From: Tony Williams
Subject: Re: current source for array of infrared leds
Date: Thu, 05 Dec 2002 00:02:32 +0000 (GMT)
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NNTP-Posting-Date: Thu, 5 Dec 2002 00:06:49 +0000 (UTC)
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In article ,
Harry Dellamano wrote:
> "Tony Williams" wrote
> > Light output Tempco= -0.55 %/K.
> > Vforward Tempco= -1.5 mV/K.
> > Theta(junct-air)= 500 K/W.
> > Radiant Flux versus Ifwd, mW/mA pairs off a lin-lin graph.
> > 0.25/12.5, 0.4/25, 0.7/37.5, 0.9/50. (very approx.)
> > Vfwd (Vf/IfmA pairs, picked off a tiny log-lin graph).
> > 1.0/1, 1.1/19, 1.2/28, 1.25/60.
> Taking a quick whack at your numbers.......
Sorry Harry.... rest of your calc snipped, because I
misread that tiny Vf/If log-lin graph. That 1.1/19mA
should have been 1.1/9mA. It does look better with
that 9mA..... a log plus low value resistive portion.
Apologies about that.
> I then calculate the current vs light power to be about
> 50mA/mW. We need a 6.05% increase in light so we need to
> increase the current by: 50mA*.065= 3.25mA.
I came at it from a different direction. The radiant
flux is 0.4mW at 25mA, so it will be about 0.32mW at 20mA
and the warmup will drop that 0.32mW by 6.05%, to 0.3mW.
We are now 0.02mW short. 0.4mW/25mA is about 0.016mW/mA,
so we need a 1.25mA increase in current.
> Frank does not have an array, just a string, so sorting (binning) is
> not necessary. Just crank up the voltage until the current is 20mA,
> vary the temp and measure light output. He could try different values
> of external resistance but until those negative puppies on back order
> come in 0.0 ohms may be the best value. Hope he shares his data.
> Then again, I don't know what those diodes are. When I did my unit
> the thresholds were about 2.15V and I forgot what the internal
> resistance was, but in the 5.0 to 12.0 ohm range.
> Bottom line, Know thy I/C curve.
That -6% came from Siemen's figure of 500K/W for
Theta(j-a). My HP data sheets don't give (j-a)
they give Theta(j-c), where the "c" is the cathode
lead, and with a typical figure of 90 C/W. If the
leads of the LED are short and go to a large area
of copper on the pcb then this should reduce that
-6% warmup drift.