Sony 20x burners $24 at Newegg

Abray said:

OH I didn't know the MXDL ran off of 2 batteries. sorry guys

but thanks for all the help on the circuit!

so If I wanted to run the diode at 420 mA, then I would want a 2.97 resistor... lets just go with 3!
Another question. I've heard of the wattage of a resistor about twice. Can somebody explain that too?

Alex

Click to expand...

1.25 x current = wattage of resistor needed:

1.25 x 420 = .525 watts

you will be fine with half watt resistors, or the resistor that comes with the MXDL

regards,

amk

Gazoo said some are starting to come with 2.4 Ohm resistors. What are the resistors that are normally in the MXDL?

Cause I wouldn't want to use a 2.4 Ohm resistor as that would be giving the diode 520mA!!!

Abray said:

Gazoo said some are starting to come with 2.4 Ohm resistors. What are the resistors that are normally in the MXDL?

Cause I wouldn't want to use a 2.4 Ohm resistor as that would be giving the diode 520mA!!!

Click to expand...

they normally come with 3 ohms, sometimes with 2.4's

when you get yours, read off the line colors and input them here: http://samengstrom.com/nxl/3660/4_band_resistor_color_code_page.en.html to see it is 3 or 2.4

regards,

amk

ok I'll do that.

I have a glass lens and the MXDL on the way. And I should be earning 70 dollars pretty soon although when I do, I won't have a blu-ray anymore

I had potentiometer problems and the diode got fried

BUT I can do the blu-ray later!

amkdeath said:

2x  3.6 batteries = 7.2 volts  :-?

Click to expand...

Yes, but when they're full it's 4.2V each, which brings you to 8.4V when full and 6V when empty, which is a perfect range for the LM317 circuit..

3.6 is just the average voltage in the middle of this range, between full and empty.


It's similiar with Ni-MHs.. They're specified as 1.2V, but have 1.4V when full. Three should give you 3.6V, the same as a Li-Po, but when full, they give you 4.2V, again, the same as a Li-Po.. So in this case, the 1.2V is again the average voltage, in the middle of the range - 1.4V when full and 1V when empty...

The shop where i ordered my NEC 7191A still didn't send anything, and only after i called them, did they tell me, that they don't have them in stock..


So i asked what they do have in the same series, and they said NEC Optiarc 7191.. Well, i ordered that, and they gave me a discount, because of the delay.. If they send it tomorrow, it should be here on saturday...


But it's better this way, since i'll have time now, to finish the SEPIC driver circuit, that can regulate the current from a single Li-Po or only three Ni-MHs... I'll finally be able to build a very small yet powerfull regulated laser...

Ok please excusse the noob question but I read all 8 pages and still have some questions.
Is the voltage of the open can LD limited to 2.8 V and sweet spot is 420mA or is it because the DDL driver it putting out the those amount? The reason why I'm asking is because I have checked my Metal Gear Dorcy 1watt under load and it is at 3.03 volts and 798 mA. I can take it down to 450 mA with a 5 ohm 1/2 watt resistor if I can leave it at 3.03 volts. So is it safe if I use this setup with a heat sink or will this kill the LD or shorten the life of it? Please note the Metal Gear Dorcy does have a built in driver.

Thanks

hydro said:

Is the voltage of the open can LD limited to 2.8 V and sweet spot is 420mA or is it because the DDL driver it putting out the those amount?  The reason why I'm asking is because I have checked my Metal Gear Dorcy 1watt under load and it is at 3.03 volts and 798 mA.  I can take it down to 450 mA with a 5 ohm 1/2 watt resistor if I can leave it at 3.03 volts.  So is it safe if I use this setup with a heat sink or will this kill the LD or shorten the life of it?  Please note the Metal Gear Dorcy does have a built in driver.  

Thanks

Click to expand...

I believe i understand how you're thinking, because i had similiar questions in the beginning..

For every voltage on the LD, there is only one current, and the other way around, of course.. (if we ignore the changes in the resistance from heat) You can not have a different voltage and the same current, unless the resistance changes.


If you have a driver, that regulates the current, it does this by adjusting the voltage, to keep the current constant..
A LED has a different forward voltage than a LD. But the same driver, set to the same current would simply adjust the voltage to get to that exact current.. (as long as it is within the drivers range)


The reason we are allways only talking about current here is, that it is much safer to keep the current constant, as opposed to the voltage.
If you keep the current constant, when the LD heats up, and it's resistance decreases, the voltage will drop a little, so the current can remain the same.

If you were to keep the voltage constant, on the other hand, the resistance of the LD would again drop, as it would heat up, but this would mean, that the current would increase.

If you're pushing a LD close to the edge, this can be dangerous, because more current means more heat, more heat means a lower resistance and again more current, which can result in a thermal runaway, killing your LD. This is why current regulation is so important.


But if your flashlight already has a current regulated driver for a LED, and you can adjust the current to a safe one for your LD, it would work with it, because it actually needs a little less voltage than a LED...

The only thing you need is an additional capacitor directly on the LD, to protect from spikes..


450mA could still be considered a safe current with enough heatsinking, but people recommend 420mA for longevity.. There wouldn't be that much visual difference anyway (i need to check the graph again)..


Or were you asking about something else?

I am not sure about that entirely,

What I grasp is that a load on a power supply draws power. In the case of a laser diode, it wants to over draw. That said, the volts should be preset as a constant and the current should be preset as a constant. The driver with an adjustable current choice allows "tweaking" to set a current that provides a desirable result. For example, the ddl driver is a current adjustable circuit, that said it does limit volts (to a respect) by using a circuit that takes a "close ratio" input 6 to 7 v and inherently drops it to ~ >3v due to circuit consumption leaving the lm317 to be "tweaked" for current limiting purposes.

Placing a laser diode (red in my case) on the circuit adds a load and as such a load tends to drop volts in any circuit (in this case the ~3+ ends up near ~2.85+) and "tweaking" the pot allows for current adjustment.

If I am wrong on this description, I am sure someone with more electronic prowess will correct my synopses.

SN

I recall that:

watts = volts x amps (less heat or waste)

as such, if your circuit runs a higher volts, then you should run it at lower amps to achieve the same watts.

This is providing that your LOAD is ok with the higher volts.

One good thing about higher volts and lower amps is smaller wires can carry it. But unless your load can tolerate the higher volts, it may not be advisable.

SN

SuperNova said:

I recall that:
watts = volts x amps (less heat or waste)

as such, if your circuit runs a higher volts, then you should run it at lower amps to achieve the same watts.  
This is providing that your LOAD is ok with the higher volts.

One good thing about higher volts and lower amps is smaller wires can carry it.  But unless your load can tolerate the higher volts, it may not be advisable.

Click to expand...

You can not drive a LD with a higher voltage and a lower current at the same time.. The current and the voltage are directly related through it's internal resistance.. U=RxI (Ohm's law) If you double the voltage, the current would also double, there is no way around it.

For every voltage, there is only one current (if the resistance doesn't change)..

Of course with laser diodes, their resistance drops, when they heat up. If the voltage was constant, the current would go up. That's why we use current regulators, which constantly adjust the voltage, to keep the current constant, even as the resistance changes.

As the LD heats up and it's internal resistance decreases, a current regulator will drop the voltage slightly, so the current can stay the same all the time.


But if a LD draws 300mA at 2.85V, you can not give it a higher voltage, but keep the current the same or make it even lower.. It's physically impossible.
Twice the voltage and half the current would mean the same power (Watts), but you can not do this with loads. That only works with transformers.

IgorT said:

[quote author=SuperNova link=1202158967/120#121 date=1203681725]I recall that:
watts = volts x amps (less heat or waste)

as such, if your circuit runs a higher volts, then you should run it at lower amps to achieve the same watts.
This is providing that your LOAD is ok with the higher volts.

One good thing about higher volts and lower amps is smaller wires can carry it. But unless your load can tolerate the higher volts, it may not be advisable.

Click to expand...

You can not drive a LD with a higher voltage and a lower current at the same time.. The current and the voltage are directly related through it's internal resistance.. U=RxI (Ohm's law) If you double the voltage, the current would also double, there is no way around it.

For every voltage, there is only one current (if the resistance doesn't change)..

Of course with laser diodes, their resistance drops, when they heat up. If the voltage was constant, the current would go up. That's why we use current regulators, which constantly adjust the voltage, to keep the current constant, even as the resistance changes.

As the LD heats up and it's internal resistance decreases, a current regulator will drop the voltage slightly, so the current can stay the same all the time.


But if a LD draws 300mA at 2.85V, you can not give it a higher voltage, but keep the current the same or make it even lower.. It's physically impossible.
Twice the voltage and half the current would mean the same power (Watts), but you can not do this with loads. That only works with transformers.[/quote]


what he said

oh and abray, don't forget to get a MXDL custom machined heat sink for the MXDL, it makes life a whole lot easier.

oh BTW, how were you measuring the current hydro? was the DMM in series or in parallel with the LED?

regards,

amk

Which Aixiz module does this diode fit? People have said the diode is 5.6mm, but how does that correspond to the module's size?

It seems the 650nm 5.0mW 12x30mm aixiz.com/shop/product_info.php?products_id=33 is the most common and is available with a glass lens for $3 more (total $7.50), will this module properly house the diode in question?

I already ordered a couple MXDLs and will be machining a heatsink/building the LM317 driver with this diode and an Aixiz module+glass lens if it all meshes nicely.

Also, what resistor value should be used to achieve 420ma from the driver? (edit: via amkdeath's post, it seems that a 3 ohm resistor (~0.5 watt) should produce 420ma with this circuit, but does anyone know the formula for determining this?)

Thanks!
-lazd

lazd said:

Which Aixiz module does this diode fit? People have said the diode is 5.6mm, but how does that correspond to the module's size?

It seems the 650nm 5.0mW 12x30mm aixiz.com/shop/product_info.php?products_id=33 is the most common and is available with a glass lens for $3 more (total $7.50), will this module properly house the diode in question?

I already ordered a couple MXDLs and will be machining a heatsink/building the LM317 driver with this diode and an Aixiz module+glass lens if it all meshes nicely.

Also, what resistor value should be used to achieve 420ma from the driver? (edit: via amkdeath's post, it seems that a 3 ohm resistor (~0.5 watt) should produce 420mz with this circuit, but does anyone know the formula for determining this?)

Thanks!
-lazd

Click to expand...

look a few posts up, the formula is:

1.25 / 3 = .420 A or 420mA

the diode itself is TO-18, or 5.6mm the module is a bit bigger than the diode itself, and the size of the module you need is 12x30mm

sales.stonetek.org sells these for 6 dollars shipped, and you don't have to worry about taking out the 5mW diode.

regards,

amk

amkdeath said:

[quote author=IgorT link=1202158967/120#122 date=1203701740][quote author=SuperNova link=1202158967/120#121 date=1203681725]I recall that:
watts = volts x amps (less heat or waste)

as such, if your circuit runs a higher volts, then you should run it at lower amps to achieve the same watts.  
This is providing that your LOAD is ok with the higher volts.

One good thing about higher volts and lower amps is smaller wires can carry it.  But unless your load can tolerate the higher volts, it may not be advisable.

Click to expand...

You can not drive a LD with a higher voltage and a lower current at the same time.. The current and the voltage are directly related through it's internal resistance.. U=RxI (Ohm's law) If you double the voltage, the current would also double, there is no way around it.

For every voltage, there is only one current (if the resistance doesn't change)..

Of course with laser diodes, their resistance drops, when they heat up. If the voltage was constant, the current would go up. That's why we use current regulators, which constantly adjust the voltage, to keep the current constant, even as the resistance changes.

As the LD heats up and it's internal resistance decreases, a current regulator will drop the voltage slightly, so the current can stay the same all the time.


But if a LD draws 300mA at 2.85V, you can not give it a higher voltage, but keep the current the same or make it even lower.. It's physically impossible.
Twice the voltage and half the current would mean the same power (Watts), but you can not do this with loads. That only works with transformers.[/quote]


what he said    

oh and abray, don't forget to get a MXDL custom machined heat sink for the MXDL, it makes life a whole lot easier.

oh BTW, how were you measuring the current hydro? was the DMM in series or in parallel with the LED?

regards,

amk[/quote]

Amk, I think in parallel.  I can't remeber, it was soo early this morning... could not sleep.  What should it be series or parallel with the LED?  I have to check it again when I get home from work.


Thanks everyone, you guys answer other questions I wanted to ask but I though it was stupid.

Thanks again, this is a great form!

In series.