Electro132
- Feb 12, 2013
- 261
- Joined
- Feb 12, 2013
- Messages
- 261
Aside from the funkyness of your drawing, it is mostly correct.
I would remove the capacitor from the output as it will cause you huge problems if the photodiode's connection ever becomes intermittent (you will cook the photodiode).
Output needs to go to the Laser/LED.
No, this is a current regulator.
ADJ goes to the load and resistance designed to drop 1.25V at the desired current goes between the output and adj.
I've noticed that the diode in the original image is superfluous too. Remove that and the capacitor.
A good test is to replace the load (the laser diode) with an ammeter and measure the current. This should be variable using the pot.
I am rather concerned that your trimpot won't be up to the current. What current do you want to use -- it would be easier to just use fixed resistors.
Using 20 ohms (both resistors in series) will give you 62mA.
Using one of these resistors (10 ohms) will give you 125mA
Using two 10 ohm resistors in parallel will give you 250mA
Start with 62mA
remove the diode and cap -- the cap just puts the laser diode at risk (I've said that earlier).
Don't hook up the laser diode, just place an ammeter in its place and measure the current. It should be within a few percent of the calculated figure.
Tell us what you get.
So, if you have the two 10 ohm resistors in parallel (giving you a max current of 250mA) and a 100 ohm pot in series with them, it's power rating needs to be 0.25*0.25*100 = 16 Watts.. It may not be practical, or if practical, it will be expensive.
When i did your calculation I got 6.25 not 16 watts. Just wanted to know if you converted it from something.
Also i was wondering what if i just applied 2x 2 pairs of darlington pairs to multiply the input power to make up for the power amount needed (e.i. 16 watts) ? The LM317 VR can go up to 35V.
Which is why I try to give the working. I calculated 1/4 * 1/4 = 1/16 then had a brain fart that 1/16*100 = 16!
I'm not sure exactly what you mean, but the answer I think is No.
Which is why I try to give the working. I calculated 1/4 * 1/4 = 1/16 then had a brain fart that 1/16*100 = 16!
I'm not sure exactly what you mean, but the answer I think is No.
Using 20 ohms (both resistors in series) will give you 62mA.
Using one of these resistors (10 ohms) will give you 125mA
Using two 10 ohm resistors in parallel will give you 250mA
Start with 62mA
remove the diode and cap -- the cap just puts the laser diode at risk (I've said that earlier).
Don't hook up the laser diode, just place an ammeter in its place and measure the current. It should be within a few percent of the calculated figure.
Tell us what you get.
Cool!Hey steve, it worked. YIppeee!! I'll load it up and give you guys a link.
Who knows? (hint -- the manufacturer, and that's why they have datasheets)In the meanwhile, just wanted to know, is it possible to go beyond 250 mA on a red LD? I was thinking of maybe using 5 ohms Res or even 1 ohm Res but was afraid that it might destroy the diode.
Cool!Who knows? (hint -- the manufacturer, and that's why they have datasheets)
Quite often you will find a graph that shows maximum current vs duty cycle (or pulse length)
Ok, since this is from a dvd burner, i'm assuming it will take 480 mA max or 430 mA at least? Just out of my head since it is no ordinary LD.
Also would you happen to know what type of lens and material the Aixiz Housing Cylinder is made out of?
It's abit difficult trying to get one here. I found one around $8 aud, but i also have to pay P &H of another $8 which is $16 altogether just for one.
I was reading that you could input some power at one end and it can be converted to a higher amount when it comes out the other end.