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

jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
LarryDFW said:
Jay;

You posted an incorrect formula. *::)

LarryDFW
Click to expand...


I just went on-line to search for an Ohms law calculator because I want to apologize if I stated the wrong information.

http://ourworld.compuserve.com/homepages/Bill_Bowden/ohmslaw.htm

I went here and I entered the first two boxes only...

1.25 Volts, and .8 Amps. Then I clicked the 'calculate' button, and it came up with 1 watt! It seems that the information that I had passed along, was correct! Gazoo must have known what he was talking about.

And somebody else must not have... again!
Jay
 
Last edited:





rkcstr

rkcstr

0
Joined
Dec 1, 2007
Messages
1,368
Points
0
Yes, power (in Watts) can be calculated with voltage mulplied by current (V * I).  Or, if you're calculating power based on resistance, it can be voltage squared divided by the resistance (V^2 / R) or current squared multiplied by resistance (I^2 * R).
 
Warske

Warske

0
Joined
Mar 10, 2009
Messages
145
Points
0
LarryDFW said:
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)
Click to expand...
Both formulas are correct. The problem is that 1.25 is the voltage, not the resistance.

The resistance required for 0.8 amps is 1.25 / 0.8 = 1.5625 ohms

So for LarryDFW to get the correct number using his formula, it would be

(1.5625 ohms x 0.8 amps x 0.8 amps) = 1 watt



If you beat a dead horse hard enough, it looks like it is running!
 
Spyderz20x6

Spyderz20x6

0
Joined
Nov 24, 2007
Messages
2,735
Points
0
LarryDFW said:
Jay;

Quit your constant whining ...

You posted an incorrect formula.  ::)

LarryDFW
Click to expand...
Trolling through the snow, in a one troll open sleigh...
O'er the fields we go, trolling all the way!
Bells on troll tail's ring!
Making spirits fight!
What fun it is to troll and sing,
A trolling song tonight!
Oh!
Trolling bells, trolling bells!
Trolling all the way!
Oh what fun, it is to troll, in a one troll open sleigh!
HEY!
Trolling bells, trolling bells!
Trolling all the way!
Oh what fun, it is to troll, in a one troll open sleigh!
HEY!


Troll, troll, troll.
STFU and get out of here.

Idiot.
 
jayrob

jayrob

0
Joined
Sep 21, 2007
Messages
9,862
Points
113
Larry, I posted in this thread trying to help with good information, and when asked, I even offered to build the driver that jake21 was looking for...

You came in not trying to help (like saying something like 'are you sure about that Jay?'), but no, you came in just trying to trash on me.

Which is fine, if I was saying something misleading, or if I was being very vague about answering questions. (like you always are in your threads) Then I would understand that.

If I am wrong about something, I admit it. And will apologize. In fact, most people here in this great forum will do the same.

Where is your acknowledgment and apology?
Jay
 
Last edited:
LarryDFW

LarryDFW

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

I apologize for my original statement.

I assumed that you were stating 1.25 ohms and that was not your intention.

LarryDFW
 
jayrob

jayrob

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

Apology accepted...

The DDL driver is talked about so much, that I guess I just took it for granted that everybody knows that the LM317 regulator is 1.25 volts between the sense pins.

The pot that I have suggested, is a 200 ohm pot. So in this circuit, it will be 6.25mA's at the most resistance. And I should add, without a series resistor, it will adjust over 800mA's, but since it is a 1 watt pot, it is only recommended for up to 800mA's in an LM317 DDL circuit...
Jay




jayrob said:
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.
Digi-Key Part Search

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!
Digi-Key Part Search

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
Click to expand...
 
Last edited:
You must log in or register to reply here.




Top