The GBA (specifically model AGB-001) uses a boost converter to power the Game Pak with a stable 3.3V from its batteries. The SP and Micro models use a buck-boost regulator to generate that same 3.3V. Presumably the NDS uses a buck-boost as well.

What is the maximum current that a Game Pak can draw? Does it depend on the current being used by the console at the time? Does it differ based on whether it's an original GBA, SP, Micro, or NDS?

In particular, I'm curious about the maximum current under the worst-case conditions so that I can ensure that the cartridge will never draw too much current in any situation.

  • 1
    I assume you mean in normal operation. Because, if it shorts for example, it will attempt to draw as much current as can be provided by the source, or before some component fuses :-)
    – paxdiablo
    Dec 22, 2022 at 20:34
  • 1
    @paxdiablo Hence "continuously draw". I'm sure a short could be delivered for a very short time!
    – forest
    Dec 22, 2022 at 23:29
  • The "AGB" in the title was not a typo. AGB-001 is the model, just as DMG is the model for the original dot matrix Game Boy systems. With that said, the new title is fine as well.
    – forest
    Jun 6, 2023 at 23:44

1 Answer 1


Note: All of this needs way more details than given. starting from model and modification all the way to cartridge used and level of backlight. So this is more a rule of thumb issue, which may need adaption to your use case.

Technically I guess it can draw as much as the source delivers - we're talking low current/power anyway. The basic GBA is powered by two (L)R6 (AA) batteries. At an internal resistance of 0.5 Ohm a current of 3A could be drawn by shortening the battery. Of course it'll be way less due additional resistance within the device. I guess the maximum ability will range around 200-300 mA (*1).

A basic 0.25 mm (10 mil) trace can carry at least 1A without being stressed, so already past whatever batter

It may be way more practical to look at two corner cases:

  • A standard GBA draws about 60-80 mA during normal operation with a basic cartridge.
  • At 200 mA the even Alkaline LR6 will be drained in less than 8 hours (*2)

This means that up to 100 mA would be a good assumption of what might be usable. Flash based cartridges with up to 80 mA consumption during operation support this.

As said, this will differ widely depending on the device and battery solution. The devices mentioned can easy span a 1:10 ratio for this topic, meaning that above 100 mA could as well be 1000 mA with the right device and battery pack/power adaptor.

If this is about construction some new device, your goal should be to draw as little as possible, no matter hoch much could be drawn. Not just to not damage the GBA or any attached device, but way more to be able to play more than a short time. This is not only die them being 'emptied', but more so due dropping output. With a higher demand the batteries will soon reach a level where they may still be way from being empty, but already insufficient from providing enough power to generate the current demanded:

Voltage drop over capacity

(Taken from Wikipedia)

So once again, it all depends on the battery you're using. Best to assume the worst...

Bottom line: There is not really a 'too much' Szenario, as the GBA will not deliver 'too much'. There is a too less szenario and its not about the GBA, but your device.

P.S.: Comments suggest that this is about some add on game module intended for as high as possible power, so why rely on the GBA power at all? Wouldn't it result in way more freedom if the cartridge brings it's own power source? That way you're able to optimize (maximize) everything from voltage to current for the need of this module but are free of all power constrains by the GBA and free of taking care for various configurations and models. They should always be able to power themself, so no need to consider that.

There are essentially three ways to do so:

  1. Add a (USB?) connector for an external power supply. Beside this being needed anyway for a SoC, the board can detect by default run in a low power more (like 200 mA max) and full power when USB is connected (500 mA at 5V (aka 600+ at 3.3) or more).

  2. Find some LiPo pack that fits the cartridge in addition to a slim line SMD PCB. ~1000 mAh at 3.7V seems possible.

  3. If power need is intermittent, a reasonable large gold cap can be used to bridge these sections of high power demand.

In fact, #2 and #3 could be combined (if the use case permits) to optimize peak performance.

Possibilities are endless :))

*1 - It's sufficient to just look at the current as the difference between 2.8..3V of the batteries and 3.3 after conversion is neglectable (<20%)

*2 - Quite visible with improved display mods

  • 1
    I'm intentionally keeping the details vague to account for the worst possible scenario (lowest source voltage so least efficiency in the boost converter, most used power by other components, etc.).
    – forest
    Dec 22, 2022 at 1:20
  • @forest Vague questions get even vaguer answers. Especially if the bandwith is that wide as with GBA variations. It's like asking for MPG numbers of classic Italian cars with respect to Fiat Topolino and Lamborghini Countach. Now, if your question is about worst possible, you need to ask for such - and making clear which setup that is for you.
    – Raffzahn
    Dec 22, 2022 at 1:24
  • I edited my question to be a bit more specific. Is it better?
    – forest
    Dec 22, 2022 at 1:34
  • I'm aware that the goal is to draw as little as possible, but in the case where more current is always better (e.g. for a co-processor that can be as fast as it can be driven), it might be a good idea to know what I can expect without having to worry about something that works fine on an NDS and AGB destroying an AGS. An answer citing an official limit given to developers by Nintendo would suffice, but a practical answer based on known limits would also be perfectly fine. If the traces can handle more than enough, then maybe the only limit is the batteries or the voltage regulator...
    – forest
    Dec 22, 2022 at 1:43
  • @forest which they are.
    – Raffzahn
    Dec 22, 2022 at 1:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .