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
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 .
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.
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.
I saw a link online that converted the display to accept 5v for power. However, the board for my display is not the same as the examples in the thread (apparently there are many different boards while the exterior looks the same).
What are your thoughts on attempting to modify the board so the display could be powered with 5 volts?
What are your thoughts on attempting to modify the board so the display could be powered with 5 volts?
I'd say if you have the knowledge to do it yourself, go ahead and do it. (i.e. if you're asking: don't)
If you don't have exactly the same screen, figuring out how to do it, probably involves some "trouble shooting" which is difficult if you don't have the appropriate skillset (and tools!).
I'm sure I have the tools, it shouldn't take more than a mulitmeter and some patience right? If we're talking oscilloscope and the like, then I'll probably go a different route.
Now this would probably be outside my skillset but one of the main reasons for doing these types of projects is to learn new skills!
So, any advice on how to do that would be greatly appreciated but if it's too complicated to explain then I understand completely!
Again, thank you so much for the replies and I hope you have a fantastic New Year.
