I'd like to start by saying I am not an electrical engineer or anything of the sort. I'm not anywhere near an expert on electronics, which is why I am here. I apologise if this is not the correct place to ask for help with such a question. That being said, hopefully it is and thank you in advance for any assistance.

So I have a device that uses a CR2032 coin style battery. It's a Pokémon Go + if anyone cares or knows what that is here. It uses just one of these batteries, and it drains them very quickly. I am tired of buying these expensive batteries every week for it, and am looking for a better solution. I briefly looked at rechargeable CR2032 batteries but they look to be more trouble than they're worth.

I am thinking of modifying it via wiring in a big rechargeable battery so I don't have to worry about it so much. I tried searching around for some website or place that might just sell a battery pack of sorts that is 3V and rechargeable. Couldn't find one so I moved onto another idea.

I am thinking of using a 5V output battery. These are very easy to find because of phones. I have a small battery pack at home that is designed for charging phones that I don't use anymore. It is too small of capacity to be much use these days for phones, but it would be great for this due to size, and would still have tons of capacity for something like this.

I am looking for the best way to go about doing this properly to avoid bad things happening to the battery or the device. Can I simply solder wires from the device to the battery with a resistor in between to reduce voltage? Should I be concerned about anything other than dropping the 5V voltage to 3V? Is there anything more I might need besides just a simple resistor?

Again thanks in advance for any insight!

 

Those "5V output" batteries are a 3.6-ish volt lithium-ion battery with a charging circuit and 3.6-to-5V boost-converter.
You could cut out some complexity by connecting a 3.6-ish volt battery to your 3.0-ish volt device.
Several (all right: two, on reddit) people claim to have done this.
Don't do it with a resistor: you don't know which resistor to use and will have satan's own time figuring it out. I'll bet that the current draw of that device varies over multiple decades

 

A linear regulator is a safer bet to drop the voltage from 5 volts to 3 volts, something like the LM317. Check the drop out voltage, voltage that is across the regulator when current is passing through it, to make sure it can reach 3 volts on the output with a 5 volt input. A LDO regulator may be needed, that is Low Drop Out regulator.

 

Okay I had to do a bit of crash course googling/wiki to understand how linear regulators and low-drop regulators work and the differences between them and resistors... I think I understand it somewhat now.

While looking into this I see that they can however be quite inefficient as they still dissipate excess power as heat. Which I'm guessing would translate to a reasonable loss in battery capacity.

I also then stumbled upon switching regulators and buck converters... I may not be understanding everything entirely right but wouldn't something like this:
https://www.amazon.com/DROK-Digital-Control-Regulator-Converter/dp/B01CE5P33M/ref=sr_1_8?ie=UTF8&qid=1491414073&sr=8-8&keywords=buck+converter&refinements=p_72:2661618011&th=1

...Be a good/easy way to do this while maintaining some additional efficiency in doing so? I mean I still have to measure the actual output of the battery charger and the device to get it right but yeah...

I'm also thinking I may want to not even disassemble the battery charger itself and just rework a USB cable that would normally plug into the charger with the regulator/device etc. That way the battery charger could be easily disconnected/separated from the device and replaced if needed in the future with a different one.

 

although this might be an extremely easy way to do it:
http://www.ebay.com/itm/like/281450833020?lpid=82&chn=ps&ul_noapp=true

 

A switching regulator would certainly do the job much more efficiently. You can reach 80 or 90% or even higher ratings rather than 60% maximum efficiency with the linear regulator.

The switching regulators are very common but more expensive.

The part you show from ebay looks like a linear regulator.

 

although this might be an extremely easy way to do it:
http://www.ebay.com/itm/like/281450833020?lpid=82&chn=ps&ul_noapp=true

I think it should work if you cannected it correctly. The chip on the breakout board is probably a LDO 3.3V regulator.

You're using it to power a pokeman game which uses a CR2032 coin cell. Your 5V battery pack would likely has 100 times more juice than the poor CR2032. So even with 60% efficiency, you won't have to change the cell once every week at least.

Allen

 

Thanks for all the help guys. I ordered the ebay one. 99 cents and free shipping from China. No idea how they can even make money on that. Probably wont be here for ages but we'll see how it goes when it does.

 

I ordered 20 pcs of SMD TL072 on 10/3 and it arrived yesterday ie 5/4. Took about 25 days and I think that's not bad.

Allen

 

If you don't want to mess with recharging, a couple of AA alkaline batteries will provide 3V and have about 11 times the life of the CR2032.

 

I

While looking into this I see that they can however be quite inefficient as they still dissipate excess power as heat. Which I'm guessing would translate to a reasonable loss in battery capacity.

wouldn't worry about inefficiency. A typical CR2032 coin cell is a ~3 - 3.4 volt 200 - 220 mAh battery so if you stuck a common 1 - 2 Ah 3.6 volt lithium battery with a common diode in series with it to bring its ~3.6 - 4 volt output down to ~3 volts it would run for 5 - 10 times as long as the coin cell did.

No need to over engineer things for a microscopic non critical power application. Especially being the diode would give you the needed voltage drop passively whereas any active voltage regulator is likely going to draw as much or more power than the device itself does and draw power even when the device is off as well.