Are there any DC power packs that will have DC out for connecting accessories? With today's LEDs built into pretty much anything, I do not think there is much need for AC terminal. Even if I were to convert the output to DC, 18V AC after rectification would give me over 25V. This is a bit too much for a few LEDs that I have. I am trying to avoid running two power supplies (one for track and one for LED accessories) and I would like to stay with one power cord for my small N gauge layout.
I'm not aware of power packs with fixed DC out terminals, but I'll bet someone does. You might want to also take a look at a "buck converter" to economically and easily produce a fixed DC output (after being rectified from your fixed AC terminals). If you shop around on the 'net, you can find a great many of these priced at around $5. Many allow you to precisely set the exact output voltage you wish and they don't produce gobs of heat. https://www.allaboutcircuits.com/technical-articles/buck-converters-and-their-cool-applications/
I use a couple of old DC power packs to operate twin coil switch machines (AC) and lighting of buildings. Using a volt meter I can control the power out. I usually run the power out between 8 to 9 volts out for conventional lighting.
Look for an old MRC Tech II 2400 (or 1500 or 2500). The 2400 has a pulse power on/off switch, which some consider an advantage. The 1500 and 2500 have a simulated momentum feature.
Thanks. Do you guys know what's the voltage output on the MRC Tech II 1500 or 2400 power pack? I can also modify the current power supply I have but after rectification, the voltage is going to be 25V which is a bit too much for few LEDs.
15V for the 1500 and 2400; 20.5V for the 2500 - as marked on the rear of the power pack. All measured under no load conditions. Note that the instruction manual for the 1500 specifies 20V. Why is 25V "a bit too much" for powering LEDs? Simply use the appropriate value (and power rating) of series resistor.
Did you try calculating the resistance and wattage? for a 20 MA LED, you need a 1,250 ohm resistor... no biggie... but how much wattage? It calculates to 1/2 watt... and hopefully you know that you need to use a resistor at least double the calculated wattage, i.e. a 1 watt resistor, and remember you are STILL dissipating 1/2 watt for every LED that is on... so 25 volts is really a lousy choice for LED power unless you can put a bunch in series... If you have the room and don't mind using larger resistors and you can dissipate the heat no problem... not good in a loco... I run G scale DCC and the track voltage is 24V, most DCC decoders provide a 5v power supply just for this reason... or we use LEDs in series, or the Suntek CL2 if we must run a single LED at that voltage. Greg
I haven't run the calculations yet but you are right, 25V for a few LEDs is a bit too much. I am working on a small inglenook layout so I may only have 2-3 LED lights total. I may just go with the Bachmann power pack that I already own and use small 3xAA battery pack for the LEDs. 3.6V worth of rechargeable batteries should be fine for LEDs and the runtime should be good.
For my LED power I have used a circuit board from a car 5V USB adapter bought from a local Poundland (Dollarstore?) and run it with a 15V input. It contains a buck convertor chip as suggested by Hardcoaler above and is very efficient with a regulated 5V output. Similar USB adapters are available that work on either 12V or truck voltages (24V). Jim
The OP isn't putting the LEDs inside a loco and there's no DCC decoder to provide 5V. Assuming a 25V input, how much power does the Suntek CL2 device dissipate and how much does it cost?
Put one of these after the full wave bridge. Cheap enough. Couple in another train forum use them. https://www.ebay.com/itm/NEW-BUCK-S...V-VOLTMETER-/351486640764?hash=item51d63c827c Rich
Jeeze.. the 1 watt resistors are still big and will get hot. That was the point. Seizing on in or out of a loco, and DCC or not is not the point. Big resistors and a lot of heat. The OP is right in not wanting to start with too high of a voltage. If this is assimilated then we can go on to other alternatives. If it gets combative then no one wins. Greg
A 1W metal-oxide resistor has approximate dimensions of 1/8" x 3/8". Is that considered big? It also costs about 10ยข. The OP plans to run 2-3 LEDs, so 1-1.5W total power dissipated by the resistors if there is a separate resistor for each LED. Is that a lot of heat? The Suntek CL2 you refer to must dissipate some power in order to regulate the current. Can you provide an estimate of that power dissipation? Also, again, what is the approximate cost of this device?
I am really leaning toward using battery pack with three rechargeable Eneloop AAAs. Voltage is going to be 3.6V so with around 2-3V voltage drop across each LED and 20mA current draw, the 800mA battery should give me hours worth of operation with just a small resistor. The layout will have open bottom so it is going to be easy to remove and recharge.
OK, Point353, no mention of cost and now that seems to be an arguing point. At 1/2 watt power, a 1 watt resistor can get warm enough to soften plastic. And heat is heat, via the laws of physics and thermodynamics. It all depends how you dissipate it. I'm done, you want to fight. Good bye. Greg p.s. you could have googled the part and done the research yourself, in common use by many model railroaders. p.p.s. Drabina, sorry to mess up your thread with contention, I think your solution is good, and you will run fewer wires.
What plastic is it that will get softened? How warm does this Suntek CL2 device get? How does it dissipate heat? Why the reluctance to provide any specifics about the cost or performance of your proposed solution?
I use an MRC Railpower 1370 connected to a Terminal Block for all of my accessories including LED lighting. Have never had an issue with it and it is fairly cheap. The 1300 model would be just as good I would think.
For powering LED's I use a cheap transformer from a model set. They are small enough to mount out of the way, and I use the DC track power connections. I use a voltmeter to set the voltage at a rate that will cause the LED's to light and lock the control. If I change the power demand of the circuit by adding or eliminating LED(s), I reset the voltage. I also don't have to include resisters in the circuits so the voltage is in the 2+V range, and heat is not a factor. Jim
