I'm trying to expand capacity for a Duracell power bank (3.7v 2600 mAh) so I can power my portable RasPi project with it. I've taken a battery (3.7v 2800 mAH) from a cheap tablet and I'm wondering if I just connect the negative of the Duracell to the positive of the Tablet battery then take the negative of the Tablet battery and connect it back to the board of the Duracell power bank, if it will increase the capacity of the power bank and function normally or will it blow up in my face melting it Raiders of Lost Ark style?I'm a novice hobbyist and any pointers would be a huge help.A little more detail about my project: I'm installing a RasPi with RetroPie into an Original Gameboy case for a pretty cool emulator. I've found a step up chip to power the new display but have run into the issue of portable power. I'd like a little more capacity than the batteries that I've seen Ben Heck and others use. Thanks for your time and I hope to hear from anyone
Hi Jack, I am not sure that this is a good idea as they are two different types of batteries and may not be compatible. Especially when you go to recharge them. You mention hooking them up in series, plus of one battery to the negative of the other. This will double your voltage to 7.4 volts but will not give you more capacity only more voltage. For more capacity the batteries would have to be hooked in parallel or plus to plus and minus to minus. There are guys on the site that know a lot more about these types of batteries that I do. I would recommend holding off until one of them looks your situation over.
The tablet cell appears to have protection, while the other is a high discharge version.
You are very likely to have different discharge rates, meaning one will go flat before the other, and with no battery management, you're likely to destroy it.
Charging is another issue.
LiPo batteries are known for catching fire when incorrectly charged, so you will need to do something.
Do yourself a favour and buy something that suits what you need, or are at least the same.
Mark
BTW its not your face (although the toxic fumes may be an issue) it's the house you should be worried about.
One of the few ways to put out a LiPo fire is smother it with sand.
The flat battery looks like a LIPO and the round one a conventional Li-Ion. As the others as said it's best not to mix the different chemistries together.
Also the charger inside the powerbank is expecting to charge a 3.7v nominal device so you will need to put two batteries of the same type in parallel (not series) also because you are effectively making 1 cell from 2 or more you need to have cells which are more or less matched in spec so that they charge and discharge in similar ways as you have no way of balancing them individually( you can get these from suppliers already made up!).
Your current power bank should power the RPI for 3-5 hours I think if you need much more then you really need a bigger bank an alternative may be a simple 12V Gel Cell and 5V regulator. I would expect 10 hours from a 5AH cell atleast depending on the regulator type usage etc.
Thanks for sending the pictures BTW as this makes helping you much easier. It's this kind of thing that gets your questions answered in the best way possible .
People are making more of a fuss about connecting different batteries together than is necessary.
For starting a boeing 787 you absolutely need balanced cells of the same chemistry: You're charging them close-to-max, you're discharging them at max when you really need them.
But for hobby purposes where you're adding an extra battery to have a larger capacity, this is less of an issue.
Worst case, what will happen is that one battery will drain from "almost full" to "almost empty" between 3.8 and 3.7V. And that the other will do so between 3.7 and 3.6V. Then during discharge you'll be (mostly) running on just one battery when the voltage is between 3.7 and 3.8, so you'll have to respect the "max discharge" for that battery (and later on in the discharge for the other). Similarly, during charge, you'll have to stick to the "max charge" of the slowest battery.
On the other hand, being extra cautious is a good idea too.
Um since my 10Ah power pack was only like £15 and you can get matched cell batteries quite easily I wouldn't even bother taking the risk of mixed chemistries.. Should be OK but since he'll probably be having it unattended then he may as well do it properly!
P.S.-never try to physically cut, bend, burn, overheat, hammer, stab or destroy a li-po battery - unless you have a controlled environment - it will either explode or burn violently in an instant - personal experiment. Have luck!
I have a 3.5in TFT LCD display (for a backup camera) and a Raspberry Pi with Retropie installed. So far the only issue I've had is portable power and charging. I was going to map the buttons to the GPIO of the RasPi until I could figure out the power.
The display is rated for 12V but I've ran it on a regular 9V battery. So, a 12V battery with the 5V regulator would be a good set up. I'm not sure if a lower voltage battery would last longer (making a battery with < 12V more intriguing) but it's something to consider.
Ultimately, I'd like to have a portable emulator that I can take on long trips to entertain myself during the dreaded wait times in airports.
Thank you everyone for all of your comments! The quick responses are such a pleasant surprise!
Those "backup camera" displays are nominal 12V, but don't really use the 12V internally for anything(*). They convert it to 5V and 3.3V first moment they get their hands on it. So those converters usually have a wide input-voltage range (but usually a maximum voltage).
"simple" (linear) voltage converters would require the input current to be equal to the output current, and waste the voltage difference as heat. Those are becoming uncommon outside hobbyist circuits. (e.g. you'll find one of these on most arduinos).
Switching converters will use less current on the input as the input voltage increases. So, to generate 5V, 500mA, from 11.1V a switching converter will use 265mA, but if you provide it with only 7.4V, it will use 397mA. To power this for a certain amount of time, you'll need say 3000mAh at 7.4V, or 2000mAh at 11.1V. So there won't be any difference. (both packs could be assembled from 6 cells with 1000mAh each).
In the calculation above, I've assumed a 85% efficiency of the switching regulator. But in fact, the efficiency goes up as the voltage difference goes down. So, you'd be better off with 3000mA, 7.4V pack.
Key is to see that a 3000mAh pack at 7.4V holds the same amount of energy and costs about the same as a 2000mAh pack at 11.1V.
(*) I have an LCD display that requires "odd" voltages like +19 and -10.4V. Those might be generated directly and at higher efficency from the 12V than from say 7.4V. However the currents are so low that having more efficiency for the 5V outweighs the advantage for these higher voltages.
