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ArcticMyst Security by Avery

TANK007 Cu H.S. GGW diode WOW

Flaminpyro

Flaminpyro

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I started this build almost 12 hours ago and now I can sleep :tired: but check this out it's a realy cool a side clicke with a tripod stand, I made a heavy copper heat sink for this one it's more than twice the weight of a aluminum one as you can emagine that translates to cooling power :yh: I have decided to start off low on the power to see if the diode will last's, then I plan on turnng it up after an hours on time is reached. I'm using a GGW-H20 driving it with an older V2 flex drive and a 18650 cell. the current is set @ 201mA with a acrilic lens it's putting out 159mW max.

Peace All..

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robjdixon

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The power seems a little low, but maybe he got a freak?
I have a GGW at 170mA that puts out 110mW with an aixiz plastic lens
 
Flaminpyro

Flaminpyro

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did a double check, the current stays at 217mA constant and Po is 152mW with a DFW lens. my meter leads are about 2 foot long each so I wonder if that could be messin with the flex drive and giving me a bogus current reading I use my meter as a current meter not a volt meter, I have been doing it like this for quite a while now and I think maby this diode's off/bad...

daguin said:
Wake up and re-check our numbers!

Peace,
dave
Click to expand...
 
dr-ebert

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Isn't the usual figure for a measly PHR given as ~100mW@120mA with Aixiz acrylic? 150mW with a 405-G-1 lens is little above that. I've read in another post that the recent available GGW batches seem to be weak, maybe they're different diodes from the original brand?
 
daguin

daguin

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WOW!

Violet diodes are variable, but . . . . "WOW!"

That's a lot of mA for only 150mW through a high output lens

Something is rotten in the state of Denmark ;)

Peace,
dave

@ Dr-Ebert -- maybe, but those diodes shouldn't survive at 215mA (at least for long AND they put out more than this before they die)
 
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robjdixon

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Are you measuring the current through the tailcap, or between the driver or diode?
 
dr-ebert

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@daguin - I wasn't suggesting that they're really PHRs being sold as GGWs. I thought about it for a moment and also decided that they'd give out higher power at that current if they survived at all.

@Flaminpyro - I don't quite see how exactly you're measuring the current. If you're measuring the current drawn from the battery, you'll be getting too low current to the diode (about 60%), a boost converter is different in that respect from a linear driver like the LM317!
 
Flaminpyro

Flaminpyro

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I just remeasured using a 1 ohm resistor and a short piece of wire to complete the circuit now I have my meter set to milivolts and am reading the milivolts across the resistor, current is being measured at the tailcap. I now get 225mA and Po @ 155mW so the longer leads of my meter did not have much if any influence on the current readings.

@ Dr. E I'm not sure what the % of effic this V2 flex drive has, I know the V4 is about 98% so if this one is only 90% that would mean it is only eating 10% of what is being taken from the battery or in this case 22.5mA which would mean that 202.5mA should be going to the diode. as I surmised in the beginning I think this is a bad/weak diode :cryyy: if I missed somethin pls point it out,thanks for lookin guys...


dr-ebert said:
@daguin - I wasn't suggesting that they're really PHRs being sold as GGWs. I thought about it for a moment and also decided that they'd give out higher power at that current if they survived at all.

@Flaminpyro - I don't quite see how exactly you're measuring the current. If you're measuring the current drawn from the battery, you'll be getting too low current to the diode (about 60%), a boost converter is different in that respect from a linear driver like the LM317!
Click to expand...
 
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daguin

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I cannot "see" anything that you are doing wrong. The thing that is "rotten" might actually be that this diode is just a very low output diode. I just have never seen such a low output from a GGW (and I've been through >40 of them) . . . .Hence the "WOW!"

I am simply amazed that it is happening. I also do not think that I would "turn it up", as you were talking about, unless you do not care about the survival of this diode.

Have you checked to see that the diode window and the lens are all clean and clear? (just grasping at ideas here)

Peace,
dave
 
dr-ebert

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There is no 98% efficiency DC-DC converter. I've read lots of datasheets and they claim up to about 85%, but that is dependent mostly on Vin which needs to be around 3.5..4V for peak efficiency (you should be close to that with your battery).

Again, if you're measuring the current between battery and driver - instead of between driver and diode - you'll get the wrong current!

Let's say your diode needs 5.5V@225mA, that's 1.25W approx. Assuming 85% efficiency of the converter, that means about 1.5W power draw from the battery, which has ~3.6V, so needs to supply about 400mA.

A linear driver, on the other hand, would draw 225mA (identical to delivery current) but require a much higher input voltage, minimum around 8.5V, for this diode, meaning a battery drain of nearly 2W. The excess power is converted into heat by the LM317 and the current sense resistor.

In short, you have to measure between driver and diode with a boost driver!
 
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Flaminpyro

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Thanks for explaining all that I'll give that a try and measure it up front when I wake up, have a good day ! Pyro...

dr-ebert said:
There is no 98% efficiency DC-DC converter. I've read lots of datasheets and they claim up to about 85%, but that is dependent mostly on Vin which needs to be around 3.5..4V for peak efficiency (you should be close to that with your battery).

Again, if you're measuring the current between battery and driver - instead of between driver and diode - you'll get the wrong current!

Let's say your diode needs 5.5V@225mA, that's 1.25W approx. Assuming 85% efficiency of the converter, that means about 1.5W power draw from the battery, which has ~3.6V, so needs to supply about 400mA.

A linear driver, on the other hand, would draw 225mA (identical to delivery current) but require a much higher input voltage, minimum around 8.5V, for this diode, meaning a battery drain of nearly 2W. The excess power is converted into heat by the LM317 and the current sense resistor.

In short, you have to measure between driver and diode with a boost driver!
Click to expand...
 
brtaman

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Yes, read dr-eberts post. Unlike a linear (LM317) driver, boost drivers such as the flexdrive drain a larger amount of current than they supply to the diode.

It is quite simple to demonstrate:

Lets say 100% efficiency to simplify:

Driver draws 215ma @3.8v this equals around 0.82W

Driver feeds diode @ 5.5v with 0.82W input...

this translates to roughly 148ma going into the diode, making the output still somewhat disappointing but nearly as bad as originally though. Now lets take into account that this particular diode has a higher forward voltage (5.8v*...141ma), then we take into account efficiency etc. There are alot of variables in this post though I think you see what I am getting at.

Like doc-ebert said meassure at the diode with resistor, if not a test load with a 1ohm resistor is the way to go.
 
Krutz

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yes, what they said, recheck the current right between the driver and diode!

besides that.. very nice work! nice host and beautiful heatsink!
how did you attach the tripod? not with a screw, like a regular cam, obviously?
what batterysize is it? AA or AAA? ah, re-checked, its a 18650.. sure looks smaller than that! :)

manuel
 
Flaminpyro

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Thanks DOC-E and everone else who helped out with advice on current mesurent :) I pulled the unit apart and installed a 1 ohm resistor in the + lead between the diode and driver, I measured the milivolts across it and found it to be 149 ma guess thoese flex drrives arn't as eff as I thought they were :yabbem: and the power with the end of the Aixiz module about 2mm away from the laserbee head no lens 146mW and voltage across just the diode was 5.42 volts after stablizing for 30 seconds. thanks again All...
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edit: now I have it set to 200ma and it's Po is 202mW no lens. Fv 5.5v after 30 sec on. now when I remove the 1 ohm resistor won't the diode get a little more power and run even harder, keep ya updated !
 
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