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

Thanks for supporting LPF!

Avery's Instagram | Open Source Cybersecurity Software by Avery | Considering selling LPF, DM if interested

The Little Diode That Could

CurtisOliver

0
LPF Site Supporter
Joined
Jun 12, 2015
Messages
7,420
Points
113
2 years 2 months roughly to go until 100,000 hours.
 



Joined
Aug 22, 2013
Messages
367
Points
43
Actually it is the current you are running it at in mA. The voltage, or Vf is not meaningful.
I guess I should clarify. These don't have driver circuits. Because they are so cheap, they only have a 92Ω resistor. The Vf of the diode is 1.834V and I'm running it from a 5V constant voltage source.

When I said the rated 5V, that's because that is how they advertised it. I don't have any real way to test the current draw without disconnecting the diode.
 
Joined
Sep 20, 2013
Messages
16,757
Points
113
I guess I should clarify. These don't have driver circuits. Because they are so cheap, they only have a 92Ω resistor. The Vf of the diode is 1.834V and I'm running it from a 5V constant voltage source.

When I said the rated 5V, that's because that is how they advertised it. I don't have any real way to test the current draw without disconnecting the diode.

You can still figure out the current coming through the diode if you know the Vf of the diode. Subtract that from the 5 volts to get the voltage drop across your resistor and use Ohm's law to get the current.

I just did it for you. I=0.034 amps or 34 mA.
 
Joined
Aug 22, 2013
Messages
367
Points
43
You can still figure out the current coming through the diode if you know the Vf of the diode. Subtract that from the 5 volts to get the voltage drop across your resistor and use Ohm's law to get the current.

I just did it for you. I=0.034 amps or 3hat is the measured Vf of the "other" diode of the same type, I'm not well versed in

You can still figure out the current coming through the diode if you know the Vf of the diode. Subtract that from the 5 volts to get the voltage drop across your resistor and use Ohm's law to get the current.

I just did it for you. I=0.034 amps or 34 mA.

That is the measured Vf of the "other" diode of the same type. And correct me if I'm wrong but won't the Vf change with temperature?

I'm admittedly not well versed in the voltage characteristics of laser diodes, ohms law I've got down, but I thought they didn't obey ohms law as they aren't linear.
 

kecked

0
Joined
Jun 18, 2012
Messages
809
Points
63
Measure the voltage drop on the current limiting resistor. You then know E and R so Calc I
 
Joined
Sep 20, 2013
Messages
16,757
Points
113
That is the measured Vf of the "other" diode of the same type. And correct me if I'm wrong but won't the Vf change with temperature?

I'm admittedly not well versed in the voltage characteristics of laser diodes, ohms law I've got down, but I thought they didn't obey ohms law as they aren't linear.

I assumed that was the Vf of this diode. If it has been running for years it is as hot as it is going to get so, no that won't be a factor. You are measuring the current through the resistor which is in series with the diode. The current will be the same through both.
 

Sowee7

Active member
Joined
Feb 1, 2021
Messages
426
Points
43
It was so little power to begin with that it's hard to really tell. In a daylight lit room, on a wall 15 feet away the dot is still easy to see.

My power meter didn't even register this thing years ago when I tried. Since quantifiable data aren't available, and given the 10 years that it's been, I would say it seems more dim,

Real time edit. I just checked it side by side with the last unused diode from this batch. Power output has reduced significantly.
Do you think it could have LED'd?
 




Top