im sure its on the forum somewhere but i cant find it.
when running more than one 445 diode off a driver what resistor should i use?
cheers
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im sure its on the forum somewhere but i cant find it.
when running more than one 445 diode off a driver what resistor should i use?
cheers
Andy dont just use a resistor use ohms law!
Rob
is it really the law, if i dont use it will i get arrested?
;)
I use 1/2 watt 1ohm. Its easy to measure the current through a 1 ohm resistor!
Given the massive tolerences of those 445s I would be more comfortable with a bit more than 1Ώ cushion between them and 1A @ 1V is 1W so a half watter will be toast
your right. If driving at 1amp need 1 watt. 3 is the most I have paralleled. If I remember right, I drove them with 2amps total. I know for fact i checked the current to each diode and they were within a few mA.
V=I X R
P=I X V or P = I^2 X R
Yeah, if the diodes a fairly well matched there will be no problem. LDs behave like zeners so if you have two that are wildly apart and you haven't got enough 'flex' between them, you could end up with one crowbarring all of the current while the other twiddles it's thumbs.
Analogy of the day: Two eggs need to be clamped to a surface with one clamp, if the clamp's face is rigid you better hope that the eggs are exactly the same size *retardo < or should that be egg-sactly :p
You could fit a spring loaded spreader to the clamp, but again, the springs need to be soft enough to give before the yolk squirts out
Lamb.Quote:
Originally Posted by Doc
Or should I say "steak and..."
:)
I used to use water analogies but they dried up, tried bread for a while but that got stale so I just stick to dairy products now. W * E + H = O
Mmmm omelette.
I would have used more than 1 ohms with these 445's, simply due to they higher Vfwd spec and variance. But then again, they are rugged little buggers.
That is the beauty of ohms law, you can't break it! You can ignore it, but it still exists and is the law.Quote:
Originally Posted by andy_con
Damn! You guys got me thinking 1 ohm may not be large enough. I am about to build a big 445 combiner. So, I figured I should do some math on it. My calculations show with a 1 ohm resistor the diode current will vary approximately 50ma for every 100mV variation in diode voltage. So, are we seeing this big a variation in diodes? I have not seen this big a variation in the diodes i have used. However, I have only paralled 2 sets of 3 diodes and they came from the same projector.
Anyone care to double check my math? It might be a good idea to find the best resistor solution as some may use this thread as reference for their projects.
That is what I was eluding to. I believe there could be 100mV or more variation in Vf, perhaps but more as a result of temperature variations between laser diodes. Is a 50mA variance at 1A average a big issue? May be not.Quote:
Originally Posted by logsquared
I think this subject requires more investigation and testing. Don't you just love the scientific process? I do.
Having said that, let me say this. If perfection is your goal; there is no substitute for individual current limiting. Series resistor current balancing is always a compromise solution. It comes down to managing the level of compromise.
One thing that sits in the back of my mind regarding this method, particularly with the more sensitive 660nm diodes is the ramifications resulting from a failed device; be it LD or resistor. The run-away destruction that could result will I am sure bring tears to even the hardest of laserists.
I had no idea that I would be needen to put resistors with the 445's shunted togeather to reduce the milli amp differances between the 445's diodes to 50ma or less.
Thanks guys great info! What math formula would I be using to figure this out?
I am a math dummy and thats why I only got a AA degree in electronics, The math realy kicked my F-en ass! The bullion algebra was easy but that was the only med to hi level math that did'nt.
Emory
For those of us who are electronic theory challenged, is there a way to make this simple? I know there should be a resistor(s) with the 445 diodes somewhere between the driver and the diode.....but where and how?
I looked online at some led resistor calculators since that is all there is. Assuming that a diode is a diode (led or laser diode), I was trying to see what resistors work and where to put them. For either a single diode or a 4 diode setup. For the single diode a calculator came up with this:
Attachment 23728
For the 4 diodes another site gave me a schematic to wire the diodes in parallel and looked like this:
Attachment 23729
That site when I choose only 1 diode still put the resistor on the - side.
One site tells me that using 1 diode the resistor goes in the + to diode and a different site tells me that for 4 diodes the resistor goes in the - sides???
I was also told that for multiple diodes in series to put the resistor across the + and - for each diode... That does not seem right to me, but the person that told me to do that is VERY good with electronics.
OK, I'm confused as hell...! I am thinking the best way to do a 4 diode setup is in series, but where and how do I put resistors in?
^^^^ +1
i know fook all about electronics
That is all well and good, however the 2nd attachment relates to direct drive from +5V. When driving LD in parallel from a constant current source (driver) the application changes considerably and the topic of current sharing requires further examination as the LED voltage drop (Vf - Volts Forward) varies slightly for each diode in the array. The above calculation assumes all diode have equal Vf, which may be good enough for LED's, however 445nm LD's do vary in Vf and if you wish to match them closely then this calculation is a generalization, particularly when driven from a constant current source driver.Quote:
Originally Posted by Phredy1
As for wiring resistors on the LED -ve or +ve (K or A) it simply does not matter. Series is series, regardless of positioning.
First of all; with a floating case diode like the 445s it makes no difference which side you connect the balance resistors, with a grounded case diode the resistors would have to go on the anode (+ve) side, else the balance resistors won't be balancing anything.
More importantly; although the schematic diagram would be the same for laser diodes as it is in your diagram, the circuit shown uses the resistors for a completely different reason.
In our application we are using a driver that controls the current (Amps) to the laser diodes. Connecting a diode in series with the LD doesn't reduce the current as the driver just increases it's output voltage to push the same set current.
In the attached diagram there is a fixed set voltage of 5v and the individual resistors are there to reduce the current to the rated diode current.
The problem is this:
A laser diode acts much like a zener diode. That is, you feed it voltage lower than it's rated forward voltage and it acts like it's open circuit, in other words the diode doesn't produce a load, it takes no current. Now if we slowly increase the voltage to the point at which it reaches it's forward voltage it very quickly starts to load up the driver and draw current.
Given the above if we have more than one LD connected in parallel and we feed them a low voltage and slowly increase the voltage we will very likely find that one of the diodes has a lower forward voltage requirement. This diode will start to conduct and take current first and in extreme circumstances it could be taking near it's rated current and lasing to it's limit before the next diode has even started to conduct.
OK the above is a theoretical example, but in the real world you could set your driver to 2A for two diodes and one diode might be taking 1.5A while the other the other takes .5A
(see how this correlates with my two eggs in a clamp analogy? :p)
What the balance resistors do is act like a soft springy face on my clamp:
The voltage increases on all diodes, diode one starts to conduct while diode two is still asleep, because there is a resistor in series with the diode and the diode is taking current there is a voltage drop across it's balance resistor, that is the voltage before the balance resistors increases relative to the voltage on the diode side of the resistor. This means that the voltage going to diode two increases as well on both sides of it's resistor (there is no load so no volt drop) so it can reach it's forward voltage. Like I said earlier; once a diode starts to cunduct, very small increases in voltage result in a very quick rise in the current it draws.
So going back to the clamp and springs analogy; clamping two different size hard and brittle things with one flat faced hard clamp can be done by fitting springs to the clamp face. Sure the largest item will be clamped first and clamped hardest, but as long as the springs are soft enough the clamp can continue forward until the second item is clamped before the first one is broken.
So we have a compromise to think about; very high value resistor will produce the most even balance but the drive voltage will shoot up, the voltage drop across the resistors will shoot up but the current will remain the same (remember LD drivers control the current), therefore the power across the resistors and therefore their required power rating will also shoot up.
I personally use 2ohm resistors in all my multi setups, probably overkill for reds but I have a big box of 'em.
I hope this all makes sense.
I would agree, 2.2ohms for 445's. 1ohm for 660's.Quote:
Originally Posted by Doc
Best bet is to put the diodes in series if at all possible. Easily done due to the isolated can. Or put a trimmer resistor on each for parallel.
Also, must consider the heat output of the resistors if they are enclosed in a module. A 6x combiner with 2ohm on each diode will give off ~ 12watts total at 1 amp current to each diode. That's a huge amount of heat to remove. Probably best to mount the resistors away from the module.
A larger resistor, say, 2-5 ohms will reduce the work of the driver pass transistor. That could be very beneficial with a set supply voltage and or multi-diode setup.
i hate electronics im so crap at it!
when to maplin today and asked fomr some 2.2ohm resistors and they gave me these -
red red black yellow brown
http://i187.photobucket.com/albums/x...0701Medium.jpg
according to this site they are 2.2ohm
http://www.hobby-hour.com/electronic...calculator.php
but wired two 445s in parallel with a single resistor on the + connected to my bench psu and nothing just dead.
doubled checked diodes and both are working fine.
please help before i go mad!!!
They're 2.2MΩ
mega ohms?
so im guessing the chap in maplin gave me the wrong ones?
Standard... *DZQuote:
Originally Posted by andy_con
http://4.bp.blogspot.com/_VPNJ1Mj-cl...n-standard.jpg
ill use ebay, you can go wrong with ebay :confused:
Don't forget the power involved @ 1Amp they will be dropping 2.2v.
2.2v x 1Amp = 2.2W dissipated.
http://cgi.ebay.co.uk/2-x-2-2-Ohm-3W...item19c0e41eba
ok cheers .
If I am understanding all this correctly, 4x 445nm diodes wired in series would need 4x 2.2 ohm 3 watt resistors. The first resistor in the + side wire going to the the first diode. The next 3 resistors inbetween each of the remaining 3 diodes, correct?
For a single 445nm diode use the same resistor?
Diodes in series don't need a resistor. The current will be the same through a series circuit. A single diode doesn't need a resistor because it is not "fighting" for current with any other diodes.Quote:
Originally Posted by Phredy1
hope this helps.
It might be one of those rare high-binned diodes that chooses the right to fight with itself. :pQuote:
Originally Posted by logsquared
Wait a minute....what?
Only multiple diodes wired in parallel do we need to put resistors in? Wired in series no resistors needed?
I'm going to borrow Andy's word. WTFook?
So, I'm going to ask the question thats just hanging out there, waiting to be asked...
If wiring in parallel is so much more fiddly, requiring more components, and greater risk of damaging all the diodes in the event of one failing (growing greater the bigger the build), why would you want to do this in the first place?
If its to keep overall currents required down, would the best compromise be pairs of diodes in series together (with appropriate resistor), and then parallel as many pairs as you need?
I think I'd be more inclined to just series wire up to the driver limit and then add another driver (and so on, etc)
Correctamundo, the current in a series circuit is the same anywhere in said circuit.Quote:
Originally Posted by Phredy1
The voltage in a parallel circuit is the same anywhere in said circuit (not counting
losses).
That's how I've done my 445s using a 12v supply to the drivers, however no resistors are required.Quote:
Originally Posted by norty303
Hi Norty. One simple answer. the choice to wire in parallel with a single driver is to avoid having to insulate the diodes, which simply moves to another set of challenges (electrical insulation while providing thermal conduction).Quote:
Originally Posted by norty303
You simply trade one set of compromises for another. That is why case isolated diodes are a win-win.
You right, the perfect solution is 1 diode per driver, although that is more $$$, more adjustment, more space ra de ra de ra.
Finally the light in my head turned on (and it was blue). Everything makes sense now. Thanks Doc and dnar for all the good advice!!!Quote:
Originally Posted by dnar
can someone help me?
i need some 1.1ohm resistors i can only seem to find 1k resistors.
can someone in the know hook me up with an ebay link?
search for 1r1 resistors sir andy
but it is not a common figure
so either go for 1r or 1r2 (1.2 ohm), something around tge 2-3 watt mark
http://cgi.ebay.co.uk/2W-Metal-Film-...#ht_1781wt_905
or http://cgi.ebay.co.uk/10-x-1R-1W-1oh...#ht_1142wt_905
if you insist on 1.1 ohm, i found some, but only at 1/2 watt http://cgi.ebay.co.uk/10pcs-1-1-OHM-...4#ht_599wt_905
umm suppose 1ohm will do but i only need 1/4 to 1/2 a watt.
will google them now, thanks for ya help
in that case, try to get metal film resistors, they are smaller than the carbon film ones. and better quality
just brought these
http://cgi.ebay.co.uk/ws/eBayISAPI.d...=STRK:MEWNX:IT
so, you won't be running the diodes at more than 0.25 amps, rite?
you are almost good to go if it is for LOCs, but not sure about the 445s...
yeah red diodesQuote:
Originally Posted by LaNeK779
Here are some really good 3W resistors http://cgi.ebay.com/Resistor-RWR89S1...item4cd9ec06e6
And 1W http://cgi.ebay.com/Resistor-RWR81S1...item4cda0fc63e
Just to complicate things I would like to know how the resistor choice might change if we want to protect the diodes (LOC's or 445's) if say a 4 diode parallel string had one diode go open circuit or a dead short. The latter is more plausible with the LOC as when a drive wire contacted the grounded case. I think the resistance may need to be significantly higher especially if we are running the diodes close to their upper limit. Any thoughts?
There is little saving grace in either failure mode.Quote:
Originally Posted by planters
O/C diode/wiring: the remaining diodes absorb the programmed current, failure highly likely.
S/C diode/wiring: the resistor on that diode will absorb the addition Vfwd of the diode (2.5 to 4.5V depending on 660/445) and could exceed the resistor rating, resulting in resistor failure, then it's the same as an O/C failure. Not to mention the sharing imbalance that will result before the resistor fails.
Rating the resistors to cope with shorts is possible, but either way the results are less than ideal. 1 driver per diode OR diodes in series really is the only safe method.
a simple yes or no will do, no techy elecy speak
when driving 4x 445 diodes with 12v and using these 3watt resistors as advised in this thread, they get mega fooking hot. they burn my fingers they are so hot, is this normal?
If you using them as balance resistors, there is no reason you need to use 1 ohm, I am using .47 ohm 1/2 watt for my LOC's @ 500ma and they barely get warm.
you could do the same with 445's by going to a lower value, which should drop the heat a lot. Depending on the current you are running through them, those 3W resistors are having to dissipate a lot of heat. ~1.5 W @ 1.2A