Cutting back to the original question, an ounce of experiment is worth a ton of, "What if."
Givens: Parallel circuit, 4 14V incandescent bulbs, 12V (nominal) power supplies.
Wanted: Voltage to achieve appropriate brightness, R value of resistors to put in series with given circuit.
Available: Various power supplies. Multimeter.
Procedure:
Converting Ohm's Law to R = E/I, subsitute 12 minus observed lamp voltage for E, observed current flow for I. The result will give you the resistance needed. Since it will probably NOT be a value you can purchase, the next higher will be slightly less bright, the next lower will be slightly brighter. OR, just dedicate an El Cheapo toy train pack to the building (aka gross overkill.)
Shifting out of instructor mode. Have fun.
Chuck (Modeling Central Japan in September, 1964 - with a bunch of 2.5V lamps in various combinations of series and parallel)
Clearly not for powering 1.5V bulbs off a 12V power supply, don't be silly. A better way for that is to use a 1.5V voltage regulator, or do like locomotive lighting does and put the 1.5V bulb parallel to a pair of diodes in series (1.4V across them). You need a load for that, in a loco the motor works, for stationary lighting you can use a taillight bulb as a ballast.
Also - that resistor calculation is PER LAMP. 10 lamps and you have 3 watts being wasted as heat across the circuit. 100 lamps and now 30 watts just goes into heating up a bunch of resistors. With diodes, you use one set of diodes and that provides the drop for up to the current limit of the diodes. Use the heavy duty 25 amp bridges for dropping and you can run 800 30ma bulbs off it, assuming the power supply is hefty enough. Even the common 1 amp diode can handle about 30 30ma bulbs, you resistos with 30 lamps would be putting out 9 watts of heat, aka wasting 9 watts of the power supply's output.
The adjustable regulators go right down to 1.2V when directly wired, which is just about perfect for 1.5V bulbs to be slightly dimmer to simulate oil lamps and also last virtually forever. With a heat sink, the TO-3 case ones can generally handle 1.5 amps, that's also a lot of light bulbs. You don't really want to feed the regulator with 12V though, the more it has to drop the more heat it has to dissipate. FFor a 1.2V regualtor, 5V is a better input source.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
rrinker Better way to drop voltage is to use diodes. No heat. Each diode will drop about .7 volt. Connect multiples in series to drop more volts. Typical silicon diodes are goot for 1 amp, some of the bigger ones are 3 amp, or you can use a bridge rectifier and get 2 drops (1.4 V) in a 25 amp package - use multiple circuits to distribute the load --Randy
Better way to drop voltage is to use diodes. No heat. Each diode will drop about .7 volt. Connect multiples in series to drop more volts. Typical silicon diodes are goot for 1 amp, some of the bigger ones are 3 amp, or you can use a bridge rectifier and get 2 drops (1.4 V) in a 25 amp package - use multiple circuits to distribute the load
So in the example earlier you would need 15 diodes wired in series or just a single 1/2 watt 350 ohm resistor. Also the heat created by the circuit is .3 watts. Not too much and even less if a higher size resistor is used.
Springfield PA
Here is a link to a on line Voltage-Current-Resistance-Power calculator I use that may be of some use.
http://www.opamplabs.com/eirp.htm
Click on the Opamps Labs Inc Web site since 1995 link for more on line calculators.
If you need a parallel resistance calculator, search Google for one.
Rich
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
SpaceMouseMy question has to do with wiring the building I'm on. I decided to go with 4 14 v grain of wheat bulbs in parallel. I want to use a resistor to dim the lighting. So I have two questions. 1) How can I use math determine the size of the resistor to use. 2) Can I run multiple lights using the same resistor?
1) How can I use math determine the size of the resistor to use. 2) Can I run multiple lights using the same resistor?
If I understand what you are asking I have say for #1 you can't. With a 12 volt power supply and 14 volt bulbs in parallel they are already going to be running in a slightly dim mode. This is a good thing because lamps run under their voltage limit will last much longer. The real issue is the dimness factor. Obviously the bigger the resistor the dimmer the bulb. How dim you want them is not something that can be put into a math formula. On the other hand if these were say 1.5V micro lamps, math could be used to get a resistor value to lower the voltage to a proper level to keep them from burning out. Then anything larger than that value would begin to dim them.
I would recommend trial and error. Get some resistors of various values and them until the desired dimmness is found.
For #2 the answer is yes. Watts=Electromotive Force * Intensity (W=EI). Simple take the current draw of a single bulb and multiply by 4 (the number of lamps in parallel). Then multiply by the voltage (12) and that should give the watts. Resistors normally come in 1/8, 1/4, 1/2, 1 watt configurations, larger wattages are available. So bulb draws 30 milliamps x 4 = 120 milliamps. 0.12 Amps x 12 V = 1.44 watts. So one would need 2 watt resistors. As one can see a single 30 ma or .03 A x 12 V = 0.36 W. In this example case, I would guess it would be cheaper to have 4 single half watt resistors instead of the single 2 watt one.
On your other musings - a computer power supply is going to be regulated. That means it will supply a constant voltage regardless of the current draw. Cheaper supplies like the old Bachmann controllers will vary in their output with the curent draw. One might have noticed that lamps hooked to the accessory output will dim at least temporarily when a train is started on the DC output.
The basic formula is E=IR (Voltage equals Current times Resistance)
Say you have a 12 volt supply and you have a bulb that is rated 1.5 volts and 30ma(miliamps)
To understand this some think of voltage as pressure and current as flow, just like a faucet.
The current will always be the same in a single wire path.
To calculate the resistor you first remove the bulb's contribution from the circuit.
There are 12 volts available and the bulb takes 1.5. 12 minus 1.5 equals 10.5 volts. This is the voltage (pressure) that the resistor has to absorb.
So now you take the 10.5 volts and divide it by the 30 milliamps which is .030amps. Remember current stays the same through the single line circuit.
10.5 divided by .030 is 350. So you use a 350 ohm resistor. If you want dimmer you raise the value of the resistor. Lower will make it brighter but could make it burn out much faster.
Good Luck
I have not decided on a power supply.
I have my choice of a 12v 5v power supply from a computer. A power supply from a Bachmann EZ track set. A few various cheapy power supplies and a Tech II.
The computer could handle all the lighting needs of the layout but does not have a rheostat. If I used several of the old PSs I could control each sector of the layout separately.
Just thinking aloud. Comments welcome.
My question has to do with wiring the building I'm on. I decided to go with 4 14 v grain of wheat bulbs in parallel. I want to use a resistor to dim the lighting. So I have two questions.
1) How can I use math determine the size of the resistor to use.
2) Can I run multiple lights using the same resistor? (I know they are cheap.)
Chip
Building the Rock Ridge Railroad with the slowest construction crew west of the Pecos.