Baldwin 2-8-0 locomotive lamp

[db22]

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.

[Yampa Bob]

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.

[Jim Banner]

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.

[pdlethbridge]

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?

[db22]

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.

[db22]

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?

[Jim Banner]

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.

[Bill Baker]

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.

[richG]

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

[pdlethbridge]

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.

[richG]

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

[mattallen37]

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

[Jim Banner]

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.

[pdlethbridge]

How does this compare, 2000 ohms 1/2 w, to a 3700ohm 1/4 w?

[Yampa Bob]

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.