I have just purchased a 6V, 4.5Ah sealed lead-acid to power my arduino-controlled bicycle LED lighting system.

The battery is to be charged by the variable power provided by a hub generator, which has a rectifier/stabilizer (diode bridge + capacitor + zener).

Currently the hub-generator/stabilizer setup is being successfully used to power the LEDS directly (arduino is not in the circuit yet). At full load I have a consumption of around 0.45A @ 6V.

My goal, after incorporating the battery and the arduino, is that the system (arduino + LEDs) works "by itself" once turned on, in the sense that:

  • The system gets its power from the battery or the hub generator, whichever is "more active" at a given moment;
  • All the excess power provided by the generator and not consumed by the system will end up recharging the battery. The battery thus would be used in "floating" mode;

I believe this is a fairly standard configuration for this type of application.

My doubt is: The battery manufacturer recommends around 6.8 Volts for a constant-voltage (floating recharging regime between 10°C and 30°C), but my system is designed to work at 6V.

So it appears that, if I used a 6.8V zener diode (1N5342) to regulate hub voltage, I would successfully float-charge the battery, but could also be "over-volting" the system.

On the other hand, if I use a 6V zener (1N5340), as I am using now, I would not have enough voltage to charge the battery.

As a last confusing point, the open-circuit voltage of a fully charged battery of this model is around 6.5V, so even without being recharged, it is expected to "present" these 6.5V to the system, isn't it?

So, my questions are:

  1. Should I be worried about this extra 0.8V if my system is expected to run on 6V? Considering that it's made up mostly of LEDs, I suspect this is not a big problem, but I'm unsure;
  2. Suppose I absolutely needed to have strict 6V at all times, how would I float-charge the battery while the system is running without exposing the system to 6.8V? Is there something I could do, such as an extra circuit? Is there a standard way to do that?
  • 1
    \$\begingroup\$ A fully-charged lead-acid cell will be about 2.2 volts, so the 6.5 volts you see with your battery is to be expected as it is made of three cells in series. (The "12 volt" electrical system in a car can be up to 14.4 volts while charging. The 12 volt battery has six cells in series.) \$\endgroup\$ Commented May 19, 2018 at 22:20
  • \$\begingroup\$ it's hard to know if the extra 0.8V is a problem or not... some systems designed for 6V operation work fine at higher voltages, some don't. if it was designed for 6V lead acid batteries it will work fine, designers know that 6V lead acid means up-to 7.2V \$\endgroup\$ Commented May 20, 2018 at 1:07
  • \$\begingroup\$ This is a good question, but the answer totally depends on how the LED current is regulated, which you haven't told us. Are you using a resistor to soak up the excess voltage (which gives you constant current only if the excess voltage is constant) or are you using a feedback-regulated current sink? \$\endgroup\$
    – Ben Voigt
    Commented May 20, 2018 at 2:03
  • 1
    \$\begingroup\$ You do need some regulation in any case, because the default behavior of a bare LED is positive feedback. At any fixed voltage, the current through a semiconductor will be an increasing function of temperature -- as temperature increases, more current leads to more heating, and very soon you can have thermal runaway. A resistor makes the overall feedback negative -- and specialized current-controller ICs are even more tolerant of temperature and supply voltage changes. \$\endgroup\$
    – Ben Voigt
    Commented May 20, 2018 at 23:13
  • 1
    \$\begingroup\$ look for a switched mode LED driver instead of wasting your battery energy in resistors. \$\endgroup\$ Commented May 21, 2018 at 2:33

1 Answer 1


Your system needs to cope with a battery voltage from about 5.5V through to 7V - these figures will depend on the exact type of battery. Short of adding buck or boost converters, there isn't really any way around this.

An Arduino is probably overkill, but you need some sort of undervoltage cutout.

  • \$\begingroup\$ Thanks for your answer. The intended function or the Arduino in this system has to do with button and light logic (blinking modes and patterns, etc.). I don't necessarily plan to use it to control the battery charging or anything. But anyway, I think your first sentence synthesizes what I am starting to figure out: The right way to solve my problem is to design the system to allow for a range of operating voltages instead of expecting a tightly controlled one. \$\endgroup\$ Commented May 20, 2018 at 22:26

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