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ArcticMyst Security by Avery

Driver test load confusion`

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Jan 4, 2011
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You said that you need more diodes in a series to make them behave like a laser diode. If I were to use higher current diodes, (say 3A) and a stronger resistor (5 or 10W) would you still want to use 6 diodes to test the current through a 445nm diode?
 
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rhd

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You said that you need more diodes in a series to make them behave like a laser diode. If I were to use higher current diodes, (say 3A) and a stronger resistor (5 or 10W) would you still want to use 6 diodes to test the current through a 445nm diode?

6 is too much. at 1.8A, that'll be like 7V.

If you're just trying to measure current, and can do it fast enough, just stick with the resistors you have on hand. If you've got 1W 1Ohm resistors, stick two of them in series, and then stick that in parallel with another two that are in series. Effectively you'll have a 1 ohm 2W resistor (EDIT: 2W right? not 4?). Then put your test probes on that resistor array, and your voltage = current (ohm's law).

Super simple, no need to buy anything you don't have on hand. As has been mentioned, with a linear driver, you don't really need the diodes (unless there is concern that you may be under-supplying the driver voltage-wise, and want to verify - in which case, 4 diodes should be fine)
 
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rhd

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So a 445nm diode has a resistance of ~2W is what you're saying?

Nope.

Just saying that if you 2S2P a set of 4 resistors that are 1W 1Ohm, you'll end up with something equivilent to a 1Ohm 2W resistor.

My slight hesitation is the result of not being 100% on that 2W figure - it might be 4W.

But it has nothing to do with the specs of a 445.
 
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Ohh ok. So I guess I'm being over cautious but you're saying that a 2W 1ohm resistor is a good dummy load for a 445nm diode on a switch driver.
 
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rhd

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Ohh ok. So I guess I'm being over cautious but you're saying that a 2W 10 ohm resistor is a good dummy load for a 445nm diode on a switch driver.

No. A 10 Ohm resistor throws off your ease of current calculation.
 

rhd

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That was a typo. =P
I just edited it to say 1ohm.
So 2W 1ohm is a good dummy load?*

Just to be really clear -

- IF you're using a linear driver
- IF you're really fast in taking your measurement
- THEN you should be ok squeezing by with just a 2W 1Ohm resistor

BUT

- You SHOULD be using a higher wattage resistor if you want to run the test load for more than just a short few-second test
- If you don't use diodes in the test load, your LM317 (or whatever linear regulator IC you're using) is going to get MUCH hotter, MUCH faster because it will be dropping almost all of the input voltage
 
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Yeah, Its not a linear driver, and this is the first time that I will ever not be using a linear driver. (Which is why I have so many questions.)
 
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rhd

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Yeah, Its not a linear driver, and this is the first time that I will ever not be using a linear driver. (Which is why I have so many questions.)

This thread was about a linear driver.

If you're not using a linear driver, than the previous is not helpful to you (would have been nice to mention that your driver was not of the type being discussed in the OP!)

If you're not going linear, then you need a proper testload.
 
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My bad. I did say that it was a switch driver earlier but it still wasn't very clear. So back to the test driver that Im trying to build; I have 3A diodes and 10W 1ohm resistors. Im mainly trying to figure out how to know how many diodes to use to replicate a 445nm diode, and if that is the incorrect way to do it, then how I would go about making a realistic dummy load. I watched bobhaha's video but unfortunately it builds the load with different components and for current up to 1A, and didn't catch any explanation of calculating how many diodes to use for different laser diodes. Thanks for all of your help so far rhd!
 
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The only part of the dummy load that I don't understand is how I'm supposed to know the number of rectifier diodes I'm supposed to use.

Or, I think, what the operating voltage of the 445nm diodes.
 
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It looks like the operating voltage is 3.7-4.5 DCV, so I don't understand why I would need any diodes on a test load since my battery only supplies 3.7 volts.
 




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