I am wiring my control panel with red and green lights. I have a 12volt DC power supply with 3amps. There are 36 wheat grain bulbs 12volts each. I want to use only 1 resitor in the positve power supply to the lights. What value of restor would be best to dim them about 1/4 of thier brightness.
How much current does each bulb draw and how will you have them wired?? Series, Parallel, both?? You would be much better off and only about $2 poorer (if that) using individual resistors…
Jeff
Sorry parallel. I’m guessing at about .5 volts.
If you have a meter, go to radio shack and get a multi-turn potentiometer. Put it in-lne with the bultbs (between them and the power supply) in the positive line. Turn it all the way down (when the screw is in all the way) turn on the power and start dialing it up. When you get the desired brightness, take it out of the wiring, and read the value on an ohm meter. The value that displays will be the size of resistor you should get.
A potentiometer won’t cost much, and it’s not nearly as hard as it sounds. Once you have the potentiometer it shouldn’t take about 3 minutes to do the rest. Feel free to e-mail me if you have any questions.
Greg
Ok but what would be wrong with leaving the potentiometer in line instead of resitor. never thought of this before. It would also give me the advantage of increasing if more lights are needed later.
Nothing wrong with it… As long as it’s big enough… BTW, Current is measured in Amps… Are you sure about that .5 number?? Sounds mighty high for a 12 volt gow bulb… Or is it 0.05 (50mA)… ?? Let’s do a little ohms law math here… 36 bulbs at 50mA per (in parallel the current draw adds), 1.8 amps… The potentiometer will drop anywhere from 1-12 volts (it’ll be in series with the lighting circuit)… At half power, 6 volts, your pot will have to dissapate (if all 36 are lit) 10.8 watts… At 11 volts (if the lamps still light), that number grows to 19.8W… 4 volts on the lamps, leaving 8 on the pot, 14.4 watts (W)… Most radio shack type pots, are around 1-5 watt rated.
Come to think of it, why not save the trouble of all of it, go to the hardware and get a rotary dimmer switch?? Rated at 600 watts, 120 volts, plenty of protection and has the on/off capability too… Yes, probably overkill but isn’t all of model railroading??
Jeff
Yup already thought of that. A friend electrian has 12 Volt dimmers. Getting him to price me one. Just thought resistor was cheaper. Thanks alot for help will probably come in handy in near future with other wiring projects.
Here is another rivet of electronics information. Remember that in parallel Voltage is common, current is additive and resistance is calculated by using this formula
1/(1/R1+1/R2+1/R3+1/R4+ however many more loads (R) you have in parallel)
In series tho resistance is additive as is current again. but in series the voltage drops over the light bulbs will leave a little less voltage for the next light making it slightly dimmer than the next.
something to cosider in when making a circuit of lights is Current will see a parallel network as one load and therefore as one resistor so to speak. dont put yourself so close to the tolerances of the bulbs you are using that losing one will lower the total resistance enough to make the current too high. Leave yourself some room there.
I think this part is a little misleading… The components in a series circuit will all see the same current… All things being equal, bulbs, each will see the same amount of current and therefore the same voltage drop (again, if equal)… The series circuit becomes a volage divider, while a Parallel circuit is a current divider…
If a 12 volt bulb draws 50mA at steady state (been on for awhile and has settled down), that’s 240 ohms… Using my numbers from the previous post, the 36 bulb parallel network becomes 6.6666 ohms, drawing 1.8A from the 12 volt source. Put those same 36 bulbs in Series, at steady state, the series circuit of bulbs becomes 8640 ohms and each will see a voltage drop of 0.3333 volts (and probably won’t light at all)… Again, Bulbs being equal. Quick crunch of the numbers, the entire circuit is going to draw 1.4mA. Higher load (more resistance) on a fixed voltage source, less current drawn. Lower load (less resistance) on a fixed voltage source, More current drawn… Ohms Law: V=IR; Power = I (squared) times R…
Jeff
since you are using bulbs and not LED’s, you may want to nix the resistor idea and step down the voltage …here is a website that with a few inexpensive electronic parts, you can do this…they will work for any incoming 12v supply and can step down the voltage to 5v or 1.5 v which ever you need… http://www.mrollins.com/circuit.html chuck