2 Questions regarding Resistors for toy train applications

[FJ and G]

The other side begs the question:

If diodes are so much more efficient than resistors at reducing voltage without building up as much heat, are there any instances in which using a resistor would be preferrable to using a diode?
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Some of the projects I'll be doing require an oscilloscope. Anyone know where I can get a low-cost but high quality one?

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Dale,

I guess similar advice applies in electronics as in doing toy train benchwork (measure twice, cut once), namely, measure twice, solder once.

[johnandjulie13]

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

[lionelsoni]

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 + k*k) * (pi / 2 - asin(k / sqr(2))) - 1.5 * k * sqr(2 - k*k)) / 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.

[johnandjulie13]

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

[Anonymous]

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