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Moped Cellphone Charger design

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qbone

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Help, Moped Cellphone Charger design

Hey guys. I am trying to built myself a battery charger for my cellphone, powered by my moped.
It is stated on the moped that the output is 6V, which is 6Vac.
So far I have rectified it by making a bridge of 4x 1N5819 so the voltage drop is as low as possible.

Now about the charging. I know that a 3.7V Lithium battery can be charged up to 4.2Volts, so I guess I should make a DC converter which gives 4.2V on the output.
After the rectifier I have about 5.35Vdc.

How do you suggest I go about designing this circuit?
I don't have time to order all sorts of fancy party like MAX1555, which I know would be the best way, but I do have a series of 78xx, LM317, LT1074 and MC34063 ICs available.
And all sorts of inductors, capacitors and resistor.

Could you help me out?
The cellphone is a Nokia 6100, and the Battery is a BL-4C rated 3.7V 820mAh.

Could I do it so that I make a voltage regulated circuit from one of the above ICs and limit the voltage to 4.2V and limit the current to something like 10c? (82mAh) - or how much do you think is safe?
One guy told me that if I charge it with something like that it wouldn't blow up the battery, and would just stop charging when the battery level reaches the output from my voltage regulator.

If someone has a diagram this would be much appreciated.
So far my diagram is very small to say the least, but I can post it if you wish. So far its just the rectifier, a capacitor and a transcient diode.
 
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P.S. I already did search the forum, but never found anything helpful.
Or maybe the issue is that this isnt even possible with 30+ views. Please let me know then :)
 

The 780x series will work but have fixed output voltages, you will find it difficult to get one with 4.2V output.
The LM317 will work, you can use two resistors to set it's output voltage to 4.2V and add a resistor in series with it's input pin to limit the current.
I do not have the data sheet for the LT1074 t hand so I can't help you with that one.
The MC34063 will work well and may be your most efficient option. Bring a switching regulator it is efficient and it has built-in current limiting. Take a look at **broken link removed** for a simple design tool that will give you the values to use.

Current always flows when two points of different potential difference are connected together, if your battery is at 4.2V and the charger is producing 4.2V no current will flow so in a sense it is self regulating. What you have to consider is the current that would flow if the battery was fully discharged or even faulty. Limiting the current will help to protect the battery and the charger circuit.

Brian.
 
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    qbone

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try first with lm317 for your concept to hardware.
mc34063 is good but you can use it after your test has given expected result.

use lm317 in constant current mode , and test that also.
 

betwixt said:
What you have to consider is the current that would flow if the battery was fully discharged or even faulty. Limiting the current will help to protect the battery and the charger circuit.

This I can regulate with a resistor in series with the output, correct?

srizbf said:
use lm317 in constant current mode , and test that also.

Before I do that I will need a little explanation, cause everywhere I have read, a lithium battery should be charged with a constant voltage, rather than a constant current.
 

Never try to limit the current with a resistor in the OUTPUT. Doing do negates the purpose of the regulator as the output voltage will drop according the the current being drawn. If using a linear regulator like the LM317 add the resistor in the INPUT side so the output stays constant until enough current forces it's input pin voltage to go below regulation threshold. This will keep the output voltage stable until overload occurs then the voltage will drop and damage will be prevented.

If you use the MC34063, the resistor works in a diffeent way but still limits the maximum current the IC will provide.

While it's true that a lithium battery is charged with a constant voltage, you should still limit the current to a safe level, not to help the charging proces but to prevent damage if the battery goes faulty.

Brian.
 
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    qbone

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@srizbf; Myeah, when programming websites I usually "just test" and see if it works out. But atm. I only have this phone and one battery, so I would rather not just test.

I did find this - rather big - circuit which looks quite promising. Only problem is that it need 12Vdc.
Now a friend of mine did some testing and measured on his moped, same type, and it gives 6 to 12Vac depending on the revolutions.

Could this circuit be modified to accept something in that range?
https://www.shdesigns.org/pdf/lionchg3.pdf

The above circuit has tons of features, but Im gonna cut away everything I dont need, so that it charges 1 cell, 4.2Volts and 650mA or something like, that would be 0,75c.

---------- Post added at 15:59 ---------- Previous post was at 15:53 ----------

Nevermind my above post. I am gonna make it from MC34063A and limit it.. my problem was, using the design tool it said it couldn't work with 5,35Vdc but it might when I get a little more.

Just a quick question. If I supply the MC34063 circuit with less than what is needed, how will it react? Will it just not let anything out, or will it just output less power and current?
 

It will probably do it's best but not manage as much output voltage. Bear in mind that your AC will be rectified and filtered (you need a reservoir capacitor) so it will be somewhat higher than the AC you are measuring. Depending on load it may be as much as 1.414 times the AC RMS you are measuring so it's doubtful the input will drop below the minimum level. The MC34063 has an advantage that it's heat dissipation will remain fairly steady even if much higher voltages are fed into it. A linear regulator on the other hand will get progressively hotter as the input voltage increases and it will also need around 3V more input than the regulated output to work properly.

Brian.
 

Oh very nice, then there is no problem.
And gaaah yeah, stupid me! i keep forgetting that it is AC Vpp times sqrt(2) and not divided :)

Now I have no issues, it should all work just fine :)
Thank you very much for your help and inputs!
 

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