Baldwin 2-8-0 locomotive lamp

[pdlethbridge]

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.

[Yampa Bob]

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. 

[Jhanecker2]

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.

[Yampa Bob]

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

[pdlethbridge]

Now thats a bright idea!

[pdlethbridge]

like this:
http://www.allelectronics.com/make-a-store/item/VTP-5K/5K-VERTICAL-TRIMPOT/-/1.html

[mattallen37]

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

[Yampa Bob]

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.

[richG]

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

[Yampa Bob]

Thanks Rich, did you say "OOPS" or something a tad stronger?   

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". 

[db22]

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?

[richG]

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