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Resistor change...Daedals 317 circuit..

LarryQ

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I just finished another burner....

I was shooting for @ 200-250Ma current..as I am well covered thermally...Huge heatsink.

This is a GB diode...

I have (2) 10 ohm 1/2 watt resistors across the LM317

I am only seeing 185 Ma current (I am using a sense resistor for accuracy)

Ok..If I use my math..I need 4 Ohms (I have 1/2 watt 3.9 Ohm resistors..but need 1 watt)

I also have 5 Ohm 1/2 watt resistors...and a TON of 1/4 watt that came in a Radioshack pack!



How can I cut the 5 Ohm total at 1 watt down by 1 ohm???


I don't have (2) 8 Ohm 1/2 watts to use....


Basic electronics...I know....but I need help!!!

Thanks!!!!

LArry
 



Gazoo

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I just noticed you have 1/2 watt 3.9 ohm resistors? One of those will work.
 

xgeek

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I have 1/2 watt 3.9 Ohm resistors..but need 1 watt
If you have 4 of them you can get back to 3.9 at 1 watt if you put 2 in series and then another 2 in series and then parallel them.

    |--3.9---3.9---|
----| |----
|--3.9---3.9---|
 

Gazoo

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Good thinking but you don't need a 1 watt resistor. A 1/2 watt resistor is fine...it does not absorb all of the current...the LM317 absorbs some of it. I forgot where I saw the calculations for this but Daedal also confirmed a 1/2 watt resistor is good enough.

Now a current sense resistor is a different matter and should have a minimum 1 watt rating.
 

LarryQ

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I was told a 1/2 watt single 3.9 was NOT going to be able to handle @225-250Ma....

Input voltage is (2) 3.0 CR123 Lithiums (6V)

THat was why I went with (2) 10 Ohm 1/2 watt for 1 watt 5 ohm total!

I don't know if (4) 3.9 1/2 watts will fit!

UUUUGGGGHHHHHHH



LarryQ

I hate radio Shack....

1 watt 8 Ohm....
Come on 8 OHM..and they don't have them!!!!!!!!!

Larry
 

Gazoo

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I found what I was looking for...the wattage of the resistor needed is calculated by taking the reference voltage and multiplying it by the current. The reference voltage is 1.25 volts. So 1.25 X .250ma's = .3125 watts!
 

LarryQ

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Thanks GAZOOO.....
3.9 Ohm 1/2 Watt installed...225Mw current!!!


I may need more as she it's even getting warm!!!!!

Too Cool!!!!!

LarryQ
 

a_pyro_is

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More ?s...
Is this to become a labby?
What heatsink?
Pics/Vids of the destruction this can cause? lol
Sorry to threadjack, but I love to see what people come up with on here for all the ideas it gives me. :D
 

LarryQ

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No prob...

It's based on an MXDL flashlight..
It is a work in progress but will have the AixiZ module pressed into a SOLID alum heatsink
1.100 Inch in dia .825 long.

I'm shooting for @300-325 ma current...hense the large heatsink.

Could I get your opinion/help on something??

Obliously I'm using an LM317 based circuit...but I don't understand what's happening.

The power source is (2) CR123 Lithiums (New).

There is 1 N14001 and 1 47Uf 16V tantalum cap across the leads of a GB diode..in an AixiZ module.

Pretty standard stuff...

I placed (2) 10 Ohm 1/2 watt resistors across the ADJ of the LM 317..

I placed a 1 Ohm as a sense resistor and powered it up.

I only got 185 Ma.....

Hmmmmm

I calculated that with 5 ohms at 185 Ma...If I dropped 1 Ohm down to 4 OHM..I should result in 222 Ma....

Lo and behold I did indeed get 225ma!!

So...My target is 300ma....

2 Ohm should give @296 ma
1 Ohm should give @333 ma

Do these numbers sound proper to you???

THe reason I ask is because If I calculate what the current SHOULD HAVE BEEN with 5 Ohm...

I Get 1.25V / .005= 250Ma

Why am I only getting 185 in the real world?

Thanks....

I'm going to tear into this mod again tonight after work!

LarryQ
 

Gazoo

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Larry...lol. To calculate the resistance needed take the reference voltage and divide it by the current you want to use.

1.25/.300 = 4.16 ohms! I don't know how much current the sense resistor is dropping but you need to figure that into your calculations. BTW, it's awesome you are using a sense resistor...it is the best way to measure current without risking damage to the diode.

I am in the process of testing the AMC7135 regulator. It is designed to regulate current at 350ma's. and has a voltage drop that is negligible..like millivolts..lol. This thing is tiny. To lower current it requires a resistor to be put in parallel across the laser diode. The resistor absorbs the current we don't want to get to the diode. Needless to say, at 350ma's it is perfect for the open can diodes as is...the only additional component needed is a capacitor. The one I am testing is this one:

http://www.dealextreme.com/details.dx/sku.3160

I have a lot more testing to do but so far it is looking very good.
 

LarryQ

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Yes..but When I actually  measure the current:

With 5 Ohms I only see 185 ma...
With 3.9 Ohm...I only see 225ma

The formula works both ways..I only use .005 to keep the decimal straight in my head... :D

1.25V/5 (ohm)=.250 or 250ma (or 1.25/.250=5 ohm)

1.25/3.9 (ohm)=.320 or 320ma (or 1.25/.320=3.9 Ohm)

Why such a diff between predicted and measured??

I'm off by 26 percent!!! predicted versus actual.

Using the 1ohm resistor as a sence resistor ....
Is it giving me THAT much current drop in the circuit??

LarryQ
 

Gazoo

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Well I assume you have the 1 ohm resistor in series with the LD...right? It should not be dropping that much current...you are right.
The only other thing I can think of is the regulator is beginning to drop out as you increase current going to the LD. What is the voltage of the batteries with the LD turned on?
 







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