That's a really useful item you have there JECS, but just because the socket worked for you doesn't mean it'll work for everybody. Just don't go around saying using a socket is a great idea, it is, but only when you're using a regulated power source like you're doing. Most of the new people are building the DDL driver, most of them don't know a lot about electronics, (I don't know a lot about electronics, all I've learned has been because of DDL's driver) they'll not know it's working for you unless you explain that you're using a regulated power source, and if you tell them to use a socket with DDL's driver it'll surely not work for obvious reasons.
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solderless ld connection
- Thread starter explsvcookie257
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phenol
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JECS said:
hm, i still dont get it - if the voltage remains the same w/ and w/o load, then this means that you are feeding your LD from a voltage source with rock-solid 3.08volts. using a current source would imply that when the load is disconnected, the voltage would increase up to its max trying to keep the current constant...
i have 2 knobs to change the voltage and amperage independently. the voltage is adjusted with a trimpot so i can have any output i want, and the amperage is controlled with a radial switch which lets me choose between 5 predefined settings. im waiting for my green and blue in the mail. i want to try and have them all connected to it at the same time. might be a good power supply to use in a scanner. not including the transformer the board is like 2x3.5 inches
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phenol said:
this is why he is not burning out his laser diode, and why if a socket is used with a DDL driver it would fry a diode, his supply is voltage clamped which prevents the voltage spike that kills the diode.
phenol
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well, it is clamped at 3 volts, which th LD seems to handle safely... what if he touches the voltage knob by accident and sets the clamping level at 5 volts for instance? if the diode is inserted in the socket, the voltage would increase to the point where the current is equal to the setting he has selected. the very moment the diode is unplugged, the voltage would hit 5 volts and charge the buffer caps that are probably present on the output of this PSU with 5 volts. when the ld isreinserted, it will suffer a brief/or not so brief if the filter cap is 2200uF/ surge.
i vary my voltage from 3 volts, for long lasing, to 4.5v, for burning. my psu does have some broken wires or something where sometimes just touching a knob will briefly cut the power.. this wasnt about the type of supply though. im just saying its possible to use a socket without frying the diode. the diode comes out all the time and i reinsert it without turning the power off or discharging the cap. i keep seeing people say not to use them under any circumstance and im just saying it can be done.
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JECS said:
it's just better to say not to, it's not real easy to break a LD by soldering so idk why so many people are paanoid about doing it. Some people may quickly scan through a thread like this not reading real closely and may end up breaking a LD because they use a socket with DDL's driver.
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just think about it, if you use an lm317 circuit with the diode in a socket, it comes unplugged for a split second causing the regulator to bring the voltage up to 6v, charging the cap up to 6v then when it conects back the capacitor discharges 6v to it all at once.. Or in short words.. socket=ld death
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roSSco
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So you're saying a socket shouldn't be used with a LD? 
