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FrozenGate by Avery

Red DIY DVD laser burner.

where are you finding a datasheet? I wasn't able to find one.  hmmmmm.  Oh I see your looking at the datasheet for the chip itself.  Ok.  I'll show you on the picture where your vout+ and vout- is.

 





yeah sorry, realized after before you posted that that was what you were looking at. So essentially that thing all by itself is regulating the current to 350ma. this could be used for any circuit as long as you add a resistor (variable) and a capacitor for spikes.
 
Yup, that's all it is..very simple. It isn't at all like daedal's circuit. This regulator does not require any external components when using it with an LED or practically anything else other than a laser diode. As far as the resistor, whether variable or not, you need to make sure to use one with a high enough wattage rating so it can handle the current.

Using ohms law: Watts = Volts X Amps. So worst case scenario..3.6 volts X .350ma's = 1.26 watts.

As you can see you would really need to play with different voltages to determine what would be best.
 
[quote author=Kenom link=1181635652/195#204 date=1191180060]oh and for anyone interested here is a picture of the original dorcy circuit.

Are there any components on the backof teh dorcy circuit board? Is this the circuit used in the dorcy that uses a cr123 battery?

This looks like an inductive based boost switcher. Has anyone taken any output current vs input voltage measurements?
 
there is nothing on the back except two circles. One outter circle for negative input and one inner circle for positive input. I've not tried any other means of powering this circuit but I can do so if you would prefer. and yes it is powered by the 1 cr123.
 
Just an off-the-cuff guess: +3 volts comes into one leg of the inductor. The other end of the inductor goes to an output transistor inside IC1 and to a schottky diode. The output of the diode goes to IC1 for monitoring and the 2 caps C5 and C6 in parallel. The voltage on the caps is regulated to ~5 volts. R330 (3.3 ohm resistor???) goes from the filtered 5 volts to the LED to limit current to the LED. So, (5 volts - Vd) / R330 = ~ 350 mA???

This is just a quick guess to stimulate thought.
 
Now that you spelled it out it makes perfect sense and R330 is the only resistor in the circuit. So replacing it with a variable resistor would be what Kenom is looking for. But it would require a pretty hefty power rating of around 1 watt, if I am understanding this correctly.
 
We need an open circuit output voltage measurement to see what the boost switcher is regulating to. Also, if the value of the R330 resistor could be measured with a DMM would help understanding the circuit.
 
well if ya want me to I'll ship you the circuit and you can play with it. All you need to do is pay shipping. I'm to broke to afford that kinda stuff.
 
I just bought a dorcy that runs off a cr123 battery. It has the same circuit as Kenom posted. The open circuit voltage is 5.09 volts and the R330 resistor is a 0.33 ohm resistor (not for current regulaton as previously thought). It looks like the LED is operated below the knee of the V versus I curve (constant current mode) such and its current is limited to ~320 mA when Vload > Vbattery. I'm working on a more complete output V versus output I for the supply.

Preliminary testing shows an operating point for the supply at 3.9 volts and being able to deliver 250 mA (i.e., voltage regulate mode) for the GB laser diode. To preliminary size the resistor to put in series with the GB LD for 250 mA you have to use this equation:

Resistance = (~3.9 - Vf@250mA) / 0.25 A

I put a regular diode (VF = 1V @ 1A) in place of the LED. The current through the diode was over 0.5 amp. I'm not sure how the circuit behaves when Vload < Vbattery.

Has anyone measured the Vf vs If for the GB LD?
 
I think were making it more complicated then we should. I think we could add a resistor inside the spring somehow, and add a cap right up near the diode.
 
Why not simply add in a 5 Ohm resistor to the leg of the diode ? The circuit is VERY stable, and produces minimal spikes at best - therefor it should work out okay with only adding the resistor.
 
OK, I guess I was making things too comlpicated, but I wanted to see for myself how the boost circuit works. The weird thing is that if your LD voltage is much less than the battery voltage you might be getting much more current than you expect. Without having the GB LD in hand, I'm guessing a ~2-3 ohm 0.5 watt resistor in series with the will get you ~250 mA based on how my dorcy supply operates.
 
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