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.
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).
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 [^]
The trick with LEDs is that the voltage across them stays the same, and you are actually using the resistor to limit the current. So you can make the LED dimmer by lowering the voltage of the power supply, which reduces the current because less voltage drops over the resistor, or by increasing the value of the resistor. LEDs are not great for dimming, because they are really on/off devices, but with some fiddling, you can get what you want.
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…
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.
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.
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).
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, 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. [:)]
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.
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
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 c
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.
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 w
Yep. You run into the same basic problem trying to control the voltage across bulbs in parallel with one resistor (the pot is immaterial, unlesss you are going to adjust it when a bulb burns out) as you do trying to control the current though parallel LEDs with one reistor. When one goes out, the load changes, and you’ve lost control, most likely reducing the life of the rest of the bulbs or LEDs. Separate resistors for each device make a lot of sense, though with LEDs the risk could be considered acceptable, since they are very unlikely to fail as long as you have things right ot start with.
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).
For a bunch of bulbs, building a little regulator would make a lot of sense. All the problems go away. In any case the lower voltage supply makes sense, reather than throwing away all the voltage across resistors. For a headlight in a loco, it’s no big deal. But for a bunch of structures the wasted power/heat becomes pretty significant, as you pointed out.
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.
