Charger Input and Output (Resistor) Question

[saumaun]

Hey everyone, I'm a new guy here =]

Well I have a few questions. I am working on making a hand crank charger for my cell phone. I have the hand crank device, and it has an output of 6 volts. It is labeled as such, and I have tested it with a volt meter.

I also have a USB wall charger (DC, right?) with and input of 100-240 volts (50-60 Herts, .2 amps) and output of 5 volts -.. 1 amp.

Basically, I want to be able to solder wires from the hand crank device to the board of the charger. The hand crank was connected to a board that had a 3.6 volt 80 mah battery. I desoldered that...

So basically, I am wondering what parts I will need to convert the wall charger board to allow 6 volt input and have a 5 volt output. I'm not sure, but I think it may involve replacing resistors.

If necessary, I can post picutres.

 

[Diver300]

The wall charger is of no use at all. It isn't possible to use a circuit designed for 100 - 240 V input and run it from 6V.

As I understand it, you want to charge a cell phone from 6 V when it was designed to run from 5V.

The first suggestion that springs to mind it just to connect it up. If you are getting 6V no load from the hand generator, what will probably happen is that the cell phone will take quite a bit of current and the 6V will drop to much nearer to 5V, and a cell phone designed to accept 5V will probably not be damaged by 6 V anyhow.

If you don't want to be that brave, you want a low-dropout 5V linear regulator.

Low-dropout means that it will work with a small voltage difference between input and output, like you have. Regulators that are not low-dropout can drop anywhere up to 2.5 V so you wouldn't get more than about 3.5 V with a 6V input.

A linear regulator is the simple type that gives out slightly less current than you put in. The alternative, call switching regulators, are needed when you increase the voltage or make it negative, and are more efficient if the input voltage is much bigger than the output. None of those apply here, so you should keep it simple.

Something like https://www.electro-tech-online.com/custompdfs/2008/03/KA78R05C.pdf is very easy to use. All you need is big capacitors to ground on both the input and ouput. The bigger the better, and not less than 100 :mu:F

 

[audioguru]

Your phone charges with 1 Amp for a few hours.
The hand crank generator came with a little 80mAh battery that probably charged in half an hour.

Then the output current of the hand crank generator is only 160mA and it will take 6.25 hours of cranking to charge your phone. Your arm will fall off in half that time.

 

[saumaun]

Thanks guys, I appreciate those responses.

@Diver - I just checked and the actual battery of the phone is 3.7V. It is a lithium ion, so we don't want to overcharge it :wink: I'm not sure if I want to be brave with my phone... It was quite expensive (Sprint Mogul). Also, it gets rather warm while charging as it is.
I am looking at the pdf you linked to, and it seems like that is what I need. Just a question though - Can I get one at radioshack? :lol:

@audioguru - I was thinking that my arm would fall off too :wink: However, if I get a 5V battery with a capacity of around 500mah, would that be better? I forgot to mention that the battery capacity of my phone is 1500mah. And like I said above, the phone's battery is 3.7V.

Another question: If I manage to get the output of the hand crank generator down to 3.6V with a linear regulator, would the phone still charge even though the standard output is 5V and the battery is 3.7V?

Again, I really appreciate your help on this matter!

 

[audioguru]

A 3.7V lithium battery charges to 4.20V. The phone has the charger circuit in it which stops the charging before the battery catches on fire. The circuit also prevents the battery from discharging too low..

There is no way that a hand crank generator can supply 1500mAh to charge the phone's battery. It is made to charge a little 80mAh battery that is 19 times smaller.

 

[saumaun]

Yay!! I found some hope!

check this out: https://geektechnique.org/projectlab/236

The flashlight is almost identicle to mine, except the DC output jack is labeled 5V on mine. (Hey! Thats what the charger output is!)

Do you think that if I replace the battery in my flashlight with one of a larger capacity, this would work? The guy in the tutorial uses a linear regulator as well, but since I already have the proper output, would it still be necessary?

@Uncle Scrooge - When you said the circuit stops the charging before the batter catches on fire, do you mean the circuit within the phone itself? (Please tell me that makes my life a bit easier)

 

[audioguru]

Your flashlight uses a low current. Your phone charges with a high current.
The hand crank generator and your arm are not strong enough to charge the phone's battery completely.
I think the battery in an ipod is much smaller than the battery in your phone and he cranked for 30 minutes and it was no where near fully charged.

Lithium batteries catch on fire if they are over-charged. So your phone has a proper lithium charger circuit built into it.

 

[saumaun]

So does that mean that I should give it up? If I can get a handcrank with a different voltage, could I do it?

Lithium batteries catch on fire if they are over-charged. So your phone has a proper lithium charger circuit built into it.

But isn't that controlled by the phone, not the charger? I thought that I wouldn't have to worry about overcharging with a charger.


I attached my pathetic attempt to further explain my (hopefully not) failed project. Tell me if what parts I need to make this work. I'm thinking of using rechargeable batteries (like those found in airsoft guns and rc cars) and a different hand crank.

Thanks

 

[audioguru]

The phone has a charger circuit inside it. Your hand crank thing is a generator, not a charger circuit. If you feed your phone 5.0V at up to 1A then the charger circuit in the phone will safely charge its battery.

Your hand crank generator doesn't provide enough current to charge the battery in your phone unless you crank for a very long time.

Rechargeble Ni-MH batteries make 4.8V, not 5V. 4.8V might be too low.
They would need to be charged by a real charger, not your hand crank generator.
Try it.

 

[saumaun]

Rechargeble Ni-MH batteries make 4.8V, not 5V. 4.8V might be too low.
They would need to be charged by a real charger, not your hand crank generator.
Try it.

Its 1.2 times the number of cells, right? So if I get 5 cells and have a 6V output, use a linear regulator and solder the charger cable to the 'creation', would this work (Im not picky about how much I have to crank YET)? Also, would the current be a high current?

I have a feeling I'm being super noobish...

I definitely appreciate your patience with my minimal knowledge of circuitry =]

 

[audioguru]

A "6V" Ni-MH battery is about 7.5V fresh out of its charger.

An ordinary 5V regulator doesn't work if its input voltage is less than about 7.5V.

NOOB!

 

[saumaun]

Ok now I'm confused. Is it a bad thing if the Ni-MH battery is 7.5V off the charger?

I really am not sure what to ask at this point. Based on the image I posted above, and assuming I can crank enough to charge the batteries, what do I need?

The output has to be 5V 1A, right? What type of batteries would do that? And how many volts for the batteries?

 

[blueroomelectronics]

I've actually seen one of these at a 5 & 10 store in Toronto for about $6.
Just Google it and order one, they probably require lots O cranking.
**broken link removed**

 

[audioguru]

Your phone's charging circuit is designed for an input voltage of 5.0V. It might fry if a Ni-MH battery presents it with 7.5V which is 50% too high.

 

[saumaun]

**broken link removed**
The output for that charger is 6.2V. How does that not fry cellphones?

For mine, if I use the 5 Ni-MH batteries @6V (7.5V off the charger lol) with this 5V regulator (**broken link removed**) could I POSSIBLY get this to work?

 

[saumaun]

bump! happy easter, btw

 

[audioguru]

**broken link removed**
The output for that charger is 6.2V. How does that not fry cellphones?

For mine, if I use the 5 Ni-MH batteries @6V (7.5V off the charger lol) with this 5V regulator (**broken link removed**) could I POSSIBLY get this to work?

6.2V is 24% higher than 5.0V. The charger circuit in some cell phones might fry if it is designed for 5.0v max.

The 7805 has a minimum supply voltage of about 7.5V. Its output voltage is lower than 5V if its input voltage is less than 7.5V.

Some 5V regulators are called "low dropout" and work fine until their input voltage is less than about 5.5V.

 

[blueroomelectronics]

Such as the LM2940

 

[saumaun]

Ah, so basically I would have to have a low-dropout 5V regulator with a supply of over 5.5V in batteries? Therefore, I could have my mystery hand-crank, 5-6 batteries and an LM2940 between the batteries and the charging cable? (well, knowing my luck, I have it completely messed up)



You guys know your stuff!