Resistor have many other application than limiting the current to the led, i.e. when there are multiple resistors it can also be for other reasons. The current limiting resistors will often be the largest ones.
Capacitors may look like resistors (They usual have another color and no letters/numbers on them), but have other functions.
When modding a driver to higher output power, it is often easier to solder a resistor “piggy back” on the current limiting resistor, than it is to replace the original resistor.
When it comes to modding a light, I am having trouble identifying a worthwhile mod, because it seems as if everything I might want already is available at the same cost or better than it would cost me. I still want to do some mods, because that has to be best way to learn how the lights work, once you have a reasonable framework for understanding. Most of these components are theoretical for me at this point. In a college lab, there was a poster: One experiment is worth one thousand theories.
I don’t think that’s true at all. I don’t have any charging curves in front of me to look at but I can say that my 1.5A li-ion charger is CONSIDERABLY faster at bringing cells from ~3v to 4.15/4.2v (measured with DMM) than the 0.5A max charger I had before.
I haven’t timed it but the roughly 3x faster it should be sounds about right.
It doesnt run the same, something will change. What changes depends on how the driver is designed. (The driver is the electronics that sit between the battery and the LED to control things.)
The 3.7v and 7.4v are supplied to the driver, the driver can control the current that it gives to the LED.
Voltage is not power, power is a combination of voltage and current (voltage multiplied by current to be precise, 4 amps at 3.7 volts has the same power as 2 amps at 7.4 volts. 4x3.7 = 2x7.4)
If the driver is designed to supply the same power to the LED regardless of the voltage it is being fed then the LED will act the same with 2 batteries as it does with 1, but 2 batteries will give twice the runtime at the same power as 1 battery. (ie 2 batteries last twice as long as 1).
If the driver is designed to give more power to the LED when the voltage it is being fed increases then the LED will be brighter with 2 batteries than with 1 for the same runtime.
In other words, with 2 batteries you would either get more runtime or more brightness than with 1 battery, and sometimes a bit of both.
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?
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.
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.
I believe you are mistaken. Current to a given LED is mainly function of the voltage (temperature has some effect but it’s less important). If you keep the current to the LED constant the voltage will remain constant as well. A regulated buck driver will keep the output current constant while varying the input current to adjust for the input voltage. In an ideal buck driver (disregarding losses) if the input voltage is doubled because of two cells, the input current will be halved thus getting about double the runtime.
Most linear drivers (which can’t use the extra voltage) will just overheat if you attempt to use two cells to drive a 3V LED.
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.
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.
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.
Because the buck driver only steps down the voltage once it passes a certain point. So for one cell it just takes the voltage, while for two it steps it down.
Batteries. Astrolux MF01 comes with a recommendation to use Samsung 30Q cells, rated for 20A.
I am wondering what to expect if I put Samsung 35E cells instead. They are rated at 8A. I am thinking the 35s will sag in voltage, making the light dimmer. I am also thinking they might consider the light that is trying to pull 20A almost a short circuit from the cells’ point of view and the cells might get really hot, vent, catch on fire, etc. Maybe not, maybe the cells run the light dim.
I don’t want to experiment to find out something bad.
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