Hello /diy/, I recently found an old MacBook Pro 17" from 2009 (I believe) that an old client of mine had destroyed by pouring coffee all over it, he practically blew the motherboard. I gave up trying to fix the mobo and decided to cannibalize the laptop for parts, most of which I have already utilized in other projects or am going too. I'm left with this here Li-ion polymer battery. I recently purchased a Raspberry Pi 2 and thought to myself, "Wouldn't it be cool if I could power my RPI 2 with this massive freaking battery? The life-span would be ridiculous." Problem is I'm not electrician and I don't know if this would cause any damage. Looking at the back of this battery these are the specs:
>7.3V
>95Wh
The Raspberri Pi 2 Model B has a:
>recommended PSU current capacity of 1.8A, >maximum total USB peripheral current draw of 600mA/1.2A(switchable)
The Raspberry Pi is powered by a 5V micro-usb connector. They say that 1.2A(1200mA) power supply will provide ample power to run the Pi, but I may want to get a 2.5A(2500mA) power supply if I utilize everything on the thing. So since the battery is 7.3V and 95Wh, I'm guessing it would fry it the Pi? Is there a way to make this work? Tips and pointing me in the right direction to resources and what not would be much appreciated. Thanks!
While we're at it, post any cool battery projects you've done.
So obviously the voltage is too high correct? So I'd damage the Pi, so I have to put some stuff in the way to lower the voltage down to 5V. Now the 95 watt hours baffles me. I have found formulas that allow you to calculate amp hours from watt hours and watt hours into amps but I'm not sure if I'm doing it right... also MORE amps isn't bad correct? More amperage doesn't damage a device its the volts from my simple understanding.
>>960013
Voltage is a fixed thing. Amperage is variable depending on how much power the device is taking.
Watts = Voltage × Amperage
With the Pi demanding 5V. I'd just buy a third market voltage regulating circuit for the 7.3V. Or, if it's going to be low power, diodes take 0.5V regularly. So you can add 4 diodes to the power and it'll work.
I personally think the charging of the battery is going to be the hardest. Li-ion cells can be a bitch, especially those witb proprietary connectors.
>>960025
Right, so I'd need 0.38ohm resistance to lower it down to 5V correct? I got pics following of the connector on the mobo, you think if I look up the connector number I may be able to get an adapter for it or something? Hmm. I need to get it from this weird pin connector down to a micro USB.
>>960025
Btw I still have the old connectors and everything, plus I have the charger for the battery.
Also I looked up the number and its a 13Ah battery (13000mAh).
This is the connector it the battery plugged into on the mobo.
>>960025
>So you can add 4 diodes to the power and it'll work.
Different anon here, Ive tried that with doides and I couldnt get more than 1v drop even with 3 of them in series. It seemed that everything past the first diode was the exact same voltage.
last pic. Also its weird, its like its soldered into the mobo, but the back little pins are spring loaded, you can push down on them. I don't understand why or their point. (pic above)
I'm interested in doing this eventually as well, but for a portable speaker.
Amp circuits are generally forgiving when it comes to input right? I forget which chip I'm going to use (probably just going to use a schematic from that frank mims3 book, I think I remember him using 741s a lot) though.
Ive looked around and cant find the pinout diagram for this damn connector any ideas how I can figure it out guys?
>>960029
No. Voltage dividers are useless under significant load. You'd be better off with a 5V regulator or (better) buck convertor.
As others have said, charging is going to be a complete bitch, because with Li-ion cells you need to monitor and control each cell separately. That brick probably has 6 cells in it.
>>960367
Correction: a voltage divider is going to be useless where one of your loads (the RPi) varies in current requirements.
>>960029
>>960025
>>960034
Don't use diodes, resistors or anything silly like that for regulating power.
Even a linear regulator would be an infinitely better choice and even then it's a very bad idea to use one.
>>960008
>Is there a way to make this work?
It's called a buck-boost converter aka switching regulator aka switching power supply aka the thing that everything uses (yes, even your PC's mains-to-3.3V PSU).
Look it up. Then find one with high enough input voltage and low enough output, read the application notes in the datasheet and you're set.
>>960374
Will do going to research now. Thank you kind ser.
>>960031
Check the power management ICs on the board. Figure out pinout using a multimeter (they even sell multimeters at lidl these days)
>>960376
>do all three require each other to work properly
All chemical batteries no matter the technology behind them can work independently or in series.
BUT you don't want to mix unmatched batteries.
There's no reason to disassemble the cell block if you're using a switching regulator.
Even better if you don't, since once you do and use some of the batteries, you shouldn't put them back together unless you have a charger that can rebalance them.
>>960377
Great! Research should be at least 2/3 of your project if you're not exactly sure what you're doing.
>>960374
>Don't use diodes
I'm the one that said I couldn't get more than a 1v drop.
I actually had that setup just to test out nigger rigging a voltage divider. Things didn't just werk.
>>960388
Thanks man, super appreciative. I found this little guy on Amazon.
http://www.amazon.com/DROK-Adjustable-Non-isolated-Buck-Boost-Transformer/dp/B00GWCC6S4/ref=sr_1_2?ie=UTF8&qid=1457749822&sr=8-2&keywords=buck+booster
Looks like exactly what I need. Now there is a 12W and a 25W version.
Since I'm going to be regulating the voltage down from 7.5V to 5V do I do the Watt to Amp conversion before the voltage regulation or after the voltage regulation?
i.e. I = 12 / 7.5 or is it going to be I = 12 / 5 ?
They say its recommended to give the PI at least 2.5A. So if the wattage is calculated prior to the voltage regulation, I'd be ending up with only 1.6A, in which case I should get the 25W version. If the wattage is calculated after the voltage regulation then I should be fine.
My assumption is that the amperes, since they are the power, are based on the ending voltage, so in that case:
12W/5V=2.4A; I should be good, yes?
>>960439
Wait fuck, the buck-booster is rated at 3A/12W or 3A/25W, so my calculations are irrelevant, its always going to be supplying 3A, even if I adjust the voltage right?
>>960439
Wait, I'm still stupid.
I need to ensure the buck-booster can handle the amperage right?
So since I'm inputing a 7.5V & 13Ah DC battery and I need to step it down to 5V & 2.5A-3.0A
this guy right here, should do the trick, yes?
http://www.amazon.com/DROK-Converter-Transformer-Synchronous-Adjustable/dp/B00C9UUFHC/ref=sr_1_12?ie=UTF8&qid=1457753712&sr=8-12&keywords=buck+converter
Or am I still being stupid? I'd safely assume the latter.
Just get a 5v 2a wall adapter and call it a wrap. Open the Mac battery to get the 18650 cells out for use in high powered flashlights and other devices.
>>960518
That defeats the purpose of tinkering though!
I'm going to purchase myself that DROK Converter with the 15A rating and give it a shot. I'm almost off work and I'll order it right after. Once I get home I'm going to open up the battery but not disconnect the cells from one another in order to see what the wires are doing and where they're going and see if there is a controller chip inside. I'm gonna bust out my multi-meter and test the 13 pins and see what they're doing as well. I'm going to then see if I can get it to charge through the mobo I have.
>>960439
>if the wattage is calculated prior to the voltage regulation
Doesn't matter much unless you're operating near the specification limit in which case you should look for a better product.
Switching regulators can be calculated similarly to a transformer (because they kind of are).
Input V*A = Output V*A * efficiency.
Efficiency is well over 0.9 for switching regulators.
Linear regulators just dump the excess power resulting from voltage difference as heat, that's why they're bad unless you need a very stable very low power souce.
>12W/5V=2.4A
Correct. That's their max current draw at that voltage. Just keep in mind that maximum doesn't necessarily mean sustainable. Oh and the higher wattage one is cheaper for whatever reason.
>I should be good, yes?
Yep, but the same modules are available cheaper, even on amazon so shop around a bit before you buy.
>>960034
>Ive tried that with doides and I couldnt get more than 1v drop even with 3 of them in series
that's coz you measured it wrong. you have to test it under a suitable load. if the load is just your voltmeter, it's meaningless.
>>960008
Get a 5v 4a UBEC from eBay. they're used for drones/model aircraft. Also find a big capacitor from somewhere, because their output is very ripply.
>>960008
Harvest the lithium and overdose one of your enemies with it.
