Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest
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While I can work out the electrical needs with some measuring and figuring, I'm curious if there is some already known wisdom on this topic. I'm setting up my Christmas diorama with a short train doing an endless loop around the room on a dogbone. Around that I'm building up some H0 scale winter dioramas, a christmas market, snowed trees, and including the Busch lighted christmas tree (5409). It comes with LEDs per-applied, and some wiring pre-installed with a per-installed, heatshrunk resistor in line. (no idea what size it is). The kit indicates it is designed to run as is on 14-16 v AC or DC. The track voltage from the trackbox/MS2 is 19V AC (ish). So my question is: is running the tree direct off of the track power (there's only the one loco and this one tree on the whole circuit) an issue with respect to the voltage? A quick test showed the tree flickered (of course), so I'm thinking adding a diode and capacitor to the tree side to smooth out the power to the tree. the LEDs wold drop the typical 2v and limited to 20mA... so if I do the math that hidden resistor would be at least (16v-2v) / 20mA = 700 ohm or larger. On the AC side, the diode isn't seeing constant power from the track voltage, but if I put in a diode rectifier and capacitor to smooth things out it might be near 19v DC, so then I get on a 700ohm resistor (19v-2v)/700 ohm = 25mA which would be too much juice possibly. The lights flikering as they are on the tree as is, are not high brightness so at present I don't see the system pushing the LEDs near limits, so I may be fine. Do I need to worry about adding another 100 ohm resistor including the capacitor/diode circuit to stop the flicker? I've got no interest in adding an additional power supply; tapping into the track power where that tree diorama will sit is just fine. The tree : http://www.busch-model.c...id=5409&sprach_id=en |
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Joined: 16/02/2004(UTC) Posts: 15,463 Location: DE-NW
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Originally Posted by: Minok  The track voltage from the trackbox/MS2 is 19V AC (ish). If you feed 18 V DC (switching-mode power supply) into the track box you will get about 16 V track voltage. If you add another bridge rectifier with capacitor, voltage will be reduced further. Most likely you won't need a resistor. |
Regards Tom --- "In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS  |
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Joined: 10/02/2006(UTC) Posts: 3,997
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Adding a rectifier and a capacitor will likely increase the voltage by a factor of 1.4
Assemble the rectifier and capacitor and don't connect it to the tree yet. Attach it to the track and then measure the DC voltage you are getting out. If it is > 16 Volts then add some diodes in series between the rectifier and the tree side. Each diode will use up 1.7 Volts.
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 1 user liked this useful post by DaleSchultz
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Joined: 16/02/2004(UTC) Posts: 15,463 Location: DE-NW
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Originally Posted by: DaleSchultz  Adding a rectifier and a capacitor will likely increase the voltage by a factor of 1.4 This applies to sine-wave AC, but not to digital track voltage. Rectifying the track voltage will not increase the voltage.. |
Regards Tom --- "In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS  |
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Joined: 10/02/2006(UTC) Posts: 3,997
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Originally Posted by: H0  Originally Posted by: DaleSchultz  Adding a rectifier and a capacitor will likely increase the voltage by a factor of 1.4 This applies to sine-wave AC, but not to digital track voltage. Rectifying the track voltage will not increase the voltage.. true, forgot he was using a digital supply... anyway, adding diodes can be used to bring it down to 16V |
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Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest
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Originally Posted by: DaleSchultz  Each diode will use up 1.7 Volts.
Diodes typically drop 0.7v for a srd silicon diode. But I get your idea. Assemble, measure and verify before connecting up the tree. |
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 1 user liked this useful post by Minok
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Joined: 16/02/2004(UTC) Posts: 15,463 Location: DE-NW
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Originally Posted by: Minok  Diodes typically drop 0.7v for a srd silicon diode. Typically and under load. With just a digital voltmeter connected, there will be just a very small voltage drop. Without load you will get a higher reading than with the tree connected. With empty track you will get a higher reading than with running trains. The voltage goes through two diodes and two transistors before it reaches the track where you have a bridge rectifier, so there will be two more diodes to pass. The output voltage of the power supply is pretty stable at 18 V. Remember that a transformer with a nominal voltage of 16 V AC will give you more than 18 V when idle. A MRR gadget specified fir 16 V AC should be able to handle 18 V. |
Regards Tom --- "In all of the gauges, we particularly emphasize a high level of quality, the best possible fidelity to the prototype, and absolute precision. You will see that in all of our products." (from Märklin New Items Brochure 2015, page 1) ROFLBTCUTS  |
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Joined: 10/02/2006(UTC) Posts: 3,997
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my brain must be frozen , yes 0.7 V drop per diode
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Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest
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In then end no further circuitry needed. connecting and running direct off of the C-track power produced no issues or flickering (other than when the train itself may have caused a brief glitch). At present I have the leads soldered to a very short c-track piece. I'll need to add the right sort of plugs on the ends of the wires.. maybe. I"m torn between having the track part be part of the diorama kit (for under the Christmas tree so its its own boxed set), vs being able to unplug it which would allow me to move it about the holiday layout more easily. |
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