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FrozenGate by Avery

MXDL 3W blu-ray

  • Thread starter Thread starter jamie.91
  • Start date Start date
Re: Is this host possible ?

sorry but i dont want to buy a driver i want to make one for two reasons

1. im a cheap lol ;D
2. i am 17 and still learning and by doing everything myself i learn quite a bit :P

and if i buy one i have to wait for postage whereas if i go to town and get all the parts i need i can build it in the postage time which gives me something to do lol

but schematics,tutorials and any other info is always appreciated  ;)

thanks jamie
 





Re: Is this host possible ?

ok sorry

There are instructions on the second link if you want to build it, but you need a boost driver of some sort
 
Re: Is this host possible ?

Would i realy need a driver ? could i not just use a reststior or at least as few parts as possible ?
 
Re: Is this host possible ?

I think you do and 90% of the LPF community would agree
 
Re: Is this host possible ?

well how can i make a driver fit in there lol
 
Re: Is this host possible ?

i knew i couldn't make one so i bought one.
If you think you can make one, try, but the other option is to buy one (if you are making a blu ray as they need higher currents than 3.6v) or risk it with a red diode (that needs fewer volts to work)
 
Re: Is this host possible ?

well i was plannig on making a red one first then a blu-ray and i think i could run this off a 6v batery woth some modifications would that be able to run a blu ray and a red with varying parts and the right resistor for each one ?
 
Re: MXDL

Got your PM... And read the thread..

So, first of all, you DO need a driver. There are MANY reasons for this..


Direct drive: (battery + resistor)
- Current starts high on a full battery, and then drops all the time, the power is never the same twice, when you turn it on
- As the diode heats up, it's internal resistance decreases, so the current actually climbs at first. If it's not limited low enough, it can climb too high, and kill the diode. More current means more heat, more heat more current, sometimes resulting in a thermal runaway.
- If you put a better battery in, or a rechargable instead of primary, with a higher voltage, your diode is dead

This means, that while it is possible to safelly drive a diode from a battery through a resistor, you have to set the current much lower, so that you are safe for sure. On a full battery, the current will have to be below the max safe current, because it will climb first (with heat). Then it will drop, and the power will be lower and lower - the battery voltage drops as it discharges.

If your diode dies, and you put another one in, you have to adjust it again, because the current will not be the same, even if it is the same diode!

Because of the properties, you have to set the current through a DMM. But as you then remove the DMM, the current actually goes up, because you remove the resistance, the DMM introduced! Unless you have a 1 Ohm 1% resistor permanently in the circuit and measure the voltage drop across it.. But this 1 Ohm resistor will become part of the limiting resistance and influence the current.


Constant voltage: (7805, Cdanjo's boost circuit, or any other CV source....)
- here the power will start at the same level every time you turn the laser on, untill the battery is empty.. BUT!
- again, as the diode heats up, the internal resistance decreases. Since the voltage stays the same, and resistance goes down, the current goes up with heat.

Again, this means you have to set the current BELOW the max safe current, because it will climb with heat. How far it will climb, depends on heatsinking, but it WILL climb in any case, even with the largest heatsink. To prevent thermal runaway, you should limit the current 20mA below the usual current, and hope, that the current will not increase more than 10mA.

It's still much better, than just a resistor, because the power doesn't drop with the battery voltage - you get the same power, every time you turn the laser on. Also, if you put a better battery in, the current will not jump up and kill your diode, like it would with a resistor. But the stronger the diode, the less suitable this option becomes, because it will heat up more..

If you put another diode into the circuit, you have to adjust it again, because the current will not be the same, even if it is the same diode!

Because of the properties, you have to set the current through a DMM. But as you then remove the DMM, the current actually goes up, because you remove the resistance, the DMM introduced! Unless you have a 1 Ohm 1% resistor permanently in the circuit and measure the voltage drop across it.. But this 1 Ohm resistor will become a part of the limiting resistance, and influence the current, so you have to take this into account.


Driver: (317, Rkcstr's micro driver, LavaDrive, any constant current source)
- The laser will again come on at the same power every time you turn it on, even when the battery voltage drops, right untill the battery is empty.
- But with a driver, the current will remain the same, no matter how much the diode heats up.
- This means, the diode will heat up, the internal resistance will decrease, and the driver will automatically LOWER the voltage just enough, so that current stays the same all the time.

So, your power is not dropping with the battery voltage, when you use a driver. And there is no risk of a thermal runaway. You can actually set the diode to the highest safe current, because you know for sure, that it won't change.
You get the same power every time you turn the laser on, right until the battery is empty.

If you put another diode in, the current will be the same, even if it is a completelly different diode!

If you set the current through the DMM, and then remove it, the change in the resistance does NOT influence the current. If you put a 1 Ohm resistor in, to measure the current, it also does not influence the current. The driver makes sure the current is always the same, regardless of the load.



With low power diodes, that don't heat up, you can use a constant voltage source if you adjust the current correctly.. The current won't change much, but it should still be set a few mA lower. (with low power diodes a few mA too much is enough to kill them)

With high power diodes, that create a lot of heat, you should always use a driver.
It can be done without a driver. But there are the above mentioned drawbacks - less power, more risk.

With a driver, you can actually set the power higher, and be safer at the same time!


Constant voltage is better, than a resistor, but again, the current will climb with heat, use a series resistor large enough, so the current ALWAYS stays below the max safe current.

With only a resistor, you have to limit the current so, that it stays quite a bit below the max safe power when the battery is COMPLETELLY FULL, and the diode hot.



Oh, and a constant voltage circuit (boost or not) is not a driver. But it is still much better, than a resistor!

A real driver is a constant current source. With a constant current source, the resistor actually SETS the current. If you know the regulator's feedback voltage, you can just calculate the resistance needed for a certain current. After you set it, the current will ALWAYS be the same, regardless of the battery voltage or small differences in the forward voltage of the diode, or even with a completelly different diode - the driver simply adjusts the voltage so that the load is always getting just enough for the current you set to flow through it (as long as thedriver is also getting what it needs for this).

With a constant voltage source, the current will be different with every load, even if it's two identical diodes from the same batch. There are always small differences in the Vf (forward voltage) of the diode, and a small difference in voltage results in a BIG difference in current.

So with constant voltage, you can't set the current, you can only limit it somewhat, with a series resistor. And you have to adjust is for each load separatelly, because of the small changes. It's actually harder to adjust correctly, and after that, the rest depends on your heatsinking..

And a resistor + capacitor is the worst of all options.



To drive a red from one Li-Ion, you could use an AMC7135. It's cheap from DX. But then you have to insulate the diode heatsink from the host body. If you can insulate the heatsink, then the AMC is very good. Otherwise you need two Li-Ions and a linear driver or one Li-Ion and a buck/boost driver.
For a blu ray, you need three Li-Ions and a linear driver, or one Li-Ion a boost driver.


A red could work from a single Li-Ion and a resistor, but you would have a lower power laser with an always changing, mostly dropping power.
A blu ray could work from two Li-Ions, with the same drawbacks.



If you want to save money on the driver, you will end up losing it on diodes. And if you don't lose them, your laser will not be what it could be with a proper driver.
 
Re: MXDL

lol its ok ;)

so IgorT would a flexdrive be the best option to run a bluray from a single cr123a battery ?

thanks jamie
 
Re: MXDL

Haha! I was just trying to explain all the options in as much detail as possible. I'm sorry if it's too long, but i wanted to make sure it would be understood in the end.

But what did you say wrong, robjdixon?



Anyway, Jaimie, yes, a flexdrive should be a good option. The only other option i see is etching and soldering your own driver.

And making your own can be hard. I've been making electronics for 10 years now, and the tiny drivers i have to make are getting on my nerves.. I actually killed like five of the tiny legless regulators, while handsoldering them. I would like to think, that's because of the large number i had to make in a very short time..  ::)  ;)


Sometimes, trying to save money can make you lose money.. And in this case, the end result could not compare to using a driver..

I mean, you can always just throw something together, but here the end results are than worth doing it properly.



I've seen some people use the unregulated boost circuits, that these flashlights often contain. Now, theoretically, you can make it work, but it would be worse, than a battery and a resistor. Those circuits are mostly completelly unregulated, and have a horribly large current ripple.. That's because LEDs are very sturdy devices, and can take the abuse without a problem. LEDs don't care much, if the current is two times too high, or if the boost circuit is putting out only spikes - they are often made to blink the LED rapidly..

LDs can be very powerful on one hand, but fragile on the other. But as long as you treat them right, they are MUCH more fun than LEDs..  ;)
 
Re: MXDL

thanks IgorT great help i think i shall invest in a flexdrive lol

thanks jamie

btw any other things i should know or any helpfull pointers your help is very much appreciated
 
Re: MXDL

Yeah, there are some things you should know about the LavaDrive..

As i said, it's a constant current driver. Some people, when they make a constant current driver, like the LM317, then power it up, to measure the voltage it is putting out, to see if it's ok for the diode.
Then they connect their DMM to the output, and measure almost 6V, and think they did something wrong, because 6V is too much for a diode, isn't it?

But the driver is just trying to push a certain current through the load.. And in this case, the load is air.. Since it's hard to put out 300mA through empty air, the driver tries raising the voltage, which then peaks slightly below the battery voltage.

The driver regulates the current, by adjusting the voltage - it always gives the load just enough for a constant current to flow through it. So the voltage will depend only on the load - it will be whatever the load needs for that current.


So don't get the idea to measure the voltage without a load! Voltage doesn't mean much here anyway. As long as the driver is regulating the current (which you test on a dummy load), it means it is taking care of the voltage..


But since this is a boost driver, it's a little different. If you powered it up without a load, it would not stop below the battery voltage - it can go higher than that! So it would go higher and higher, then you would hear it whining (yes, whining - as it is changing the modulation of the switching into the coil), and it could actually hurt itself!

Do not power a LavaDrive up without a load! Use a dummy load for testing!


Another thing about the LavaDrive is, that it can't regulate the current if the forward voltage of the load is under 2.7V. So you have to make sure, that you use enough diodes in your dummy load! 1N4007 has a Vf of 0.7V or so, and it changes with the current. So use enough, that their combined Vf is over 3V, but under 5.5V (LavaDrive can't boost above 5.5V, well it probably can a little, but it shouldn't)


Also make sure, you discharge the output caps on the driver BEFORE even thinking of putting a diode on there. The caps could contain a high charge (especially when powered without a load), and they could then discharge through your LD the moment you would connect it.

This can also happen, if you don't have permanent connections between the LavaDrive and the diode.. If it's just clipped on, it can lose contact for a moment, then the capacitors charge up, and kill your diode, when it reconnects a moment later..


So permanent connections, and always short the output, before attaching the diode..


You also have to set the current before attaching the diode. And you should not be adjusting it with the diode connected! If i'm not mistaken, the low range driver can go much higher, than the specified range, the low range is just the part that is easily adjustable. After that it becomes exponential, and a small change in the pot position results in a huge change in the current.


So all the current adjustments should be done on a dummy load. You can put the DMM in between the driver and the dummy load, or you can solder a 1 Ohm resistor in series with the diodes, and measure the current on that resistor.

If you measure the voltage drop across a 1 Ohm 1% resistor, in the 2000mV DMM range, the voltage drop in mV ACROSS it, is the same as the current in mA THROUGH it (Ohm's Law - U = R x I, if R = 1, then U = I)...


Otherwise, Lava already put a 0.1 Ohm resistor on the output instead of 1 Ohm.. 0.1 Ohm loses less power, if it has to be there permanently (on the dummy load it doesn't matter) and you can measure the voltage drop across the 0.1 Ohm, and simply multiply it by 10, to get the current...


Hope that helps.. ;)
 
Re: MXDL

yes this helps a lot but as this would be my first experience with the flexdrive there is a lot to remember lol

you say the dummy load needs more than one diode how many would be a good amout between the 2.5V and 5.5V ?
btw could i not use an LED ?

i have built an lm317 circuit but this sounds toatally different lmao :D

thanks jamie
 
Re: MXDL

Hehe, it does the same thing as a 317.. But in a different way.


As i said, use 6 or 7 diodes for a dummy load. You could also use a high current LED.. I use a CREE LED for a dummy. It's slightly above a red LD in Vf, and with three 1N4007 diodes, it's close to the PHR diode in Vf, so it's a good load, and also gives visual feedback. Of course this feedback is useless, you still need a DMM.


But as long as the Vf of your load is between 3 - 5.5V, the current you set there will be the same as the current once you put a diode in.


This is another good thing about constant current drivers - you can't use a dummy load with a constant voltage source or with a resistor, because no dummy load will behave exactly like the LD.
 





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