RichG,
This is a belated THANK YOU for your LM- voltage regulator circuit posted on Aug 8. I have been looking for one of those circuits for literally, months! Just a simple, adjustable relatively low-current power source.
I rarely visit this page, but I think that's gonna change.
Les W.
richg1998 wrote: I would use a LM317 voltage regulator and set the output for 1.4 volts to prolong lamp life. The LM317 will handle the current. No large wattage resistors are needed. Just bolt the metal tab on the LM317 to a piece of aluminum as a heat sink. Just make sure the aluminum does not contact anything electrical. Radio Shack should have the components. No large wattage resistors needed. Here is a link to the circuit. Very simple. I have used the LM317 in similar situations. Our local club has a 12 volt DC buss for turnouts. I just tap off the 12 volts and put in a LM317 for 5 volt logic circuits. I have the calculated resistor instead of the pot. I made about six of these for various parts of the layout. http://www.national.com/mpf/LM/LM317.htmlAdjust the potentiometer until you get the voltage you want. Measure the pot and put in the nearest standard value or lock the pot. down and out of the way. Rich
I would use a LM317 voltage regulator and set the output for 1.4 volts to prolong lamp life. The LM317 will handle the current. No large wattage resistors are needed. Just bolt the metal tab on the LM317 to a piece of aluminum as a heat sink. Just make sure the aluminum does not contact anything electrical. Radio Shack should have the components. No large wattage resistors needed.
Here is a link to the circuit. Very simple. I have used the LM317 in similar situations. Our local club has a 12 volt DC buss for turnouts. I just tap off the 12 volts and put in a LM317 for 5 volt logic circuits. I have the calculated resistor instead of the pot. I made about six of these for various parts of the layout.
http://www.national.com/mpf/LM/LM317.html
Adjust the potentiometer until you get the voltage you want. Measure the pot and put in the nearest standard value or lock the pot. down and out of the way.
Rich
With multiple lamps in parallel, this is the best approach. If you lose one of the lamps, the circuit resistance changes, and the amount of current through each lamp will INCREASE. This is not good. The LM317 (if you know my posts, I discussed this device in the LZ version several times) is very useful to modelers because the output voltage will remain constant; this will also allow a variance in the supply voltage to the regulator without affecting the output.
Resistors are easier to install but for lamps regulators (current and / or voltage) give you better results in the long term.
I know many of you read MR and in the articles, resistors are used. This is because they follow the keep-it-basic approach to any electronic projects. They do not want to discourage anyone from the project with may appear to be intimidating parts, like a voltage regulator. Let me tell you, laying track, scratchbuilding a project or weathering a boxcar is harder than wiring three leads to a VR.
I realize some readers become anxious with just the mere mention of electronics; a little patience and help from those of us who combine our vocation with our avocation will get you through it.
Happy Railroading!
Carl in Florida - - - - - - - - - - We need an HO Amtrak SDP40F and GE U36B oh wait- We GOT THEM!
Karle,
A pair of 50 ohm resistors in series would be 100 ohms - too much. Putting the same pair in parallel would give you 25 ohms. I would just use a small 'wall wart' power supply - you can get them for a $1 a piece at most Salvation Army thrift shops. lots of voltages/amperages to choose from! The problem here is that you are trying to drive a 'wire brad' with a 16 lb sledge hammer!
A lot of cell phone wall warts are 3.7v with about 300 ma output. You can use smaller 4-5 ohm resister with a 1 watt rating....
Jim
Modeling BNSF and Milwaukee Road in SW Wisconsin
bnsf76 wrote: You could put a resistor in series with each bulb.bnsf76
You could put a resistor in series with each bulb.
bnsf76
Correct. I've explained here numerous times the danger of the single resistor and what happens when a bulb or two burns out and you run the risk of losing them all. Folks can search on prior threads on this topic.
Engineer Jeff NS Nut Visit my layout at: http://www.thebinks.com/trains/
Rich beat me to it ....
If the bulbs are not replaceable, I would seriously be looking for a better matched power source. You might even want to check out the "wall-warts" - many of them are low voltage in the 1/2 amp rating.
Mark.
¡ uʍop ǝpısdn sı ǝɹnʇɐuƃıs ʎɯ 'dlǝɥ
Miniatronics has a 1.5 volt 1.5 amp transformer that will work. $25.98
Cir-Kit Concepts has one also. There are a lot of lighting options in dollhouse lighting.
Transformers CK1009G 1.5 volt 3 watts 2 amps $10.95. Shipping is about $4.85. You cannot go wrong on this one.
GRS Micro Liting has voltage regulators specifically for low voltage lamps. They are part of Cir-Kit Concepts.l
If you ever fall over in public, pick yourself up and say “sorry it’s been a while since I inhabited a body.” And just walk away.
Thanks for the suggestions. Here is some additional info and questions:
The ore dock was built about 7 years ago before I learned about the advantages of LEDs. I am now using LEDs for all my structure lighting. But for the dock, I scratch built over-head lights with what I had on hand at the time, 1.5V bulbs. I don't want to scrap these lights due to the lost labor, and because I wired them in parallel to a bus inside the dock, it would be very difficult to put a resistor in series with every bulb at this point, although I appreciate the suggestion. Hindsight is 20/20!
I was hoping I could just wire a resistor in the hot output to the bus, but apparently, assuming the previuos calc.s are correct, I need about a 37 ohm 11.48 watt resistor. I can't easily find a resistor with that capacity, the biggest I find is 10 watts. Could I use (2) 50 ohm 10 watt resistors wired in series? Would my current drop to 20.5/100 or .205 amps, making the bulbs dim (which would probably be OK). Will these resistors get so hot I could cause a fire?
If I burn out these bulbs it will be a shame. (if I burn up the house it will be more than a shame)...maybe I should stick with a 1.5V D battery!
I checked my Sams Handbook of Electronics Tables and Formulas, and you're correct about the wattage.
The DC Power formula is P=EI or P=I2R or P=E2/R, so using the first formula 20.5 Volts x .56 Amps = 11.48 Watts. A 20 Watt resistor should be used.
I'm gonna play devil's advocate here ....
You're drawing 1/2amp and require a 10 watt resistor to run only 7 bulbs !?!
You would be much further ahead to use warm white LEDs ! The lighting itself is much more realistically scaled, they will literally last forever, all you would need is a 1000 ohm - 1/2 watt resistor on each LED and for that same 1/2 amp draw, you could run four or five times as many LEDs.
Just MY thought ....
cacole wrote:No, your formula is off. R=E/I or Resistance = Voltage divided by Current. Starting with a 22 Volt supply, subtract the voltage of the bulb, 1.5. With the bulbs wired in parallel, you're still drawing only 1.5 Volts. You must drop 20.5 Volts at 560 mA, or 20.5 divided by 0.560 = 30.6 Ohms. 560 mA is just a little over 1/2 Amp, so a 1/4 Watt carbon film resistor is plenty big. The next highest standard value of resistor would be 33 Ohms. Personally, I would use a 47 Ohm resistor to insure that the bulbs last longer.
Sorry, although I am neither an electrician nor a physicist, but I think that the numbers given by the original poster are largely correct. You both agree that the resistor has to have somewhere around 30-40 Ohm. As to the wattage rating, the voltage drop that has to be covered by the resistor is about 20 V; that means that his resistor has to have a rating of about 10 W. I don't know how your calculations end up with a 1/4 W resistor.
Just my 2 cents....
Hope someone can advise: I have (7) 1.5 volt grain of wheat bulbs I am using for ore dock lights. I want to power them from a 22 VDC power supply from a spare model rectifier power pack, with all the bulbs wired in parallel. According to my VOM, when powering a single bulb with a 1.5V D size battery, it draws 80 mA. Is the following correct:
The total amp draw will be 560 mA, since each bulb is wired in parallel, and each draws 80mA.
Since the power supply is 22V, 22/0.56 = 39 ohm resistor required wired in series with the power supply output.
Since each bulb drops 1.5V, the voltage drop across the 39 ohm resistor will = 20.5 V.
20.5V * 0.56 A = 11.48 watts. So I need a 39 ohm resistor rated at least 11.48 watts.
Is all the above correct? In looking at Radio Shack, I can find a 50 ohm "wire wound" (?) resistor rated at 10 watts, but nothing of larger wattage. What is a wire wound resistor? Do I need to decrease the number of bulbs to 6, which will decrease the required resistor to 45.8 ohms and 9.84 watts, so I can use the radio shack resistor? Is there a better way to do this? I want to use my 22V supply and not wire the bulbs in series. Wish I had not fabricated my dock lights from 1.5 V bulbs! THANKS!