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.
Thanks in advance
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.
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
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
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.
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.)
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
* - I think the fraction John/Sexed appeared in it somewhere :)
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?
(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
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.
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.