From: "Harry Dellamano"
Subject: Re: current source for array of infrared leds
Date: Wed, 4 Dec 2002 14:10:51 -0800
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"Tony Williams" wrote in message
> In article ,
> Harry Dellamano wrote:
> [lots of snippage Harry, to the bits I can remark on.]
> > Frank did not tell us how many in his array.
> I think he has 7x IR LEDs in series.
> Flipping through a few Siemens data sheets, their
> GaAS IREDs have the following typical characteristics;
> Light output Tempco= -0.55 %/K.
> Vforward Tempco= -1.5 mV/K.
> Theta(junct-air)= 500 K/W.
> Vfwd (Vf/IfmA pairs, picked off a tiny log-lin graph).
> 1.0/1, 1.1/19, 1.2/28, 1.25/60.
> 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.)
> [snip again]
> > Bottom line, know your LED's I/V curve over it's temp range
> > and a little math will result in a good nights sleep.
> Frank wants a constant light output, which is probably
> going to be obtained from some optimum voltage value
> and optimum series-R. He is operating at around 20mA.
> So how do I thrash around some sums here.... like this?
> Vf/20mA= about 1.11V, so P(20mA) = 22.2mW.
> At 22.2mW and 500K/W, Temp rise= 11 degrees C.
> Warmup change in light output= -6.05%.
> From the radiant flux versus If, at 20mA, that -6.5%
> can be recovered by increasing the current by about 1.1mA.
> For an 11C temp rise, Vf falls by 16.5mV, and for const
> light output that should cause an increase in If of 1.1mA.
> ie Rsource= 16.5/1.1 = 15 ohms.
> So the final circuit for 1 LED at 20mA is a voltage
> source of 1.41V, and a 15 ohm series-R.
> For 7 LEDs it would be 8.9V and 105 ohms.
> Does that look a reasonable way to do it, do the final
> results look reasonable?
> Tony Williams.
Taking a quick whack at your numbers and saying that the original temp was
25C I then get a model of a 1.0V threshold, 8.0 ohm internal resistance and
1.160V supply to create 20mA current flow at 25C. We then get self heating
to 36C causing the diode's threshold to drop to 0.9835V and the current to
increase to 22mA. 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. We got a 2mA increase but 3.25mA is needed so
put in a negative resistor. Digi-key #P-125RW2R2-ND.
An 11C change is not very much and with ambient temp at 45C and self
heating we could be at 70C. Do it for the end point temps then measure how
many bumps you have in the middle.
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.