Yes, but you will have to put a rectifier upstream of the e-unit so that you will be running it on DC. The good news is that this will also quiet the usual buzzing. The bad news is that the e-unit will run hotter than normal because the RMS voltage will increase by 40 percent compared to AC operation, doubling the power dissipation, and further because the coil's inductive reactance will no longer be part of its impedance.
Nevertheless, you may find this solution satisfactory. But, if the coil gets too hot, you may have to put some resistance in series with it, particularly if you run dual motors in series. If so, I suggest using a lamp rather than a fixed resistor.
Bob Nelson
Bob,
Normal 0
You never fail to amaze me with what solutions you come up with. If you haven't written a book on model electronic solutions and tricks, you should. I just purchased the Vol II Wiring for Model RR and it is great, but not nearly as clever as many of your answers to the posts here.
Terry Thomann Fredericksburg, Virginia That is me on the left. My brother got the train TCA 09-64381
How does a rectifier increase the RMS voltage by 40 percent?
You convert AC to DC. This is done with the bridge rectifier.
Read the definition of AC and DC and RMS. I sort of understand, but not enough to explain it. It's safer that way.
An ideal full-wave rectifier would cause no change in the RMS voltage.
Adding a capacitor, however, has the effect of making the voltage equal to, more or less, the peak voltage of the waveform. During a cycle, the capacitor charges to the point where its potential difference is equal to the peak potential difference of the source. As the voltage of the source falls off, the capacitor then steps in to supply power.
Provided that the capacitor is large enough and the current draw is not too large, the voltage will not fall much below that of the peak voltage.
For a simple capacitance filter, the voltage output is equal to sqrt(2)Vrms(1-(1/4RCf), where Vrms is the RMS voltage of the source, R is the load resistance, C is the capacitance of the filter capacitor, and f is the frequency of the source.
(Bob, feel free to add to or correct anything that's wrong. I hope that I've done the answer justice).
I think you've got it right. Note that the expression 4RCf is all meant to be in the denominator of the fraction 1/(4RCf) . However, with a time constant long enough to delay e-unit stepping, that term (which expresses the effect of ripple) shouldn't have much effect.
A bridge rectifier may not be needed, which can greatly simplify the wiring, since the e-unit case is normally connected, electrically and mechanically, to the locomotive frame and is difficult to insulate when you want to use a bridge. Nevertheless, some modern "transformers" are sensitive to unbalanced loads and may switch off if you use a simple 1-diode half-wave rectifier. A real transformer won't be bothered by that.
ben10ben Adding a capacitor, however, has the effect of making the voltage equal to, more or less, the peak voltage of the waveform. During a cycle, the capacitor charges to the point where its potential difference is equal to the peak potential difference of the source. As the voltage of the source falls off, the capacitor then steps in to supply power. Provided that the capacitor is large enough and the current draw is not too large, the voltage will not fall much below that of the peak voltage.
Given the load of an e-unit coil, I'd guess that would have to be one big capacitor
. According to a quick look on the web, RMS voltage is approximately 70 percent of peak, and most voltage measuring devices are reporting the RMS value, not the peak.
I just measured one at 12 ohms. With a (not-too-big) 10 millifarad capacitor, that would be a 120 millisecond time constant. So you could probably run through a 200-millisecond gap without stepping. I think that would be noticeable.
What you say about meters and RMS voltage is true; but I don't understand your point. Do you understand that the capacitor charges to a DC voltage that is equal to the peak of the sine wave, 40 percent higher than the sine wave's RMS voltage, and that the RMS voltage of a DC voltage is equal to the DC voltage itself?
lionelsoni Do you understand that the capacitor charges to a DC voltage that is equal to the peak of the sine wave, 40 percent higher than the sine wave's RMS voltage, and that the RMS voltage of a DC voltage is equal to the DC voltage itself?
I understand that the RMS voltage is equal to the DC voltage itself for a constant DC power source such as a battery.. But the RMS voltage of the DC put out by connecting a bridge recitifer to an AC source is still going to have the voltage vary over time, so isn't the RMS voltage still about 70% if the peak?
Yes, the RMS voltage is unchanged by simple full-wave rectification (neglecting the diode drops). But here we're talking about rectifying into a capacitor large enough to hold a constant voltage between cycles. It will charge to the peak of the AC waveform; and that DC voltage will be 40 percent higher than the RMS value of the AC waveform and will therefore have its own RMS voltage 40 percent higher than the original AC RMS voltage.
This is getting complicated for a simple-minded non-electrical engineer. I thought caps. block DC but allow AC to pass, so how could they output any DC voltage?
I sure don't want to burn out any of my E-units, so I guess I'll have to keep them switched out. On a "permanent" layout, does anyone solder wires to assure conductivity between rail sections? I don't have too much luck with just the rail pins.
I operate with rusty track. I do not solder connections. To compensate I keep the top free of corrosion. I use track pliers to keep the pins tight. On really bad connections you can feel the heat generated. For good track you may need separate feeds every 20 feet or more. Mine are sometimes 8 feet. If the engine slows on the far side, I try another feed. If the track is room temperature and the engine doesn't slow you are good to go.
A capacitor in series does block DC; but we're talking here about a capacitor in parallel, which tends to block AC.
I solder the rails themselves together, without track pins. I started doing this not for conductivity but to make it easier to remove and replace track, for repair and modification. But the reliability of the connections is a great bonus. Some folks do solder short pieces of bare wire across the joints, since it can be tricky to flow solder into a butt joint; but I don't find it too difficult.
Guys: especially Bob and Ben, thank you for answering my questions. I don't have any formal schooling in electronics, just what I have picked up by reading, and from folks like you.
I think Bob is reincarnated. No one should retain that much knowledge from once around the block. He guided me through an Atlas slip switch problem a year ago and the thing is still working flawlessly. Everyone that has seen the switch work is amazed at how completely, and with a nice audible
" Click " it throws. It connects a double loop and sends each train on a different loop automatically, set up by the touching of the locomotives wheels as it approches the slip switch from either loop. I bought stuff, wired it and I still have NO idea of what I did, but it works like a charm. Bob is a huge help on this board as many of the other guys/gals are. Jake
(Blush)
While testing my new Williams trainmaster, I noticed that its reversing unit didn't respond to quick pushes of the reversing button on my KW-had to hold the button down for ~half a second.
Looks like this engine has some sort of delay built in. I can also run it at very low speeds with no tendency to stall. I like it a lot.
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