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Old 08-03-2012, 05:23 AM #1
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Default Test load 445.. Problem

Hey Guys,

Working on a build and something is amiss.

Flashlight host with 2x18650's, fully charged. (trustfire, cheapos)

Properly heat sinked 1.8A Jib driver that I bought from Mohrenburg months ago.

3A Test load from Jufran. Jumper set to pin 4.

Reading the voltage across the resistor and I get 1.5A, and then slowly after about 20 seconds climbs to 1.67A... however after 30 seconds the test load is getting way too hot, the diodes and resistor can't be touched. So I turn it off.

Should the driver be going straight to 1.8A? Is the driver bad?
Are my batteries not able to provide enough current? I have two good sanyo's I bought, however their contact points are shallow so I'd need to rig them up somehow else..

Thanks!


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Old 08-03-2012, 02:41 PM #2
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Default Re: Test load 445.. Problem



Who is the bastard that keeps telling people 4 diodes is a good approximation for a blue diode? It drops over 5V!

Use three instead, mister bobo.
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Old 08-03-2012, 02:50 PM #3
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Default Re: Test load 445.. Problem

Quote:
Originally Posted by Cyparagon View Post
Who is the bastard that keeps telling people 4 diodes is a good approximation for a blue diode? It drops over 5V!

Use three instead, mister bobo.
Doh! I thought Jufran put in his instructions to use 3 for red, 4 for 445, and 5 or 6 I think for violet.

Grande, if you happen to look at this thread, this might be our answer we had been looking for.

Cheers!
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Old 08-03-2012, 03:13 PM #4
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Default Re: Test load 445.. Problem

The IDEA of the test load is to use a cheaper diode with the same electrical properties. Since it's hard to find another diode that drops that many volts, we use a bunch of regular, cheap diodes in series, and add up their Vf (@ whatever current you will be running them at) to as close an approximation of the Vf of the laser diode at that current. The resistor is put in place simply to get a reading of teh current without affecting the circuit too much. We use 1 Ohm resistors because everybody is dumb and can't do multiplication or division.
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Old 08-04-2012, 07:05 AM #5
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Default Re: Test load 445.. Problem

It all depends on the current that goes throught the diodes. Larger currents will result in larger voltage drop. It's not three for 445nm or four for 445nm...

It can be 5 or 6 for 445nm running at 200mA for example. The more current the more voltage drop across each diode. Do some experimenting reading the voltage with different currents and diodes. It's easy stuff
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Old 08-23-2012, 01:54 AM #6
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Default Re: Test load 445.. Problem

Thanks for the responses!

I think I'm still a little thick on something, so if someone could please correct me.

The forward voltage (Vf) is the voltage a component uses to operate. So my 445 diode has a different Vf for a different current. So lets say I have an A140(more on this later), I would need to find from a chart what the Vf is for a given current setting. This will give me the Vf I want to "simulate". Lets say its 5V. Jufrans' 3A selectable test load uses 1N5404 diodes which have a drop of 1.2V. So to get as close as possible, I want to set it to pin 4.
Was all that right? I hope? Does the source voltage play a roll in these calculations? For example if I'm using a linear driver, I have an 8.4V source, or if I have a buck-boost then source is always what the diode wants...

Just a little more clarification perhaps!

Thanks

p.s- the diode I am working with is the 445 diode that was in survival lasers original laser, in the C6 host... anyone know what diode that is ?
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Old 08-23-2012, 03:25 AM #7
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Default Re: Test load 445.. Problem

How do you figure 1.2V? It's closer to 0.9V according to the datasheet. Don't forget the current sense resistor drops 1V for every ampere of current through it, so you need 3 diodes. If I assume you're over-driving the piss out of it like everyone else, that's 1.8V + 3*0.9V = 4.5V

The source voltage does not play a role. However, it must be in the operating range of the driver.

You can't call your source 8.4V if it's two lithium ion cells. That's the full voltage, yes, but it falls very quickly to ~7.2V, and if you plan on utilizing the full capacity of the cells, you should design for a lower limit of 5.6V.
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Old 08-23-2012, 04:01 AM #8
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Default Re: Test load 445.. Problem

Thanks for the reply! I was looking at the "instantaneous forward voltage" @ 3A... I still can't find 0.9V anywhere.

The 1.8V is because I'm doing 1.8A (from linear driver)? And the 4.5 is roughly the Vf of the laser diode, the number we obtained from what shiny graph/datasheet?
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Old 08-23-2012, 10:06 AM #9
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Default Re: Test load 445.. Problem

At 1.8A the testload will heat up very quickly. At 20-30 seconds it is already blistering hot.
Unless you heatsink the testload and remove the heat buildup quickly, you can't make long testing on it. It is meant for testing if the current is correct. Not for duration testing of the driver.
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Old 08-23-2012, 10:27 AM #10
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Default Re: Test load 445.. Problem

Quote:
Originally Posted by Cyparagon View Post


Who is the bastard that keeps telling people 4 diodes is a good approximation for a blue diode? It drops over 5V!

Use three instead, mister bobo.


Made my day haha
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Old 08-23-2012, 01:40 PM #11
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Default Re: Test load 445.. Problem

The reason I was testing so long is being the current wouldn't stabilize... because of the heat on the diodes or a bad driver ?
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Old 09-07-2012, 12:26 AM #12
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Default Re: Test load 445.. Problem

So JUST to make sure I have this down... before I blow another diode

Assuming I'm doing a test for a BDR-S06J ...
According to this site: http://diy-lasers.com/blog/2010/11/i...f-bw512-sleds/

The Vf@500mA=6.1V

0.5 + X*0.9=6.1 Number of diodes to use: 6

Success?
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Old 09-07-2012, 12:32 AM #13
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Default Re: Test load 445.. Problem

The voltage drop of your components depends on the current flowing through them. For half amp, you can expect closer to 0.8V for diodes and 0.5V for the resistor. So maybe 7 ideally? 6 is fine. So is 5. It doesn't matter much if it's a linear driver and if it is heat sinked well.
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Old 09-07-2012, 05:25 AM #14
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Default Re: Test load 445.. Problem

Quote:
Originally Posted by Cyparagon View Post
The voltage drop of your components depends on the current flowing through them. For half amp, you can expect closer to 0.8V for diodes and 0.5V for the resistor. So maybe 7 ideally? 6 is fine. So is 5. It doesn't matter much if it's a linear driver and if it is heat sinked well.
This is something I am trying to explain to everyone (that depending on the current different voltage drops occur and you have to use that many diodes depending on the current and voltage you are trying to simulate) So far I can't see if anyone actually understood me
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Old 09-07-2012, 07:31 PM #15
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Default Re: Test load 445.. Problem

Maybe if we say it differently.

Just like you checked to see what the Vf was at the current you will be running it at, you need to do the same for the diodes in your test load to know how many to use. Remember, they're both diodes.

Some people respond positively to snide questions that simultaneously insult them:
Why would you go through all this work to make sure you know the right Vf at that current, and then just assume a Vf for your test load, unless you are some kind of n00b?
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Old 09-07-2012, 07:53 PM #16
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Default Re: Test load 445.. Problem

Not offended

But that wording hit the proverbial nail on the head. Each component has its own Vf at a different current. Crystal clear.

So looking at the data sheet here:http://www.datasheetcatalog.org/data...nts/1N5404.pdf

Using the chart at the end we just read off the Vf at a given current setting. Use that Vf.
The resistor voltage drop will just be whatever current we are working at, because its a 1ohm resistor. (0.5A=0.5V.. etc)

Thanks to all! Quite clear now!
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5 mW 532 nm x 4
300mW DD Elite 532 nm
275mw Small Police Host 650nm from lazerer!
DIY 800 mA 635 nm http://laserpointerforums.com/f40/2n...avy-70708.html

Last edited by bobo99; 09-07-2012 at 07:53 PM.
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