# Help with Resistors ?

I am trying to reduce the DC transformer 14v down to 3v. I went to Radio Shack
and started looking at resistors. Found out very quickly I did not know what I
was looking at. I would find something for example that read 470-OHM 1/2 watt
5% tolerance. Nothing about volts. So what do I need? I am trying to equal two
double A batteries.
Chris
ChrisGW spake thus:
I think your problem is going to be that because you're trying to set up such a large voltage drop (14-3 = 11 volts), you're going to need a fairly large resistor (power-dissipation-wise, not resistance) to dissipate all the power. Otherwise you'll burn up the resistor.
Better to find one of them wall warts close to the voltage you need. I see them in second-hand stores all the time for cheap.
By the way, resistors don't "come in volts". They don't make resistors to drop x volts to y volts. No getting away from using Ohm's Law here.
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Here's a site with info about resistors:

History of N Scale:
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Bad Boys Rape Our Young Girls, But Violet Gives Willingly
On Fri, 26 Jan 2007 17:21:04 -0800, I said, "Pick a card, any card" and Steve Caple instead replied:
Jpegs, please. Of Violet. -- Ray
On Fri, 26 Jan 2007 19:17:02 -0500, I said, "Pick a card, any card" and ChrisGW instead replied:
Resistance is feudal. -- Ray
You need Ohms Law:
R = V / I
(Resistance (in Ohms) equals Voltage (in Volts) over Current (in Amps)).
You want to drop 14-3 == 11 Volts. At how many amps? What is your load? A white LED (which are 3V devices) draw about 20Miliamps (.020Amps), so:
11 /
.020 = 550 Ohms. The stock value available would be 560 Ohms (Green-Blue-Brown-Gold). It will need to disipate 220miliwatts, so a 1/4 watt will do just fine.
Robert Heller spake thus:
(aka "R = E/I")
So, to generalize this for the OP:
* To compute the value of the resistor you need, divide the voltage drop (11) by the current (in amperes), as per the example above.
*
To compute the *power dissipation* of the resistor in watts, multiply the voltage drop by the current in amperes. Make sure you use a resistor large enough not to get too hot.
A resistor is not the way to go. What are you trying to power? How much current does it draw? I would suggest using a voltage regulator IC, something like the LM317T here:
The datasheet describing its use is here:
Here's power supply circuit using this:
This latter circuit shows a formula for calculating a fixed resistor value for R2, but if you use a 5Kohm potentiometer instead you can get an adjustable output range from about 1.25V to 18V. A similar circuit showing this was just published in the January issue of Scale Rails (a monthly magazine sent to NMRA members). As shown, a few additional components are required to make it work properly, but it will be far more reliable than just using a resistor to drop the voltage.
For some applications a resistor is the way to go. For LEDs that's pretty much the universal solution up to 120 volts.
She's only an 8 on a scale of 10, with a tolerant heart of gold.
Steve Caple spake thus:
So you musta seen that old one I saw inscribed on the bathroom wall at the College of Engineering where I went to school:
b(4) i(4)q, qt(insert symbol for pi here), ru/18?
(Doesn't work so well in ASCII.)
lotsa snipping
There were several variants on that - my fav "equation" - and I forgot the first line*, had as the next two (pronounce the "/" as "over"):
John/Joan + Joan/Fertile = Joan/Due
Joan/Due + \$5000 = Joan(squared) + John/Seas
Steve * - I think the fraction John/Sexed appeared in it somewhere :)
What you need is a quick course in electricity. Really. :-)
I see that other posters have kindly provided you with links to suitable sources.
Have fun!
Nope. I learned the resistor color code mnemonic from a family friend (who used to work at White Sands, and later at the Indianapolic Naval Ordnance Lab), and picked up other mnemonics ("IDA CAD" = Inverter, DC to AC; Converter, AC to DC) in Navy airborne electronics school at Memphis. Where one famous instructor taught that cigars were training devices for c--ksuckers, a piece of wisdom I remember every time I see our Gropenator Ahhhhnoald sucking on one.
Wolf spake thus:
No; what he *needs* is the right value resistor. Really. :-)
That's what he asked for, remember? Why is it that every query of this type is an invitation to subtly belittle the asker and suggest that they get their PhD. in electronics before deigning to ask such a question? Sheesh.
(Sorry if I seem a bit snarky about this: this newsgroup isn't as bad as some others, but I'm absolutely fed up with the rampant elitism about these kinds of questions I see here and elsewhere.)
No!!! What he doesn't need, is for folks like you to imply to him that his original request was complete enough for an intelligent answer. It wasn't. The replies have been as gentle as possible for a basically NONSENSE question.
Chuck Davis
The OP did not give enough information to give him an answer. We (the group) still don't know how much current is involved. Without that, all we can do is suggest how to compute the value, that is how to use Ohms Law.
True, hence the question above about what he is trying to power.
Charles Davis spake thus:
OK, so it's true he didn't give us enough information. Are you suggesting a) beating him over the head for being an idiot or b) asking him for more information? Most of the responses I've seen so far tend towards a) is all I'm saying.

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