New to the forum. Finally got a house with space in the basement for my postwar Lionel trains that have been in storage since I was a teenager almost 5 decades ago. My “layout” back then was an annual Christmas setup in my parents’ basement on a ping pong table and old kitchen table with a couple of 8 ft 2x6’s bridging the gap.
Do you know listed forward voltage of your diodes? Did they come with a spec sheet? Not all diodes drop the same voltage.
"MIC" is probably "Master Instrument Company", a manufacturer of many types of diodes. So you may indeed be using something like a Schottky diode with a very low forward voltage. For the voltage-dropping purpose you need ordinary run-of-the-mill silicon diodes.
You should also be aware that you need a true-RMS voltmeter to measure the voltage reduction accurately. An ordinary VOM (volt-ohm-milliammeter) measures AC voltage accurately only if the waveform is sinusoidal, which is not the case with this diode circuit.
Bob Nelson
Bob,
First of all, I have read many of your posts over the past few months and every time I have learned something useful.
I got the diodes from Amazon. They are listed as "general purpose plastic rectifier diodes". There is no spec sheet listing a forward voltage but I am pretty sure they are not Schottky diodes...I made that mistake with an original purchase and found out that they had a very small voltage drop. I will try a couple of those again and see how they compare.
I'm pretty sure you hit the nail on the head with your statement about the voltmeter. It was fairly inexpensive so I'm guessing it doesn't read true-RMS voltage (a concept which I vaguely remember from the one EE course I took in college). I have another voltmeter which I will also try but my guess is it is the same type.
Would a simple VOM measuring a non-sinusoidal waveform be expected to read less voltage drop (a higher resulting voltage) than is actually occurring? If that's the case then I'm not too concerned about knowing the actual drop if it's likely to be closer to 0.5v than 0.3v. I know that deterimining the total reduction I will ultimately need on each of my three downhills (which are a long way from finished) is is going to be a trial and error process anyway. Just trying to plan ahead on the numder of diode pairs I will need.
Thanks.
Here is an example, assuming 16 volts RMS from the transformer, a single anti-parallel diode pair, and a .65 forward-voltage drop: The RMS voltage to the locomotive is 15.42, that is, a drop of .58 volt. A simple rectifying meter would read 15.28, an apparent drop of .72 volt.
By the way, there is an alternative to individual diodes. You can make the equivalent of two anti-parallel pairs in series from a bridge-rectifier module just by connecting the + and - terminals together. The ~ terminals are the ends of the series string and the +- combined terminal is a center tap, useful for fine adjustment.
Here's a little explanation about RMS--"root-mean-square". It's the square-root of the mean (average) of the square of an AC waveform. It's the value of a DC voltage or current that would produce the same heating in a resistive load. For a sinusoidal waveform, that works out to be the peak value of the waveform divided by the square-root of 2.
This means that the 120-volt waveform that you get from the outlet is actually rising to a peak of 170 volts, 120 times per second, to make up for the fact that it spends half its time below 120 volts.
A meter that measures the average of the absolute value of the waveform will show a value that is about 90 percent of the RMS value for a sinusoid. So cheap multimeters that do that have a built in 11-percent boost to make them useful--but only for sinusoidal waveforms.
> assuming 16 volts RMS from the transformer, a single anti-parallel diode pair, and a .65 forward-voltage drop: The RMS voltage to the locomotive is 15.42, that is, a drop of .58 volt. A simple rectifying meter would read 15.28, an apparent drop of .72 volt.
Am I correct that a "simple recifying meter"** is more or less the same thing as a "cheap multimeter"? If that's the case then my cheap VOM will be displaying a lower voltage downstream of my diode pairs than the actual RMS voltage and consequently the voltage drop for each diode pair is actually less than the 0.3 volts I mentioned in my original post.
I'm either misunderstanding something you said or I must have gotten the wrong type of diodes (something with a forward voltage drop much lower than 0.65v). The other ones I got that were Shottky diodes measure a drop of 0.1v per pair. I had no idea...thought that was a brand name. I have since learned they are made with a different material. Fortunately I've bought small quantities since I figured there would be a learning curve. My experimenting thus far has cost less than a couple of sections of Gargraves track.
> By the way, there istive to individual diodes.
Yes, I've read your (and others') posts about that on the forum. It seemed to me that the price of a bunch of 4-diode modules would be much greater than the price of 4 times that many diodes plus the terminal blocks to hook them together. Seems like I need to do more research.
Where do you buy diodes (or bridge-rectifier modules) when you need them? Since Radio Shack more or less cratered I don't know of any local electronics parts stores. What specs should I look for - amps, volts, etc? (One of my transformers is a ZW, so 20v max.) Is the forward-voltage drop of a diode always called that or is some other nomenclature used by different manufacturers.
**I Googled "rectifying meter" and if I understood it correcly it seems to be what you described in your last paragraph: In order to make it possible for a multimeter to measure AC & DC with a single (DC) meter a full-wave recifying bridge is added to the AC measuring function to convert te AC to DC which is then displayed as AC voltage on the meter. And it concludes with this statement: "The AC voltmeter produces an output voltage, which is equal to 0.9 times the rms value of the sinusoidal (AC) input voltage signal." Which is what you say in your last paragraph.
Thank you so much for helping me with my rookie questions. Last time I used my trains it was: set them up, put a lockon or two in each block, make sure the transformers are in-phase (I have taken note of your serious cautions on that), set up a couple 153C (?) controllers to keep the trains from hitttng each other and let 'em go. Can't believe how much good info is available on the forum.
> By the way, there is an alternative to individual diodes.
Yes, I've read your (and others') posts about that on the forum. It seemed to me that the price of a bunch of 4-diode modules was much greater than the price of 4 times that many diodes plus the terminal blocks to hook them together. I need to do more research.
It does seem likely that you have some sort of low-forward-voltage diodes, like Schottky or germanium diodes.
"Cheap" and "rectifying" indeed have the same connotation here.
In this application, each diode is protected by the forward voltage of its anti-parallel companion. So you could probably get away with a peak-inverse-voltage rating of only 1 volt. However, you are unlikely to find silicon-rectifier diodes rated less than about 50 volts. But note that, for other applications, the transformer voltage peaks are the RMS voltage multiplied by the square-root of 2.
I would get diodes with a current rating about the same as that of the breaker protecting the circuit. Since each diode of each pair conducts only half the time, that should provide a decent margin of error.
I just found the Amazon page that you bought from:
https://www.amazon.com/General-Purpose-Plastic-Rectifier-Diodes/dp/B072KKYBMD
And that led me to this:
https://www.top500.de/lexikon/unspsc_code_32111504.php
where they are described as Schottky diodes.
I would buy diodes from a distributor like Mouser.
I plan to protect each track power output with a 10A breaker but am going to wire it all with 14ga. (Smaller wire and breakers for switch motors, etc.) Everything I've got except for the Gargraves track and turnouts is Lionel postwar from the mid-1950's.
One final set of questions related to that:
Am I correct in thinking that I don't need a breaker in the common? An entire circuit from power terminal out to the track center rail and back to the common terminal is protected by the single breaker in the power wire, right?
Even in the unlikely event of simultaneous shorts in more than one circuit I assume the individual breakers will trip before the common bus/wire is overloaded. Does the often-mentioned issue about the ZW not having internal protection against shorts between pairs of power terminals impact any of this?
Sorry for posting my last reply twice. The first time I clicked "submit" there was no indication that anything happened so I submitted again after 15-20 minutes.
You don't need to add a circuit breaker in the common wire, but you do need one, because the total current from all 4 circuits protected by 10-ampere breakers can easily exceed the 15 amperes that your common wiring and the transformer can handle. But, as you say, the ZW's internal breaker can handle that.
Using individual breakers on the 4 circuits takes care of any circuit-to-circuit faults that may occur.
You may already know that there is no appreciable electrical difference between stranded and solid wire, and both are generally available in a variety of colors at home-improvement stores in gauges 14 AWG and larger.
https://www.mouser.com/datasheet/2/345/10a05-10a10-16323.pdf
How about the 10a05 (10 amps, 50 volts) version of these? As I'm sure you are aware there are literally thousands of different diodes on Mouser. Applying search filters these are the first ones I found with the right specs. I assume the max Vf of 1.1v is somehow related to the ~0.7 volt drop for an anti-parallel pair?
It doesn't matter whether the diodes are in a pair or not: whichever one is conducting will have the same drop as if it were by itself.
The 1.1-volt rating at 10 amperes is the worst case that the manufacturer is claiming for his product. The forward voltage for a real diode is not truly constant for all forward currents. It starts at zero for zero current and increases as the current increases, but in a way that doesn't stray far from about .6 or .7 volts for most of the diode's useful current range. Close to the diode's specified maximum current (10 amperes for this one), the bulk resistance of the silicon also begins to exacerbate the forward drop above what the semiconductor physics predicts. But diodes used to slow a descending train will probably be passing far less than their rated 10 amperes, so a model that assumes a constant drop of about .65 volts is appropriate.
Personally, I think this whole conversation has gone way off the rails, (pun intended), and this is a classic case of overthinking.
Lets assume you are using a ZW transformer or multiple phased xW transformers. I assume you would have one ZW output set at 18v and another set for 10v. In both cases the return is via the U connection on the transformer(s). There is a myth that if you short the 18v output with the 10v output that you are placing an 8v direct short in play. That would only happen on the bench if you are ignoring the U connection. That is not the case on a layout where U is the return and connected to the outer rail. Because current flow on the layout is referenced to the U connection there is no path from 10-18v as both are +V referenced and you will not have a short.
To solve your problem you can Power your main with one output of a ZW set at 18v for the level and upgrades, and the downgrades can be powered off another control of the ZW set for 10V or less. No Diodes or resistors required. In fact, Lionel was building layouts like this 70-80 years ago before semiconductors were commonly available.
So what happens when a locomotive stradles the 10 & 18v blocks? Nothing really, the 18v block momentarily drops to 10v any current flow is though the transformer winding, the motor, then to U (outside rail).
One of these days I'm going to do a youtube lecture series on this. However in the meanwhile I've wired my layouts like this for over 50 years and never experienced an electrical issue.
BigAl956 - EE Chicago 1978.
aboard!
Trains, trains, wonderful trains. The more you get, the more you toot!
Rather than repeat this old argument with Al, I'll just post a link to a previous version of it: http://cs.trains.com/ctt/f/95/p/255980/2867653.aspx#2867653
OMG. Maybe Roger Carp can look into the error of not following Bob Nelson's recommendations on using multiple taps of a high power transformer and instead paralleling them set at different voltages.
CLICKABLE LINK to Bob's reference above.
Rob
I decided to use the diode method out of an abundance (or over-abundance?) of caution, to learn to do something new, and also because my layout will probably have more up and down sections than I have throttles. I had some questions about how to do it and Bob Nelson answered my questions. I'm sorry that this has re-opened an apparantly old debate but it has made me consider another question.
I am not qualifed to say which position is correct as far as the potential danger (or not) of something accidentally and non-instantaneously bridging the gap between blocks powered by different outputs of different voltages But, it seems to me that if one uses "the traditional" method and adds an individual breaker (approriately sized for the wire size) in the wiring from every transformer output terminal that would provide protection against the worst case scenario. Yes or no?
bgreen...adds an individual breaker (approriately sized for the wire size) in the wiring from every transformer output terminal that would provide protection against the worst case scenario. Yes or no?
Maybe. The fault current I have measured is small, under 4-5 amps at a few volts difference, but quickly jumps to 20-30 amps at 6-7 volts difference in throttle settings. This was with a ZW.
If Big Al ever wants to test this with a ZW, he can slowly roll a Polar Express passenger car(or the whole train) from one block set at 6 volts through to another block set at 18 volts and report back.
With respect to fire safety, I would say yes.
But Rob's mention of passenger cars reminds me that the wiring between the trucks of some lighted cars may be too light to carry across the gap the fault current that the circuit breakers will allow. I recall seeing a picture posted on this forum years ago that showed such a wire buried in melted plastic.
lionelsoni...I recall seeing a picture posted on this forum years ago that showed such a wire buried in melted plastic.
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