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A constant current device without a load will ramp up to a very high voltage and charge up the output capacitor.  If an led is connected to this open circuit voltage it will burn out.

This is not a problem for voltage regulating circuits.

bikeNomad
Flashaholic
Member # 686
  posted 02-06-2002 10:33 AM
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When you're developing driver circuits for the LS lamps, you don't want to blow up the precious LED when you make a mistake.
A 10 ohm 2W resistor isn't a good choice, because its resistance is constant regardless of voltage. This is different from the LS, which have a non-linear resistance curve.

A LS will show 0.1V change in forward voltage for a 100mA change in current, above (say) 100mA. The resistor only shows a 0.01V change for the same change in current.

What to do?

My solution: I built a dummy load out of 5 1N4004 power diodes (you could also use 1N4001, etc.) in series.

I have it set up so I can use either 4 or 5 of them to simulate the range of LS voltages.

The forward voltage and dynamic resistance is almost exactly that of a LS, and the total cost is less than $0.50.
 

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Posts: 121 | From: Stanwood WA | Registered: Sep 2001  |  IP: Logged

MrAl
Flashaholic
Member # 689
  posted 02-08-2002 01:23 AM
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Hi there,
I've found that 4 1N4001 diodes in series with the base of a FTZ651 transistor to be
so very close to the dynamic curve of the LS
from about 50ma to 500ma. The collector is
connected to the anode of the first diode
in the series string. The emitter goes to
ground. You can add Schottky's in series
with the 1N4001 diodes to simulate other
LS's.

This is the closest simulator i have found
so far. The diodes alone arent too bad
though.

--Al

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LED's vs Bulb's, the battle is on.
 

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Posts: 396 | From: New Jersey | Registered: Sep 2001  |  IP: Logged