wooooooolazer said:Oops i meant voltage, the regulator will never work right, it will always be dropped out because the batteries cant give 6v since ni-mh have a normal 1.2v charge while alkalines have a normal 1.5v
flogged said:[quote author=wooooooolazer link=1185701612/690#702 date=1199485647]Thats not a good idea, it wont give nearly enough current, 4aa ni-mh wont even be 6v when fully charged, just use 6 of them and measure the current when you adjust it.
...lazer.... ;D ;D ;D
Gazoo said:No it is not preferable...the reason being is because your regulator is dropping out and not maintaining a constant current to the diode. If for some reason you use alkaline or lithium batteries, your current going to the diode will spike and possibly kill the diode. If you measure your voltage between the output and adjust it should always be 1.25 volts. If it is lower then your regulator is dropping out.
It is much better to run more voltage to the regulator and have it set at whatever current you want to drive your diode with, than to have the regulator dropping out. So for example if you want to drive your diode with ~160ma's, you would put an 8 ohm resistor between the output and adjust. The current going to the diode would stay the same no matter what your voltage going to the regulator is....as long as it is 6 volts or more. This is what is meant by constant current regulation.
Daedal said:I can see that too much confusion is being had at this circuit. Let me try and clarify a couple of things as I see them:
Third, the capacitor is charged to what the circuit is providing. The capacitor is being charged UP TO 3V and while it is connected to the LD, it is dissipating 3V. This is because the LM317 is BEFORE the capacitor, so it will only charge to what the LM317 is supplying. When there is a spike, all this does is take the first little jolt by charging up in a split second and not getting the full jolt from the battery spike to the LD immediately, as would be the case WITHOUT a capacitor. Therefore, and let me take it slowly here, LM317 is putting out a total voltage of (Battery - 3V), if the battery spikes up, the voltage is being dropped by 3V by the LM317, then the capacitor is a selfish SOB and will suck up all the current before it lets it pass to the LD. Once the capacitor is fully charged, then the LD starts seeing some current and starts lighting up.
Fourth, the 1N4001 is being used for two reasons. One is to stop the diode from overkilling itself when you connect the battery the wrong way around by passing all the current through the 1N4001 and not through the LD. Second, when the battery is connected properly, there is a drop of 0.7V across the 1N4001. Therefore, 6V - 3V = 3V to the diode. With most cases I have tried this I got a little past 3V and closer to 3.5V to the LD. The diodes we are getting from the SenKat group buy, and that is what this circuit is designed for, and most other laser diodes even, require between 2.5 and 3V to lase. With a 0.7V drop off the 3-3.5V supply form the LM317, you have a margin of 2.3-2.8V going into the diode. Which is perfect! You can take it out if you want, again, this is a personal choice.
Abray said:hey, I'm back...
just 2 questiong this time though, you solder the silicon diode in "backwards" right? so that if the batteries were placed in the wrong way, it would be correct for the silicon diode?
and also. blu-rays need ~40 miliAmps, and GB diodes need ~250 (or something like that). would you just use different resistors and keep everything else in the circuit the same to make that change? or do you just need to adjust the pot?