Electronic – Zener over voltage protection from DC hand crank generator

If you are using a 6volt battery then for most of the time you need a boost switching regulator except for those periods when the generator might be producing voltages at about 6volts. The problem here is that the generator is about the same voltage as the battery and this effectively means using a boost switching regulator generating maybe 8volts followed by a buck regulator that takes the 8volts and charges the battery thereon.

The alternative is to use a 12volt battery and avoid the buck regulator. In effect you just have a boost regulator and it will be about 90% efficient.

Ensuring the battery does not lose energy back into the charger when the generator is stopped is trivial so you have to decide on your battery voltage and your battery technology. This, might warrant a fresh question should you struggle to decide.

For any application, this very basic (and not much reliable) "dc-dc step down", could be done using a resistor between your 6V output and Vin. Considering:

Vout = 6V
Pout = 3W
Iout = Pout/Vout = 0.5A

for a total voltage drop of Vr = 1V, the resistor value should be:

R <= Vr/Iout = 2 Ohm

Higher value of R won't let you use the total 3W of your power source. If value of R is too low, your circuit will have high Ibias when in standby mode (arduino not connected). Consider the case where R-->0, and see that your 5V zenner will drain more current, and also make your 6V power source more unstable.

At the same time, you should take into consideration the resistor's dissipation power:

Pr = ((Vr)^2)/R = ((1V)^2)/(2ohm) =~ 500mW

where Vr = Vout-Vin.

Finally, calculate the zener's dissipation power.

edit: for a 500mW resistor, it would be safer to use a 1W resistor. This way it won't overheat, reducing it's variation and hence the output stability.