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ArcticMyst Security by Avery

20w coherent Will computer PSU work?

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Purchased a 20w coherent array from ebay, with crystals. I have read elsewhere some people using comp PSU to power them. How is this accomplished? Will the PSU need to be modded? The research I've done shows there to be quite a few PSU's with 12v, 5v, and 3v outs. How do I isolate the 3v out?? Thank you for any insight. Also, will a 1w 808 produce anything with the crystals?
 

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Thank you. That looks to be very helpful. So it seems I just need to find one w/ the right wattage? What would that be for my diode? All the info I got with it is in the 3rd pic. Also would I be able to regulate the power to it or would it just be full power when switched on?
 
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You might want to wait for someone who knows about this to answer but based on the Operating Current of 32A and you using the 3V rail that would be equal to 96 Watts so I would say 150-200 Watt supply would do fine... I don't think you will find much under 350 though.
 
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So I dont want one over, say, 200watts? Good to know, as I was researching the 600+ watt psus.
 
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So I dont want one over, say, 200watts? Good to know, as I was researching the 600+ watt psus.

You _shouldn't_ NEED one over 200 Watts based on the spec sheet unless I am reading it wrong.. Adam had a 60 Watt Coherent and he had a fancy power supply, he was running i think 2.3 volts @ 60 amps...

A 200 Watt supply should theoretically be able to handle that Amperage at that Voltage level however I am no expert on this subject so you might want to wait for someone who knows more than me to respond before you buy anything.

I have never worked with any CO2 lasers or anything even close to a Watt.
I have worked with PC power supplies before and used them to run high current chargers however with no problems.
 
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The wattage rating of an ATX power supply is based on all rails at max load. You need one that can do 32A on ONE RAIL (preferably the 3.3V rail). It varies between models, but it appears to me you'll need at least 650W.

On the other hand, you don't NEED to power the diode at max.
 
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The wattage rating of an ATX power supply is based on all rails at max load. You need one that can do 32A on ONE RAIL (preferably the 3.3V rail). It varies between models, but it appears to me you'll need at least 650W.

On the other hand, you don't NEED to power the diode at max.

Just out of curiosity what calculation did you use for that?

You might also need to beef up the wires a bit, I don't know if it's built to have 32 Amps on that one wire or not...
 
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No need to calculate that, it's been common for years that a computer psu's rated power is it's combined power over all voltages and rails and in case of cheap brands the peak power.

Get a quality power supply and also put some load on the unused lines. Build a good current driver capable of handling any noisy or spiking PSU output. Add a thermistor and a few TEC's to regulate the temperature. It only needs to be stable enough to hold to roughly tune the diode bars wavelength. With 19 emitters it can possibly be right for all of them at once so it's no supercritical. Don't forget the crystal's cooling. be carefull with static electricity, it can also kill these big diode bars.

I take it you have appropriate safety eyewear?
 
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I am aware the rating is based on peak and over all rails but the other rails wouldn't be loaded.. I was just wondering where the 650 watt number came from...
 
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So that will or is going to be 1 beast of a laser doing ~4-5 watt of 532
 
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Build a good current driver capable of handling any noisy or spiking PSU output. Add a thermistor and a few TEC's to regulate the temperature. It only needs to be stable enough to hold to roughly tune the diode bars wavelength. With 19 emitters it can possibly be right for all of them at once so it's no supercritical. Don't forget the crystal's cooling. be carefull with static electricity, it can also kill these big diode bars.

I take it you have appropriate safety eyewear?

Anything like building small drivers like ddl etc?? Yes, I purchased googles for this for sure;)
 




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