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.
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
Springfield PA
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
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
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.
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.
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
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.
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.
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)
Thanks everyone. This has been very helpful.
I'd like to hijack this thread a bit. As you know I'm starting a new layout, and what I need is a lighting strategy. I listed a few options for power. Sounds like the computer power supply is the most viable option.
More background. The layout is set in 1905 which means that some structures will use early electricity and others will burn kerosene. Either way, the lights will need to be dim.
I used grain of wheat bulbs in my current structure only because that is what was available locally and I wanted to keep moving forward with my project. but bulbs burn out and most of my structures will be craftsmen kits and replacing bulbs will not be an option.
So I want to find a good (& cheap) source of suitable warm white LEDs.
So with the idea that I will have at least 3 grain of wheat bulbs at 14V and a crapload of LEDs, how would you set-up a good lighting system?
Cool idea Chuck.
rrinkerBetter 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
I would quibble with the "No Heat" statement. Any circuit element which has a current flow and a voltage drop will consume power, measured in Watts and equal to the product of the voltage and current. That power has to go somewhere, and most circuit elements give it up as heat. So, while a diode with a voltage drop of only 0.7 volts doesn't produce much heat, it does produce some.
It takes an iron man to play with a toy iron horse.
I like 12 volt supplies. I have mostly 16-volt lighting on my layout, and the 12 volt supply runs them well below their rated voltage, which should prolong bulb life and also gives me a nicer, warmer glow.
A few weeks back, I burned out a supply. I had inadvertently wired a hard short while repairing a loose wire under my layout. Before installing a new supply, I went to Radio $hack and bought a fuse block and some fuses. Now, I've got the supply protected so this won't happen again. A word to the wise.
SpaceMouse1) How can I use math determine the size of the resistor to use.
There is a nifty little formula called "Ohm's Law" that will do the trick quite nicely. The formula is expressed:
e = i r
Where:
e = volts
i = current, in amps
r = resistance, in ohms
You can use this formula to determine one of: volts, current, or resistance, where the other two are known. Thus you can rearrange this formula like so:
r = e / i
to solve for resistance where volts and current are known.
Or:
i = e / r
to solve for current where volts and resistance are known.
In your case you want to solve for resistance. Therefore you will need to know the amount of voltage you'll use, and the amount of current to permit through the circuit.
Just for an example, let's assume your going to use 5 volts and permit 25 mA through your circuit. First we'll need to adjust for amps, because that's what our formula uses. So:
25 mA equals 0.025 Amps
So if we plug those values in we get:
r = 5 / 0.25
Which equals 20, which we know from above is in units "Ohms", so the result is a 20 Ohm resistor.
Technically speaking, the volts value is supposed to be DC and if you use AC, you should technically use a formula called "Root Mean Square" to approximate a DC value. For reasonably small values and non "electronic" loads, you can probably ignore that difference.
Many people ask how in heck can I remember how to flip those formula bits around? Here's an easy way to remember...
We start with "e = i r". Think of "e" as "Eagle", "i" as "Indian", and "r" as "River". So using that nomenclature, let's rewrite the formula:
Eagle = Indian River
Rearranged it becomes:
Indian = Eagle over River
Rearranged again it becomes:
River = Eagle over Indian
Just remember the Eagle is always over everything, and the Indian is always over the River.
You'll just have to remember though which one doesn't get the dividing slash-- but I think its easy if you just remember "e = i r", the other two forms get the slash.
Also its interesting to note that if you remember about the "Eagle", "Indian" and "River" you can rearrange any triad equation. Just substitute the letters "e", "i", and "r" for whatever's in the formula and you'll be able to figure it out with the "Eagle-Indian-River" trick.
John
SpaceMouse Thanks everyone. This has been very helpful. I'd like to hijack this thread a bit. As you know I'm starting a new layout, and what I need is a lighting strategy. I listed a few options for power. Sounds like the computer power supply is the most viable option. More background. The layout is set in 1905 which means that some structures will use early electricity and others will burn kerosene. Either way, the lights will need to be dim. I used grain of wheat bulbs in my current structure only because that is what was available locally and I wanted to keep moving forward with my project. but bulbs burn out and most of my structures will be craftsmen kits and replacing bulbs will not be an option. So I want to find a good (& cheap) source of suitable warm white LEDs. So with the idea that I will have at least 3 grain of wheat bulbs at 14V and a crapload of LEDs, how would you set-up a good lighting system?
Minatronics has 14V 30ma bulbs that have a 16,000 hour lifespan. I use them for my structures. Also, if you are looking for general lighting in your structures, here's a trick I use. I drill a 1/4" hole under the structure through whatever material the structure is sitting on. Then I take a 1/4" wooden dowel pin and drill a 3/32" hole lengthwise down the middle. Then I take one of the 14V bulbs and thread the wires through the hole and put a drop of glue under the bulb to glue it to the end of the dowel pin. Then I insert the dowel pin up through the base of the structure and through the 1/4" hole I drilled to start. This makes changing bulbs easy, although with 16,000 hour bulbs, I haven't had to change one yet. This obviously won't work for certain types of lighting approaches.
Engineer Jeff NS Nut Visit my layout at: http://www.thebinks.com/trains/
Jeff, I like that idea. I don't have a place for my current structure, but once I start placing structures that will work. 16,000 hours is a lot. I don't know if I will even get that much total running time, let alone night running--which just might be relegated to official ops sessions.
SpaceMouseI used grain of wheat bulbs in my current structure only because that is what was available locally and I wanted to keep moving forward with my project. but bulbs burn out and most of my structures will be craftsmen kits and replacing bulbs will not be an option.
I would think you might want to consider a lightly "flickering" look for the kerosene lanterns. Real bulbs would work better for that. Simple electronics could do that. One could even make several "flickering" buses so fewer flicker circuits would be needed.
Are your structures going to have interior with floors or just glowing windows? I asked because it might be better to just mount the bulb/LED on the layout and set the structure down over it. That way if one ever did burn out you could replace it from below.
I would NOT run just one big 12V bus from the computer power supply - they're not great at handling accidentle shorts, and if it's a fairly recent one you're talking 20+ amps - that's a lot of potential. Think house wiring, and divide the power into multiple branch circuits each with its own fuse at a safer level, like 1-1.5 amps. For LEDs, calculate the current based on the resistor used, for lamps use the specifications, ie 15ma, 30ma, etc. For each lamp or LED on a circuit, just add the current and keep it below the limit of the fuse for the branch. ie, 10x 30ma lamps = 300ma of current. 20 would be 600ma, still good on a 1 amp fuse. 30 would be 900ma and pushing it. 25 bulbs at 30ma is probably the limit per 1 amp fuse. White LEDs with a 1K resistor results in about 10ma, so 75 of them would be 750ma, fine with a 1 amp fuse. Flicker effects though will almost always work better with lamps. That's not to say a circuit couldn;t be built that does work with LEDs - kind of like decoder flashing effects, some like TCS can be adjusted so they work well with LEDs and lamps, other like Digitrax work much better with real lamps than with LEDs.
rrinkerthey're not great at handling accidentle shorts, and if it's a fairly recent one you're talking 20+ amps - that's a lot of potential. Think house wiring, and divide the power into multiple branch circuits each with its own fuse at a safer level, like 1-1.5 amps.
JUST TO BE REALLY CLEAR -- Randy is NOT suggesting you USE house wiring to power the lights on your layout, but rather the STYLE of wiring used in typical house wiring, where you have the MAIN POWER routed through a POWER DISTRIBUTION PANEL with CIRCUIT BREAKERS (fuses) and each "CIRCUIT" (wiring branch) carries only a SMALL PORTION of the overall power through the circuit.
"House Power", "Mains Power", "Line Power", all mean the same thing-- SERIOUS INJURY or DEATH if handled incorrectly !!!
So I should take out the 8 ga wires I stuck in the dryer outlet?
Yes, you need #2 ga wire for that.
Considering the sentence before that one was "I would not run a single heavy 12V bus around the layout" I think it was fairly clear that I was talking about dividing up said 12V lotsa amps supply and not suggesting running house current through the layout. I'm pretty much riddled with ADD myself and I don;t think I'd miss that connection, but I suppose YMMV. Slow down, sit back, and read - reading is fun. Don't skip steps because it would be "too much reading".
Hmm, This thread is getting scary. Anyone have a fire extinguisher? Quick call Murphy from the other thread.
Can anyone think of a reason why I shouldn't wire my lighting with Cat 5 twisted pairs? I got tons of the stuff.
Cat 5 would work fine but you'd have a lot of unused conductors. You wouldn't want to put too much load on the wiring if it was too much distance. The voltage drop would dim the lights. But for a few lights it would be fine.
You could easily get away with 10 or so lights/LEDs per circuit, and with 4 pairs you get 4 circuits on one cable. 40x15ma bulbs is 600ma, fuse that with a 1 amp fuse, then connect to your big power supply.
SpaceMouse Can anyone think of a reason why I shouldn't wire my lighting with Cat 5 twisted pairs? I got tons of the stuff.
I run 18ga thermostat wire for lighting buses. I get the 7 conductor stuff at Home Depot and run three buses (1 eatra wire). I do this so I have an interior structure lighting bus (with variable voltage supply to adjust the brightness of buildings), a switched on/off 12V bus for exterior structure lights and street lights, and a 12V always on bus to power accessories.
rrinkerYes, you need #2 ga wire for that. Considering the sentence before that one was "I would not run a single heavy 12V bus around the layout" I think it was fairly clear that I was talking about dividing up said 12V lotsa amps supply and not suggesting running house current through the layout. I'm pretty much riddled with ADD myself and I don;t think I'd miss that connection, but I suppose YMMV. Slow down, sit back, and read - reading is fun. Don't skip steps because it would be "too much reading". --Randy
It was reasonably clear, but I've seen people take the most innocuous looking directions and analyze them with their keen, legal-lawyerly like minds-- looking for loopholes-- and then decide that something that seems obvious to *everybody else* is really *ambiguous* and then use it to rig up something incredibly stupid and fry themselves. And then, technically, because you said it, you could be construed as legally liable for it. Sad, but true in these overly-litigious times we live in.
So anyway, its all good-- I got yer back!
jwhitten There is a nifty little formula called "Ohm's Law" that will do the trick quite nicely. The formula is expressed: e = i rWhere: e = volts i = current, in amps r = resistance, in ohmsYou can use this formula to determine one of: volts, current, or resistance, where the other two are known. Thus you can rearrange this formula like so: r = e / ito solve for resistance where volts and current are known. Or: i = e / rto solve for current where volts and resistance are known. ... Many people ask how in heck can I remember how to flip those formula bits around? Here's an easy way to remember...We start with "e = i r". Think of "e" as "Eagle", "i" as "Indian", and "r" as "River". So using that nomenclature, let's rewrite the formula: Eagle = Indian RiverRearranged it becomes: Indian = Eagle over RiverRearranged again it becomes: River = Eagle over IndianJust remember the Eagle is always over everything, and the Indian is always over the River.You'll just have to remember though which one doesn't get the dividing slash-- but I think its easy if you just remember "e = i r", the other two forms get the slash. Also its interesting to note that if you remember about the "Eagle", "Indian" and "River" you can rearrange any triad equation. Just substitute the letters "e", "i", and "r" for whatever's in the formula and you'll be able to figure it out with the "Eagle-Indian-River" trick. John
...
That Eagle River Indian process is something new. I find it's easier to just apply basic algebra to the equation for whichever part you want to solve for. (remember solving for X in math class?)
You have
If you want to solve for anything else you just divide each side by the what you want to get rid of (always remember to "Do onto one side of the equation that you do to the other"). For instance, if you want to solve for 'r' then divide each side by 'i' to move it from one side to the other.
e/i = (i r) / i
Dividing 'i' by 'i' cancels that out and you're left with
e/i = r OR r = e/i
Apply the same to solve for 'i' and you'd have to divide by 'r' on each side to move it.
e/r = (i r) / r
e/r = i OR i =e/r
D94RThat Eagle River Indian process is something new. I find it's easier to just apply basic algebra to the equation for whichever part you want to solve for. (remember solving for X in math class?)
Sure, I know, but not everybody is into math, and that's a neat mnemonic for remembering how to arrange the terms.
Here is a good reason to not use one resistor, besides the cost of a 2 watt part. The voltage "dropped" across the resistor, is dependent on the current thru the resistor. When all 4 lamps are burning, they pull enough to drop the voltage at the point where they connect to the resistor. Now, when one lamp burns out, there is less current through the resistor so less voltage wasted which means more voltage to the remaining bulbs. Now they will in short order, burn out one by one, each lasting a shorter time.
Better to use a cheap 1/4 or 1/2 watt resistor in series with each lamp. Others have given the formulas in their replies and seem to be valid.
Same thing applies to the LEDs, but the first one has to fail, so it may not be so likely. As I have stated before, we are not making life dependent devices.
Re: Indian , River and Eagle. Years ago it was a Rabbit in place of the River. The rabbit was on the bottom of any division it was a part of. Never could remember algebra manipulations, but the E/R/I scene was easy remember. Just like the learning color codes, these "crutches" can be very handy when you are with out references.
Bob T
From what you describe, when one bulb burns out they all extinguish. It's called a series circuit. They don't burn out, they just stop working because the open circuit created by the burned bulb opens the circuit.