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Re: 638 700mw build beam shots added
Yes that last photo answers the question I PMd you about, thanks.
Alan
Yes that last photo answers the question I PMd you about, thanks.
Alan
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638 700mw build MORE PICTURES
- Thread starter APEX1
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Re: 638 700mw build beam shots added
O. so 1.1A*4.7V=4.07W
O. so 1.1A*4.7V=4.07W
APEX1
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No dude! Are you trying to troll?
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Re: 638 700mw build beam shots added
I thought current times voltage is watts
I thought current times voltage is watts
APEX1
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Re: 638 700mw build beam shots added
Not the way you are thinking for the power of the beam though. The diode I used in this build will hit 700mw with the right optics. Now if I had a bigger and better host with correct heatsinking and a driver that hit 1.4 amps It could hit over the 700mw mark its rated for. You have the theory correct to a degree but not for output power of a Laser "Beam"
Maybe someone else could help you understand it better cause I honestly don't know how else to explain it to you. :beer:
Not the way you are thinking for the power of the beam though. The diode I used in this build will hit 700mw with the right optics. Now if I had a bigger and better host with correct heatsinking and a driver that hit 1.4 amps It could hit over the 700mw mark its rated for. You have the theory correct to a degree but not for output power of a Laser "Beam"
Maybe someone else could help you understand it better cause I honestly don't know how else to explain it to you. :beer:
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Re: 638 700mw build beam shots added
I kinda understand it better now because the actual beam can be affected by other factors, sucha as a different lens
I kinda understand it better now because the actual beam can be affected by other factors, sucha as a different lens
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Re: 638 700mw build beam shots added
1.1*4.7=5.17 I don't know where you get the 4.7, I think you meant 3.7, the 3.7v batteries charge to 4.2v but under load will start out at more like 4v and drop from there. You are correct that P=IV but if you want to know the power output of a laser it can't be calculated. You are just talking about the power drawn from the battery. There is much power used in the driver, the diode, and optics that is just dissipated as heat, only a fraction of it is laser output.
Alan
1.1*4.7=5.17 I don't know where you get the 4.7, I think you meant 3.7, the 3.7v batteries charge to 4.2v but under load will start out at more like 4v and drop from there. You are correct that P=IV but if you want to know the power output of a laser it can't be calculated. You are just talking about the power drawn from the battery. There is much power used in the driver, the diode, and optics that is just dissipated as heat, only a fraction of it is laser output.
Alan
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Re: 638 700mw build beam shots added
It makes more sense now. I thought that the power in the circuit had to be the same power rating of the diode. and im terrible at math
It makes more sense now. I thought that the power in the circuit had to be the same power rating of the diode. and im terrible at math
APEX1
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Re: 638 700mw build beam shots added
That still doesn't make sense bud. Just curious as to what made you think that lol I'm not being mean or a smart a*# I just don't understand what you mean by this. Alan just explained it in a non mess up or confusing way. If you ever need help with anything shoot me a pm, I'd be glad to help.. :beer:
That still doesn't make sense bud. Just curious as to what made you think that lol I'm not being mean or a smart a*# I just don't understand what you mean by this. Alan just explained it in a non mess up or confusing way. If you ever need help with anything shoot me a pm, I'd be glad to help.. :beer:
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ok. thanks for your help though
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Re: 638 700mw build beam shots added
Nope. That would be right if diodes were 100% efficient, but diodes aren't anywhere near that
Nope. That would be right if diodes were 100% efficient, but diodes aren't anywhere near that
DTR
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Re: 638 700mw build beam shots added
The Oclaro diode is consuming about 2.4V when driven @ 1.1A so 1.1A*2.4V=2.64W of energy beeing feed to the diode. If the diode is about average it will have around .86W of optical output with a G-2 lens(about 20-25% back of that with a three element lens). So 2.64W-0.86W leaves around 1.78W of energy being dumped as waste heat due to the efficiency of the conversion process.
The Oclaro diode is consuming about 2.4V when driven @ 1.1A so 1.1A*2.4V=2.64W of energy beeing feed to the diode. If the diode is about average it will have around .86W of optical output with a G-2 lens(about 20-25% back of that with a three element lens). So 2.64W-0.86W leaves around 1.78W of energy being dumped as waste heat due to the efficiency of the conversion process.
Gadget
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Re: 638 700mw build beam shots added
It does, but only in the electrical side of the system.
You can only supply the diode with a current up to its
Imax, go beyond this rating and you risk ruining the diode.
Think of it this way. This is a crude representation of a
handheld laser system.
(Batteries) - (Driver) * (Laser Diode) - (Beam)
Everything left of the asterisk is electrical power.
Everything right of the asterisk is optical power.
These are two totally different types of energy.
Values on the left of the asterisk can be pre-calculated with ohm's law in volts, amps, and ohms.
Values on the right of the asterisk CANNOT be pre-calculated with ohm's law, or any other
formulas for that matter. They must be determined by hard measurements.
(Edit: Well, they can to an extent, but it requires access to ALOT more data than the average hobbyist has)
Hope this clears things up for you in terms of the power types found in a handheld laser system.
-G
It does, but only in the electrical side of the system.
You can only supply the diode with a current up to its
Imax, go beyond this rating and you risk ruining the diode.
Think of it this way. This is a crude representation of a
handheld laser system.
(Batteries) - (Driver) * (Laser Diode) - (Beam)
Everything left of the asterisk is electrical power.
Everything right of the asterisk is optical power.
These are two totally different types of energy.
Values on the left of the asterisk can be pre-calculated with ohm's law in volts, amps, and ohms.
Values on the right of the asterisk CANNOT be pre-calculated with ohm's law, or any other
formulas for that matter. They must be determined by hard measurements.
(Edit: Well, they can to an extent, but it requires access to ALOT more data than the average hobbyist has)
Hope this clears things up for you in terms of the power types found in a handheld laser system.
-G
Last edited:
APEX1
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I doesn't get any better than that!!!!
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thanks for the help guys. now I have a better understanding of how light energy and electrical energy work together in the system
APEX1
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Yeah man, I hope you don't think I was being mean or anything I just assumed that you new diodes are never 100% efficient
