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I'm trying to monitor the health of my duracell consumer battery.

Is there an IC that can do this? I've come across the MAX6775, but it appears that you have to use a voltage divider. Isn't this drawing excessive current? This is for a ultra low power application.

Ideally, I'd like to switch on and off the monitor IC every 10 minutes, take a reading and turn it off again so as to minimise power.

Can someone please suggest how to do this/suggest a suitable low-power IC?

Many thanks in advance,

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In the data sheet of the MAX6775, the device itself draws 3-7 uA, plus your voltage divider for the LBI input. The input has very low (5 nA) LBI bias current, so you could get away with some very high-valued resistors for your divider (for reference, if you have 5 nA bias, you would still be ~1% accurate with a divider that draws 500 nA, or 0.5uA, adding little to the chip itself.) And in the 5 uA range for the monitor circuit, the time it would take for your circuit to discharge the batteries would be on the same order as the self-discharge/shelf-life of the batteries themselves (using AA's for example, and assuming we aren't talking coin cells, but you didn't specify).

If you need to get it lower, you could use a high-side MOSFET to disconnect the battery monitor from the batteries, but then again, your micro is likely going to be drawing several microamps of its own in sleep at the very least.

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  • \$\begingroup\$ Hi, many thanks for the reply. Using four AA batteries. I think I'll go with high resistance resistors. \$\endgroup\$ – Eamorr Jun 4 '12 at 4:38
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Voltage divider as such is not inefficient, it all depends on resistor values - you can use resistors in megaohm range, it would take ages to drain battery via such voltage divider.

On contrary note - you can't get to very high values - it all depends on current requirements of the circuit you are connecting your voltage divider to - you have to choose your resistors big enough as to provide enough current, but not too big to conserve power.

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