I have reached a point in a project where I am stumped and need some guidance from some of the more seasoned experts out there. I'm not a 100% newbie as I've been playing with Arduinos for a couple of years now, but I'm not an expert either so please go easy on me! Project diagram and datasheets attached.
I'm working on a high power LED flashlight project. The Flashlight is controlled by an Arduino Pro Mini 3.3 v (for dimming, strobe settings, etc.). I'm using a 3400 MA 18650 3.7v LI-ION battery to power the project.
I have run into a couple of problems.
Problem 1: I have set the RSET using external resistors R3 and R4 to a total of 549 ohms. According to the CAT4101 datasheet, that should set the current limiter to constantly sink current right around 1 amp. When I turn the flashlight on and energize the PWM pin on the CAT4101 for full brightness, the LED does come on, but my ammeter shows a variation of current between 187 MA and 350 MA. It never gets up to an amp. The LED is super bright, so part of me wonders if my ammeter is broken or if there is something obvious going on.
Problem 2: When I leave the LED on full brightness (100% duty cycle), the LED slowly gains in current. It usually starts around 250 MA. Then slowly climbs to around 400MA. Once this happens the LED starts to cook. This lead me down a long learning path and I finally realized the importance of thermal design! I had no heat sink on the LED. Still, is the growing current slowly as the thing burns up normal? I am in the process of getting a heat sink solution but not sure if there is something else I should be concerned about. (I have other LEDs as replacement as I'm sure I damaged the original one!)
The LED has a Vf of 3.5volt typical, and 3.9volt max.
The LED driver has a dropout voltage of 500mV@1A.
You need a minimum battery voltage of 3.9+0.5 = 4.4volt.
That's more than a single LiPo battery can provide.
Vf (working voltage) of a LED drops when the chip gets hotter, therefore you see a current increase.
It won't reach 1Amp, because battery voltage is too low.
A ~3.5volt/1Amp LED generates ~3.5watt of heat.
Temp must kept under control with a heatsink.
They are usually mounted on an aluminium star base, but that alone might not be enough.
Don't see any image.
I think it's only visible if you have a Google account.
Leo..
Thank you so much Leo! I had a suspicion about that "dropout voltage" rating but could not quite bring myself to understand what it meant.
I still can't quite visualize what dropout voltage is. Is it happening on the LED due to limitations in the driver? Or is it happening in the driver itself? I do see what you mean by the math you spelled out in terms of how to get parts that work better together so I am finally unblocked in terms of moving on to plan B.
To see my circuit you can view the attachment on my original post labeled "screenshot". That's the same image as the google one. I wasn't sure if that would not show up or not. Sorry about that!
Is there a much easier way to accomplish what I am trying to do? Maybe without an LED driver? Since the source voltage is so close to the Vf on the LED can I just use a resistor and run it through a MOSFET or something? Just trying to think of other ways to make this work.
morrisdirector:
I still can't quite visualize what dropout voltage is. Is it happening on the LED due to limitations in the driver? Or is it happening in the driver itself?
That chip is a "dumb" linear transistor "switch" with fixed current limiting that you can turn on/off with PWM.
The "switch" itself is not ideal, and drops (looses) 0.5volt@1Amp minimum.
Not sure how to fix the problem. I think flashlights are using switching regulators to overcome those problems.
Try "flashlight board" on ebay.
Leo..
The CREE XP-L has a Vf of 2.95 typical and 3.2v maximum rating. Datasheet says that Vf should be right around 2.95 @ 1 amp.
So if the LED driver's dropout is 500mv at 1A, that means 2.95v + .5v = 3.45v typical. Or 3.2v + .5v = 3.7 max. Am I doing my math right?
Lastly, if I had more than one of these drivers running in parallel (driver datasheet says this is common application) to boost the current driven to two LEDs instead of one (2 amps), does that mean that voltage dropout is now 1 volt for both LEDs? Or does each LED/Driver combination follow the same calculations we've been discussing separately? I guess what I am asking is does voltage dropout effect the available voltage in everything in the entire circuit or just the LED that is being synced?
No. Most power LEDs are around 3.3volt, except for the red ones (~2.4volt).
That LED might have been selected for a slightly lower Vf, but I don't think it will make a big difference.
morrisdirector:
So if the LED driver's dropout is 500mv at 1A, that means 2.95v + .5v = 3.45v typical. Or 3.2v + .5v = 3.7 max. Am I doing my math right?
You forgot the wiring and internal resistance of the battery. That could drop another 0.2volt.
And at 3.7volt the battery is still almost half full.
Yes, two in parallel with half the current each might help a bit.
Not sure if it will be enough.
Leo..
