Testing DDL Circuit with multimeter

... but the KES-400 can emit 2 wavelenght at the same time, which give you magenta !  

See these great post   :
http://www.laserpointerforums.com/forums/YaBB.pl?num=1213303871

http://www.laserpointerforums.com/forums/YaBB.pl?num=1215859801

and Milos's analyse :
http://www.laserpointerforums.com/forums/YaBB.pl?num=1209527536/0#0

check this out:
http://www.digikey.com/scripts/US/DKSUS.dll?Detail?name=425-2697-ND

I have another Q ;D
what do I do when I want to set permanent current levels?
I should remove the POT and use just resistors correct?
But what if its hard to match the needed resistance? In that situation would I have to use a POT?
For example, Im interested in making the magenta laser out of a KES diode and 2 DDL drivers combined on one board to run the 650nm and 405nm laser at the same time.
I already tried running them in parallel with 1 DDL driver, but just the 650nm diode lit up.

So Im expecting to get it working with 2 DDL drivers in Parallel.
My plan is to make a double DDL driver and find the best current levels, then make new double DDL drivers with the current levels preset.
What im wondering is, if the power level requirements for magenta are like im guestimating... ill need about 10-15mw for the 650nm and about 30-35mw for the 405nm.

so for:
10mw = 125 ohms
15mw = 83.3~ohms
20mw = 62.5 ohms
25mw = 50 ohms
30mw = 41.6~ohms
35mw = 31.71 ohms

How do I match the needed resistance? Seems I would need either ALOT of resistors to get the needed ohms... Or a POT.
I would much rather use resistors to reduce the workload.
Ever run into this problem before?

viroy said:

How do I match the needed resistance? Seems I would need either ALOT of resistors to get the needed ohms... Or a POT.
I would much rather use resistors to reduce the workload.
Ever run into this problem before?

Click to expand...

Personally, I don't use pots, I always used fixed value resistors, I try and keep to 2 esistors and find that you can make almost any combination you need. You can use a 40 ohm and 1.6ohm in series to get 41.6. you can use 2 125ohm resistors in parallel to get 62.5. Just look at the resistors you have available to you (at the shop, online etc.) and work out the math. Ty looking for the exact value you want, if that doesn't work then use 2 resistors double the original value, or half the value to use in series, or try just as close as you can ie 40 + 1.6 example. Hope this helps  

hmm i got some odd results today.
I wanted to limit the current to 160ma.
At the local store I was limited on resistor choices so I got (6) 47ohm resistors.
47 divided by 6 = 7.833ohms = 159ma correct?
after soldering it all up and setting the pot to min resistance... I only get 90ma max.
not sure why, any ideas?

update: it appears that the dummy load i was using did not have enough current draw... which led to even more odd results... maybe they are right though, hopefully someone can help clear this up.

Dummy loads:
(1) 1ohm resistor and 4 green LED's = MM shows 90ma max draw with POT at min resistance on DDL circuit with 8 ohms.
(1) 1ohm resistor and 4 1n4001 diodes in series = MM Shows 160ma max draw with POT at min resistance on DDL circuit with 8 ohms.  

So I used the 2nd dummy load and preset the current to 120ma.
I then connected the real PHR diode in series with the 1ohm resistor and MM still attached... It shows only about 95ma on the MM across the 1ohm resistor.

I dont understand why I get the difference between the 2nd dummy load and the PHR diode.
*Most importantly: Is that ma reading correct across the 1ohm resistor with the PHR diode? So is 95ma the true reading?
If so, Do I need a better dummy load that more accurately mimics a PHR diode? Any solutions for this already?

Please help  :-/