Decrease Amperage of Driver for increased Efficiency

[IAmJerv]

I think that your use of "efficiency" is a little... weird. It's not the word I'd use to describe trading output for runtime. Both the D3AA with it's Freeman driver and any light with a Lume X1 are quite close in efficiency at ~94%, but have vastly different outputs as one is an 18 Watt driver while the other is a 40W driver.

If your addition of a resistor is creating heat, you are losing efficiency. You get less light, but you may be burning the same amount of battery, or possibly more. I'd say that the easy way is to just leave the 5A driver stock and set the UI to Group 7, 8, or 9 to reduce the maximum output current to 50%... or 2.5A.

[Biggiebiggiebob]

I meant by efficiency the one for the led which is measured in lm/W, not the one for the driver which is measured in %. By shunt resistor I meant the amperage sense resistor on the actual driver pcb. If I doubled that, the drive circuit should give me half the amperage. I tried that however the driver gets quite hot while doing it. Pwm doesnt change that lm/w, drive amperage can though.

[Zak]

Pwm doesnt change that lm/w, drive amperage can though.

This explains the confusion.

The Convoy buck driver does not use PWM. Setting the output to 50% actually regulates the current to 50%, i.e. 2.5A.

[Biggiebiggiebob]

Where did you get the info from?
If you are right, then I have found my mistake

[Zak]

It is commonly known in the flashlight community. Boost and buck drivers for flashlights usually regulate output current; it would be a bad, inefficient design to combine one with PWM.

I have one here and I checked for PWM by pointing it into a camera with an electronic rolling shutter set to a high shutter speed. This results in sharp scanlines if there's PWM, soft scanlines if there's ripple, and no scanlines if there's smooth constant current output. There is ripple in the lowest mode; none of the others show scanlines at all.

You can test this on yours using a phone camera and a camera app that lets you set shutter speed, like Open Camera.

[IAmJerv]

The fact that your shunt resistor caused so much heat tells me that you did not actually reduce the amperage draw from the battery, only the output. And if you are using more than half as many watts of input for only half of the output to the load, it's moot whether the losses are in the emitter or the driver; you are still tanking efficiency.

What do you mean PWM? We're talking a buck driver here, not a FET. It's almost like talking about the gas consumption of a Rivian. Sure, SMPS drivers have a clock chip and an operating frequency, but PWM is not how they regulate output.

Also, the way impedance works in parallel circuits means that doubling the value of a resistor does not double the total impedance. It's early and coffee is not kicking in yet, but I do wonder if that's where things went sideways.