I see lots written on the lamp being too dim but I still do not know how to fix it. This loco came in the NY Central set and is DCC. I can turn it on and off with F10 but I am gussing that the wires go from the tender directly to the LED? If that is the case then the resistor on the PC board needs to be changed or how do I know if the CV is set for full brightness? Bachmann don't give you any information with the set as to the specs, CV's or the model numbers of what's included. Sorry if this has been answered before but I really cannt afford 5 hours a day to read hundreds of posts. It would be so much easier if Bachmann had a knowledge base and gave the wiring specs and the PC board layout and schematics.
Jim Banner has successfully converted the headlight to an LED on several locomotives. I'm not sure if he has written a step by step tutorial, but I will do a search and try to locate some of his notes.
I have 4 Spectrum 2-8-0 and I agree the lamp is puny, even on DCC. I plan on changing to LEDs, just a low priority for now.
I'm sure Jim will be in to comment on this matter, so stay tuned.
Thanks, Bob.
I have converted several Spectrum 2-8-0s to LED headlights. I use 3 mm warm white LEDs which fit right in place of the original grain of wheat bulb and install a 1000 ohm, 1/4 watt resistor right behind it. Alternately, you could install the resistor inside the tender but then you might run the risk of blowing the LED if its leads accidentally short to the motor or pickup wires (not very likely but possible.) On the other hand, a resistor in the tender is easy to change if you want a different brightness (I wouldn't go below 560 ohms.)
The relatively high efficiency warm white LEDs I use give a nice electric headlight effect with a 1000 ohms resistor at 6 volts on the track. For use on DCC, I would try about 2700 or 3300 ohms. Your results will vary depending on the LEDs you use.
If you need more details, please let me know. I have two more 2-8-0s that I need to service and upgrade in the next week or so and will take photos if anyone is interested.
If you put 2 1000 ohm 1/4w resistors in series would that be like a 2000 ohm 1/4w resistor or 1000 ohm 1/2 w?
Quote from: pdlethbridge on December 02, 2008, 04:26:16 AM
If you put 2 1000 ohm 1/4w resistors in series would that be like a 2000 ohm 1/4w resistor or 1000 ohm 1/2 w?
1/4 watt. Parallel the R's to half the Total R and double the wattage.
Thanks all for responding and Jim: I still have no idea how to take the tank off of the loco to access the lamp? My loco looks like an LED but you say it's incandesant? 1000 ohms at 6v - I'm assuming DC? For DCC you say use approx 3K ohms - why?
Quote from: pdlethbridge on December 02, 2008, 04:26:16 AM
If you put 2 1000 ohm 1/4w resistors in series would that be like a 2000 ohm 1/4w resistor or 1000 ohm 1/2 w?
Neither. In series, the resistances add. So it would be 2000 ohms. Each resistor could safely dissipate 1/4 watt so between them they could dissipate 1/2 watt. Correct answer - 2000 ohms, 1/2 watt.
If you put them in parallel, they would form a 500 ohm, 1/2 watt resistor.
Doubling the power handling works only if the resistors are equal.
====================================================
1000 ohms for 6 volts on dc, 3000 ohms for 12 volts on DCC does look a little odd doesn't it. But not if you figure it this way:
At 6 volts, the resistor has to drop 3 volts which is the difference between the supply voltage and the voltage dropped across a typical white LED. With 3 volts across the 1000 ohm resistor, the current through the resistor and through the LED is 3/1000 = .003 amps or 3 milliamps.
At 12 volts, the resistor has to drop 9 volts while the LED still drops its required 3 volts. With 9 volts across a 3000 ohm resistor, the current through the resistor and the LED is 9/3000 = 3 milliamps.
The exact value of the resistor you choose will depend on several things:
- efficiency of the LEDs you use
- how the LED is mounted in the locomotive (direct view, light pipe, lenses, etc.
- operating voltage, which depends on DCC track voltage
- background lighting of the layout
- personal preference
The minimum resistance of about 560 ohms is to safely limit the current to about 20 milliamps even if your DCC system puts 16 volts on the rails, giving 14 volts out of your decoders. With the high efficiency LEDs I have been using, 20 milliamps is way to bright. In some applications, for example an oil burning headlight, even with 3000 ohms they are too bright. A coat of Tamiya clear yellow paint tones them down and tints them to a better colour, but they still need a resistor of 10,000 ohms or so.
Jim,
Wow, I feel like Plato sitting at the feet of Aristotle...or was it the other way around? Anyway, I've printed your response out and am starting a Jim Banner technical book.
Quote from: db22 on December 02, 2008, 09:24:54 AM
Thanks all for responding and Jim: I still have no idea how to take the tank off of the loco to access the lamp?
Look here.
http://members.shaw.ca/sask.rail/dcc/2-8-0/index.html
Rich
This might work if your resistor leaves a bright bulb in the loco after you put in a new bulb. Adding resistors in the tender in series is easier than taking the loco apart again.
If you model 1900 a bright headlight would be ok. Carbon arc headlights were being installed in 1897 through maybe 1905. Do not know when the arc lamps was done away with. They were so bright, oncoming loco crews could be blinded. Some times crews of the loco with an arc lamps could temporarily be blinded my fog, rain reflecting light back. Some railroads installed a low wattage filament type lamp for this issue. The crew would turn off the arc lamp when approaching another loco or in a station area. It took a few years for rugged filaments to be developed. Some railroads started using acetylene lamps which provided a light that was not as bright as the arc lamps.
Rich
Quote from: Jim Banner on December 02, 2008, 10:51:16 AM
Neither. In series, the resistances add. So it would be 2000 ohms. Each resistor could safely dissipate 1/4 watt so between them they could dissipate 1/2 watt. Correct answer - 2000 ohms, 1/2 watt.
I think it would be 1/4 watt 2000 ohms, not 1/2 watt 2000.
Matt
Sorry, Matt, but it is 1/2 watt. Let's look at it this way. We take two 1000 ohm, 1/4 watt resistors and put them in series. Then we connect a 31.6 volt power supply across the two of them. What is the voltage across the first resistor?
31.6 / 2 = 15.8 volts
What power is the first resistor dissipating? Power = voltage squared / resistance. In this case, that would be
15.8 x 15.8 / 1000 = 1/4 watts
Likewise, the voltage across the second resistor is also 15.8 volts and the power it dissipates is 1/4 watts.
Now we both agree that the two resistors in series make 2000 ohms. So how much power does that dissipate at 31.6 volts?
31.6 x 31.6 / 2000 = 1/2 watts
From this we can conclude that two equal resistors working at full power can be connected in series to form a resistor with double the resistance and capable of working at double the power of each single resistor.
How does this compare, 2000 ohms 1/2 w, to a 3700ohm 1/4 w?
Sorta like comparing apples to oranges. Do you mean putting the two in series?
As Jim noted: "Doubling the power handling works only if the resistors are equal".
Four basic rules for resistors in series.
1. The total resistance in a SERIES circuit equals the sum of the individual resistances.
2. The sum of voltage drops in a series circuit will equal the voltage source.
3. The voltage drop across a resistor in a series circuit is directly proportional to the size (Ohm rating) of the resistor.
4. The total power consumed in a series circuit is equal to the sum of the individual powers consumed by each circuit component.
For dissimilar resistors, we can't halve the voltage, you have to calculate for each one. If we put the 2 you noted in series, and assume 45 volts applied, then the total current would be 45/5700 (total resistance) = .0079 amp.
For the 2000 ohm, voltage drop is .0079 X 2000 = ~16 volts. Power consumed is 16 X .0079 = .12 watt.
For the 3700 ohm, voltage drop is .0079 X 3700 = ~29 volts. Power consumed is
29 X .0079 = .23 watt.
Total power in the circuit is .12 + .23 = .35 watt. To verify 45 volts X .0079 amp = .35 watt.
Due to the higher voltage drop across the 3700 resistor, it's dissipating close to its rating, the other one is overrated. Notice all 4 rules above have been observed.
Now I'm really confused. My thought was this, if you change a bulb to a different one in the heavy mountain and use a resistor ( 1/4w 1000ohm ) with it in the loco but get too bright a bulb, would adding a resistor in series in the tender lower the brightness. My thinking is if you take an engine apart once and got it back together you were lucky the first time, the second time maybe not so lucky so put the second resistor in the tender.
Don't let the power doubling confuse you, just add another 1/4 watt resistor in series, in the tender, say a 500 ohm.
Your original 1000 ohm at 12 volts = 12 milliamps. Add the 500, so 1500 ohms total at 12 volts = 8 milliamps.
Remember the bulb has a voltage drop, we're just adding resistance to create more drops over and above what the bulb needs. (or can stand)
Now I have a headache, g'night. 8)
To Bill Baker : Aristotle was the student of Plato , Plato was the student of Socrates . Interestingly Aristotle was one of the teachers of Alexander the Great. J2.
If you have an ohmmeter, here's an easy way to determine the resistor value you need. Pick up a 5,000 ohm, 1/4 watt, linear taper miniature potentiometer. Radio Shack has them, about $3. Hook it up in series with small jumpers to the lamp or LED and the headlight wire. Dial in the brightness you want then check the value of the pot with the ohmmeter.
They are small, might even have room in the tender to mount it permanently.
http://en.wikipedia.org/wiki/Potentiometer
Now thats a bright idea! 8)
like this:
http://www.allelectronics.com/make-a-store/item/VTP-5K/5K-VERTICAL-TRIMPOT/-/1.html (http://www.allelectronics.com/make-a-store/item/VTP-5K/5K-VERTICAL-TRIMPOT/-/1.html)
(http://www.allelectronics.com/mas_assets/image_cache/image_product.height.500;width.500;modified.1207856984.VTP-5K.jpg)
Quote from: Jim Banner on December 03, 2008, 11:06:16 PM
Sorry, Matt, but it is 1/2 watt. Let's look at it this way. We take two 1000 ohm, 1/4 watt resistors and put them in series. Then we connect a 31.6 volt power supply across the two of them. What is the voltage across the first resistor?
31.6 / 2 = 15.8 volts
What power is the first resistor dissipating? Power = voltage squared / resistance. In this case, that would be
15.8 x 15.8 / 1000 = 1/4 watts
Likewise, the voltage across the second resistor is also 15.8 volts and the power it dissipates is 1/4 watts.
Now we both agree that the two resistors in series make 2000 ohms. So how much power does that dissipate at 31.6 volts?
31.6 x 31.6 / 2000 = 1/2 watts
From this we can conclude that two equal resistors working at full power can be connected in series to form a resistor with double the resistance and capable of working at double the power of each single resistor.
That seems logical, but, if you get the power from the middle of them (making a voltage divider) then essentually they are not in series. what i was talking about, is when they are in series (without center tap). for example if you have a water hoses that has X resistance (at one cubic foot per minute max speed) then you would be able to put one cubic foot of water through in one minute with X resistance. Next, suppose you have another hose exactally the same, if you put the two in parallel, you could get two cubic feet of water per minute max, at .5X resistance. Or, if you put the two hoses is series, you could only get one cubic foot of water through it (excluding(because it does not apply to the electrical resistors) the fact that the first hose would have more pressure than the second, thus reducing the amount even further) and it would have a resistance of 2x. Or what if you have a transformer, that is 10:1 CT (12v) rated 1 amp, if you ran one amp off the center tap, it would be maxed out at 1 amp 6 volt, but, if you ran one amp off it at 12 volts, it would still be maxed out, even though it can output twice as much power. I am still learning, so if I am wrong, please correct me. My uncle is a computer/robot builder/programer, so i'll call him to see what he says.
Matt
Perhaps we can de-mystify LEDs a bit by stating some constants, and eliminate some variables. This has been hashed before, and my apologies to others who have presented the same information, but allow me to present it in a slightly different way. (the way it makes sense to this tired brain.)
First, regular bulbs are resistive and are voltage driven. You buy a 3 volt, 12 volt, 110 volt, etc A 100 watt light bulb will draw about 1 amp. More specific, 100 watts / 110 volts = .9 amp.
If you want to use a 1.5 volt bulb on 12 volts, then 10.5 volts must be dropped by using a resistor. The bulb is usually designed for 15 milliamp (.015 amp). 10.5 volts / .015 amp = 700 ohm resistor or higher. ( 10.5 X .015 = .16 watt..use 1/4 watt. )
LEDs are current driven devices, not voltage driven. Generally LEDs are designed to operate at a conservative 20 milliamps (.020 amp) or less for long life.
LED voltage drops vary from about 2 volts for red/green and about 4 for most blues and warm whites. Let's use 3 volts as an average, as Jim mentioned.
For DCC we'll use 12 volts for applied voltage. Now that we have a few constants, we only have one variable to consider, the resistor. The goal is to limit current to .020 amp.
Applied voltage (12) - LED voltage (3) = 9 volt drop required. Voltage drop divided by current = resistance.
9 / .020 = 450 ohms. (9 X .020 = .18 watt...use a 1/4 watt)
Note this is for maximum brightness, which is probably too bright for our locomotives, so let's cut the current in half.
9 / .010 = 900 ohms.
Cut the current in half again..... 9 / .005 = 1800 ohms. Plug any current value from .005 to .020 into the simple formula to find the resistor. (use the next higher standard or available value)
A tiny 5k, 1/4 watt trim pot installed in the tender will let you fine tune the brightness, but you must install at least a 470 ohm 1/4 watt fixed resistor for safety in case the trim pot is shorted or turned to the zero ohm position, or just use the pot to determine a fixed resistor. Whether you install a 470 to 1000 or any in between in the locomotive is up to you, just remember the minimum is 470 ohms for 12 volts, some suggest 560-700 for added safety when operating DCC. For example at 14 volts...14-3= 11/.020 = 550. Since Jim has a lot more experience with LEDs than I do, I would bow to his judgment about minimums.
There are many LED/resistor calculators on the web, but as you can see it's very simple and I suggest doing the calculations for better learning.
One important thing to remember, watch how you set up the trim pot. When set for maximum resistance between one side contact and the center contact, between the center and other side will be zero ohms. If you inadvertantly flip the trim pot over, you could hook up the the wrong contacts and blow the LED. Been there, done that.
Rich
Thanks Rich, did you say "OOPS" or something a tad stronger? :D
I forgot to include that point in my post, so I edited and changed the "suggest installing" a fixed resistor to "must install" a fixed resistor.
The safest type of pot to use is a multiturn preset. It uses a worm gear for fine adjustment, and not likely to get accidentally changed. Once set, mark the side terminal to be used, and cut off the other one.
http://www.allelectronics.com/make-a-store/item/RTP-5K/5K-MULTITURN-TRIMPOT/-/1.html
The site has lots of other "goodies". 8)
I started this thread and some great information has come out of it but I still have some unanswered questions: 1. nobody has stated if this loco is bulb or LED. 2. I still do not know how to access the bulb to replace it or change it. 3. where is the official Bachmann database of disassembly, maintenance, parts etc?
Secondly - I looked inside my diesel and there is a LED at both ends of the PC board. (front and rear lights) The resistor is on the board but how do I find out which one it is? This engine has a LED that also is only just visible if you look straight at it.
I have 2 steam locos from Bachmann, one lights up the room and the other is barely visible. Is this intentional or did the designer not ever check the resistor values?
It is easy to figure out, Use the ohm meter portion of a multimeter and trace the circuit. Follow the PC board traces also. That is what I do. A LED will have no continuity. Lamps will have some comtinuity, i.e. resistance. If you understand electronics, it will be easy to do.
Rich