I know I have a mental block the size of the Chinese border wall. Give me multi-dimensional forecasting or emotional logic, and I can get it. Heck, I had two years of Calculus and a year of Calculus Physics. But give me a linear equation like Ohm's Law and my brain shuts down.
Okay, I understand in concept V=IR.
But when I have a power pack, a resistor and an LED, I don't know how to apply it.
The resistor is R (and to a minor extent the LED)
But what is V and what is A.
I'm using a Bachmann power pack and a 3v LED. What I'm trying to do is dim the LED so that it gives off the effect of a kerosene lamp.
Chip
Building the Rock Ridge Railroad with the slowest construction crew west of the Pecos.
If your power pack puts out 12 volts.
Your LED uses 3 volts.
You want to drop 9 volts across a resistor (assuming constant voltage from the power).
IIRC, LEDs use about 20mA or 0.020 Amps (that's the I part in V=IR) Verify that before going furuther.
If ALL that are in your circuit are the power pack, the LED, and the Resistor, then the resistor needs to be, R=V/I = 9/0.020 = 450 ohms. Typically you would use a 470 in this case. Now, The resistor is going to disappate heat or power. Power (P) = VI = 9 x 0.020 = 0.18 watts. A quarter watt resistor should be large enough.
I don't know that you are going to be able to dim the LED but to try, Increase the voltage across the resistor by 1.5 volts and redo the math (leaving that to you).
Okay, I think this sinking in.
I am not concerned with the voltage of the power pack in the equation rather I'm trying to calculate the size of the resistor based upon the difference in the voltage of the power pack and the LED. Assuming that the pack is 12v and the LED is 3v, then I use 9v as value of V. This has always been the missing part of the formula for me I think.
So how do I dim the LED, decrease the voltage in the Power pack with the arrow pointy knobby thingy?
SpaceMouse wrote: Okay, I think this sinking in. I am not concerned with the voltage of the power pack in the equation rather I'm trying to calculate the size of the resistor based upon the difference in the voltage of the power pack and the LED. Assuming that the pack is 12v and the LED is 3v, then I use 9v as value of V. This has always been the missing part of the formula for me I think. So how do I dim the LED, decrease the voltage in the Power pack with the arrow pointy knobby thingy?
If you are using the variable outputs on the power pack for the 12V then "yes". Between off and 12V the LED will get brighter as you raise the output voltage.
If all of you folks keep writing so many threads on building lighting and 12V power supplies I may be forced to find some company to build my DCC controlled power supply design so that everyone can have an easy to use system to power building lights, control panels, switch machines and more
Engineer Jeff NS Nut Visit my layout at: http://www.thebinks.com/trains/
Jeff But it's a dry heat!
Exactly, Chip, As long as you know two of your variables, voltage {V} requred for the LED or lamp, and the current {A or I} you can figure out how to avoid letting the smoke out of the device.
This also works for lights in DCC circuits as well. Oft times the lamps have a current rating along with a max voltage. In this case, ALWAYS err on the heavier side (as we did in my example). If you put too small a resistor in (47 ohm) for instance, you could draw too much current and burn the output of the decoder as it's likely to go before the resistor smokes. I'll Splain some day the expensive way I know that.
I know you didn't ask for it but someone may find it useful..
Resistor color codes (along with a couple limricks to remember it by)
Just don't forget about that power thing P = VI (or VA). Many times, in my case anyway, resistor required for my DCC lights is actually 1/2 watt because the voltage is usually higher than 12v (around 17 in my case). 1/4 would probably do but after awhile it'd get warm. I always tend to err on the heavy side when figuring these things. If you put too small a resistor (power-wise) in, interesting things will happen for about 1 second or 2.
Jeff, et al,
Would it be better to use a grain or wheat bulb if I want to dim the lights? These suckers are going to be buried inside craftsman kits and I figure if the lights go out, well, that structure will be dark. I'm thinking an LED is a little less likely to give up the ghost. But if I can't dim them, then they don't fit the bill.
Now I can use a 5v or 12v power supply from a computer. But I figure there may be only 20-30 lights powered from it.
I'm also planning to run the lights in parallel in the form a bus under the town.
I myself would be hesitant putting any kind of light bulb or LED in a building that I couldn't replace. Have you considered fiber optics, with the light source external of the building? Just another option to think about.
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.
Chip,
A one amp 5 or 12 volt non-switching wall wart will power a lot of lights. If you have the computer power supply on hand then by all means go for it.
For interior lighting, I have to agree with the others who've stated that the incandecant looks better and is more controllable. If you are looking at single point lights and just want a light showing in a particular scene, Gandydancer's suggestion of fiberoptic is worth considering. You can still dim it's source. What exactly I do would depend on how easy it is to replace a light/LED.
After thinking about this a minute, and doing a couple drawings (below), I'd go with 12 volt bulbs and a 12 volt supply. You can go with LEDs or smaller bulbs if you wish (the 12v ones are pretty large), but things get much easier using 12 volt bulbs.
The above circuit has a single limiting resistor and a bank of parallel lights. EAch light is going to draw the same amount of current and when connected like this, it is additive. So, If each of these 4 draw 20mA like your 3 volt LEDs, then your circuit current will be 80mA (or 0.080A) give or take (close enough). Each time you add another parallel lamp or LED, you add another 20mA to that. Depending on the bulbs you use, the resistor has to be able to drop and dissapate the heat of the excess voltage. Let's assume you are using 1.5 volt bulbs and the 12 volt supply (just to exagerate this a bit). You need to drop 10.5 volts across the single limit resistor in an 80mA circuit. Works out to 131 ohms (and change). The physical size of the resistor becomes, 1 watt (need to disappate 0.84W).
If you're using 12 volt bulbs and a 12v supply, then knock out the limit resisitor (in both circuits). For light bulbs, the above circuit is fine. If you are using LEDs however, the below one is better (though the top will work).
The limit resistors in each branch are identical. But to calculate and make sure, another example.
20mA 1.5V bulbs. Still need to drop 10.5 on the resistor. 525 ohms, go for a 560, 1/4Watt should be fine. Again, if using 12/12, then knock out the limit resistors.
The Dimmer is a simple Poteniometer. 5 watt 1000 ohm LINEAR taper should do. They usually come with 3 tabs and some also have a switch, you connect the power supply to one of the end ones, and connect your lights to the center one (usually). Leave the other tab hang empty. Use the tab to the right of center however instead of left as I've drawn it. The way I have it drawn, full brightness will be when the pot is turned all the way down (sorry bout that)..
PS. Excuse the sloppiness of the drawings, still trying to get use to writing in photoshop with my tablet pen.
I think I would do some experimenting to figure what works best for you, Chip. I think that you could use cheap potentiometers and be able to control the brightness of each LED individually, if you wanted to. I'd either put a resistor in series, or use a pot that didn't go below a "safe" value to make sure you never gaev one too much current. The effect of dimming an LED may well vary from one type to another, that's why I'd experiment.
SpaceMouse wrote:I know I have a mental block the size of the Chinese border wall. Give me multi-dimensional forecasting or emotional logic, and I can get it. Heck, I had two years of Calculus and a year of Calculus Physics. But give me a linear equation like Ohm's Law and my brain shuts down. Okay, I understand in concept V=IR. But when I have a power pack, a resistor and an LED, I don't know how to apply it. The resistor is R (and to a minor extent the LED)But what is V and what is A. I'm using a Bachmann power pack and a 3v LED. What I'm trying to do is dim the LED so that it gives off the effect of a kerosene lamp.
Spacemouse, we must have gone to the same school. LOL! Between this thread and my thread "simple question" we're working Engineer Jeff and the guys pretty hard. If you figure it out, come over to my thread and explain it to me.
- Harry
SpaceMouse wrote: Jeff, et al,Would it be better to use a grain or wheat bulb if I want to dim the lights? These suckers are going to be buried inside craftsman kits and I figure if the lights go out, well, that structure will be dark. I'm thinking an LED is a little less likely to give up the ghost. But if I can't dim them, then they don't fit the bill.Now I can use a 5v or 12v power supply from a computer. But I figure there may be only 20-30 lights powered from it. I'm also planning to run the lights in parallel in the form a bus under the town.
Actually LEDs will work you just might not have as much control between totally dark and full brightness. I suspect you are thinking of LEDs due to longevity. I use LEDs when I want pure white light, like for a soda machine.
You can see how white it is. I use a resistor to adjust the brightness leave it fixed. Here's the same thing with a GOW bulb.
Big difference. For inside buildings I use GOW bulbs because most homes and such use incandescant lighting. Office buildings and such use flourescant so one could argue that you could go either way, white LED or GOW. In order not to worry about bulb life I use Minatronics 14V bulbs that have a 16,000 hr rated life.
http://www.miniatronics.com/Merchant2/merchant.mvc?Screen=PROD&Store_Code=M&Product_Code=18-014-10&Category_Code=1_1&Product_Count=5
I run them at around 12.5 - 13V max. At that voltage they will last forever.
Also a trick I use for lighting structures, so I can replace a bulb, if necessary is that I drill a 1/4" hole into the area I want to light from underneath the layout. This obviously only works on some structures and ground floors. Then I take a 1/4" dowel pin and drill a 1/8" hole lengthwise through it. Then I take one of the bulbs above and slip the wires through the hole in the dowel pin. I place a drop of glue on the base of the bulb to hold it to the dowel pin (keeps it from moving). Then I stick the dowel pin up through the hole into the bottom of the structure. I leave enough sticking through the layout so I can pull it back out if I ever need to. You could do the same with brass or plastic tubing for thicker layouts.
The math isn't so easy because an LED is a non-linear device. Once a voltage drop threshold is achieved the current (and light) flow will increase to the maximum available with almost no increase in voltage drop. If the maximum available current is greater than the diode's rating, the magic smoke gets let out. So a current limiting device - the resistor - is inserted in the circuit.
Most LEDs have a max current rating of 20ma and will start lighting at less than 10ma, so I am personally reluctant to design for more than about 15ma. So I tend to use larger resistors than the typical 470 ohms calculated for 12 volt power supply. If I need brighter, then I decrease resistance.
FWIW, the LED voltage drop varies with the color of the LED. Red and green tend to be down around 2.5V, with the blue-white being slightly above 3V. For your purposes, the golden white LEDs are probably best.
Others have given you the formulas and calculations so I won't repeat them. Design the circuit so that you can't blow out the LED with the power pack at max. Your Bachmann power pack likely has a 50 to 70 ohm pot as the speed control, and could put out as much as 14 volts with little to no load (transformers are not totally linear, and the rectifiers are just 2-4 diodes). The amount of voltage dropped by the pot is going to depend upon the load current. With a single LED load, the pot/speed control is going to be pretty useless for controlling brightness. The variable output of the power pack will be much more controllable with a load of 100ma or more.
Final point to consider - LEDs are quite directional compared to incadescent bulbs.
my thoughts, your choices
Fred W
rolleiman wrote: Chip, A one amp 5 or 12 volt non-switching wall wart will power a lot of lights. If you have the computer power supply on hand then by all means go for it. For interior lighting, I have to agree with the others who've stated that the incandecant looks better and is more controllable. If you are looking at single point lights and just want a light showing in a particular scene, Gandydancer's suggestion of fiberoptic is worth considering. You can still dim it's source. What exactly I do would depend on how easy it is to replace a light/LED. After thinking about this a minute, and doing a couple drawings (below), I'd go with 12 volt bulbs and a 12 volt supply. You can go with LEDs or smaller bulbs if you wish (the 12v ones are pretty large), but things get much easier using 12 volt bulbs.The above circuit has a single limiting resistor and a bank of parallel lights. EAch light is going to draw the same amount of current and when connected like this, it is additive. So, If each of these 4 draw 20mA like your 3 volt LEDs, then your circuit current will be 80mA (or 0.080A) give or take (close enough). Each time you add another parallel lamp or LED, you add another 20mA to that. Depending on the bulbs you use, the resistor has to be able to drop and dissapate the heat of the excess voltage. Let's assume you are using 1.5 volt bulbs and the 12 volt supply (just to exagerate this a bit). You need to drop 10.5 volts across the single limit resistor in an 80mA circuit. Works out to 131 ohms (and change). The physical size of the resistor becomes, 1 watt (need to disappate 0.84W).
One of the problems with this type of design is if a bulb burns out. If a bulb burns out more current will flow through the other bulbs in order to equalize the remaining current flows of the filaments against the resistor value. This could cause enough current through the other filaments to cause them to burn out also. Your resistor per leg design is more predicatable and reliable.
jbinkley60 wrote:One of the problems with this type of design is if a bulb burns out. If a bulb burns out more current will flow through the other bulbs in order to equalize the remaining current flows of the filaments against the resistor value. This could cause enough current through the other filaments to cause them to burn out also.
One of the problems with this type of design is if a bulb burns out. If a bulb burns out more current will flow through the other bulbs in order to equalize the remaining current flows of the filaments against the resistor value. This could cause enough current through the other filaments to cause them to burn out also.
Ohm's law says you're right, to a point, if the pot wasn't in the circuit. Remember that it is going to drop voltage as well. The pot, the resistor, and the lamp bank make up a 3 part voltage divider. With 3 bulbs, the current draw of the bank will be 60mA (assuming 20 each). If only the limiting resister were there, it would need to be 175 instead of the 131. Of course, what exactly happens will depend on where the pot is set. The load, is only going to draw what it needs. We aren't forcing it to remain at 80mA (in the example).
I do agree, the other circuit (of the two) is more foolproof.
rolleiman wrote: jbinkley60 wrote: One of the problems with this type of design is if a bulb burns out. If a bulb burns out more current will flow through the other bulbs in order to equalize the remaining current flows of the filaments against the resistor value. This could cause enough current through the other filaments to cause them to burn out also. Ohm's law says you're right, to a point, if the pot wasn't in the circuit. Remember that it is going to drop voltage as well. The pot, the resistor, and the lamp bank make up a 3 part voltage divider. With 3 bulbs, the current draw of the bank will be 60mA (assuming 20 each). If only the limiting resister were there, it would need to be 175 instead of the 131. Of course, what exactly happens will depend on where the pot is set. The load, is only going to draw what it needs. We aren't forcing it to remain at 80mA (in the example). I do agree, the other circuit (of the two) is more foolproof.
jbinkley60 wrote: One of the problems with this type of design is if a bulb burns out. If a bulb burns out more current will flow through the other bulbs in order to equalize the remaining current flows of the filaments against the resistor value. This could cause enough current through the other filaments to cause them to burn out also.
Incandescant bulbs are dynamic and not fixed load devices. For my point, the pot doesn't matter much in the burnt out bulb scenario. Let's assume things are at the design point where each bulb has 20ma flowing through it. As you indicate the resistor/pot combination would be at 131 ohms. Now if we lose a bulb (and we assume for an instant that the bulbs are fixed load devices) then the current draw of the remaining bulbs will drop to 60ma. This would change the voltage drop across the resistor/pot to .06 * 131 = 7.86V and then 4.14V across the bulbs. So what would happen next is that with the additional voltage on them the bulbs would draw more current. They'd all have to draw 26.7ma each in order to get back to 80ma to bring the voltage back down to 1.5V across the bulbs. We don't know if they can withstand that or not. Suppose two of the three could withstand it. Another bulb would blow and now the remaining two would have to sink 40ma each.
Where all of this might not occur is if a bulb burns out and the pot is adjusted so that less than 20ma is flowing through each bulb. That was your point about the pot setting. That would mean the combination of the pot/resistor was higher than 131 ohms. If it was set so 40ma was flowing through the 4 bulbs then you could lose a bulb and be fine until you tried to turn up the brightness and then you could still exceed 20ma per bulb.
Interesting.. 3 Jeffs discussing Chip's lighting..
I will (and did) concede that the single resistor solution isn't the best. We could get a little fancy and turn this into a fixed voltage supply but it's a little more involved than Chip likely wants to get into (All he wanted to know at first was how Ohm's Law works with respect to a single LED circuit).
The largest thing I didn't really like about the circuit was the physical size of the limiting resistor. For just a 4 bulb bank, the limit resistor needed to be a 1 watt. If we went to 8 bulbs, now it needs to be a 5 watt (actual disappation is 3.35w).
A simple solution for using one device to limit current is a 10V 1W zener diode in place of the resistor. A 1N4740A would work fine. Just use it in place of the resistor. You'd be limited to 5 bulbs in parallel and at that point the diode would be at its maximum rating. A 1N5347 is a 5W version which could be used for up to 250 20ma bulbs.
Well, you can try a 10,000 ohm resistor with/for the white LED's. And then paint the LED's with yellow stain glass paint a time or two. That is what I am doing with Lunar light signals.
10,000 = 10K = (brown, black, orange)
Jeffs,
This has been a very enlightening and educational thread for me. When you finish, I'd like to ask you a few questions about quantum mechanics. Just kidding, but it really has been very helpful.
Let's see we take 1 part theoretical physics, 2 parts superstring theory, 1.5 parts of mathematical renormalization and a full 3 parts of quantum field theory and we'll have us a whale of a discussion underway Glad I could help with the electronics stuff...
jbinkley60 wrote: Let's see we take 1 part theoretical physics, 2 parts superstring theory, 1.5 parts of mathematical renormalization and a full 3 parts of quantum field theory and we'll have us a whale of a discussion underway Glad I could help with the electronis stuff...
Let's see we take 1 part theoretical physics, 2 parts superstring theory, 1.5 parts of mathematical renormalization and a full 3 parts of quantum field theory and we'll have us a whale of a discussion underway Glad I could help with the electronis stuff...
Ooh, ooh, I know this one. Superstring is where you use your lights frm your Christmas tree and run them from house to house on your layout. I think setting them to "chase" would be really cool.
Ha!
My quantum physics is 25 years old. They probably teach it in kindergarten now! Or more likely, it's become a graduate course!
Ah, Grasshopper, you have come very far on your journey. The final lesson you must learn is:
One wire hook 'em up......Two wire screw 'em up!
jc5729 John Colley, Port Townsend, WA