Most efficient way to achieve 3.6V or less from NiMH or Li-Ion batteries (when fully charged)

Question

I have a micro and RF module both which I'd like to power from either 3 AA NiMH cells or a single 18650 Li-ion cell. Haven't decided which I'll use yet. My current requirements aren't that much (~2mA or so) but I'm looking to achieve a fairly long runtime (30+ days between recharging).

Both the micro and the RF module have a fairly wide input voltage, however the RF module has a max input of 3.6V. So the issue is with fully-charged batteries either of these set-ups would easily supply more than the 3.6V max: more like 3.8-4.2V I'm guessing.

My gut tells me that the drop from, say, a 1N4001 diode (~0.6V) should do the trick, at least initially but as the cells discharge I'll no longer need that 0.6V drop and will then just be a waste.

My other thought was use a high-efficiency buck/boost converter (such as this one) with Vout set to 3V or so. I don't necessarily need a fixed 3V output since both the micro and RF module will work down to 1.9V.

What's the most efficient way to drop the output so that either of these battery set-ups will provide 3.6V or less?

Answer 1

There are several things here...

The NiMH cells you chose could, in theory, support 2 mA current draw for 30+ days based only on the mAH rating. However, there is self-discharge to consider. This Wikipedia article says that a typical NiMH battery will loose 5-10% of it's charge the first day and about another 1% every day after that. My experience is that it is true for new batteries only, and older batteries will self-discharge much quicker. My cordless drill, for example, would be useless 6 weeks after a full charge when new but now is useless after only a week-- when not being used at all during that time.

Lithium rechargeable batteries, however, have a much lower self-discharge rate. Which is why I'll be buying a lithium based cordless drill real soon now!

As for your question... The problem with using a diode is that the voltage drop is not always 0.6v. At low currents it could be much less. The datasheet for the diode will contain a graph showing the voltage drop vs. current. You say 2 mA, but it will more likely be less than 0.5 mA most of the time with pulses greater than 4 mA (just a guess). So when it is taking 0.5 mA the voltage to the module might be too high.

The ideal device is a Low Dropout Linear Regulator (LDO) that is designed specifically with this in mind. Sorry, but I don't have time to find one right now. But there are LDO's designed for running off of a battery. When the battery is new it regulates like normal, but as Vin drops below the ideal Vout the regulator will stop regulating and Vout will basically equal Vin. Not every LDO will do this! If it doesn't say in the datasheet then you must assume that it won't do this.

Another idea would be to create your own "lame switching buck converter". The business end of the converter is a simple MOSFET feeding Vbat to a storage cap. The output of the cap feeds your module. If the cap voltage drops below, say, 2.5v then the MOSFET turns on. When the cap voltage reaches 3.3v the cap turns off. A simple voltage comparator with hysteresis can do this. Look for a super low power comparator, since you don't want this thing to consume more power than your module. Make the cap large enough that the turn on/off rate of the MOSFET isn't too fast. Below 10 KHz, or maybe even below 1 KHz. An inductor in series with the MOSFET might not be a bad idea, to reduce the peak current from the battery. I would rather use an LDO than this, however, as it would be much easier to get right the first time.

Answer 2

Use a low dropout regulator that's well below the point at which the batteries become useless to you - something like 2.5V (or even 2V) should do the trick if both your devices will work at that voltage. They're cheap, easy to use and only need a couple of decoupling capacitors. Generally your batteries aren't going to go below 3V before they need charging.

Don't waste money on a switching regulator, it's just extra hassle and components.

E.g. MCP1825S you can get a 2.5V fixed version with a dropout of around 2.7V ish.

Answer 3

Good solution:

2 x AA NimH AA from reputable maker (NOT 3).

Better still use low self discharge NimH AA.

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2 mA x 24 hours x 30 days =~ 1500 mA hours. A modern good quality 2500 mAh AA NimH cell has a genuine capacity approaching 2500 mAh (say 2400 mAh to be safe). In fact the capacity will be higher at the very low discharge rate of 2 mA. A Sanyo Eneloop AA has about 2000 mAh capacity.

To supply >= 1500 mAh the 2400 mAh call would need to lose no more than (2400-1500)/2400 =~37% capacity. Any reputable NimH cell will have less than that much loss in one month.
An Eneloop or any other reputable low self discharge NimH call will lose far less than that in a month.

As you can tolerate a 1.9V supply voltage, using 2 x NimH makes sense. At 2 mA an AA Nimh will supply 1.2V+ over the very large part of it's discharge cycle.

With two cells and 3.6V max allowed voltage you can have up to 1.8V/cell - which you will never reach. Max ever is say 1.5V/cell on charge abs max, 1.35 V/cell just off charge dropping rapidly to under 1.3V/ cell.

Discharge should not go below say 1.05 - 1.1V/cell at very low discharge rates. (Lower at higher rates).