Lighting Math

Hmm, This thread is getting scary. Anyone have a fire extinguisher? Quick call Murphy from the other thread.[:D]

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.

–Randy

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.

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! [swg]

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 cla

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.

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

–Randy

{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.

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

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.

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.

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!

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