Need help with power supply for 500mw 635 C-mount

[Jmillerdoc]

Hi,
I recently bought a new C-mount 635nm 500mw diode, new, from another forum member. I have not yet worked with a c-mount and the power supply needs to be 2.8v at 1.6a according to the package insert for the max 1000mw output it claims to produce. I bought one of those c-mount lens modules on eBay and hav it all mounted and ready to fire up.

Question...what's the best way to test my power supply with a dummy load....I know with the 445's you use three diodes in series to mimick the load. For the 635 that's rated at 2.8v would you use two? What's the best way for me to do this? I have a great dummy load with a precision 1R resistor that has been very accurate for my 445's.

I want to build a DDL driver for this project so I was thinking of using two 18650's in series with a nominal 7.4v (6.1v after the driver drop). Again, is this too much voltage or will the diode handle that much voltage across it?

I do have some Micro Flex drives....was thinking of possibly just setting one to the max 1.5a output and use a single 18650...

Any help here will be appreciated....

Thanks,
Jeff

 

[Sigurthr]

Now, I've never built a dummy load, but if I understand it's construction correctly than this might work. I advise you to confirm it with someone experienced though: Two forward biased 0.7V Vf silicon diodes in series will yield a 1.4V Vf, add that to the 1.6V voltage drop across a 1Ohm resistor and you'll have 3V, which is pretty close. You could do one 0.7V Vf silicon diode and one 0.3V Vf Schottky diode for 2.6V Vf too.

Two 18650s in series will likely kill it with a LDO linear. Keep in mind 18650s are 4.2V not 3.7V when fully charged. You could do two 18650s in PARALLEL using that 1.3V drop LDO - that would yield 2.9V, but as soon as the cells voltage dropped below 4V you'd lose output power. A Buck driver would probably work best. I think RHD just developed a new one, should ask him.

 

[Fiddy]

In theory, Vdrop of silicon diode = 0.7V (or 700mV)

2800mV / 700mV = 4 diodes

At 1.6A temperature will affect your test load diodes, which will drop like 1V instead.

You can quiet easily use 1 flexdrive, it wont hit 1.5A off 4.2V tho, they usually make 1.2 ~1.3A on 4.2V input. You could use 2x flexdrive at 800mA each to hit 1600mA tho!

You could use 4x AMC7135 chips to hit 1400mA, each additional AMC chips adds 350mA.

Fiddy.

 

[Sigurthr]

In theory, Vdrop of silicon diode = 0.7V (or 700mV)

2800mV / 700mV = 4 diodes
Fiddy.

I was thinking the same thing, but then I remembered there will be a voltage drop across the 1 ohm resistor in addition to the forward voltages of the silicon diodes. I could be wrong of course, but it just seems to me that if I was; then using 3 silicon diodes for a 445/450nm diode's dummy load would be insufficient.

 

[Fiddy]

I was thinking the same thing, but then I remembered there will be a voltage drop across the 1 ohm resistor in addition to the forward voltages of the silicon diodes. I could be wrong of course, but it just seems to me that if I was; then using 3 silicon diodes for a 445/450nm diode's dummy load would be insufficient.

You measure the voltage drop across the 1 ohm resistor...

why would you use 3 diodes for a 445 or 405nm?

 

[Jmillerdoc]

I have always used three diodes for my dummy loads on a 445, learned that here on the forum....

I happen to have two flex drives...if I were to set them for 800 ma each what type of dummy load should I use to setitfor the 635?

 

[Fiddy]

Umm ok, i use 5 for 445 and 6 for 405.. use 4 diodes for 635

 

[Jmillerdoc]

Strange...I tried five at first but kept getting the wrong readings on a known driver output...meaning I had a DDL with a 0.68r set resistor and when I tested the voltage across the resistor I kept getting around 1a with five diodes...someone here told me to use three...when I did I got the 1.8a I expected...I have used three ever since and set all my DDLs withit for the 445's I have built. The optical output also falls in line with what I would expect from the current measurements I get.....so what's up with that? I hope someone here can clarify, verify what I should be doing and furthermore, what I have done....

 

[Sigurthr]

As I said, I've never worked with these dummy loads. I wasn't disagreeing with you at all, simply stating my thought process and why I gave the answer I gave (and why i gave the warning that it may be incorrect).

Placing a resistor in series with a load creates a voltage drop across the resistor based on the current. You measure it to determine the current because at 1 ohm the current is equal to the voltage drop, but a voltage drop is there as a fundamental aspect of series resistance. Should not it be taken in to account as well? Just as a diode's Vf increases as current increases the voltage drop across the 1 ohm resistor increases with the increase in current.

As for the three diodes part:

....I know with the 445's you use three diodes in series to mimick the load.

I'm not sure how many are actually used on proper dummy loads. I would guess that you would need six diodes for a 405nm build (6x 0.7 = 4.2V, the addtional drop provided by the 1R resistor based on the current setting).

 

[Fiddy]

what are your diodes rated for?

 

[Bionic-Badger]

Some of the higher current 1N540x rectifier diodes (which I bought specifically for diodes like the 445nm laser diodes) drop 1.2V instead of 0.7V, so using 3x 1.2V = 3.6V in series with a resistor at > 1.0A currents might be appropriate.

Don't just take the number of devices at face value. Make sure they make sense!