Dang, this is going to make things a pain in the rear
I definitely can't get away with using a flexdrive using an 18650, right? Too much current for the flex to handle...
Edit:
I was looking at the flexdrive manual and looking at the formula.
(Vout (Volts) * Iout (Amps) ) / Vin (Volts) < 1.2
The only part I don't understand is the end of the formula, < 1.2
I plugged the settings I'd use for this diode into the formula, and came up with this.
(4.2) * (.750) / (4.2) =.75
So, does this mean it would work with a 4.2V Li-Ion since my answe is .74, which is less than 1.2? :thinking:
@Dave, now it makes perfect sense, I totally forgot about the shortcuts lol. Also, thanks Mario & Jay.
Double Edit:
If this is the case, then we can definitely power this thing with a decent capacity Li-Ion such as the 18650. Once it drops down to about 3 volts, that's when problems will arise I guess? What would happen? Would it just cut power to the diode? Would it kill the diode? Would the driver just generate an insane amount of heat?