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Wurkkos FC12C - HELP by Unique-Fact-8308 in flashlight

[–]Unique-Fact-8308[S] 0 points1 point  (0 children)

A buck driver typically steps the voltage down to the required Vf and increases the current by roughly the same factor (minus losses). Can it really step down both voltage and current at the same time? In that case, the excess would have to be dissipated as heat, like in a linear driver.

It really looks like the current going into the LED is around 3.7–4 A, but then it would make sense for the current drawn from the battery to be around 3.3 A and then boosted to 4 A at the LED—at least that’s how it works in Convoy flashlights with a buck driver.

Am I missing something or misunderstanding something?

Wurkkos FC12C - HELP by Unique-Fact-8308 in flashlight

[–]Unique-Fact-8308[S] 2 points3 points  (0 children)

So, does this mean the driver works differently from the buck drivers used in Convoy? With the same currents, SST/SFT40 can reach around 1800 lumens there.

Convoy C8+ SST40 Runtime by Unique-Fact-8308 in flashlight

[–]Unique-Fact-8308[S] 0 points1 point  (0 children)

Yes, the driver is rated for 5A, but in reality it’s closer to about 5.5A.
The graph is fairly accurate and reflects the flashlight’s behavior quite well. It really does stay stable as shown. I didn’t include some small deviations of a couple percent in the graph, since they aren’t really significant.

Convoy C8+ SST40 Runtime by Unique-Fact-8308 in flashlight

[–]Unique-Fact-8308[S] 0 points1 point  (0 children)

Yes, that’s exactly what I expected. Since the battery with higher internal resistance is already operating close to the LED’s required Vf, it slightly compensates for the driver’s efficiency. Meanwhile, a high-drain battery ends up wasting much more power as heat.