Measuring peakcurrent

[koala]

Hello cpfers,

As usual, programming my atmel microcontroller do pwm then I thought of something regarding the peak current running through the Luxeon. Sorry, I don't have a scope /ubbthreads/images/graemlins/frown.gif.

Consider the highly sophisticated circuit above /ubbthreads/images/graemlins/naughty.gif put it in words, that's direct drive(do not implement!).
The white Luxeon in this circuit has a vf of 3.6v, direct driven with 4.2v.

There's no current limiting resistor so let the resistance of the circuit be 0.1ohm. The current going through the Luxeon when the switch is toggled will be...

0.1ohm = (4.2-3.6) / ?Amp /ubbthreads/images/graemlins/confused.gif

6Amps? Direct driven? I might be wrong here...
So, back to the question. If I got really fast fingers /ubbthreads/images/graemlins/ooo.gif that can toggle the switch on and off(PWM) at 1khz will the peak current be 6Amps?

Vince.

 

[MrAl]

Hi again koala,

In a word, "Yes" /ubbthreads/images/graemlins/smile.gif

If you wanted to take a look at the heating in the LED
it would be a little different, but i think for the
purpose of understanding this mechanism you dont have to
worry about that.
This is basically why LED's cant be driven from a voltage
source without a series resistance or other current limiting.
For a 1 watt device they spec 500ma max, so you see how
constraining this can be. Some people just about ignore
this spec, but i dont.

Take care,
Al

 

[NaCl]

You need to put an inductor in series, clamp Diode in front of inductor and Capacitor acress LED.

If your inductance are suitable value and Controller fast enough, you can measure current rising through inductor as linear ramp that start from zero ampare until it constant at 6 A (then Boommm if you not switch it off).


Controller may check for switch current and turn off when current go up to level you programmed. You can turn it off for fixed interval and turn on again or measure Inductor current look for power stored in Inductor go down to some level and then switch on again.


It' s easy to implement you can try to do this.

But better result if you havs a Scope.

 

[koala]

That sounds pretty bad, considering the spec for a Luxeon III peak current is 1000ma. Is there a study on long term out of spec pulsing on Luxeons? Would be very interesting.

If my lithium ion supplies an average of 3.9volts then the peak current will be around 3000ma provided that there is really 0.1ohm resistance in the circuit and the battery can manage to supply such current. That's still pretty out of spec.

How do I solve this without much toll on the efficiency? A small resistor say 1ohm in series, that will consume around 0.09watts(300ma x 0.3v)? Or a capacitor parallel to the Luxeon to suck up the "ripple"? /ubbthreads/images/graemlins/confused.gif

NaCl - sounds complicated to me but I'll try it. How many uH inductor should I use? Does it matter? What about the capacitor value?

EDIT: HAH I totally forgot about the LDO circuit floating aroudn lately!

vince.

 

[NaCl]

 

[koala]

Woohooo /ubbthreads/images/graemlins/buttrock.gif /ubbthreads/images/graemlins/buttrock.gif rock: thanks man! I will try it tommorow afternoon.

 

[evan9162]

BTW, the current through a Luxeon I to cause a 4.2V Vf is really around 2A. The current for 3.9V is around 1.2A. I have tested this experimentally. It will, of course, vary on an individual basis, but this is the ballpark range to be looking in.

A li-ion cell's voltage will drop dramatically under that load. I doubt you'll get 2A through the luxeon, 1.2A is a more likely starting point (depending on the Li-ion cell type/size)

 

[koala]

evan9162 are you talking about peak or avg current?

 

[evan9162]

Um, it doesn't matter. Since your battery will never exceed 4.2V, then you will never exceed ~2A of current in direct drive.