Welcome to Laser Pointer Forums - discuss green laser pointers, blue laser pointers, and all types of lasers

How to Register on LPF | LPF Donations

Pondering Laser Efficiency

TZ250

TZ250

0
Joined
Aug 30, 2012
Messages
8
Points
0
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.
 
Last edited:





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%.
 
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.
 
Last edited:
Cyparagon said:
That's the efficiency of the entire system - the "wall plug efficiency," although you're not accounting for cell charger or charge cycle losses.
Click to expand...
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". :)




DrSid said:
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.
Click to expand...
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.




Cel said:
15% is quite good. Just look at gas lasers...
Click to expand...
Thank you. I will research gas lasers. Prior to this experiment, I could not imagine calling 15% efficiency 'good'. :thanks:
 
Did you measured the battery voltage under load ? It could sag a lot under 1.7Amps.
 
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.
 
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:
 
Did you measure current draw at the tailcap? It should be closer to 1.37-1.40A.
 
Garoq said:
Did you measure current draw at the tailcap? It should be closer to 1.37-1.40A.
Click to expand...

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.
 
TZ250 said:
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.
Click to expand...

I should say that I usually measure 1.4A at 8.4V, so your number isn't that different.
 
You must log in or register to reply here.


Back
Top