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04-08-2011, 12:28 PM #17
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Xplorer877
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Quote:
 Originally Posted by Toke Yes, diodes have a voltage drop that does not increase much with current.
I disagree with this. But I'd suppose you'd have to specify what kinds of currents were talking about.

But consider diode dummy loads, for 405nm diodes at currents <500mA you'd use 6 rectifiers. Each has about .7v Vdrop, simulating a 4.2v Vdrop for the driver. But when the 445nm lasers first came out there was alot of confusion as to how many rectifiers to use. Dr. Lava said to use 4 even though 445nm diodes also run at ~4v. At high currents, >800mW, their Vdrop is more like 1v. So if you used 6, you'd really be stressing the driver to provide an amp of current at 6v!

Anyway, Mich what you gotta understand is how electricity works. Trying to apply these formulas without understanding what is actually going on is very hard, even though most of us find this stuff to be trivial.

What is electricity? It's the flow of electrons from atom to atom in a conditioning material, like copper aluminum silver.. ect.

Voltage is the force pushing the electrons along, or rather it's how much they are attracted to + end. They physically flow from negative to positive. It's counter intuitive, you can thank Benjamin Franklin for that. Anyways, think of voltage as water pressure. It's "how badly" the water/electrons want to flow. High voltage is harder to resist, low voltage is easier to resist.

Current is the amount of flow. Going back to the flowing water analogy higher current would be a wider pipe with more water passing by any one point at a given time. Current is measured in Amps. One amp is actually kind of alot of current for most circuits so the standard unit is the mA (milli amp, 1/1000th of an amp). 1000mA = 1A.

Resistance, measured in ohms, is the restive property of a material which only slightly conducts. Think of it as a kink in the hose. If you kink it, depending on the water pressure (voltage), you'll slow the current down considerably.

Ohms Law is the relationship between these.
1 ohm of resistance with 1v of voltage presented across it will cause 1 amp of current to flow.

Batteries have a fixed voltage (more or less), and the current that is drawn from them will depend on the resistance presented across them. They do not just output 30mA or whatever automatically.

Your LEDs have a voltage of 3v and a current of 20mA. That means that if you put 3v across the LED 20mA of current will flow. So, using Ohms Law, you can figure out that the LEDs essentially have a resistance of 150ohms each (3v / .02A = 150 ohms). At these low currents that resistance wont change much. So we can assume that won't change.

So as Toke pointed out, you usually don't want to wire LEDs in parallel (all + connected together, all - connected together) although I've never had a problem doing that in the my little projects over the years. Anyway, it's best to wire them in series (the + of one LED connects to the - of the other).

Now since each LED is 150ohm, two in series is 300ohms. Two kinks in the hose, double the resistance, so they're added together. Now your battery is 9v and you want 20mA for your LEDs. Even though you've got two in series you still only need 20mA. This is because they're connected together. When there is only one single loop for the electricity to flow through the current remain constant at all points. Remember that current is how much charge (electrons) are passing by any one point. Since there is only one path for them to take, the current must remain even.

So you want 20mA from your 9v, using ohms law... 9/.02 = 450 ohms. Your two LEDs already make up 300ohms so you only need a 150 ohm resistor in series with your LEDs.

battery+ ------- (150ohms) --- switch --- +LED1- ------ +LED2- ----- - battery

I hope this helps a bit.

-Tony

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Last edited by Xplorer877; 04-08-2011 at 12:35 PM.

04-08-2011, 02:17 PM #18
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123splat
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A+ XPLORER877
The only area where I might disagree is your calculation of the resistance of the LED and using that as a basis for calculation of the total series resistance. The 20mA is only the limit of what the manufacturer says the diode can stand without damage. It can and will pass much much more if allowed to do so. However, since weare now talking about two LEDs in series, I must agree that the 150 Ohms is most probablly adequate, so you win!
Outstanding tutorial! and my thanks for mentioning Dr.Lava's short dissertation on the test loads at higher currents (so, I do remember correctly that Dr. Lava suggests using a 4 diode stack rather than 6 when approaching 1A, or better?? Please respond to that one, I've been looking everywhere, but cannot seem to find Dr. Lava's quote).

04-08-2011, 02:26 PM #19
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anselm
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Quote:
 Outstanding tutorial! and my thanks for mentioning Dr.Lava's short dissertation on the test loads at higher currents (so, I do remember correctly that Dr. Lava suggests using a 4 diode stack rather than 6 when approaching 1A, or better?? Please respond to that one, I've been looking everywhere, but cannot seem to find Dr. Lava's quote).
It's easily backed up by looking at the 1N4XXX datasheet. At higher currents like used
for those 455nm diodes, say 1-1.5A , the silicon diodes voltage drop is closer to 1V each than 0.7V.
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04-08-2011, 03:15 PM #20
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123splat
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Thanks anselm,
That's what the logic tells me,,, BUT, what about Laser diodes? Does the Vf increase with forward current in Galium diodes as it does in Silicon? If so, then it would seem that you would also need to increase the diodes in the test load. Nobody seems to want to say. From what little is available in the way of datasheets for LD's, I have not been able to confirm ( and I not ready to start sacraficing 445 1+Watters to find out. I blow enough cheapies through stupidity).
So, what I get out of this is: when setting up drivers for 445's at 800mA or above, use the RED test load (four diode stack). I use 1N5401's in my homebrew test load.

Michealoorah, Sorry for highjacking yout thread.

Last edited by 123splat; 04-08-2011 at 03:17 PM. Reason: Apologies to Michaeloorah for highjack

04-08-2011, 07:18 PM #21
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michaeloohra
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lol NP I foud this site, and it helps a lot. It really breaks it down for me:
LED series parallel array wizard

And within that site are these 2 links that helped with the lable power ratings:
http://led.linear1.org/id-help.php
http://led.linear1.org/vf-help.php
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532nm -- 230mW Untested
532nm -- 200mW QT
532nm -- 200mW kaleidoscope
532nm -- 100mW QT
532nm -- 5mW -- QT
405nm -- 200mW Waiting for shipment
405nm -- 100mW QT
650nm -- 150mW Untested
650nm -- 250mW Waiting for shipment

Last edited by michaeloohra; 04-08-2011 at 07:20 PM.

04-09-2011, 02:48 AM #22
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Xplorer877
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Quote:
 Originally Posted by 123splat Thanks anselm, That's what the logic tells me,,, BUT, what about Laser diodes? Does the Vf increase with forward current in Galium diodes as it does in Silicon? If so, then it would seem that you would also need to increase the diodes in the test load. Nobody seems to want to say. From what little is available in the way of datasheets for LD's, I have not been able to confirm ( and I not ready to start sacraficing 445 1+Watters to find out. I blow enough cheapies through stupidity). So, what I get out of this is: when setting up drivers for 445's at 800mA or above, use the RED test load (four diode stack). I use 1N5401's in my homebrew test load. Michealoorah, Sorry for highjacking yout thread.
The drivers we use are current regulated, not voltage regulated. So it doesn't matter if the dummy load perfectly emulates the voltage drop. You probably could use 5 or even 6 rectifiers for a 445nm dummy load, but you'll just be stressing the driver as you set the current. But it has been done before. Like I said, when the diodes first came out people just used 6 diodes for a 4.2v Vd, thinking they were similar to 405nm diode, (since thats about what the 445nm use ~4.0v) but those who weren't electronically inclined didn't know that rectifiers drop more voltage at high currents. You set the driver to whatever current, say 800mA, and regardless of the Vd of the diode (within limits of course) the driver will adjust the voltage so that 800mA always flows.

I cant remember what thread it was that Dr Lava advised that only 4 diodes will suffice but as Anselm pointed out the proof can be found in the data sheet.

-Tony
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04-09-2011, 11:28 AM #23
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anselm
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Quote:
 Originally Posted by 123splat BUT, what about Laser diodes? Does the Vf increase with forward current in Galium diodes as it does in Silicon?
http://laserpointerforums.com/f65/3x...lot-53927.html

Offtopic: I think this is the key to understand how apparently some people like LarryDFW
have gotten away with DIRECTdriving (no driver at all) a 445nm diode from a single
18650 battery. The diodes's forward voltage (at high currents) is higher than the
4.2V the 18650 can provide (really not even 4.2, under load it's less), so unless
you have a "freak diode" with very low Vf, it seems to me that a 445nm diode would be
safe from a 18650, in that the battery can only push so much current through the
diode, because it almost drops all of the voltage the battery can provide (at a certain current).
In this case it would seem to me that the 445nm would be running at arounf 300-500mA
when direct driven from a single LiIon cell.....
NO GUARANTEE, PROCEED AT YOUR OWN RISK, THIS IS JUST ME HAVING IDEAS!
I CAN'T BACK IT UP WITH EMPIRICAL DATA FROM ACTUAL EXPERIMENTS (yet)!
__________________
I'm away from home and lasers until late July!
Kryton Groove switch workaround
Get your warning label stickers 0.90\$ or less, depending on qty
The INDEX to the LPF
Beware! Scams a-plenty, don't be a sucker!
___________________________
ATTENTION
This forum is fullfilled mit special electronische equippment. Fingergrabbing and pressing the cnoeppkes from the lasers is
allowed for die experts only! So all the “lefthanders” stay away and do not disturben the brainstorming von here working
intelligencies. Otherwise you will be out thrown and kicked anderswhere! Also: please keep still and only watchen astaunished the blinkenlights.

04-10-2011, 01:02 AM #24
 Class 3R Laser Join Date: Oct 2008 Location: So. Cal. Posts: 1,059 Rep Power: 1985
Xplorer877
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^ Yeah I'd agree with that.
The bottom line is that a 18650 just doesn't have the voltage to make these diodes want to draw lethal currents. However, that is not the risk IMO. If I was going to direct drive a diode I'd at least protect it with a 10uf capacitor and reverse protection diode too.

-Tony
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300mW 808nm IR Burner
250mW 660nm LOC Ehgemus Torch
5mW 635nm Z-Bolt Pointer
160mW 532nm CNI PGL-III-A
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300mW 445nm Ehgemus CR123 Build
260mW 405nm 6x Ehgemus Torch
20mW Magenta 405nm + 660nm PS3 Hybird Aixiz Pointer

04-10-2011, 05:34 AM #25
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LarryDFW
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Quote:
 Originally Posted by anselm Offtopic: I think this is the key to understand how apparently some people like LarryDFW have gotten away with DIRECTdriving (no driver at all) a 445nm diode from a single 18650 battery. The diodes's forward voltage (at high currents) is higher than the 4.2V the 18650 can provide (really not even 4.2, under load it's less), so unless you have a "freak diode" with very low Vf, it seems to me that a 445nm diode would be safe from a 18650, in that the battery can only push so much current through the diode, because it almost drops all of the voltage the battery can provide (at a certain current). In this case it would seem to me that the 445nm would be running at arounf 300-500mA when direct driven from a single LiIon cell..... (yet)!
The lithium-ion cell I am using, has a charging voltage of 4.35 VDC.

This higher voltage, combined with low internal resistance,

outputs over 600 mw from my 445 LD.

LarryDFW

P.S. On my Red's, I use a Schottky diode for ESD protection.
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