1.5volt bulbs on 12volts (resistor)

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
--
Redline Race Controls
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Western Pacific Model Railways
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Reply to
Nigel Nichols
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With DCC, the track voltage is indeed fairly constant and thus won't affect 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.
-- Bob May Losing weight is easy! If you ever want to lose weight, eat and drink less. Works every time it is tried!
Reply to
Bob May
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
Reply to
David J. Starr
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
-- Redline Race Controls
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Western Pacific Model Railways
Reply to
Nigel Nichols
1/2 watt resistors are big; you only need 1/4 watt (.17 < .25) and they are significantly smaller.
Ed.
in article bup8od$v7o$ snipped-for-privacy@news.wave.co.nz, Nigel Nichols at snipped-for-privacy@wave.co.nz wrote on 1/23/04 9:05 AM:
Reply to
Edward A. Oates

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