803T...15min use=LED...back from the dead!?!?!

[IgorT]

He said 9V battery.. As in no driver.

If you have a driver, input voltage doesn't matter (as long as it is what the driver needs to work), because the driver gives the diode the voltage it needs for the current to always be the same, and the rest is converted to heat. You can give the driver as much voltage you want, and the voltage on the diode won't change. Only the driver will get hotter and eventually shut down.


Unless Jake was talking about connecting the driver to a 9V? In that case it would make sense, because linear drivers need much more voltage for blu rays, than for reds..

But i think it was hooked up to a battery directly.

 

[brtaman]

Ohh then I misinterpreted the statement, I thought jake was using something along the lines of a lm317/ddl driver?

Either way too bad for your diode man :-/


brtaman

 

[jake21]

i used a adjustable rkcstr driver i had it set at 70ma but was dimm but not dead, so i cranked it up to 100ma and still nothing, so i bought 2 new cr123 batteries still nothing, so i got a 9v batt and hooked it to rkcstr driver and it started to work really good so im confussed why it works only with a 9v batt

 

[IgorT]

Oh, ok, that's easy Jake...

A rkcstr driver needs at least 2.25 - 2.5V more, than what the diode needs.

A PHR diode at >100mA needs 5.3-5.5V. So the minimum voltage for the driver to regulate for sure is >8V.


Two CR2 batteries do not supply enough voltage for a rkcstr driver to regulate the current through a blu ray diode. Especialy if they are only 3V. Two 3V batteries are not even enough for a red!

You always need to select your batteries so, that the voltage is ABOVE the minimum, even when the batteries are empty.

For a red, that needs 3V with a driver that needs 2.5V, that means the minimum voltage is 5.5V, so two Li-Ions at 3.6V each will have 8.4V when full and 6V when empty, and the current through the diode will be the same all the way down, as the battery voltage drops.

For a blu ray, that needs almost 5.5V (depending on the current), with a driver, that needs 2.5V, that means your minimum voltage should be above 8V. Two Li-Ions can not do this. You need at least three Li-Ions.

A 9V can work for a while, untill the voltage drops to 8V, then the current starts dropping.


Only if the driver is getting MORE than the minimum is your current really what you set it to. If the driver is not getting more than the minimum voltage, the current is NOT what you set it to.

That's why it didn't work untill you used a 9V.

Also, if you don't make sure, the voltage is above the minimum with the batteries empty, the power of the laser will drop as the batteries go empty. If you do make sure the voltage is above the minimum with the batteries empty, the laser will have the same power EVERY SINGLE TIME you turn it on. That's the whole point of drivers - so that the power can't escape up nor down!


You should really know this by now:

For a boost driver the voltage just has to be between two values, like between 2.5 and 5.5V, but for a linear driver, the voltage has to be a certain value PLUS what the diode needs, like 2.5V + 5.5V for rkcstr's driver or 3V + 5.5V for a 317 driver. And this is the minimum! You need more than the minimum!

Also, increasing the voltage into a linear driver WILL NOT increase the current through the diode. If you use more batteries, or a battery with a higher voltage, it will just stay in regulation for longer.


You actually did the right thing, hooking up the driver to a 9V battery. But that is only practical with a PS3 diode. For a PHR with higher requirements, three Li-Ions or a boost driver is the only way to go.

P.S. And when i say boost driver i'm not talking about unregulated LED abuser circuits.

 

[IgorT]

glad u got it   my 803t diode was working but then went dimm  so i tryed many diffrent things but would not work, so i hooked it up to a 9v batterie and now it works fine  :-/

Now i also understand the bolded part..

You obviously did used two 3.6V Li-Ions (and not two 3V), and two 3.6V batteries have 8.4V when completelly full. This means, they were able to discharge down to 8V or maybe even 7.7V before the current through your diode started dropping, because the driver didn't get enough voltage.

The full charge voltage above the battery nominal voltage sags very quickly under load, so your diode looked as if it went dim.
But it was just your driver dropping out of regulation, and the current getting lower and lower. As soon as you gave it enough voltage, the current was once again what you set it to.

There was nothing wrong with the diode, or the driver. You just didn't give the driver what it needs.

 

[jake21]

ok now i no lol thanks Igort

 

[IgorT]

Well, if you remember it this time, it won't lead to "misteriously dead and revived" diodes..