Agony of Confusion About Batteries and Current

[danallen]

@ggf31416

The buck drivers have a small ‘transformer’ that reduces the voltage from whatever the input is to the 3.x V needed by the led. As for how it works, an electrical engineering background is needed to really understand them and I don’t have such background.

Edit: Read Switched-mode power supply - Wikipedia and Buck converter - Wikipedia . Don’t expect that it will be easy to understand.

Thank you, that article is exactly what I needed. For the flashlight to produce the same output across a range of input voltages that varies from one cell to two cells in series, it uses a buck driver that varies the duty cycle with the input voltage, and has enough inductance or capacitance to handle an input voltage twice as high as needed to run the LED. When only one cell is used and its voltage is low enough, the switch controlling the duty cycle can stay closed, because bucking is not needed at that level. I don’t know the details of how to make the switching vary the duty cycle based on input voltage, but I sure do want to know.

@BlueSwordM

It’s using a buck driver, a driver that steps down the high DC voltage to a usable lower DC voltage.

Let’s start with an example: with a 2S battery pack, you get a maximum battery voltage of 8,4V.

However, in this case, you have a 3V LED driven at 3A. You would burn it in direct drive since the delta in voltage is massive!

So you use a buck driver to step down the voltage. As the battery voltage goes down, to sustain the same power level, the current goes up.

The part of this I don’t understand is how the driver can handle the voltage from two batteries in series and also handle running only one battery.

@LouieAtienza

@danallen

With this light, the runtime doubles with 2 cells vs. 1 cell, led power appears the same, might be slightly different if precise measurements are taken.

How does a driver output the same power to the led over such a wide range of input voltages? That is the part I don’t understand.

The driver (in your case) does not output power to the LED, it supplies a_ constant current_ to the LED, and the voltage will vary with a specific range to maintain the current. Power is the energy used by the LED (over time), and the wattage (power) rating of an LED does not necessarily determine its brightness. This is because certain LEDs are more or less efficient than others.

When running two (identical) cells in series the voltage doubles but the capacity rating stays the same. When running two cells in parallel the voltage remains the same but the capacity doubles. Running two cells in your light, in your case, would not seem to make sense, since the driver does not seem to take advantage of the extra voltage.

The advantage of two cells is a result of less current out of the batteries when the voltage is higher. The two cells in series make the voltage into the driver higher than when there is only one battery. The output voltage and current both are the same whether there are two batteries or one. Lower current from the batteries means longer run time. To make the work, the driver has to use transistors to route the electricity into inductors or capacitors and control the current from the batteries. It is the combination of power switching and energy storage of the inductor or capacitor that makes this work efficiently. I want to find out the details of how these components perform.