickly recovering to 2.8-2.9V, does this suggest I am drawing to much current out of the battery? or that the battery is damaged or simply running dry?

I assume my design is damaging or stressing the batteries somehow, are there any standard practices to be used in this scenario?

Based on a constant discharge test with reference to the battery manufacturer I estimate that the batteries that drop off in voltage drastically after a period of time being used are less than 1/3 discharged. I.e with a load drawing a defined constant current and measuring teh voltage across the terminals and comparign this with a the discharge curve (http://data.energizer.com/pdfs/l91.pdf)

Unfortunately I can't upload schematics due to confidentiality agreements with the customer, however I'll provide whatever additional information I can.

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  • \$\begingroup\$ What kind of lifetime are you seeing with the batteries you are using and what is the required / expected lifetime? \$\endgroup\$
    – Arsenal
    Sep 16, 2019 at 13:02
  • \$\begingroup\$ Pls show current waveforms on battery. Any imbalance on battery ESR, Ah results in the weaker battery draining faster. \$\endgroup\$ Sep 16, 2019 at 19:44

1 Answer 1

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Batteries have different capacities and the one with the lesser capacity will define how much capacity you get from a series connection of multiple cells.

As you can see in the datasheet you linked, the cells will hold their voltage pretty constant until they drop off a cliff at the end. So I don't think you have a strange behaviour with one cell at 1.47 V and the other at 0.2 V. Both are actually discharged. Here is a relevant document explaining that cells which have an open circuit voltage of less than 1.7 V are to be considered discharged when they had the chance to rest (and recover their open circuit voltage)

But there is a second effect happening which might be the troublesome part for you. As the cell gets discharged, the internal resistance rises. So if one cell is drained the internal resistance might be too high for you to draw all the current you need.

Third thing is that your current requirement will rise during discharge. While it is 1 A at 3 V at the beginning it will be 2 A at 1.5 V when the batteries are almost dead which is overstressing them.

The effect you see in 2. is the battery relaxation effect. I think that the batteries are running dry and you are in the region where the internal resistance is rising. In that state you can extract energy of the battery only with a smaller current.

Depending on your SMPS design, it might draw peaks which are a lot higher than your average current, which will stress the batteries even more. You could try to reduce the problem with a really low ESR capacitor bank which provides the peak currents while the batteries then provide the average current.

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  • \$\begingroup\$ Thanks very much for this information \$\endgroup\$ Sep 17, 2019 at 9:04

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