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I’ve got an 850mAh LiPo (https://www.sparkfun.com/products/13854) hooked up to a charger circuit (https://www.sparkfun.com/products/10217) and 3V voltage regulator (https://www.sparkfun.com/products/526) to power a Gameboy Advance (AGB-001). I don’t quite understand why the battery drains so quickly when the unit is off (seems to last only a few days to a week maybe). I’m able to power on other natively LiPo-powered units like the Gameboy Advance SP or Gameboy Advance Micro just fine after months of being stored. Any thoughts?

Is there something about the voltage regulator that would cause this? I’ve seen other Gameboy battery packs for sale (e.g. https://www.retromodding.com/collections/gameboy-advance/products/gameboy-advance-rechargeable-battery-pack?variant=2565317361697) that have a switch to flip when storing for long periods of time, though I’m not sure if it’s to prevent drain.

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2 Answers 2

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The parameter you should be looking at is called quiescent current (Iq). This is the current consumed by the regulator when the output current is zero. Both linear regulators (e.g., LDO's) and switching regulators (DC-DC converters) can possibly have high Iq. In general it is harder to make a DC-DC with very low Iq.

If you need to leave a regulator connected to a battery at all times, you should be sure to use one with a low Iq. To figure out how long the battery will last, divide capacity in mAh by Iq in mA. That will tell you how long the battery will last.

In your case the Iq is 5mA. For some reason the datasheet uses the symbol Id instead of Iq. But it is the same thing. Anyway, 850 / 5 = 170 hours or 7 days from full charge to flat empty.

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  • \$\begingroup\$ Thanks everyone for the insightful answers. @mkeith, Are there any ways to make the 3.7V LiPo around 3V and have zero draw when the power is off? I’ve heard of using diodes but am not positive. The obvious but likely naive approach is to use a resistor, but that doesn’t “regulate”. \$\endgroup\$
    – Andrew M.
    Commented Apr 7, 2018 at 20:52
  • \$\begingroup\$ Also, I’m guessing that if I put a switch between the regulator and the battery, I could eliminate the lq altogether and be only subject to the natural drainage of the battery. It seems that’s what handhelds that are built around being rechargeable behave like. Is this because they effectively have the regulator on the “other” side of the unit’s power switch? (Something I can’t achieve trivially because I’m retrofitting a AA-powered device and can’t easily modify that part of the PCB circuit). \$\endgroup\$
    – Andrew M.
    Commented Apr 7, 2018 at 20:59
  • \$\begingroup\$ I have designed a lot of handheld battery powered products. Most of them have at least one regulator that is on all the time. The power button typically just sends a signal to the processor telling it to wake up. Also, there is some type of low-voltage cutoff that takes standby current down to less than 10 uA (unless the normal standby current is less than 10 uA already, which it can be sometimes). The low voltage cutoff kicks in at around 0.8V to 1V per cell for alkaline batteries. This helps prevent the batteries from leaking electrolyte. Protects NiMh from over-discharge destruction. \$\endgroup\$
    – user57037
    Commented Apr 7, 2018 at 21:31
  • \$\begingroup\$ In your case, the best bet might be to just use a good LDO with only a few uA quiescent current. \$\endgroup\$
    – user57037
    Commented Apr 7, 2018 at 21:32
  • \$\begingroup\$ Great, thanks. Would a voltage reference do the same? What do you think of this guy? sparkfun.com/products/11078. Seems like it could do what’s needed but I can’t find the Iq for it (if that concept exists for one of these). \$\endgroup\$
    – Andrew M.
    Commented Apr 8, 2018 at 1:59
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Voltage regulators, like the one you're using, dispense a lot of excess power in the form of heat (therefore the presence of a metallic heatsink). If the GBA is off, but the regulator still "regulates", then you lose a lot of power this way.

For battery supplied applications, one should opt for DC-DC converters, which have greater efficiency and don't heat up as much. Or, when turning the GBA off, flip a switch that disconnects the battery from the circuit, just like described in the question.

I would go for the converter option AND the switch, as it would save the most power when the GBA is both powered on and off.

Not a product recomendation, just an illustrative example, something like this guy would get the job done.

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  • \$\begingroup\$ The average efficiency of an LDO running from a lithium ion battery is 3.3V / 3.7V = 89%. Most of the time it is not worth it to use a DC-DC to get 3.3V. You will only gain a few percent in efficiency. \$\endgroup\$
    – user57037
    Commented Apr 7, 2018 at 6:34
  • \$\begingroup\$ I disagree, maximum Iq of the example component is 40uA. Compared to 5mA, that's over 100-fold power saving. \$\endgroup\$
    – Vicente Cunha
    Commented Apr 7, 2018 at 6:42
  • \$\begingroup\$ Also, efficiency of LDO is not given by voltage ratio, but by output power over input power - which can be quite different. \$\endgroup\$
    – Vicente Cunha
    Commented Apr 7, 2018 at 6:43
  • \$\begingroup\$ The regulator the OP chose has a very high quiescent current. It is not a good choice. A good LDO will usually have lower quiescent current than a good DC-DC and also be cheaper. When quiescent current is small compared to load, the efficiency for an LDO is very closely approximated by Vout/Vin (because Iin = Iout + Iq). \$\endgroup\$
    – user57037
    Commented Apr 7, 2018 at 6:52
  • \$\begingroup\$ There is nothing wrong with using a DC-DC converter, and the one you chose has nice low current. But if it is a design which will go into volume production, then the cost of the DC-DC usually cannot be justified. A DC-DC may cost US$ 0.2, and then you need a diode and inductor. The LDO may only be $0.06 and needs no inductor or diode. \$\endgroup\$
    – user57037
    Commented Apr 7, 2018 at 6:57

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