Replacing battery with a capacitor...

[Anonymous No.1216324]

Hi.

I have a Samsung Young 2 (G130H) that i plan have permanently connected to a power supply. It's manly connected to a car so it powers on when the car is on and powers off when the car is off.

However in the long run this will not be good for the original Li-ION battery, so i am thinking in replace it for an alternative.

I only need that the phone runs at least 2 minutes after the power goes off so a capacitor should do the trick.

However i am not sure what kind capacitor should i use to replace the 3.7V Li-ion (4.81Wh): charge voltage 4.2v

Can someone help?

Thank you

[Anonymous No.1216327]

BTW i've mooded the phone, so now he can power on when the power is connected and turns down 1 min after the power is unconnected.

The thing here is to give the phone the ability to safely shutdown using other power source.

[Anonymous No.1216329]

>>1216327
The battery is always ready to supply power, retard. Turn off the phone, then turn off your car. Fucking easy.

[Anonymous No.1216331]

With battery everything runs perfect. However in the long run (and due to the fact that will be placed inside of a car) the weather conditions - like a hot day - can damage the battery. So a best solution should be replace it for a capacitor.

[Anonymous No.1216339]

Hey, tell me something /diy/ why do we rate batteries in such retarded ways? Like mAh or Ah or even Wh

I mean, if they're rated for a specific voktage, why not just use Joules@Xvoltage? It seems much clearer.

Like why we use kWh for homes, that's just retarded. Just stick with joules and watts.

[Anonymous No.1216349]

Because that's the information that the battery has on it...

Personally i think it's simpler to have only one measuring unit.

[Anonymous No.1216547]

Anyone?

[Anonymous No.1216562]

>>1216324

Lets say 1W, for 120 seconds is 120 Joule. 1/2C*V^2 = 120, C*12=20 , so you need about 1F at 12V.

There are balance boards for capacitors on ebay/aliexpress which prevent series strings of supercapacitors overcharging one of the capacitors.

[Anonymous No.1216566]

>>1216562

PS. the individual capacitors of course need to be larger, 6 10F capacitors in series should do it.

[Anonymous No.1216568]

>>1216566
60 farads? Easy

[Anonymous No.1216569]

>>1216566

All that said, just connect it directly to the car battery and use a delay switch (modules also available on ebay/aliexpress).

The only battery you need is the car battery.

[Anonymous No.1216629]

>>1216324
Caps bleed out their power constantly. If you don't charge it up all the time or don't have access to charge it then it will be useless in a dozen or so hours. Super capacitors are really bad about that. Also, most super caps that give you the range you want (90F or so) and are small in size are actually a bunch of super caps with heat shrink plastic over them without any balancing circuitry.

[Anonymous No.1216679]

>>1216349
I like it. We should have one unit for everything. Voltage, watt-hours, volt-ampere and C rating. Just call it 1 Battery. 1B is the equivalent of "that shit you put into a phone"

[Anonymous No.1216682]

Just hook up about a dozen of these. 6 series, 2 parallel. Your phone will feel the power

[Anonymous No.1216688]

>>1216339
Because you can ask 'how long will this battery run my x amp device. Let's see y amp hours divided by x Amps equals time'

[Anonymous No.1217480]

>>1216339
>I mean, if they're rated for a specific voktage, why not just use Joules@Xvoltage? It seems much clearer.

Because that's a near-useless measurement for everything except comparing energy density.

[Anonymous No.1217580]

>>1216339
mAh is retarded, but Wh i'd like to see. You instantly know how long you can power your shit.

[Anonymous No.1217596]

>>1217580
>mAh is retarded, but Wh i'd like to see. You instantly know how long you can power your shit.

What's the difference? The voltage is specified. I realize that the voltage is not exactly constant as you drain the battery, but for a 3.6v battery you know that 1000 mAh is half as much as 2000 mAh.

I don't see how you "instantly" know any more information from Wh than you do from voltage plus mAh.

[Anonymous No.1217840]

>>1217596
>assuming any particular discharge curve

[Anonymous No.1218165]

>>1217840

[Anonymous No.1218172]

>>1218165

>no sensible chuckle.gif on this computer

[Anonymous No.1218193]

>>1218172
I got u senpai

[Anonymous No.1218196]

>>1218165
I deserved that

[Anonymous No.1218259]

>>1216562
that's not how it works

the phone will never use all the energy since it will stop working after the capacitor goes below a certain voltage (maybe 3V or smth).

this is the problem with replacing batteries with supercaps
while a Li-Ion battery supplies almost all of its rated energy within a small voltage range, with a capacitor you need a very low cut-off voltage otherwise you're wasting a lot of energy

[Anonymous No.1218281]

im not too well educated in regards to electrics, but as a phone doesnt take a lot of power, you don't need a huge capacitor. this site states, that a smartphone will probably never use more than 6W (but most likely only 2-3) https://petewarden.com/2015/10/08/smartphone-energy-consumption/

this site is also interesting, although on a different scale than your project
https://www.wired.com/2013/05/can-you-power-a-phone-with-a-capacitor/

2 minutes of 6W = 720J
C= 1440/16 = 90F at 4V

should be doable, but I cant tell you, if you need any exta parts inbetween

[Anonymous No.1218327]

>>1218281
see
>>1218259
you're charging your capacitor from 0 to 4V and your phone will stop working once it's a bit below 3,7V i think.

The only way this will work is with an aditional circuit that reduces the capacitor's voltage to 3,7V, so you can charge for example to 8-12V for the energy recuired to power the phone
Even just a zener diode might do the trick, but i believe there are more elegant solutions
electronics-tutorials (dot) ws/diode/diode_7.html

I'm by no means an expert by the way

[Anonymous No.1218415]

>>1218327
>your phone will stop working once it's a bit below 3,7V
Please review actual LiPo discharge curves and mobile PMIC datasheets.

[Anonymous No.1218423]

>>1218259

Stored energy grows with V^2, so with 12V storage by the time your converter cuts out due to low voltage of the capacitor nearly all the energy will have been removed.

>>1218327

I assumed he'd charge a capacitor at 12V to run a converter to power the phone.

But a delay switch still makes more sense regardless.

[Anonymous No.1218429]

>>1218415
I haven't said anything about batteries m8, i just assumed a phone stops working once it's fed with a voltage a bit below 3,7V
No idea how much "a bit" exactly is, but that's beside the point

[Anonymous No.1218791]

>>1216324
These guys are all overthinking it. Just run a DC-DC converter set to ~4 volts to the battery contacts. Then you don't have to worry about charging or capacity, and it can just run off the car battery all the time.

[Anonymous No.1218958]

>>1218791
Mostly this. The phone probably doesn't use more than 100mA anyway. Choose an especially efficient dc-dc converter with undervoltage lockout.
>or, just do the same thing with the usb inlet

[Anonymous No.1218981]

>>1218791

Then you don't drive for a week while the phone for some reason is stuck in a loop consuming peak power and your battery is dead.

[Anonymous No.1219105]

>>1218981
Yeah, like 2 watts... I think it would take longer than that anyway, and how often has your phone locked up so that it won't turn off? Probably almost never.

>>1218958
I thought of that, but then I remembered that my phone doesn't like it when you power it from USB with no battery, so I figured some others might give trouble as well.