1.5volt bulbs on 12volts (resistor)
Nigel Nichols
I want to run some 1.5volt bulbs in my locos on my DCC system. The track
voltage is 12volts. What is a simple formula to work
out what ohm resistor I will need to reduce the 12volts to 1.5volts.
I want to run the 1.5volt bulbs instead of replacing them with 12volt bulbs
in order to keep the heat down around the loco plastic body. This being of
course that my assumption that a 1.5volt bulb on 1.5volts operate cooler
that a 12volt bulb on 12volts
Many thanks
Nigel
--
Redline Race Controls
http://www.wave.co.nz/~lakewood/Redline2.htm
Western Pacific Model Railways
http://www.wave.co.nz/~lakewood/MyWP.htm
Bill Sohl
"Nigel Nichols" <lake...@wave.co.nz> wrote in message
news:bumlrq$aed$1...@news.wave.co.nz...
> Hi people
>
> I want to run some 1.5volt bulbs in my locos on my DCC system. The track
> voltage is 12volts. What is a simple formula to work
> out what ohm resistor I will need to reduce the 12volts to 1.5volts.
> I want to run the 1.5volt bulbs instead of replacing them with 12volt
bulbs
> in order to keep the heat down around the loco plastic body. This being of
> course that my assumption that a 1.5volt bulb on 1.5volts operate cooler
> that a 12volt bulb on 12volts
> Many thanks
> Nigel
Nigel, the heat generated is a function of the bulb's wattage rating..not
simply
the voltage. Just as we have different wattage bulbs for 120v AC lamps, you
can
and do have different wattage bulbs for DC at 12 or 1.5v. One could
have a 1 watt bulb for 1.5v as well as a 1 watt bulb for 12v. The bulb for
12 volts would have a current flow of 1/12 amp while the bulb for 1.5v
would have a current flow 2/3 amp. The formula is W (wattage) - E (volts)
times I (current),
Based on that and assuming you don't know the current for the 1.5v
bulbs you have, you can find out by measuring actual current flow
with the lamp connected to a 1.5v source. Once you know the current flow
you can calculate the resistance value needed using Ohm's law as
follows: You need the resistor to drop 10.5v (12v - 1.5v) for
a certain current that you have determined by measurement. The
formula is R = E (volts) divided by I (the current measured). If,
for example, the current flow was .050 amps (50 milliamps) then
the formula becomes: R = 10.5/.050 which calculates to 210 ohms.
Next you must calculate the wattage of the resistor needed. That is,
using the wattage formula above: W = 10.5 * .050 which equals .525 watts.
For safety, you would then want to use a 210 ohm, one (1) watt resistor.
In doing these calculations, you'll find the calulated resistance needed
may not be exactly available. Don't worry, use the next highest
value for one that is. Again, by example, you won't find a 210 ohm
1 watt resistor, but you will find a 220 ohm 1 watt.
Hope that helps,
Cheers and happy calculating.
Bill Sohl
Bob May
do a diode arrangement to provide the voltage level for the 1.5V bulbs.
This will keep the bulbs at a constant level of brightness over the majority
of the range of the track voltage. The 12V bulb can be inside somewhere
that won't matter if the light exists and the surface of the bulb can be
painted black to help keep the light in.
I've done this technique for both passenger car lights and for locos that
have micromotors in that don't draw enough current to light the lamps.
--
Bob May
Losing weight is easy! If you ever want to lose weight, eat and drink less.
Works every time it is tried!
Mark Mathu
news:10747383...@news-1.nethere.net...
> This will keep the bulbs at a constant level of brightness over the
majority
> of the range of the track voltage.
With DCC, won't the voltage to the bulb be pretty much constant? And isn't
the track voltage constant?
Peter W.
if you are worried about heat you might consider using White LEDs instead of bulbs.
It appears that there are now at least 3 different types of White LED
available from various Model Train vendors. In the USA that is, but I'm sure
you can mail-order them.
There is the original Bluish-White LED, then there is Golden White and
Sunny White from Richmond Controls. Miniatrinics also sells Yeloglow White
LED (I think it is similar to the Sunny White). I find the last 2 closest
in color to incadescent Bulbs.LEDs are very low power so they don't generate
heat. The current limiting resistor also generates very little heat.
http://www.miniatronics.com
http://www.richmondcontrols.com/
Other responders provided you with a good way of calcualting resistor values
and power rating. Same equation applies here. Only difference is that
the voltage across the white LED will be around 3.7V (not 1.5 like the bulb).
The current will be in a range from 1mA (low brightness) to 20mA (full
brightness). You would have to decide what looks good for your application.
Getting back to bulbs, many micro-bulbs are 1.5V 15mA or 30mA. Using that
type of a bulb with a resistor will generate the least amount of heat.
Any bulb which uses more current or operates at higher voltage will produce more
heat.
Somebody else mentioned using a "ballast" bulb of a higher wattage with a
diode circuit to power the 1.5V bulb. That circuit generates more heat
than the one with just a single resistor.
But either will work.
And as you probably guessed - I prefer LEDs ....
Peteski
"Nigel Nichols" <lake...@wave.co.nz> wrote in message news:<bumlrq$aed$1...@news.wave.co.nz>...
Peter W.
Peter W.
Peter W.
Bob May
the brightness by that much. I do tend to prefer to put the diodes across
the 1.5V bulbs just to insure that the bulbs don't get overvoltaged in the
first place. Besides, I'm still running DC and thus need the diodes to
control the voltage to the bulbs.
Nigel Nichols
I,m going to do some experiments with the LED's Peter.
Lets check I got this right then.
The LED's that are available at the local electronics shop are 3.5v - 20ma.
So 12v track power minus 3.5v = 8.5v
8.5v divided by .020 = 425.
So a 425k resistor is needed ( I can get a 470k)
Resistor wattage will need to be 8.5v*.020 = .17w
If I get a 470k 1/2w resistor then I'm onto something.... true???
Nigel
--
Redline Race Controls
http://www.wave.co.nz/~lakewood/Redline2.htm
Western Pacific Model Railways
http://www.wave.co.nz/~lakewood/MyWP.htm
Peter W. <pet...@my-deja.com> wrote in message
news:b2412277.04012...@posting.google.com...
David J. Starr
Nigel Nichols wrote:
>
> Thanks to all
>
> I,m going to do some experiments with the LED's Peter.
>
> Lets check I got this right then.
> The LED's that are available at the local electronics shop are 3.5v - 20ma.
> So 12v track power minus 3.5v = 8.5v
> 8.5v divided by .020 = 425.
> So a 425k resistor is needed ( I can get a 470k)
>
> Resistor wattage will need to be 8.5v*.020 = .17w
>
> If I get a 470k 1/2w resistor then I'm onto something.... true???
>
> Nigel
>
Your math is good so far. But don't stick a "k" on the 425. "k" is
the electronic guys way of saying "one thousand". You want a 470 ohm
resistor, not a 470,000 ohm one.
The other thing you want to consider is polarity. LEDs only light up
when forward biased. Reverse bias turns them off, and enough reverse
bias destroys them. "Enough" is variable, some LED's will withstand 12
V reverse bias, others won't.
When you flip the reversing switch on a standard power pack, you are
reversing track polarity, and changing the bias on the LEDs from forward
to reverse (or vice versa). Standard solution is to get a bridge
rectifier (cheap four pin part from Radio Shack) and wire it to supply
DC of the right polarity to the LED's no matter what the track polarity
is. Put the track power pickups to the AC pins on the bridge, put the
LED and it's dropping resistor across the DC pins. Double check LED
polarity, make sure you have got it forward biased. Then you want to
account for the 1.4 volt drop going thru the bridge rectifier.
So,
Let R = 12 V - 3.5 V LED bias - 1.4 V FW bridge drop / 0.02 A = 355
ohms.
Nearest standard values are 330 ohms and 390 ohms.
Good Luck
David J. Starr
Edward A. Oates
significantly smaller.
Ed.
in article bup8od$v7o$1...@news.wave.co.nz, Nigel Nichols at
lake...@wave.co.nz wrote on 1/23/04 9:05 AM:
--
Ed Oates
http://home.earthlink.net/~edoates
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