Test Load Diagram's for all laser diodes!

Hey All

Made a one stop thread for all diode laser test loads.

Inform me if they are wrong and i will fix.

also the .900VDC is just an example voltage/current

As you know 1mV = 1mA







Cheers!

Fiddy said:

Hey All

Made a one stop thread for all diode laser test loads.

Inform me if they are wrong and i will fix.

also the .900VDC is just an example voltage/current

As you know 1mV = 1mA

Cheers!

Click to expand...

Why do all your test loads show a reading of .900VDC with
no load or power supply...:thinking:

There are tons of pictures of all different tests loads...
All you needed to do is post links to those...


Jerry

I also dont think your first diagram works for high power 445nm. You only want to use 4 diodes. For some reason the voltage drop of the diodes changes at higher currents. (I am not an electronic expert)

Heres a quote from drlava

drlava said:

you're over-volting the output of the flexdrive. at 1A those 6 diodes are 0.77V each and your 1 ohm load is 1V so that's 5.62V. Take 2 diodes out to get to 4V.

Click to expand...

I'm not entirely sure myself but while trying ot source the best diodes for a load, I kept on seeing ratings like " Vd of 1v @ 2A" or such, when in fact almost all diodes have a Vd of .5-.8V, at best I'm guessing its like a transistor at some point you start saturating the PN junction and since the diode has a limited current path it may just start clamping the voltage off as the current getts bigger, plus if you look at the LD's themselves you'll notice the have an almost identical current to Vd ratio as a regular diode with a .5V swing or flucuation.

Ideally you should use a voltage controlled electronic load combined with a current limiter... That way its not a constant guessing game of what diodes to use and how many diodes to use at certain currents.

Ideally all that matters is the vpeak at Ipeak.

So for a 445nm you have vpeak =5V with Ipeak = 1.8A... not sure about the rest. Anyone know?

I've never used test loads. Using a power resistor or just shorting the output with a multimeter have always worked for me. Sure, the IV curve isn't the same and it can put more/less stress on the driver, but it is only a test... to set the current. The driver will output the same current no matter the load (within reason), so I've never seen the point of providing it with such a precise load.

Cyparagon said:

I've never used test loads. Using a power resistor or just shorting the output with a multimeter have always worked for me. Sure, the IV curve isn't the same and it can put more/less stress on the driver, but it is only a test... to set the current. The driver will output the same current no matter the load (within reason), so I've never seen the point of providing it with such a precise load.

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within reason isn't exactly a fair statement for the drivers that are most commonly used here. Many boost drivers and or linear supplies are rated for a max current such as 1.5A ..ect but that doesn't mean they can maintain the Vmax at that current. Its critical that you ensure that both your voltage and your current match actual diode peak output. If I direct drive a 445nm with a 18650, i'll pump 3A+ through it but the vdrop will only be 4.2-3.7V as opposed to 5V. This applies to all drivers.

To my knowledge a diode isn't going to magically start limiting the current simply because the powersupply does not have the ability to reach a max vdrop. If this were the case there would be no danger in direct driving 445nm because the output is only 4.2V and would never reach 5V. I would assume the same thing applies with test loads.

Again, its best to create a electronic load that is both voltage and current dependent.

random person said:

Many boost drivers and or linear supplies are rated for a max current such as 1.5A ..ect but that doesn't mean they can maintain the Vmax at that current.

Click to expand...

The limitations of common drivers are well known. The vast majority of users (as far as I know) only use test loads to set current. Maybe if it was an experimental driver, I could see the use - in which case I'd just use a dead diode instead of a test load.

random person said:

If I direct drive a 445nm with a 18650, i'll pump 3A+ through it but the vdrop will only be 4.2-3.7V as opposed to 5V.

Click to expand...

How do you figure? For that to be true, you'd need to find a diode that has a Vf of 4.2V at 3A.

random person said:

If this were the case there would be no danger in direct driving 445nm because the output is only 4.2V and would never reach 5V.

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There isn't any danger of powering them directly off 4.2V. That isn't enough voltage to push a lethal current through them according to the PIV graphs I've seen.

The main problem with direct driving (in this case) is the current falls quickly as the battery discharges. You'd only get to use 10-20% of battery capacity before the diode doesn't get enough current to lase.

The main problem with direct driving (in this case) is the current falls quickly as the battery discharges. You'd only get to use 10-20% of battery capacity before the diode doesn't get enough current to lase.

Click to expand...

Well... that would make alot more sense if that was the case. Hmm...my simulation software has been throwing me off and I'm still new to laser diodes so are probably correct. I was under the impression that direct driving would kill it but that idea never made sense to me. However, something isn't adding up still. Direct driving is the most efficient way to power anything. So, unless your pumping excess current, your going to burn your battery out far slower than if you were using a boost drive set to 4.2V.

Do you mean that the Vmax of the battery drops quickly and consequently the current output is decreased? I guess that would make alot more sense.

Yup. Match up a 445 IV graph with a Li-ion discharge graph. You might get 100% "driver efficiency", but also constantly declining light output and a run-time of only a few minutes.

With other direct-drive cases like the infamous kipkay video, the batteries did have the voltage required to possibly ruin the diode.