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Author Topic: 12v down to 5.8 volts at 1A? *updated x2 - last post  (Read 7036 times)

Daneel

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #10 on: December 28, 2004, 05:24:31 AM »

yes, but are they what i want for 12v to 5.8 volts?
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viridius

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #11 on: December 28, 2004, 02:30:06 PM »

Yeah.
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Daneel

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #12 on: December 29, 2004, 07:52:49 AM »

ok, so how do i wire one of these suckers up to give me 5.8 volts?  i would like to use a pot to fine tune the output voltage.  i'm guessing it'd have something to do with caps and resistors as per bottom left of page 30 of this.

i had initially thought that there would be a dial on the regulator for output voltage or something (in hindsight i realise that this is stupid).
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Skylined

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #13 on: December 29, 2004, 09:28:44 AM »

If you want you can use a LM317 together with a pot, but you'll have to use a multimeter to fine tune the output voltage.
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Daneel

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #14 on: December 29, 2004, 10:41:16 AM »

i have my multimeter handy at all times, just speak ..err..  post the magic explanation of where to put the pot for adjustable voltage regulation assuming i'm using the circuit in the bottom left corner of page two of that jameco pdf file..


*edit, i think i've figured it out now, actually..

i guess that the 0.1uF cap can go either way (ie, is not polar) but the 1uF cap is polar and the neg has to go away from the output line..

is that right?

i'm also assuming that that circuit would work with an input voltage of 12v instead of >= 28v..

will it still work?
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viridius

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #15 on: December 30, 2004, 06:39:12 AM »

I would use the "Adjustable Regulator with Improved Ripple Rejection" schematic on page 12.  And yes, 12 volts will work.
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Daneel

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #16 on: December 30, 2004, 10:37:58 AM »

ahh, didn't spot that one..

thanks to all that helped, esp viridius, i can get on with the buying and building now.  see my project log for the buildign details.

log title: H2Own Water-cooled Briefcase PC


*edit - oops, one final question, how do i wire up the pot?  i can't seem to find a pinout, but there must be some form of standard..

*edit edit - i guess i could just play around with my multimeter and find the two contacts that give me 5k, couldn't i?

*edit edit edit - and will this resistor: "240ohm 1/2 Watt 1% Metal Film Resistor" be ebough to cope as the 240ohm resistor in that circuit.

thanks


i do know a bit about circuitry, but this simple circuit is completely new for me.  hell, i soldered up an intercomm system for my house but i can't do this, oh my.
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viridius

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #17 on: December 30, 2004, 07:07:50 PM »

Find the two legs that give you 5K and the other one is the wiper.  That resistor should be fine.
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Daneel

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #18 on: December 30, 2004, 10:24:01 PM »

i don't know what the wiper is supposed to do but i guess i don't need it.

thanks again
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viridius

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12v down to 5.8 volts at 1A? *updated x2 - last post
« Reply #19 on: December 31, 2004, 05:50:25 AM »

The two legs that give you 5K are the ends of the resistive element within the potentiometer.  The wiper slides along the element so the resistance between one leg and the wiper decreases and the resistance between the other leg and the wiper increases as the knob is turned.  In the circuit you're using, you will use the wiper and one of the legs (which one doesn't really matter, it just determines whether or not the resistance increases or decreases when you turn the knob clockwise).
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