Dale:
Thank you for the additional analysis. I am slightly confused by one thing: If putting the diodes in causes a voltage drop, how does this reconcile with your comments yesterday in setting up a sound activation circuit. In that response you said that jumping the diodes boosts voltage by 1.75 volts RMS.
Maybe I don’t understand the term “jumping.” Could you please clarify? I feel that I am close to understanding this, but I am not quite there yet.
Thanks,
JO
The 1033, 1044, and others of their ilk will go all the way down to zero if you use the option of the B terminal as common. The maximum is then 11 volts.
JO
I was speaking of 12 diodes in series,6 in each direction in series to the track. All 12 would block about 4.2 volts AC. Now when you jump or connect 5 on one side together with a jumper wire via a switch.,The sine wave would be 3.5 volts higher on one side than the other. The whistle relay (or bell sensor depending on polarity) will pick up this imbalance, energize and blow the whistle. That is 3.5 more volts going to the track on the top or bottom of the sine wave. This would be an equivalent of about 1.75 volts RMS in potential.
An air whistle would probably use more than this and may slow the train down a bit but solid state electronics whistles would not use much and not slow… You could make the diode string longer and leave more voltage in reserve to compensate for an air whistle,just as locos with cruise control hold voltage in reserve for speed regulation. You could then give a bigger boost to an air whistle. With a string of 8 pairs of diodes for example using a 2 pull pushbutton or relay you could jump 5 on one side plus 2 pairs of diodes. That would give the equivalent potential of about 3 volts AC just a little less than a ZW booster winding.,which has a very inefficient selinum rectifier in it.
My layout is automated and I use a long string of diode pairs coupled with relays and timers to produce a soft start circuit when the locos start. My locos start in 9 steps, first ringing the bell then slowly increase speed to full in about a minute. Not only is this more realisitc but it saves mechanical wear from jackrabbit starting.
Dale hz
Gee, this will take a whole mess of time to digest. I won’t have any more questions for a while, as I’m studying this and continuing my dummies book, even tho it is imperfect.
Hmmm. “Automating a layout with a long string of diode pairs coupled with relays and timers to produce a soft start circuit…”
The circuits in my brain are about fried. Ask a simple question and …
Thanks, tho, this is very interesting stuff. I didn’t mean to waste so many people’s time on this but I’m grateful. If I ever learn this stuff, perhaps I can help others someday as well.
Dale:
Thank you for the clarification. So, when you “jump” the diodes, you are really short circuiting them. Is that accurate? This then results in a potential difference with the diodes on the other side, thus creating the DC component for the engine’s relay to recognize. Is that right? The math seems to work: .7V RMS for each diode, times 6 diodes in series, gives a total of 4.2V RMS. Jumping all but one on the other side, would leave one diode at .7V RMS and a resulting difference of 3.5V.
So, in response to Bob Nelson’s comment, this setup reduces the voltage to the track, resulting in a lower top speed for the engine. On the other hand, activating the button will not cause the engine to surge ahead like it would with a whistle throttle postwar transformer. Is that right?
I feel like one of the monkeys in 2001: A Space Odyssey, that has just touched the monolith.
Regards,
JO
Jo
Yes,You are short circuiting the diodes if you just look at the diode portion of the circuit. A better term may be you are bypassing them ,taking them out of the circuit…The total circuit is not shorted.
Yes the to speed of the engine is limited but in most applications 20 volts is too much anyway. If you really needed a voltage boost you could bypass the whole thing with a switch or relay if you needed extra voltage for climbing a grade.
Dale Hz
The little bit of extra voltage should not cause a surge. Even an electronic whistle uses a bit of power. With the 6 pair of diode an air whistle would actually slow the train down a bit. You can customize the diodes for the type of equipment you run. Postwar and modern locos run differently and characteristics vary by manufacturer.
Dale Hz
Of course, when we talk about the effects on the track voltage from various arrangements of diodes, we are only making rough estimates. I thought it would be interesting to see what the exact values are, assuming perfect diodes with a forward drop of exactly .7 volts.
For the 6-diode strings most often discussed and a ZW with 21 volts RMS, the diodes drop the RMS voltage by 3.700 volts. This is the voltage of interest for lamps and for universal motors. For DC “can” motors, I assumed 2 additional diode drops in each direction for a bridge rectifier in the locomotive and calculated the difference between the rectified DC motor voltage with no other diodes and with the 6 pairs, which is 3.884 volts.
When the whistle is blown by shunting 5 diodes in one string, the DC component becomes 1.658 volts. The RMS is dropped 2.101 compared to no diodes, which amounts to a rise of 1.6 volts. The rectified DC is dropped 2.279, which amounts to a rise of 1.605 volts.
However, these numbers are unrealistic in that they assume that the ZW is turned all the way up. For comparison, I computed the values for a ZW voltage of 15 volts, which might be closer to normal. I got an RMS drop of 3.666 and a rectified DC drop of 3.756 volts, which is not much different from the 21-volt values. When the whistle blows, the RMS drop becomes 2.057, the rectified DC 2.210, and the DC component 1.621 volts.
As I have said before, for small reductions, the RMS voltage goes down by about 90 percent of the diode drop. It does look like that is a good rule of thumb for all these kinds of estimates.
Well, I downloaded this entire file and it will take some reading and a lot of digesting to do.
I found out more and wonderous things about resistors, btw (into Chapter 4 now):
The makers of Resistors take meticulous care to stamp the number or use color rings to indicate their Ohms ratings, as well as their tolerances + or -. But, they fail to rate the power in watts! Too many watts could literally fry your little baby resistor! So, you must either use a multimeter to find it out yourself or judge the value of the power by the size of the resistor!
There’s more than 1 resistor too!
a. There’s precision resistors
b. there’s HIGH-precision resistors with lower tolerances
c. there’s variable resistors called Potentiometers (which you can use to control the TV on your Lionel loco-scope)
Well, as you can tell, it’s fun being an electronics Dummy and learning all of this Alice in Wonderland stuff.
Most wire wound resistors give wattage value on them . A thing to remember about adjustable wirewounds is that if you adjust the resistance you change the capacity. Lets say you have a 25 watt wirewound 10 ohm resistor and you adjust it by a third to slow something down. You reduce the wattage capacity also by a third to 17 watts because you are not using the full lenth of the coil. If you cut it by 2/3 you have an 8 watt resistor. It is important therefore to properly size resistors for their ohm rating as well as wattage. In this case if you had to cut it by 2/3 you would be better off with a 5 ohm adjustable.
As usual Bob is right on his calculations. As a rule of thumb diodes drop between the .6 and .7 range RMS because of the complication of the sinewave. Did not want to bring it up and confuse people. Most textbooks do not bring this up and some even give the .6 figure. In reality you make adjustments under field conditions. As long as the diodes are the same in each pair no imbalance should result.
Dale H
The other side begs the question:
Some of the projects I’ll be doing require an oscilloscope. Anyone know where I can get a low-cost but high quality one?
Dale,
I guess similar advice applies in electronics as in doing toy train benchwork (measure twice, cut once), namely, measure twice, solder once.
Hello Bob and Dale:
Thanks again for the great information! I am wondering how Bob derived the actual values. Are these mathematically calculated? Or, are they measured? If they are calculated, could you share the underlying mathematical formulas?
While David is making good progress with “Electronics for Dummies” I have been making decent progress with “Teach Yourself Electricity and Electronics.” I have just about completed the first section which focuses on DC circuits and general principles. I am looking forward to the second section of the book which gets into AC and various components (capicitors, inductors, DIODES, transistors, etc.)
Also, I have purchased one of those Lionel sound activation units. I can’t wait to get it home to see what the circuit looks like.
Thanks again for all of the input.
Regards,
JO
Dave, as I said on the previous page, “Using diodes instead of resistors in a DC circuit does not save power. The power dissipated by the diode is exactly the same as that dissipated by the resistor that produces the same voltage drop.”
In an AC circuit, a half-wave rectifier can be used to reduce the RMS voltage by about 30 percent. In this case, the diode does run cooler because it simply shuts off the current half the time rather than trying to carry the current and create a voltage drop, which is the situation we’re dealing with here.
JO, I integrated the waveforms analytically. I derived two expressions:
halfavg = -k / 2 + (sqr(2 - k*k) + k * asin(k / sqr(2))) / pi
The first one gives the average voltage, relative to the RMS voltage of the original sinewave, for each half wave, as a function of k, the ratio of the total diode drop to the original RMS voltage. Differencing two halfavg values, one for each polarity, gives the overall DC output. When k is the same for each half wave, that difference is obviously zero–no DC. But with two different values of k, corresponding to different numbers of diodes, a non-zero DC component emerges.
halfms = ((1 + kk) * (pi / 2 - asin(k / sqr(2))) - 1.5 * k * sqr(2 - kk)) / pi
The second one gives the average of the square, relative to the square of the RMS voltage of the original sinewave, for each half wave, as a function of the same k. Adding two halfms values, one for each polarity, gives the square of the overall RMS output. The square-root of that sum, multiplied by the original RMS voltage, is the RMS result.
For the DC rectifier drop, I used the first equation, adding the two terms instead of differencing, to account for the rectification, and using 2 extra diodes in each direction.
Hello Bob:
Ewww! I am almost sorry that I asked. Now I have to dust off one of my college text books and become reacquainted with sine waves. In any event your explanation as to how the “positive” wave will compare to the “negative” wave based on the relative number of diodes helps clarify for me how the net RMS voltage is calculated. I will have to chew on this over the weekend.
Thanks again for your help. Before this is over, you are going to have to add “Tutor” to your list of interests under your profile [;)].
Regards,
John
Dave
In many or most applications resistors have an advantage. A simple example would be a volume control,you simply adjust it to a desired setting. Using Ohms law,if the variables are known resistors are more precise. Powering LEDs is another example as well as RC circuits in a radio.The problem with resistors regulating a model train is that the variables change when dfferent models are placed on the track. Just when you think you have the answer,the questions keep changing when you employ resistors.This really has no effect on diodes as the drop is predictable regardless of the load. A good book with basic imformation is one Radio Shack used to carry Getting Started In Electronics by Forest Mims. It pretty much gets into the basics without being overly technical. Unfortunatly Radio Shack seems to have less and less for people making their own circuits.
Dale Hz