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Running a laser pointer? Answered

For a school project, I am running a 4.5 volt, 0.03 amp laser pointer. I need to run it from a 24v power supple. I have tried 1/4 watt resistors, and they ended up smoking. I don't know how to determine what components I need between the + power and the +lead of the laser pointer. Any assistance would be appreciated.

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liquidhandwash
liquidhandwash

8 years ago

I think you should have a look at a regulator like an LM317 they are cheap and available at any good electronic hobby shop and you can adjust the voltage to whatever you want. The are also good to 1.5 amps.
have a look at the pdf.

Screen Shot 2012-11-24 at 4.18.01 PM.png
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dkop1
dkop1

Answer 8 years ago

Thank you. I should have thought of this, as I am using (2)-317s in a separate project.

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iceng
iceng

Answer 8 years ago

While the 317 is a very excellent fast analog voltage or current regulator it
must follow ohms law.

A

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frollard
frollard

8 years ago

You need to try and find a lower power supply, or run more than one laser diode in series.

The laser drops 4.5 volts, leaving 20 volts of potential difference to reduce. that's 80% of your power burning up in a resistor.

Resistor: 30mA at 19.5v is 0.585 watts -- a quarter watt resistor will not be able to dissipate that much power.

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dkop1
dkop1

Answer 8 years ago

This is being run from a nanoline plc controller, and will be sent to Germany in the spring. I am trying to avoid an alternate power source, as converting to their outlets is sketchy at best. I don't mind burning up power in resistors, though it'll be inefficient. Would higher wattage resistors be my solution?

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frollard
frollard

Answer 8 years ago

Yes.

As iceng says use a 1 watt resistor, if you absolutely must. It will get warm/hot.

A cheap dealextreme dc:dc buck converter will do the job for less than 5 dollars.
http://dx.com/p/lm2596-dc-dc-1-25-27v-step-down-adjustable-power-supply-module-149205

Feed it anything and it switch-mode regulates it down to what you need. 80-95% efficient

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iceng
iceng

Best Answer 8 years ago

frollard is very correct in everything ;
  • You are wasting over a half watt  o.585W  in a resistor.
  • ( this means a 1 Watt resistor should be used )
  • A lower source voltage is an obvious improvement to pursue.
  • Using an LM317 will still dissipate the same analog power.
  • Until a lower voltage source like 12 or 6 VDC are used even in Germany.
  • A current regulation 317 mode is the best method of driving a laser.

If you want a circuit to do that just ask :-)

A
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dkop1
dkop1

8 years ago

As well, Iceng; In using a 317, I should still use my 0.03 amps in my P=IE calculation? If I have a 4.5-4.6 volt output, *0.03 to get my wattage?

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iceng
iceng

Answer 8 years ago

If you are still using 24VDC as the source there is still a 19.5 volt
drop to 4.5 volts at 30 ma heating something to over a half watt.

BTW current control is preferred to voltage control.

317constLASERgen.JPG