Jay
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Arjay1969 wrote: This is for structure lighting? And you have a 12v source? IMO, your best bet would be to use a voltage regulator (LM317T) with the resistors to set the voltage to 1.4 volts. The 317T can handle up to 1A of current (1000mA), so with one regulator, you could put about 20 x 40mA bulbs on it safely (1000/40=25, but to avoid burning out the regulator you don't really want to go over about 80% of the rated capacity). Then it's just a matter of running a buss line for the lights to connect to.
This is for structure lighting? And you have a 12v source?
IMO, your best bet would be to use a voltage regulator (LM317T) with the resistors to set the voltage to 1.4 volts. The 317T can handle up to 1A of current (1000mA), so with one regulator, you could put about 20 x 40mA bulbs on it safely (1000/40=25, but to avoid burning out the regulator you don't really want to go over about 80% of the rated capacity). Then it's just a matter of running a buss line for the lights to connect to.
You would need a heat sink on the 317 at that current. Also, the bigger the difference between the input and output voltages, the more heat the 317 will produce.
..... Bob
Beam me up, Scotty, there's no intelligent life down here. (Captain Kirk)
I reject your reality and substitute my own. (Adam Savage)
Resistance is not futile--it is voltage divided by current.
Robert Beaty
The Laughing Hippie
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Then it comes to be that the soothing light at the
end of your tunnel, Was just a freight train coming
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nedthomas wrote: Better to use a resistor for each lamp. Using only one resistor is good until one or more lamps burn out. This lowers the current the dropping resistor is seeing and decreases the voltage drop. The remaining lamps a see a higher voltage and will burn out faster.
Better to use a resistor for each lamp. Using only one resistor is good until one or more lamps burn out. This lowers the current the dropping resistor is seeing and decreases the voltage drop. The remaining lamps a see a higher voltage and will burn out faster.
I agree. The problem is that I want to install 11 lamps in one large building; a 4 stall round house. 2 banks of 4 and 1 bank of 3. I think hiding/disguising the resistors inside the building would be a challenge. I figured that in parallel, I would only have to worry about 3 pairs of wires leading down from the support beams. Wiring each lamp individually is going to produce a lot of wires.
The mind is like a parachute. It works better when it's open. www.stremy.net
Ned, you are right. The wattage of the resistor will be less as well.
Just use the same calculations for one bulb, and put the same value resistor in series with each bulb. Then connect each bulb-resistor combo in parallel.
Elmer.
The above is my opinion, from an active and experienced Model Railroader in N scale and HO since 1961.
(Modeling Freelance, Eastern US, HO scale, in 1962, with NCE DCC for locomotive control and a stand alone LocoNet for block detection and signals.) http://waynes-trains.com/ at home, and N scale at the Club.
modelmaker51 wrote:He didn't ask about LEDs, he asked about light bulbs. Try to read the OP.
The math works for either LEDs or bulbs.
TONY
"If we never take the time, how can we ever have the time." - Merovingian (Matrix Reloaded)
Gandydancer19,
That was a very thorough and excellent breakdown on how to figure out resistance values.
Anyway, my little contribution to the OP's question: LED Calculator
Very useful for the lot of us that don't want to crunch numbers and need a fast answer. In the middle of the page you can choose single LED, series LEDs, or Parallel LEDs.
OK, here you go.
Since the bulbs are in parallel, the end voltage needs to be 1.5 volts to the bulbs. The variable is the current required, and that depends on the number of bulbs you will connect in parallel. You need the total current. So with four bulbs and each drawing 40ma, total current is 160ma. (4 x 40ma) I total = I1+I2+I3+I4 etc.
The voltage that you need to drop or reduce is 12v minus 1.5v (or whatever voltage the bulbs use) so 12volts - 1.5volts = 10.5 volts. (E supply - E bulbs = E drop ) So the actual size of the resistor in ohms depends on the current that will be going through it and the voltage that it has to drop or absorb. Now that you have two out of three numbers (E resistor and I total), you can find the third by using the formula R=E/I. E drop =10.5 and I total =160ma or 0.160 amps. So, 10.5 / 0.160 = 65.625 ohms. I think 68 ohms is going to be the closest standard value.
Now to the wattage of the resistor: P=IxE is the formula. So we need the power value for the resistor, so we use the values going through the resistor. P = 0.160amps x 10.5volts. P in Watts = 1.68watts. So the closest standard value is going to be either 2 watts or 5 watts. So the resistor you would be looking for would be 68 ohms at 2 watts. A higher wattage resistor will not hurt anything, but don't go lower or the resistor will overheat and may burn out, and on rare occasions start a fire.
I want to connect n lights (say 4) in parallel. I need the equation(s) for the resister to match the power supply. I know the voltage of the lights (say 1.5) and the current draw (say 40 ma). The power supply is 12 volts. How do I calculate the resister? Bonus question: How do I determine if a 1/4 watt resister is sufficient or if I need a larger one (say 1/2 watt)?
Also, if I determine that lights are too bright, can I recalculate the resister by lowering the input voltage of the lights (say from 1.5 to 1.0) in the equation(s)?
Thanks...