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

LPF Donation via Stripe | LPF Donation - Other Methods

Links below open in new window

ArcticMyst Security by Avery

WTB: 0-700ma+ driver closed found one

J

jake21

0
Joined
Dec 2, 2007
Messages
3,189
Points
0
im looking to buy a ddl driver that can go up to 700ma if passable pm me if anyone can make one thanks

closed thanks for the help ;)
 





jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
Re: WTB: 0-700ma+ driver

Per ohms law, using an LM317 circuit, you would need a 1 watt resistor to go to 700 or 800mA's. (1.25 X .800mA = 1watt)

A 2 ohm 1 watt resistor would give you about 625mA's.
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=CF12JR-ND

If you want it adjustable, you will need to use a 1 watt pot.

I have some nice small pots, but they are 1/2 watt. They are 100 ohm 4 turn pots that make a nice DDL driver with a range from 12.5mA's to 400mA's.

Here's a 25 turn 1 watt pot that will work for you!
http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=3250W-201-ND

It is 200ohms, which means at the most resistance, you will have 6.25mA's. With 25 turns, you will have very nice adjustment. It should be used at 800mA's max in a DDL driver. I'm sure the adjustment will become more and more touchy as the current gets higher. But the adjustment should be very nice! ;)

Only problem is, the pot alone would cost over $20 bucks after shipping and stuff... (Digikey charges a $5 dollar processing fee for small orders)
Jay
 
Last edited:
J

jake21

0
Joined
Dec 2, 2007
Messages
3,189
Points
0
Re: WTB: 0-700ma+ driver

yea thats why i barley order from there ::) well if any one has the parts or can order them ill pay for it to be built :cool:
 
jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
I have all parts except the pot...

If I were to build it for you, I would rather build it into a host, such as the 3405, like I did for daguin's 8X shown here:
http://laserpointerforums.com/laser_pointer_forums_3/forum/showthread.php?t=36790&page=5#113http://www.laserpointerforums.com/forums/YaBB.pl?num=1236464366/96#112

I would charge:

* Host/heatsink combo - $38
* Custom driver installed with lead wires out the front as shown in Dave's 8X thread linked above - $50 (it is this price, because the pot alone is $20)
* Shipping - $6

Total - $94

Then you would use 3 X 10440 batteries for the supply. (not included)
Jay
 
Last edited:
J

jake21

0
Joined
Dec 2, 2007
Messages
3,189
Points
0
sorry jay this will never fit in a host ;D ill find the link and post it here
 
jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
Ahh... I see said the blind man.

Cool! :cool:

Show that baby when you get it going!
Jay
 
Last edited:
Warske

Warske

0
Joined
Mar 10, 2009
Messages
145
Points
0
jake21 said:
0-700ma+ driver closed found one
Click to expand...
I'm glad you found one.  If its readily available and not too expensive, I would be interested too (if not a secret)...
 
LarryDFW

LarryDFW

0
Joined
Dec 6, 2008
Messages
1,581
Points
63
Re: WTB: 0-700ma+ driver

Per ohms law, using an LM317 circuit, you would need a 1 watt resistor to go to 700 or 800mA's. (1.25 X .800mA = 1watt)
Jay
Click to expand...

The correct formula for Power is I x I x R (current squared times resistance)
(1.25 x 0.8 amps x 0.8 amps) = 0.8 watts  (20% less)

I hope this might keep a few builders from using the wrong specifications on components.

Edited- as I misunderstood the original statement, I thought the 1.25 was ohms
    Sorry Jay

LarryDFW
 
jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
Well then, I stand corrected! (if your right, I'm not sure I trust what you say) I got my information from Gazoo, and I trust him.

The fact remains, that for this circuit, you will need a 1 watt pot. You cannot buy a .8 watt pot. Not that I know of...

LarryDFW said:
Inaccurate info from jayrob ... again.

The correct formula for Power is I x I xR
(1.25 x 0.8 amps x 0.8 amps) = 0.8 watts *(20% less)

I hope this might keep a few builders from using the wrong specifications on components.

LarryDFW
Click to expand...

I'm not the one trying to say that I have a lens that is more power with red than a Meredith glass lens when I have not even tried a Meredith glass lens!!!

I make a statement that you think I am wrong about, (although I am correct, as shown below), and you say this? I think maybe you have been sniffing too much of that glue that you use to glue your lens assembly's together... :crackup:
Jay
 
Last edited:
LarryDFW

LarryDFW

0
Joined
Dec 6, 2008
Messages
1,581
Points
63
Jerry;

Your formula is correct.

Resistors have to be large enough to dissipate the power they consume.

LarryDFW  
 
lasersbee

lasersbee

0
Joined
Sep 20, 2008
Messages
17,622
Points
113
Agreed... isn't that why resistors are rated in watts...

A 1 watt resistor is physically large enough to dissipate 1 Watt of power...
(as far as I know)... I've only been using resistors for about 25 years... ;D ;D

I could be wrong... ::)


Jerry
 
jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
lasersbee said:
Not to stir up any crap... but.... I have always used Volts X Amps = Watts :-/

Here are a few examples.... ::)

http://ebtx.com/mech/ampvolt.htm

At 1:10 in the video....
http://www.youtube.com/watch?v=SAB4YKfIGQM

http://van.physics.illinois.edu/qa/listing.php?id=2332


Jerry
Click to expand...

I've always believed the same...

Thus, 1.25 volts at the LM317 sense pins X 800mA's = 1 watt.

But I must say, that I do not claim to know all about electronics. I got this information from Gazoo long ago. And I have always believed that it was accurate information.

But I am all for understanding and correcting my thinking if I have had this wrong all this time.
Jay
 
You must log in or register to reply here.




Top