what resistor for 445 diodes in parallel?

[logsquared]

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.

[dnar]

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.

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.

[beamann]

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

[Phredy1]

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:



For the 4 diodes another site gave me a schematic to wire the diodes in parallel and looked like this:



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?

[andy_con]

^^^^ +1

i know fook all about electronics

[dnar]

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:



For the 4 diodes another site gave me a schematic to wire the diodes in parallel and looked like this:



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?

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.

As for wiring resistors on the LED -ve or +ve (K or A) it simply does not matter. Series is series, regardless of positioning.

[Doc]

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? )

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.

[dnar]

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.

[logsquared]

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.

[andy_con]

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



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!!!