No, the lamps are not in series, just the one resistor. The lamps are in parallel. If one opens, the current goes down thru the resistor resulting in more voltage to the lamps. They will continue to burn but because of the higher voltage now applied, one of the 3 will open after some time, and now the remaining 2 lamps have even more voltage on them.
This won't be such an issue in this case because of the 14 volt lamps and 12 volt REGULATED supply. But the effect is there. Any way the remaining lamps will burn brighter while they last.
Bob T
My approach is a little different than those here. I watch and collect old 5V and 9V wall warts and they seem to be everywhere at pretty decent 1+ amps. Instead of building a single big power supply and running 'big wire/power' around, I get a power strip and plug in these little guys around the layout. I then run 5 or 9 v lines to my buildings. In the buildings, I use two 1N4000-series diodes in series and wire my 1.5v lamps or LEDs in parallel. The 1N4000 diodes are super cheap on-line and make it easy to just put in the building.
On the base of the building, I write how many lamps I have. Now when I put the building in place, I figure the resistor I need for the number of lamps and the small voltage drop I have from these lower voltage power supplies. If I have a 5v wart, then I only have to drop 3.5v. I put this resistor in series with the line running to the building. This has worked well for me as I have been able to 'rescue' these little power supplies for free.
Everyone has done a great job in explaining how to figure the resistors and their wattage elsewhere so I won't repeat that info.
In the few places that I have used the grain of wheat bulbs, I used a 12 V wart which gives reasonable brightness. If the grain of wheat bulbs are used with the grain of rice bulbs, I still need only 3 wires running to the building as I can wire a common line between the two warts and the other 2 lines will give me 5 and 12 V. Color code the lines in the building so you don't connect a grain of rice bulb to the 12 V line. You will get to see that bulb one time and one time only - Flash!
You could do that if all the bulbs you are using are the same ratings. You typically would use a constant voltage power supply though so that you know what's feeding the circuit. Dialing in the voltage using brightness is blind and could result in blowing them, especially when a guest thinks it's a controller for a train and turns it up.
Springfield PA
Okay, to everyone I admit I'm not the brightest bulb (pun intended) but please tell me why I can't just take one of those cheap Bachmann, Life Like, etc. power packs (that seem to reproduce all by themselves), hook it up to a strip (or whatever it's called) and hook up lights to the other side of the strip? Then turn the rheostat to whatever brightness I want? By now you've correctly guessed that I am not an electronic wizard, so please be gentle.
Haven't tried any of the newest type computer supplies (they're all in use in running computers), but the slightly older ones I did mess around with don't need a minimum load to turn on. They DO need a minimum load to maintain decent regulation. If turned on with no load, I usually measure the $12 down to 11 volts or so, the +5 comes out high at almost 6 volts sometimes. Adding a load gets the +12 up where it should be, and drops the +5.
I have one I permanently modified with a big sandbar resistor (you need a high wattage resistor for this - 1/4 and 1/2 watt types are NOT suitable) heatsinked to the power supply case, and added binding posts for +12, +5, and ground. All other wires were cut back and insulated. I also added an on/off switch and a power LED on the Power Good lead. However - DO NOT attempt this if you are unfamilair with these types of power supplies. COmputer power supplies are switching power supplies with large capacitors that hold a large jolt long after you've pulled the plug. Best way to go about this is to get some Molex connectors that match the power supply conectors and locate the switch, power LED, load resistor, and binding posts on a small circuit board that the power supply simply plugs in to. No need to open up the power supply in this case. Plus if one power supply dies you can just plug another in its place.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
{So I want to find a good (& cheap) source of suitable warm white LEDs.]
I picked up two boxes of warm white christmas lights (50 light strings) at a (national chain) pharmacy last christmas for $5.00 each. One regular price with 5mm LEDs (These have a cone shaped concave end that diffuses most of the light radially instead of directing it in a beam), one on sale at half (of $10.00) price using 3mm LEDs and a faceted plastic "bulb" (the plastic bulb pops off the led with pliers). The "bulb's of both strings are made to be removed from the socket. That works out to ten cents per LED.
[I have my choice of a 12v 5v power supply from a computer.]
If you choose a computer power supply, use a fuse or circuit breaker in the output! A computer power supply requires a MINIMUM load before it will turn on. Usually one half to one amp on either the +5 or +12 volt supply will work. The newer supplies (2x10 or 2x12 pin plugs) require a load from pin 14 to ground to turn on (unless it is a name brand computer with non-standard wiring). I find a 100 ohm resister in series with a spst switch works. These also supply +3.3 volts. The older type with a push on-push off switch needs only a minimum load, but does not have the 3.3 volt output. All the output voltages have a Common Ground.
The electrical math has already been well covered.
No, he's describing a parallel circuit with one monster resistor providing the voltage drop. Loads in parallel add current, so 10x 30ma lamps in parallel draws 300ma. Now, cut off 2 of those lights, the circuit now draws 8x30ma, 240ma. Assuming a reasonably regulated power supply putting out a constant 12 volts, and also the fact that loads in series get equal current - the parallel bunch of lamps is in series with the reistor - if the load drops to 240ma, the current across the resistor is also 240ma. The resistor value didn't change, so the voltage dropped across the resistor MUST change. Make it simple, 100 ohm resistor. At 300ma (all 10 bulbs working), the resistor drops 30 volts (see this isn't even PRACTICAL). So say we has a 31.5 volt power supply that lets 1.5 volts get to the bulbs - perfect. Now at 240ma, the resistor drops only 24 volts. 31.5 volts - 24 volts is 7.5 volts to the bulbs - instant poof!
It's really the same with LEDs - 1 resistor per LED, not one massive resistor to be the limiter for a whole string of LEDs.
Diodes for reducing the voltage do not have this issue, at least with the voltages and current levels we are talking about. No matter if the load is 500ma or 200ma, the dide will drop .6 volt per.
This resistor problem is also why an old HO rheostat power pack has no control over an N scale loco or even many modern HO locos with more efficient motors. The less current flows through the rheostate, the less voltage it drops
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.
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.
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.
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
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.
...
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
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
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
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".
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!
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.
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.
Engineer Jeff NS Nut Visit my layout at: http://www.thebinks.com/trains/
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.
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.
Chip
Building the Rock Ridge Railroad with the slowest construction crew west of the Pecos.
Hmm, This thread is getting scary. Anyone have a fire extinguisher? Quick call Murphy from the other thread.
So I should take out the 8 ga wires I stuck in the dryer outlet?
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 !!!
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.
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.
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?
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.
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.
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?
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.
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.
SpaceMouse1) How can I use math determine the size of the resistor to use.
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.
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.
It takes an iron man to play with a toy iron horse.
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.
Cool idea Chuck.