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ArcticMyst Security by Avery

Need help finding PHR-803T diodes for NASA

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Well we all know k-shell's real plan here is to have NASA order 200 PHR-803Ts, use about five, and then "scrap" the rest, which k-shell can kindly "recycle" for free. Government supplies for big-ticket agencies like the Dept. of Defense or NASA has so much waste involved. Universities too. I picked up three spools of 1000ft of high quality Belden signal wire, that goes for about $0.70/ft (that's the BULK rate for 1000ft-at-a-time purchases) for about $30 each because it wasn't needed anymore. I can bet they ordered three, used 10ft or something, and didn't have the room to store it.

Lucky me, and lucky k-shell!
 





ReNNo

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Sorry, but i have to correct you.

Leds and LDs acts in the same way (not for the emission, ofcourse :p) ..... both them are light emitting diodes, both of them are, basically, diodes junctions with a voltage threshold (VF), both of them works in current ..... the reason for which you put a resistor in serie to a led, is for give it a certain current, based from Vpower and the resistor value ..... otherwise, you just burn them, like LDs ..... only, LDs are much more sensitive and delicates ..... i had high power IR leds from HP, that i overdrived til the small filament that connect anode pin to the chip becomed dark orange incandescent, and when i left cool them, some of them was still giving out IR ..... LDs just die much much before this, if you try the same :p

That all is true, and I'm saying what I experienced in practical use.. (not by regulating current, but voltage)

We can easily give 1.5V to IR LED and it will work on nominal current...no need to use any current limiter.
For example, if we have 5V source, and IR LED (1.5 FV, 100mA), we can add 36Ohm resistor that would make 3.5 Volt drop, and voltage across diode will be 1.5V, and that diode will work on nominal current (100mA). Or we can say it differently....
If diode works 100mA, voltage drop of diode will be 1.5V, and on 36ohm resistor will be 3.5V.
(But as we cannot regulate current from batteries or constant voltage sources it's easier to say what voltage drop we need to make to deliver nominal voltage on LED)

But if we give 3V from constant voltage source to Red laser diode, this diode will die, doesn't matter if that is nominal voltage for red diode. That is why we can't use resistors to limit current.

Even If we try to use resistor to limit current, we will need to use big resistance resistor that would make lot of voltage drop, and not enough voltage would be delivered to LD.
 

HIMNL9

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Again, sorry, is the concept that is expressed in wrong way..... i try to explain better what i mean, sorry if my english is not the better.

You don't add a resistor to a led for cause a voltage drop, you add a resistor for limit the current.

Example: your 1,5V FV IR diode ..... if it's rated for 20mA, you need to add a resistor that limit the current to 20mA, and this is calculated dividing the power supply, minus the FV of the led ofcourse, for the current ..... so, if you want to use 12V, you must add (10,5/.02)=525ohm resistor ..... this resistor don't cause a drop of 10,5V, it just limit the current to 20mA ..... where instead, if your example was good, and the 525ohm resistor was used for cause a voltage drop of 10,5, then passing the power supply from 12 to 10 volt, your diode must turn off, cause the voltage drop was higher than the power supply ..... instead, cause the led works in current, it simply become more dim, cause the current decrease, but still work.

A LD die if you give it the FV directly, just cause it's thousands of times more sensitive and delicate than a led.

I hope this explanation is more comprehensible than the previous one, sorry.
 

ReNNo

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You don't add a resistor to a led for cause a voltage drop, you add a resistor for limit the current.
What is the different?
If you add resistor, there will be voltage droop on that resistor, than means less voltage for LED. (Less voltage = lower current)

We can say it in two ways....
- We add 525ohm resistor to make 10.5V voltage drop .... and 1.5 would be delivered to LED (Why I say that.... Because we know that LED will work on nominal current when we connect it to nominal voltage)

- We can add 525ohm resistor to lower down current from very high current to nominal current. (What I did above is just more detailed description what would happen if we add resistor)

this resistor don't cause a drop of 10,5V, it just limit the current to 20mA
Not true.
525Ohm resistor will make 10.5V if current is 20mA (and we know it is, because it is stated in LEDs datasheet)

if your example was good, and the 525ohm resistor was used for cause a voltage drop of 10,5, then passing the power supply from 12 to 10 volt, your diode must turn off, cause the voltage drop was higher than the power supply
If you switch power supply to 10V that means lower current, and a bit lower voltage across LED. (Not 0)
 
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renno, your end result is getting the LED to dim, yes... but the wording is incorrect.. a resistors job is to limit current. if we took your example and limited the voltage, why don't you try your idea with a voltage regulator. like the LM317.. it will give you the 1.5 volts needed for your LED but it will be fully bright. i hope this expands a little on what himnl9 said

michael
 
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You guys are both missing some things, sorry.

You can absolutely drive a laser diode with either a current source or a voltage source, they both work equally well, especially when driven with a bench pulse generator.

Now in specific applications, either current source or voltage source can work better. When pulsing a diode, using a voltage source works better for several reasons. When running a diode CW in a portable host where consistency is an issue, a current source is much better.

But there is a set ratio: all else being equal, the same current will always give you the same voltage, and vice versa. The problem with portables is that all else is NOT equal, so current sources are safer. But in a lab with lots of consistency, driving at a high pulse rate, a voltage source is better.
 
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You guys are both missing some things, sorry.

You can absolutely drive a laser diode with either a current source or a voltage source, they both work equally well, especially when driven with a bench pulse generator.

Now in specific applications, either current source or voltage source can work better. When pulsing a diode, using a voltage source works better for several reasons. When running a diode CW in a portable host where consistency is an issue, a current source is much better.

But there is a set ratio: all else being equal, the same current will always give you the same voltage, and vice versa. The problem with portables is that all else is NOT equal, so current sources are safer. But in a lab with lots of consistency, driving at a high pulse rate, a voltage source is better.

yes, the pulse generator im using has the ability to limit the voltage but not the current. and i am using an oscilloscope to measure the peak voltages to the diode.
so in this configuration i can only see the voltages.. im sure there would be a way to configure the scope to measure the voltage drop across a resistor and solve for amps, but at the moment im not too concerned.

[edit] we just ordered a handful of LDs, i have one here already but would prefer not to push it until the others arrive. as soon as they do i will sacrifice one diode to see just how quickly and brightly we can pulse it. and ill let you all know what we got.
 
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yes, the pulse generator im using has the ability to limit the voltage but not the current. and i am using an oscilloscope to measure the peak voltages to the diode.
so in this configuration i can only see the voltages.. im sure there would be a way to configure the scope to measure across a resistor and solve for amps, but at the moment im not too concerned.

That's the exact basis for our set-up, a pulse-generator running as voltage source. We ramp up voltage from zero, and watch the current, voltage, and power on a digital o-scope.

As long as your o-scope has another input and you're running in pulsed mode anyway, get an inductive current probe. Just a loop around the wire, direct input into the o-scope. It doesn't interfere with your circuit as much as if you put a sense load in series with the diode, and easily puts your current and voltage on the same screen on your digital o-scope. Put a photodiode on another channel, and you get all three waves on your scope for very little money.

Be warned though, you're going to get ugly waveforms for the current and such. Current can overshoot, and it can overshoot more the faster you drive it. Using a current source makes it worse than using a voltage source, but if you want clean waveforms, then you'll have to start thinking about things like impedance matching and being careful about your transmission lines. I don't think you care too much about a clean waveform, right?

Also if you're looking at power output, with a photodiode, start thinking about things like photodiode risetime vs the length of the pulse you want to look at. Lots of things to think about if you want to make more precise measurements, but it doesn't sound like you need a lot of precision at this point.
 

ReNNo

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renno, your end result is getting the LED to dim, yes... but the wording is incorrect.. a resistors job is to limit current. if we took your example and limited the voltage, why don't you try your idea with a voltage regulator. like the LM317.. it will give you the 1.5 volts needed for your LED but it will be fully bright. i hope this expands a little on what himnl9 said

michael
Voltage and current are related values.
You can watch it from two angles, and with good calculations both results would be the same.

What resistor does?
It limits current or we can say it differently...it makes voltage drop.
Both is correct. We can look it from both angles.

It's doesn't matter how we say it...
We can say.... LED can work on its nominal power when we connect it to 1.5V voltage source. (Current will automatically be 20mA)
Or we can say....LED can work on its nominal power when we connect it to 20mA current source (that means that voltage across LED is 1.5V)

As you can see, these two values are very related.

It's not incorrect. Numbers don't lie. But someone just don't understand that these two values are related.

why don't you try your idea with a voltage regulator. like the LM317.. it will give you the 1.5 volts needed for your LED but it will be fully bright. i hope this expands a little on what himnl9 said

michael
If you connect IR LED to 1.5V using LM317, it will work on nominal power.
When IR LED works on 1.5V, trough LED will flow nominal current.
Because 1.5V is nominal voltage for IR LED (most of IR LEDs)
 
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Voltage and current are related values.
You can watch it from two angles, and with good calculations both results would be the same.

What resistor does?
It limits current or we can say it differently...it makes voltage drop.
Both is correct. We can look it from both angles.

It's doesn't matter how we say it...
We can say.... LED can work on its nominal power when we connect it to 1.5V voltage source. (Current will automatically be 20mA)
Or we can say....LED can work on its nominal power when we connect it to 20mA current source (that means that voltage across LED is 1.5V)

As you can see, these two values are very related.

It's not incorrect. Numbers don't lie. But someone just don't understand that these two values are related.


If you connect IR LED to 1.5V using LM317, it will work on nominal power.
When IR LED works on 1.5V, trough LED will flow nominal current.
Because 1.5V is nominal voltage for IR LED (most of IR LEDs)
Knock it off you two, we all know how stuff works, no need to explain it to the people who already also know it!

You two just gonna end up in a flame war or something.

ReNNo, you know the old saying "Ne valja sipati iz šupljeg u prazno?" :D
It applies here.

You are both right, the language barries makes it difficult to express one's opinion to the other , hence the misunderstanding.

As I said - you both know how it works, why explain it one to another?
 

ReNNo

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Yeah, we all know it... but we cannot deny something that is proved by numbers.
OK...it's not thread about electronic basics, but about good use of PHR diode.
 
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i have a few more questions,

like pullbangdead said, using an inductive current probe would be a great idea. i would like to test this and was wondering if i might be able to make this probe myself? ive used them before and i dont think their very complicated, looks like just a piece of iron or something that makes a loop around a wire..

---

and second question, is there a difference in the design of diodes that are meant for CW and ones that are meant to pulse (say a bluray burner vs. just a bluray player)? im looking for a good pulsing diode and it could be any wavelength from 405 and shorter.

---

naturally this entire xray source is sealed in a vacuum. in the name of weight reduction we might want to eliminate the aperture window and install the diode inside of that vacuum.
i beleive de-canning could become an option due to the extremely clean vacuum environment. do you agree?
my only concern is any thermal issues caused due to the insulation of the vacuum or any other issues it may cause. but im almost sure proper heat sinking would be more than adequate).
 
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well around here there are few threads that test the limits of some blu rays. but to say just blu ray is a bit vague... there are phr's, 6x, 8x, 10x and 12x.... they have more or less been tested here in cw to see what is the safest and longest they will last. pulsed usually isn't used. as for burner vs reader... the burners are of higher cw power.

michael
 
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i have a few more questions,

like pullbangdead said, using an inductive current probe would be a great idea. i would like to test this and was wondering if i might be able to make this probe myself? ive used them before and i dont think their very complicated, looks like just a piece of iron or something that makes a loop around a wire..

---

and second question, is there a difference in the design of diodes that are meant for CW and ones that are meant to pulse (say a bluray burner vs. just a bluray player)? im looking for a good pulsing diode and it could be any wavelength from 405 and shorter.

---

naturally this entire xray source is sealed in a vacuum. in the name of weight reduction we might want to eliminate the aperture window and install the diode inside of that vacuum.
i beleive de-canning could become an option due to the extremely clean vacuum environment. do you agree?
my only concern is any thermal issues caused due to the insulation of the vacuum or any other issues it may cause. but im almost sure proper heat sinking would be more than adequate).

The inductive current probe, as far as I understand it, is a coil of some sort, probable several loops, and a transmission line. It may have some other components to help with noise on the line, but I have no idea. I'm sure you could make one and calibrate it, then it just becomes a question of if your time to do so is worth the price of just buying one instead. I don't think they're expensive at all, is it should essentially just be a coil with a transmission line that can be hooked to the o-scope, where the o-scope does all the work. The ones we use are dang small, I can't imagine them costing much at all.

As far as diode design for CW vs. pulsed: in a word, no. It's more complicated, naturally, but for all intents and purposes you can think of these high-power violet diodes as all being able to run pulsed very well. They like running pulsed, they all like pulsed better than CW as far as their raw output. Not all pulses are created equal, especially if you care about waveforms, signal-to-noise, etc, but pulsed is easier on a diode than running CW. So all diodes will run pulsed better, and then the best diodes will also be able to run CW. When we make diodes, we start at like .1 or .01% duty cycle, and see how they do. If they do well at that, then we gradually increase duty cycle. Only the very best diodes ever get up to true CW (100% duty cycle). So you already know if the diodes will run CW, they'll have no problem running pulsed whatsoever as far as raw performance of the diode. Again, clean signal output is trickier, but just getting it to run isn't a problem. Ramping from zero and watching the current waveform (and voltage) to see how much it overshoots is a good idea, but nothing to worry about as far as if the diode will break or not unless you're running way out there on the ragged edge.

Running in a vacuum shouldn't be a problem. Diodes go through multiple steps in vacuum during processing, and I don't foresee any problems with running a diode in vacuum. Heatsink it well, naturally, but I'm sure you'd do that anyway in order to get the thing mounted securely in the chamber in the first place.
 

HIMNL9

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Well, speaking only about fast pulsed operations, then yes, a simple resistor can be enough for limit the current in the LD, and you can set the voltage of the pulses from the pulse generator, for match the LD current that you want .....

@ pullbangdead: sorry, i was talking basically for CW operations, as we use them in hobby field ;)

BTW, i don't suggest you to use any inductive probe, at the frequencies that you want to use ..... i mean, i suppose you know that, more the frequency is high, less a coil is efficent, for this, if is not matched and tuned for that frequency ..... and with your frequencies, it may be just one or two loops of copper wire, that gives you microvolts output range (or less), that require ultra-low noise amplification and all shielded like fort knox :p ..... and also after this, it gives you just pulses, not real-time reading (if you don't use a special type of hall sensor)

IMHO, much better use a known resistor and take the voltage from it with an high-speed op-amp ..... IF your pulse generator have a voltage regulation very stable and with low tolerance, you can also just place an 1 ohm in serie with the LD, and read directly the curent on the resistor leads, as 1mV=1mA ;)
 




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