Although I havn’t got my MTH SD70ACe yet, I have gotten my DCS and am wondering if this wiring scheme will work. Because I wont be seperating the mainlines, and using just 1 KW, I wired both hots to the (+) post on the TIU. Since I believe like the ZW, the KW has internally connected ground posts, I just connected the 20 volt ground to the ground post on the TIU. Will this work?
Once also thing I am wondering is is that when I turn on the KW’s throttles, can I set the both to 18 volts or do I need to set them each at 9 volts?
Grayson:
That is a good question and I am not sure of the answer. However, why not use the second tranformer output and run it through the Fixed 2 of the TIU. I believe the TIU commons are internally connected (which is why my TMCC can work on my entire layout with only the TMCC wire connected to the Fixed 1 output of my TIU).
Regards,
John
Grayson,
First of all ABSOLUTELY DO NOT CONNECT YOUR KW IN THE MANNER YOU’VE ILLUSTRATED.
As Bob Nelson has told us many times, unless the EMFs of the two outputs are exactly matched(which would be very difficult, if not impossible), you will effectively be short circuiting the transformer. Furthermore, you will be short circuiting the transformer in such a way that the circuit breaker WILL NOT trip-meaning that you’re going to have a fried secondary coil.
What you need to do is pick one of the variable outputs of your KW(I like to use B, as it’s easier to turn), and connect that output to the TIU. Then, connect one of the U posts(they’re connected internally) to the black input on the TIU. Then, connect the black output of the TIU to the outer rail, and the red output to the center rail.
Of course, the Legacy base gets connected to outer rail somewhere, but it shouldn’t matter where. It coud be the other U post on the KW, one of the black posts on the TIU, or even connected at the opposite end of the layout of where the transformer initially connects(how my TMCC base is set up).
That’s it, and you should be ready to go.
First of all ABSOLUTELY DO NOT CONNECT YOUR KW IN THE MANNER YOU’VE ILLUSTRATED.
Great advice Ben !!!
Grayson - With all the knowledge regarding trains that I’ve seen you demonstrate. If I may a suggestion… I’m not sure what grade your in, but IF you could find a basic electroncis course at a summer school, or anywhere. It would benefit you greatly! I think you would find it very intriguing and it would open a lot of doors of posibility that you havn’t even thought of yet. (BTW When I said “Benefit you Greatly” I did NOT mean for that to sound insulting at all. I just mean it would be extremely rewarding.)
I’m always asking myself when I get something hooked up or working, “Why didn’t I do this that way?” I could then do THIS and Also Do That, and it bugs me until I change it. I promise you a basic electronics course will ALWAYS be rewarding to you.
Grayson, what they’re telling you is absolutely true. In fact, as I have mentioned before, there is a fine-print note on the KW schematic diagram that says, “Note that the circuit breaker does not protect binding post combinations A-B, B-D, and C-U.”
If it’s any consolation, you can get as much power as the KW has to offer out of any one of A-U, B-U, or D-U pairs, with the other terminals not connected.
Bob,
If I may please ask you a question along these same lines-
Would it even be possible to match the EMFs of two different outputs from a KW? My inclination is that it would not, since the rollers are physically about a half inch apart, and thus even when resting on the same turn of the coil, the EMFs would still be slightly different.
It also seems to me as though it would be even worse when trying to match the A or B output of a Z or ZW with the C or D output, since they’re physically separated by close to a half of a turn.
Of course, this is all moot for me, since I now have my four separate power districts wired using a block system as you have described. I’ve been curious about it, though, and you seem to be the one who would know.
Thanks,
Ben
That’s a good question, Ben. The answer is that the voltage is pretty much the same even though the rollers tap the same turn at different points. Here’s why: The changing magnetic flux in the iron core is what induces voltage in the secondary winding. The voltage induced in each turn of that winding depends on how much flux the winding encircles. Since the flux is confined almost entirely to the iron, because of its very high magnetic permeability compared to air, it matters very little how loose each winding turn is. Of course, most of them are wound pretty tightly on the core; but you can think of one, the one including the wiper arm, the wires inside and outside the transformer, and the track, as being a very loose turn. If you were to move the roller along the exposed length of the secondary wire, you would change the shape of that last turn, but not the amount of flux that it encircles. Therefore the voltage would remain the same.
Even if the rollers are on opposite sides of the coil, as in a Z or ZW, there is no voltage difference as long as the (integer) number of encirclements is the same for both secondary circuits. In other words, there’s no such thing as half a turn when the circuit is complete.
Bob,
Thanks for the explanation. I can see how it’s true, however I need to digest it a little more for it to make complete sense in my mind.
Thanks again for answering this, and your general willingness to answer any and all electrical questions.
Ben
I’m glad I waited on yalls opinion before I powered it up! Thanks for the warning!