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Old 08-13-2008, 07:48 AM #1
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Default Build my own test load

I was told to use 4 diode in series and a resistor.

(Looking at rkcstr's test load, I see his load use 3 diodes and 1 resistor, and 1 resistor for discharge.)

So that lead leads me to wonder which diode (not laser) to use, and should I use 3 or 4 diodes? Resistor I'll use a 1 Ohm to keep it easy..

Thanks


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Old 08-13-2008, 08:07 AM #2
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Default Re: Build my own test load

Basically what you're doing with the diodes is mimicking the voltage drop the laser diode would cause (about 2.5V for red, 4.5V for blu-ray), and then measuring the voltage across the 1-ohm resistor in series with those diodes to determine the amperage (get a power resistor so it doesn't burn up). *So just find out what the forward voltage drop of your diodes are (usually about 0.7V each), and determine how many of them you need to be roughly the same as the laser diode's operating voltage. *

So if you want to make a dummy (test) load for a red laser, you'd put (2.5 / 0.7 = ~3.5) = 3 diodes in series with your 1 ohm resistor. *It's not an exact number and that's okay, because you'll still be able to measure that current on the resistor.
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Old 08-14-2008, 02:47 AM #3
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Default Re: Build my own test load

I chose 3 diodes in series because of the forward voltage of the specific diodes I use across the current range I wanted, which came about to about 2.3 to 2.9V over my standard red current range of 150-430mA. *Its not ideally matching a red DVD burner diode, but good enough. *A 4th diode would have made the peak current around 3.7V, way higher than a regular red diode.

The standard 1N00x series diodes typically drop 0.7V each (but will vary with current), so most use 4 in series for about 2.5 to 3.0V, not to mention the voltage dropped by any resistor you have in series as well.

But, if you're just measuring current output of a constant current driver, the voltage of the diodes won't matter since current will be the same regardless.
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