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Old 10-01-2007, 03:46 PM #1
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Default reducing voltage via potentiometer

hi
i made my homemade laser and want to change the power supply to two cr123a batteries
that would be 6Volts because they will be rechargables

i will use a pot to reduce the current but if i want the voltage to be at around say 3 volts, would just using the pot work
SO it would be
6V-3=3
3V/300ma=0.01k or 10 ohms resistor/pot

So would that be safe,
also i would rather do this than the lm117 coz it seems complicated and takes space

thanks


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Old 10-01-2007, 04:35 PM #2
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Default Re: reducing voltage via potentiometer


You can try a pot but if you should set it wrongly = expensive LED.

Quote:
also i would rather do this than the lm117 coz it seems complicated and takes space
I would strongly recommend the LM317 route, have a look around this section of the forum, there are pictures of cut down reg's that take very little space even with a resistor of the correct value.

You are still in danger of spikes at switch on, that is the reason for the capacitor, it will protect your diode...... Regulator circuit cost cents, LD costs $$$ (more or less)

Deadals circuit is well tested, easy and cheap to make, and can be made small enough for most host torches.

At the end of the day the choice is yours, we can only advise....

Regards rog8811
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Old 10-01-2007, 04:41 PM #3
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Default Re: reducing voltage via potentiometer

Also...I have had a heck of a time finding a 10 Ohm 10% tolerance 1 WATT Pot!

If you do find a source...Please let me know where you found them..

I need a Top Adjust model...and so far have found SQUAT at Digikey.

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Old 10-01-2007, 04:42 PM #4
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Default Re: reducing voltage via potentiometer

Quote:
Originally Posted by rog8811
You can try a pot but if you should set it wrongly = expensive LED.

Quote:
also i would rather do this than the lm117 coz it seems complicated and takes space
I would strongly recommend the LM317 route, have a look around this section of the forum, there are pictures of cut down reg's that take very little space even with a resistor of the correct value.

You are still in danger of spikes at switch on, that is the reason for the capacitor, it will protect your diode...... Regulator circuit cost cents, LD costs $$$ (more or less)

Deadals circuit is well tested, easy and cheap to make, and can be made small enough for most host torches.

At the end of the day the choice is yours, we can only advise....

Regards rog8811
ive read around and basically i want to know do you have to have the lm317 to drop the voltage or can it just be done with a pot

or if i was to use it, how would i go about connecting it,
6v input and i want the output voltage to be 3v so i just put in a pot between the adj and out pins of the lm317?
so would it be like this
input 6V-----------------------------------to laser+
|adj |Out pin
| |
------POT------
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Old 10-01-2007, 05:05 PM #5
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Default Re: reducing voltage via potentiometer

I don't really know what to say to you on this....
Yes you can use a pot, it will work, but set it wrongly and bang goes your LD.
The LM317 drops the voltage to what you need without the danger of getting it wrong.

The circuit as you have drawn it is incorrect I have attached the simplest circuit that will work, you can always go for maybe 2 ohm fixed resistor and a pot, in series with each other, in place of the 5 ohm if you so desire.

I hope that helps....
Regards rog8811
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reducing voltage via potentiometer-lm317_basic.jpg  
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Old 10-01-2007, 05:15 PM #6
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Default Re: reducing voltage via potentiometer

ok thanks for that
so basically could you plz explain how you get to the 3v coz i cant seem to figure it out with the ohms law

so the 5ohm resistor also regulates the current but to 600ma is it? 3/.005

thanks
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Old 10-01-2007, 05:21 PM #7
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Default Re: reducing voltage via potentiometer

Rog is doing a great job with explaining this, so I am not gonna jump in and bail him out :-) I do have one comment to make, though - you CAN cut corners in electronic circuits, it just depends on how deep your wallet is. The LM317 circuit that is drawn out on here many times, and to original concept is in the datasheet for the LM317 chip itself - and it works VERY well. By trying to reinvent it, or cut corners for one reason or another, you are comprimising hte integrity of the circuit, and increasing your chances of ending up with nothing but smoke, and a cool looking red LED. To each their own - but the circuit is REALLY easy to make, and if you have come this far by asking questions about it, and trying to reverse engineer it to suit your wants a little better, I have a pretty good hunch that you could build this circuit (as advertised) with no problems. I have always gone about stuff like this a little odd (well, odd for some people) I build it the way it is supposed to be built, and THEN messwith perfection in an attempt to make it suit my needs a little better :-)
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Old 10-01-2007, 05:27 PM #8
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Default Re: reducing voltage via potentiometer

You can't use ohms law on the output voltage of the Lm317 I am afraid, as I understand it, it is just a funcion of how the regulator works in this application. (Help! DDL)

I know it takes a bit of reading but it is well worth going through the thread on this driver, many people have posted output figures and recommendations for resistor values.

Regards rog8811
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Old 10-01-2007, 05:32 PM #9
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Default Re: reducing voltage via potentiometer

Feel free to jump in SenKat, I only learnt some of this stuff over the couple of months that I have been here.
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Old 10-01-2007, 05:34 PM #10
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Default Re: reducing voltage via potentiometer

If you setup the LM317 as a current regulator, the resistor between the ADJ and VOUT terminals limits the current to 1.25V / resistor.

If its setup as the standard voltage regulator, with R1 as the resistance between ADJ and VOUT, and R2 as the other resistor (check out the spec sheet to see diagrams since I don't think I'm explaining that too well), the output voltage is ~ 1.25V (1+R2/R1)

If you're using a regulator other than the LM317, replace the "1.25V" in those equations with the reference voltage for the regulator
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Old 10-01-2007, 05:42 PM #11
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Default Re: reducing voltage via potentiometer

Thanks Pseudo
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Old 10-01-2007, 05:44 PM #12
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Default Re: reducing voltage via potentiometer

ok now i understand (at least i tink :

so basically i put a 4ohm resistor in between adj
which makes it 1.25v /0.004=313ma roughly

if i put a pot after that then to get 3v it hast to be 3ohm to get around 2.91v
so (0.004/0.003)+1) * 1.25 =2.91v
is that correct

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Old 10-01-2007, 06:50 PM #13
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Default Re: reducing voltage via potentiometer

alright now i get it after reading ddl's posts on homemade laser minimags lights etc.

basically the lm317 will limit output voltage to 1.25 and then the rest of the current depends on the resistor.

so if i put a 3 ohm resistor it will be 1.25/.003=416ma etc.
i think i should put a 3 ohm resistor and a pot after it so i can turn it down if i want.
can you plz clarify

also how about this minimag
dealextreme.com/details.dx/sku.2025

or would it just be better to use one cr123 battery and have 3v to start of with. but it would mean the battery would need to be replaced often.
dealextreme.com/details.dx/sku.5977
thanks
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Old 10-01-2007, 10:02 PM #14
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Default Re: reducing voltage via potentiometer

To figure the desired resistor, take the reference voltage and divide it by the current you want!

For example, if you want 250ma's of current going to the diode.....1.25 divided by .250 = 5. You would use a 5 ohm resistor.

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Old 10-02-2007, 02:31 PM #15
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Default Re: reducing voltage via potentiometer

yeah thanks
thats what i eventually figured out :P

so i will just add a 3ohm resistor and a pot. 1.25/.003=416
but one more thing, the voltage will then be 1.25v output from the lm317 wont it?
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Old 10-02-2007, 03:21 PM #16
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Default Re: reducing voltage via potentiometer

Quote:
Originally Posted by izzy007
but one more thing, the voltage will then be 1.25v output from the lm317 wont it?
No.
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