Reducing Voltage from 5V Power Bank to 3.7V for Bicycle Lamp: DIY Cable Connection

Hello, I have a bicycle lamp powered by a 3.7V battery.


I would like to be able to power this lamp from the power bank. Just how to change the voltage from 5V to 3.7V? I would like to make a cable for connecting a powerbank - a flashlight that will also reduce the voltage.

Insert two IN4007 diodes in series in the forward direction

Hello,

pawel16150 wrote:

Insert two IN4007 diodes in series in the forward direction

probably only one diode is enough.
My guess is that the "3.7V battery" is simply a lithium-ion cell that is marked that way, that is 3.7V is the "standard" voltage of such a cell. A fully charged lithium-ion cell obtains a voltage of 4.2 V, this voltage must rather withstand this lamp, unless it is written in the instructions "under no circumstances should the battery be charged to 100%". Thus, 5 V - 4.2 V = 0.8 V, i.e. more or less one "normal" diode or two Schottky diodes connected in series.
Voltage breaking with a diode has a disadvantage, however, some power is released on the diode, which is lost, i.e. converted into heat. If you want the smallest possible power loss, look for an impulse lowering converter, with an output voltage of 4 V, which can be supplied with voltage from 5 V, having the highest possible efficiency at the current consumed by your lamp.

best regards

The lowest energy loss will be if it discharges the voltage directly from the powerbank's cells (I assume that the cells are connected in parallel and you need to interfere with the powebank's housing) If the lamp works with a 3.7V battery it should have protection against excessive discharge of this type of cells. Of course, a suitable fuse should also be used.