In this past weekend's WPF thread (Dec 5,6,7 2014) I posted a picture of one of a fleet of cabooses that I am working on. JaBear asked me to start a thread showing the wiring diagram for the working marker lights and interior light.
I cannot take credit for the original wiring scheme. That goes to Mark R. I have modified it a bit to adjust the brilliance of the interior light down quite a bit, and I have added in a magnetic off/on switch.
Here is Mark R.'s original diagram:
EDIT: For those not sure of the electronic symbols, The round circle on the lower left with the + and - and two sqwiggles is the bridge rectifier. The rectifier is marked for output polarity. The markings may require a magnifying glass to see. Track power polarity doesn't matter. Note that the right side wire going from the track power to the top of the rectifier does NOT connect to the wire coming off the negative side of the rectifier!
The black bar at the top is the resistor. Polarity doesn't matter.
The center component with the flat line over the curved line is the capacitor. The capacitor will have the polarity marked on it but they don't always use + and - signs. Sometimes there is a black stripe down the side of the case adjacent to the negative post.
The LEDs are on the right of the diagram. They will have one lead longer than the other. The long lead is +.
To quote Mark R. : "The resistor is 1000 ohms and the capacitor is a supercap with a rating of 0.10 farads (not micro-farads) and a voltage rating of 5 volts. I can get away with the 5 volt rating on the capacitor because the resistor is in line BEFORE the capacitor lowering the voltage to acceptable levels for both the cap and the LEDs at the same time. The physical size of this capacitor is about the same as four dimes stacked together. Be sure to observe the polarity..."
I added an 847 ohm 1/4 watt resistor in front of the LED which will be used for interior lighting. I want it to suggest an oil lamp over the conductor's desk only. I put the interior LED in a light box so the light only shines out of one window.
I also added in a magnetic reed switch so the lighting can be shut off when the caboose is not in service. It is wired between one side of the track power and one input to the bridge recitifier.
Here is a picture of the turned down LEDs compliments of Mark R. They fit quite nicely into Utah Pacific marker light castings. I fill the castings with gel CA. The CA then holds the LED in place and also serves as a light conduit: Second EDIT: The CA doesn't appear to work as consistantly as a light conduit as I had hoped. When I tested the first caboose everything seemed fine. The marker lights were bright enough. However, I just tested the other seven cabeese and the results are hit and miss. I will have to re-do the marker lights/LED arrangement to get more consistent results. Please stay tuned.
Note that the capacitor takes 10-15 seconds to charge because of the resistor, so when you first test the circuit you have to wait before you will see any light.
Here are the caps on eBay:
http://www.ebay.com/itm/5-pk-Panasonic-Coin-Electrolytic-Capacitor-5-5V-0-1F-75OHM-Super-Ultracapacitor-/121448968111?pt=LH_DefaultDomain_0&hash=item1c46ec2faf
Bridge rectifiers:
http://www.digikey.com/product-search/en?pv721=12&pv69=3&FV=fff40015%2Cfff8052c&k=bridge+rectifier&mnonly=0&newproducts=0&ColumnSort=0&page=1&quantity=0&ptm=0&fid=0&pageSize=25
The magnetic reed switch is available from this gentleman. (The reed switch is optional):
http://www.trainelectronics.com/LED_Articles_2007/LED_103/index.htm
The marker lights are from Tomar Industries under their Utah Pacific brand. Unfortunately their website is down so I can't post a link.
Dave
I'm just a dude with a bad back having a lot of fun with model trains, and finally building a layout!
Thanks, Dave and Mark...
Very clear and consice information!
I wonder if Aileen's clear tacky gel would work better as a cement? I have had quite a few LEDs get fogged by ACC and maybe the tacky glue will allow them to be removed if needed? It dries water clear.
Ed
As an addendum to the original circuit, there actually should be a resistor between the capacitor and the LEDs, otherwise, the LEDs would just get a single burst of power from the capacitor in the event of a power loss. I used a 470 ohm resistor on mine.
In this configuration, the total resistance from track power to the LEDs is 1470 ohms, which still lights the LEDs just fine on track power. The 470 ohm resistor buffers the output of the capacitor causing a slow drain to the LEDs - my two LEDs will stay lit for more than a minute.
Here's the revised diagram ....
Sorry for any problems the original diagram may have caused. Dave, your adding the 847 ohm resistor in front of the LED solved my error either by common sense or pure luck !
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
A couple of things to note about these circuits:
First, when you have LEDs wired in parallel, best practice is to have a seperate resistor for each LED. If the LEDs have different electrical characteristics, they will not be illuminated evenly using a common resistor. For example, if you have a white LED with a 3.5 volt forward voltage and a red LED with a 2 volt forward voltage then the white LED most likely won't even light because it will only see 2 volts. Even with LEDs with the "same" electrical characteristics, you can possibly have enough variation form one to the next to have a visible difference.
The other thing is to pay attention to the voltage your applying to that cap. In the original circuit the resistor and the LED work to limit the voltage the cap sees to the forward voltage of the LED. In the revised circuit, the two resistors will work as a voltage divider between the source and the LED, so now the cap will see more than the voltage drop of the LED. As an example, if we have 12 volts input (after the rectifier), a 1000 ohm resistor before the cap, a 470 ohm resistor between th ecap and LED, and one LED with a 2 volt forward voltage, the resistor will act as a voltage divider between 12 volts and 2 volts. Your resulting voltage on the cap would be ((12-2) x 470/(1000 + 470)) + 2 = 5.197 volts, which would be OK; however, if you use a white LED with a 3.5 volt forward voltage you wind up with ((12-3.5) x 470/(1000 + 470)) + 3.5 = 6.218, which is too high for a 5.5 volt cap. Likewise, if you had a 14 volt input but still had the 2 volt LED ((14-2) x 470/(1000 + 470)) + 2 = 5.837.
Thanks for the technical side of things Robert. Why or how it works, I honestly don't know. I came up with that circuit over fifteen years ago, long before I had any real handle on how much of anything electronic worked !
It was probably (based on your calculations) a fluke that it even worked at all, but it did .... and still does to this day, keeping the lights shining in my cabooses !
MR had an article many years ago about these "new fangled" super caps, with a few wiring suggestions. My design was basically a modified version of what they were suggesting.
My logic at the time (again, not knowing any better) was that I always used a 1000 ohm resistor to drop the voltage down to around around 2 volts for the yellow LEDs I was using, the same resistor would be fine for dropping the voltage low enough to use the 5 volt cap. If I would have known then, what I know now, I would have probably thrown my hands in the air and forgot about even trying !
It's only been within the last year I actually took one caboose apart to see what I actually did do when the topic came up regarding constant lighting. Again, my cabooses have been rolling along fully lit for years, so I figured it was a viable solution. I'd probably question my methods today (as you have).
George at TVW Miniatures has a circuit very similar to what I had devised, what is the big difference between what he has and what I did ? ....
https://beta.groups.yahoo.com/neo/groups/soundtraxx/files/TVW%20Miniatures/
CSX Robert A couple of things to note about these circuits: First, when you have LEDs wired in parallel, best practice is to have a seperate resistor for each LED. If the LEDs have different electrical characteristics, they will not be illuminated evenly using a common resistor. For example, if you have a white LED with a 3.5 volt forward voltage and a red LED with a 2 volt forward voltage then the white LED most likely won't even light because it will only see 2 volts. Even with LEDs with the "same" electrical characteristics, you can possibly have enough variation form one to the next to have a visible difference. The other thing is to pay attention to the voltage your applying to that cap. In the original circuit the resistor and the LED work to limit the voltage the cap sees to the forward voltage of the LED. In the revised circuit, the two resistors will work as a voltage divider between the source and the LED, so now the cap will see more than the voltage drop of the LED. As an example, if we have 12 volts input (after the rectifier), a 1000 ohm resistor before the cap, a 470 ohm resistor between th ecap and LED, and one LED with a 2 volt forward voltage, the resistor will act as a voltage divider between 12 volts and 2 volts. Your resulting voltage on the cap would be ((12-2) x 470/(1000 + 470)) + 2 = 5.197 volts, which would be OK; however, if you use a white LED with a 3.5 volt forward voltage you wind up with ((12-3.5) x 470/(1000 + 470)) + 3.5 = 6.218, which is too high for a 5.5 volt cap. Likewise, if you had a 14 volt input but still had the 2 volt LED ((14-2) x 470/(1000 + 470)) + 2 = 5.837.
Thanks for the equation. A few of questions on it.
In the numerator(normal volts area) you multiply the voltage drop by the 470 Ohms(I think this is normal for divider calcs?). Then divide by combined R, then add the load V to get a Voltage where normally V/R equals I?
Richard
CSX Robert:
You are proving to me that my little electrical knowledge just might be dangerous. Others be warned!
If I can ask, would you do a calculation for the circuit as I have it installed now. That is: A 1000 ohm resistor between the bridge rectifier and the capacitor, 5.5V 0.10F capacitor, three white LEDs in parallel with one LED having an 870 ohm resistor in series. I have not added in the 470 ohm resistor that Mark suggests.
Am I exceeding the maximum voltage for the cap?
What happens when I add Mark's suggested 470 ohm resistor to the above?
If I am exceeding the maximum voltage, what would you suggest as a solution?
Thank you very much for sharing your detailed knowledge. I owe you an apology for not responding to one of your posts in the other thread which cautioned about overloading the capacitor. My bad!
I'm thinking you should be OK as long as at least one of the LEDs is directly across the capacitor. Sources or loads in parallel share the voltage, add the current - so an LED with a 3.5V voltage rating in parallel witht he capacitor means the capacitor also gets 3.5V maximum. If you put the resisot between the cap and ALL the LEDs, then you have the voltage divider effect and will be putting excess voltage to the capacitor.
Note that the whole reason this works has nothing to do with the 1K resistor - the capacitor gets less thn its rated voltage because of the parallel diode. Assuming you had an LED that could stand a lot more than the 20-25ma typical of white LEDs, and eliminated the 1K resistor, the cap would still only see 3.5V. What the 1K resistor does is reduce the CURRENT going to the LED (and by definition, since it is in parallel) and the capacitor. I'm surprised any of them is too bright - with a 1K resistor you are supplying about 12ma to the circuit at 12V. Divided by 4 until the cap charges, then sort of (close enough) by 3. Much lower and I'd be suprised if the LED even lights.
--Randy
Modeling the Reading Railroad in the 1950's
Visit my web site at www.readingeastpenn.com for construction updates, DCC Info, and more.
I'd be interested to know how / why my circuit works as well as it does .... it theory, apparently it shouldn't. For the record, I'm using two yellow LEDs in my cabooses.
After doing some searching online, I found a couple diagrams by other people that are very similar (if not identical) to what I came up with. So what I have must have some kind of merit to its design ....
Mark:
Maybe we could persuade CSX Robert to comment on the top diagram with the zenar diode. Too bad for him that he can't charge a consulting fee.
Mark R.After doing some searching online, I found a couple diagrams by other people that are very similar (if not identical) to what I came up with. So what I have must have some kind of merit to its design ....
Texas Zepher Mark R. After doing some searching online, I found a couple diagrams by other people that are very similar (if not identical) to what I came up with. So what I have must have some kind of merit to its design .... Yes it does have merit. I joined a club that used command control in 1984 and they were melting down their caboose. So I cobbled up a similar scheme using the trial and error method just to reduce the heat. Those cabooses that have it were still working fine when I left the club in 2006.
Mark R. After doing some searching online, I found a couple diagrams by other people that are very similar (if not identical) to what I came up with. So what I have must have some kind of merit to its design ....
Yes it does have merit. I joined a club that used command control in 1984 and they were melting down their caboose. So I cobbled up a similar scheme using the trial and error method just to reduce the heat. Those cabooses that have it were still working fine when I left the club in 2006.
That's funny .... I firmly believe that things sometimes work because we don't know they shouldn't !
Same goes for my DCC buss. When I installed it twenty years ago, I ran it around the room and tied it back in where I started (because I didn't know better). The "rules" say you should not do that. Well, guess what ? It's still tied back into a loop to this day and I've never had any trouble due to it !
Mark R.That's funny .... I firmly believe that things sometimes work because we don't know they shouldn't !
And another quote...
hon30critterJaBear asked me to start a thread showing the wiring diagram for the working marker lights and interior light.
"One difference between pessimists and optimists is that while pessimists are more often right, optimists have far more fun."
Mark R. I'd be interested to know how / why my circuit works as well as it does .... it theory, apparently it shouldn't. For the record, I'm using two yellow LEDs in my cabooses. After doing some searching online, I found a couple diagrams by other people that are very similar (if not identical) to what I came up with. So what I have must have some kind of merit to its design .... Mark.
Top one works fine because the Zener limits voltage to the capacitors. A Zener goes to short once the voltage is exceeded, but the 1K resistor limits the current to 10-15ma depending on track voltage.
The bottom one may or may not work, depending on the resistor values chosen. That is a voltage divider with the slight complication of loosing the forward voltage of the LED in the right-hand side. Don't have time to fully calculate it all to see if there are resistor values that will work that will both allow the LED to light AND keep the voltage across the cap safe. If that capacitor is not a super-cap, then there is no problem - merely choose a capacitor with a high enough working voltage.
Train Modeler Thanks for the equation. A few of questions on it. In the numerator(normal volts area) you multiply the voltage drop by the 470 Ohms(I think this is normal for divider calcs?). Then divide by combined R, then add the load V to get a Voltage where normally V/R equals I? Richard
The normal formula to find the output voltage of a two resistor voltage divider circuit is Vout= (Vin * R2)/(R1 + R2). This assumes the low end of the divider is at 0 volts. In our case, the low end of the divider will be at the forward voltage of the LED, so you have to subtract that voltage from the overall input voltage to get the input to the divider circuit, and then add that voltage back to the dvider circuit output to get the total voltage at the capacitor.
I wasn't using V/R=I for my calculations, just the voltage divider equation; although you could calculate it using V/R=I as well. Using my first example (12 volt input, 2 volt LED): For the overall circuit I = V/R = (input voltage - LED voltage) / (R1 + R2) = (12-2)/(1000+470) = .0068 amps. For the resistor before the cap I = V/R, V=IR = 0.068 x 1000 = 6.8. So the voltage drop across the 1000 ohm resistor is 6.8 volts resulting in 12-6.8=5.2 volts at the cap.
Mark R.I'd be interested to know how / why my circuit works as well as it does .... it theory, apparently it shouldn't.
Actually, your circuit works just fine in theory, as long as your choice of LEDs and resistors results in a voltage on the cap that is less than it's rated voltage. The zener diode circuit does have some advantages. It limits the voltage on the cap no matter what resistors and LEDs you have, so you can make changes without having to worry about causing to high of a voltage on the cap. Additionally, you could a swicth between the zener diode and the LED to be able to turn it off if you want (without the zener diode breaking the circuit on the LED side of the cap will result in full voltage on the cap).
You may be interested to know that the zener diode circuit can actually end up working exactly like your circuit - if the choice of resistors and LEDs results in a voltage at the cap that is LESS than the zener diode voltage then the circuit works the same with or without the diode.
One thing I have issue with in the diode circuit is the wording of the choice of zener voltage - "equal to or close" to cap. Saying "close" allows for a zener voltage above the cap voltage, and I would not even allow for equal, it really needs to be less.
CSX Robert Train Modeler Thanks for the equation. A few of questions on it. In the numerator(normal volts area) you multiply the voltage drop by the 470 Ohms(I think this is normal for divider calcs?). Then divide by combined R, then add the load V to get a Voltage where normally V/R equals I? Richard The normal formula to find the output voltage of a two resistor voltage divider circuit is Vout= (Vin * R2)/(R1 + R2). This assumes the low end of the divider is at 0 volts. In our case, the low end of the divider will be at the forward voltage of the LED, so you have to subtract that voltage from the overall input voltage to get the input to the divider circuit, and then add that voltage back to the dvider circuit output to get the total voltage at the capacitor. I wasn't using V/R=I for my calculations, just the voltage divider equation; although you could calculate it using V/R=I as well. Using my first example (12 volt input, 2 volt LED): For the overall circuit I = V/R = (input voltage - LED voltage) / (R1 + R2) = (12-2)/(1000+470) = .0068 amps. For the resistor before the cap I = V/R, V=IR = 0.068 x 1000 = 6.8. So the voltage drop across the 1000 ohm resistor is 6.8 volts resulting in 12-6.8=5.2 volts at the cap.
Thanks, that clears it up very nicely. And you've shown a good way to check my work.