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# [SOLVED]Charging a phone on a 6v battery

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#### larsthiesson

##### Newbie level 4
I have a bicycle with a 6 volt system. The system is powered by NiMH (4,5 Ah 6V(when charging the voltage goes up to 7,2-7,4v and I guess it goes down to 5vt)).The system has lights, horn etc. I want charge my smartphone (HTC One S). Which option should I go for?
1. Assume that the smartphone could be charged by 6v instead of the required 5v.
2. As above but put some electronic to protect the phone against spikes from the other parts (lights, horn etc.)
3. Make a DC-DC converter
4. Or some other solution?
As I am not an electronics expert I would appreciate direction about where to finde informations about how to make the solution(2 and 3).

Can you determine what volt level comes from the phone's normal charger? That will tell you something about what it likes.

An adjustable voltage regulator could do the job, but the incoming level must be 2 or 3 V higher than what you want coming out.

Maybe you only need to drop the 6-7 V down a bit. A diode in series will do the job. It will also prevent your phone from trying to power the bicycle accessories when their supply goes low.

A smoothing capacitor should take care of spikes, etc.

It's important to find out how fast your bicycle system can charge the phone. Ideally you will halt charging when done. You ought to discover whether the normal charger stops charging when the battery is full.

Does anyone know if a smartphone has a voltage regulater build in so the phone don't need exact 5,2 volt?

I ask because I have discoverede that a normal 12v usb charger will charge the phone with 6v when delivered 6v. I guess it needs the 7v or above to regulated the voltage. When below it apparently just lets the voltage "passes through".

The normal charger delivers 5,2 volt. I guess my battery will deliver 5-7,2 volt (7,2 volt when charging).

Homemade battery chargers are always a bit tricky. It is hard to determine exactly what volt level is safe for your batteries, and for how long.

If it were me, I would try installing a resistor in between the bicycle system and my smartphone. I would make the ohm value high enough so that current from the bicycle system never exceeds the safe charge rate for the phone. (This could be 50 to 100 mA, but don't quote me.)

Also I would make sure the bicycle system never gets less than the phone voltage. If it does, then a diode is needed.

hi larsthiesson
your phone is powered by a 3.7v battery and to charge the battery u have to give 5v at the terminals of the battery .for this u can make a 5volt vlotage regulator circuit using lm7805.LM7805 can give a constant 5v output and it can also provide a current upto 1ampere at its output ,which is sufficient to charge your battery.since you are charging your smart phone so you have to make your circuit protective and robust,for this you have to use a comparator circuit with LM7805 using OPamp. the function of this comparator circuit is as follows-
when u have less than 5 volts at the input of LM7805,the output of the Lm7805 will be LOW(0 volt),and when the input voltage at the LM7805 is greater than 5v the output of the LM7805 is constant is HIGH(5volts),
the circuit will also have low battery indicator and deep discharge protection.
If u are intrested in this circuit ,i can design the schematic and working on a paper and i will send u the photographs of that paper.

Hope this helps
--BawA--

anand146.bit

### anand146.bit

Points: 2
If you have 6V you can use one diode to lower voltage by 0,7V, and give 5,3V. I think that phone will work even on 6V, because charging controller IC is in phone, and have tolerance at least 1V. I will try first with one diode for 6V to get 5,3V, and if you have 7,5V use multiple diodes in serie, each diode for lowering 0,7V.

Diode can be 1N4007. In SMD can be very small.

For usage of 7805 voltage regulators you need at least 3V Vi-Vo difference, to get 5V on output. You can check LDO they requast lower Vi-Vo difference.

One of reason for factory 5V needs for phones chargers, is to keep charging IC power dissipation in ok ranges.

I'm curious, that voltages what you mention, 6V, 7V, thise voltages are constant, or voltage varies ?

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Better search solar case for phone, that can be more practical solution.

The voltage varies over the charge/decharge cyclus of the batteries. I think fully charge batteries are 7,2 volt and discharged batteries are 5 volt (5 NiMH cells). Thats why I think the diode solution is not ideal. From what I have read on the web the 7805 solution will not work because of the voltage drop is not suficient (as tpetar wrights).

The voltage varies over the charge/decharge cyclus of the batteries. I think fully charge batteries are 7,2 volt and discharged batteries are 5 volt (5 NiMH cells). Thats why I think the diode solution is not ideal.

I mean on power source voltage stability, not on battery voltage oscillations. If we have contstant voltage of power source, lets say 7V, you can use diodes to adjust appropriate voltage what you need. Because of that I ask that, and I mean on voltage of power source. Battery like battery will have voltage oscillations from full to empty.

Try solution with LDO voltage regulators, such as MAX603. This regulator have 320mV dropout, and can deliver 500mA of current. Also have selectable 3,3V or 5V.

https://datasheets.maximintegrated.com/en/ds/MAX603-MAX604.pdf

Try solution with LDO voltage regulators, such as MAX603. This regulator have 320mV dropout, and can deliver 500mA of current. Also have selectable 3,3V or 5V.
Will this also take care of any "spikes comming from turning on the lights - which the battery also i providing power to?

Will this also take care of any "spikes comming from turning on the lights - which the battery also i providing power to?

You should not have spikes.

Will this also take care of any "spikes comming from turning on the lights - which the battery also i providing power to?

There might be a very small time during which there will be a current surge due to turning on the lights. Of course, this isn't really a surge. It's just current increasing from a lower level to a higher level in a short time and then staying at the higher level. If this causes voltage of the battery to momentarily drop, you should tackle this by using a bulk capacitor.

Hope this helps.
Tahmid.

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