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Pondering Laser Efficiency

TZ250

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Everyone,
My new Survival Laser laser is a lot of fun. Soon, I will test the output and report the results here.

The expected output is 2W. So, I did some quick checking of the battery voltage and current. My two batteries measured 7.81V. The current measured 1.7A.

Since P=VI, Watts=7.81V X 1.7A = 13.277W.

Assuming a 2W output, the efficiency is 2W/13.277W = 15.06% efficiency. :cryyy:

I was hoping for 50% efficiency, but that is not going to happen.

Does the laser really loose that much energy due to heat and resistance (which is turned to heat)?

Am I missing something?
P.S. I searched for that information, but did not find the answer. I must have used the wrong search terms.
 
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Cyparagon

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That's the efficiency of the entire system - the "wall plug efficiency," although you're not accounting for cell charger or charge cycle losses.

You have heat lost in the cells, driver, diode, and optical assembly. Diode efficiency is much better. It's like 4.5V drop at 1.7A so 26% or so. Beefy IR diodes can be up to 60%.
 

DrSid

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Yep .. 445nm diodes 20%-30%. But the driver burns another 50%. If you want to be effective, discard the driver ! :D Check direct drive build in my sig. But then you are limited to what batteries you have at hand, and there are also other problems, like output stability.
 
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TZ250

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That's the efficiency of the entire system - the "wall plug efficiency," although you're not accounting for cell charger or charge cycle losses.
Cyparagon, you are correct. It is daunting to think of trying to calculate total efficiency. We would have to include the efficiencies of the power plant, the losses in the lines, the efficiency of the coal transportation system....

That is a long way to say, that I am only interested in the "wall plug efficiency". :)




Yep .. 445nm diodes 20%-30%. But the driver burns another 50%. If you want to be effective, discard the driver ! Check direct drive build in my sig. But then you are limited to what batteries you have at hand, and there are also other problems, like output stability.
Thanks for the information. I'm a newbie to laser construction, so I'll stay with a driver for a while. I did not expect the driver to be so inefficient.




15% is quite good. Just look at gas lasers...
Thank you. I will research gas lasers. Prior to this experiment, I could not imagine calling 15% efficiency 'good'. :thanks:
 

Blord

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Did you measured the battery voltage under load ? It could sag a lot under 1.7Amps.
 

Cyparagon

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If you see the cells sag to 6V for example, you can't take that new value as the supply voltage (unless you're just calculating driver efficiency). The other 1.8V is dropped in the cell itself because of a parasitic series resistance inside the cell. Therefore, the voltage sagging is actually another source of heat and subsequent inefficiency.
 

TZ250

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Thank you all for helping me to better understand the losses in this system. I did not measure battery voltage under load.

I appreciate the helpful inputs! :beer:
 

Garoq

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Did you measure current draw at the tailcap? It should be closer to 1.37-1.40A.
 

TZ250

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Did you measure current draw at the tailcap? It should be closer to 1.37-1.40A.
Gary, I removed the tailcap and connected one probe to the battery and the other to the host.

I hope to go to the optics lab next week. I will report the results.
 

Garoq

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Gary, I removed the tailcap and connected one probe to the battery and the other to the host.

I hope to go to the optics lab next week. I will report the results.
I should say that I usually measure 1.4A at 8.4V, so your number isn't that different.
 




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