2 Questions regarding Resistors for toy train applications

[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

[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

[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 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

[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?
--------------

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.

[Anonymous]

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

[FJ and G]

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):

1. 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!

2. 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)

3. Some resistors rated for high wattages build up so much heat, in fact, that they often come encased in flameproof coatings and have their own metal heat sinks with fins that draw heat away from the resistor!

Well, as you can tell, it's fun being an electronics Dummy and learning all of this Alice in Wonderland stuff.

[lionelsoni]

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.

[Anonymous]

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

[Anonymous]

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

[johnandjulie13]

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

[FJ and G]

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.

[Anonymous]

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

[lionelsoni]

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.

[johnandjulie13]

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

[Anonymous]

QUOTE: Originally posted by johnandjulie13
I would assume the 1.5V bulb stays at a constant brightness because it is actually deriving its voltage from the voltage drop of the bridge rectifier. Is that correct?

You got it!!!

QUOTE: Originally posted by johnandjulie13
How does the light stay on even if the train is stopped? Is there still current running through a HO system when the train is not moving?

You'd have to turn the X-former on just enough to get ~1.5 volts out of it. This is enough to light the bulb, but after the voltage drop, there's nothing left to turn the motor. Most of the older and/or cheaper transformers wouldn't be able to give that small a voltage, but I built my own and was able to deliver anything from 0 to ~13 volts. There's no way you could get an old prewar or postwar transformer to do this!

QUOTE: Originally posted by lionelsoni13
Hank, you need to have the cathodes pointing toward the + terminal.

The way you drew it, it works only for one polarity of supply voltage. With the other polarity, the lamp burns out. I would rotate the bridge 90 degrees and connect the bridge's + and - terminals together.

You're right! My mistake. It's been a lot of years since I last drew out a schematic! [8)]

Oh, and just to make sure my drawing didn't give the wrong idea, you'd probably want to use a premade bridge rectifier for this instead of 4 separate diodes. IIRC, I bought the ones I used at Radio Shack for about $1 each. Don't remember the part number, but they're flat and about 3/8" x 1/2" x1/8" thick and have a pin sticking out of each corner.

Have fun with it!!! [:D] [:D] [:D]

[FJ and G]

Thanks. This will take some time to digest. Especially the forumulas. Thank God for online calculators that can do exponents and weird stuff like that.

Nothing appears to be as simple as it first seems. It's no wonder there are so few electronics specialists!

[johnandjulie13]

Hello Bob:

Thanks for the analysis! I recently converted an old LGB locomotive to MTS (LGB's DCC system) and I had to take the diode out that was connected to the front headlight. Now I know why it was there!

Thanks again,

JO

[lionelsoni]

Hank, you need to have the cathodes pointing toward the + terminal.

The way you drew it, it works only for one polarity of supply voltage. With the other polarity, the lamp burns out. I would rotate the bridge 90 degrees and connect the bridge's + and - terminals together.

[johnandjulie13]

DDHank:

This is great stuff! I would assume the 1.5V bulb stays at a constant brightness because it is actually deriving its voltgage from the voltage drop of the bridge rectifier. Is that correct? How does the light stay on even if the train is stopped? Is there still current running through a HO system when the train is not moving?

I believe I have seen bridge rectifiers used in LGB (G Gauge) for reversing loops.

Thank you for the information.

Regards,

JO

[lionelsoni]

Dave, that example from your book is terrible. I would get a new book. The resistance of an incandescent lamp varies enormously with temperature, being roughly 10 times as high when lit as when not lit. Measuring the resistance with an ohmmeter is useless. And, even if the 9 ohms were the "on" resistance, assuming that it remains at 9 ohms when you dim the lamp is wrong. If they wanted to illustrate Ohm's law, they should have selected another load for the example, a resistive one with a constant resistance.

But, let's assume that the 9 ohms is the resistance at 12 volts, even though that makes for an unbelieveably bright headlight. The rule for incandescent lamps is that, even though in the short term they behave as resistors, the RMS current varies as the .55 power of voltage in the long term. So

I / (12 V / (9 ohms)) = (E / 12 V)^.55, or

I = (12 V / (9 ohms)) * (E / 12 V)^.55

where ^ means exponentiation, is the equation for the lamp current. When we stick a 5-ohm resistor in series, the lamp voltage E drops by I*5 ohms:

E = 12 V - I * 5 ohms

There is probably no analytic solution for these non-linear equations. But they can be solved iteratively, by repeatedly evaluating E and I, using the latest value of one in the next evaluation of the other. This gives a value for E of 7.031 volts. The simplistic solution from the book would be 7.714 volts. The current is .994 amperes.

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.

The constant-voltage circuit is simply two diodes connected in parallel, anode-to-cathode, with the pair put in series with the motor. The lamp is connected in parallel with the diodes for a nearly constant voltage of .7 volts, or thereabouts. If you put two diode pairs in series, you get double the voltage and you can use a convenient bridge-rectifier module, which contains 4 diodes, instead of individual components.

If the current is AC, its polarity just switches back and forth every 1/120 of a second. The lamp doesn't care. If it is DC, there is a further possibility for directional lighting. Use the 4-diode bridge-rectifier package. Then put a single additional diode in series with the lamp, which will light only for one polarity, that is, one direction. You have to use a .7-volt lamp for this, because that fifth diode eats up half the 1.4 from the bridge rectifier diodes. These lamps are easy to find, however, since HO modelers use them this way.

When you use pairs of diodes to reduce an AC voltage, the RMS voltage drops by only about 90 percent of the diodes' forward voltage. This is because of the changes that occur in the waveform with this method.

[Anonymous]

DOH!!! I drew that wrong. Ignore my earlier drawing. (Sorry, it's been a long time since I made one of these.)

DO NOT hook that earlier circuit up! It would have put a dead short in paralell with the motor!!!

All 3 components are NOT in paralell. The rectifier & bulb should be in paralell with eachother, and that assembly should be in SERIES with the motor.

There's a much better diagram of a bridge rectifier about half way down this page: http://www.kpsec.freeuk.com/components/diode.htm (It's even animated to show how it works in an AC to DC circuit!)

I edited my earlier post to fix the diagram...

Oh, and I did use it with a DC motor. The internal circuitry of the bridge rectifier allows current to flow either way and drops 1.4 volts in either direction. I've never tried it with an AC motor, but I don't see any reason why it wouldn't work.

Once again,: DO NOT hook that circuit up as I drew it the first time!!! It would have put a dead short in paralell with the motor!!!

[FJ and G]

ddhank,

I've drawn the components here (lightbulb is at the top of the diagram)

Using your schematic, I've indicated the direction of the diodes, which has a colored ring around one end of the cylindrical type component.

I'm unsure how to determine which side of the motor is positive and which is negative (same with lightbulb). If the motor wires are yellow and blue, that probably determines which is which?

Also, I'm assuming this is a DC circuit and that the power source (from the rails or in my case from the battery pack) would also connect to the motor (in addition to the 2 wires pictured)

[Anonymous]

<EDITED to fix drawing error>

Okay, here's a VERY rough drawing of the circuit I was talking about:
(Ignore all the periods. I had to put them in to get the text to space properly.)

--|<-- = Diode

-->|-- = Diode

--|-- = Connection point

|---|
-- | --- = NO Connection point
|---|

(+), (-), & (AC) = markings on bridge rectifier



............................./---------.............................| LIGHT |
.............................| BULB |
.............................\---------/
................................|....|
................................|....|
......|-------------------------|....|-------------------------|
......|........................................................|
......|........................................................|
......|........................................................|
......|.................(This tap not used)............|
......|.............................| (AC)...................|
......|.........|--------|<--------|---------|<--------|.......|
......|.........|.......................................|.......|
......|.........|.......................................|.......|
......|---------|.(-)..............................(+).|-------|------------|
......|..........|......................................|....................|
......|..........|......................................|............|---------------|
......|..........|--------|<--------|-------|<---------|...........|. MOTOR .|
......|..............................|.(AC).......................|---------------|
......|.................(This tap not used)..........................|
......|......................................................................|
......|......................................................................|
......|..........................................................(To power suply)
......|............................................................(either + or - )
(To power suply)
(either + or - )

[FJ and G]

Thanks. A wealth of info here.

So then, this nicely sums up the value of diodes over resistors for the 2 applications I mention:

"You get a precise drop in voltage (per diode) and generate almost no heat in the process."

And furthermore, that advantage extends to AC as well as DC and just requires hooking up the diodes in the correct direction, sequence, number and amps.

Terrific~

[Anonymous]

John,
Since Dale hasn't been back, I'll take a stab at answering your questions.

QUOTE: Originally posted by johnandjulie13
Could you explain why a diode reduces the voltage .7 volts? Is that the general characteristic of a diode?

I don't remember exactly how diodes create this 0.7V drop, but yes, it's a common characteristic of diodes. They block current in the "reverse" direction and allow current to flow in the "forward" direction, but produce a 0.7 drop in voltage.

QUOTE: Originally posted by johnandjulie13
Do different diodes have different voltage consumptions?

If I remember correctly, Yes.

QUOTE: Originally posted by johnandjulie13
In DC, I would assume current would just flow through the diode normally, but would use up some voltage just like a resistor.

Not quite. A diode does drop the voltage, but not in the same way that a resistor does. The voltage drop in a resistor will be different depending on what other components are in the circuit. A diode drops the same amount regardless of other factors. (Assuming, of course, that you don't let the magic smoke out.) [:D]

QUOTE: Originally posted by johnandjulie13
Therefore, why would a diode be preferable in a DC circuit over a resistor?

You get a precise drop in voltage (per diode) and generate almost no heat in the process.


Here's another useful thing that I used do with diodes was when I ran HO & N scale trains (If you're not familiar with them, they run on DC.)
One of the most annoying things about most toy trains (AC or DC) is that the brightness of the headlight bulb varies with how fast the train is going. With a DC motor, you can put a small bridge rectifier (4 diodes in a single 4-pin chip) in series with the motor. I don't think I can draw the circuit out here (with just text to work with), but you wind up with a 1.4V drop in voltage in either direction. Put a little 1.5V "grain-of-wheat" bulb in parallel with the rectifier and you get a nice, bright light that stays at full brightness at any engine speed - even at a dead stop!

Good luck!
Doug

[Anonymous]

Jo

An AC sine wave has 2 components,the positive and the negative. A single diode inserted in a 12VAC circuit for example would totally block half the sinewave,either the positive or negative half depending upon the direction it is inserted. So you would block say the negative totally and the positive half would be reduced by .7 volts. The .7 drop is just the property of a common general purpose diode. A light bulb in the circuit would be a little less than half as bright. The half RMS voltage would be 11.4 but you would only get the top half of the wave. The bulb would get the equivalent of a little less than 6VAC as a pulsed DC current.

Now lets take the same 12VAC circuit and install 2 diodes in series,one in each direction. Both the positive and negative part of the sine wave would have a path and both sides would be reduced by .7 volts leaving 11.4 VAC to the bulb. Each pair of diodes inserted would further reduce the voltage by .7.

Now lets say we have a 12 volt DC circuit powered by a battery source lighting a bulb. Inserting a diode in one direction would totally block current flow and the bulb would go out. Reversing the diode would allow the forward current but reduce the DC current by .7 volts. Each diode in series in that direction would further reduce voltage by .7 So 4 diodes in a string all in the same direction would leave you with about 9 volts.

In summary dropping AC cuurent requires paired diodes in opposite directions while dropping DC current requires single diodes in the direction of the current flow.

The diodes themselves use little power,they do become warm because they are resisting current flow. Resistors get hot. If you mount wirewounds make sure you mount them with the proper clips away from combustibles on the board. They do the job but are wasteful electrically.

For lightbulbs and such 1amp diodes such as 1n4001 are fine. To Block track power 3 or 6 amp ones are required. Bridge rectifiers can also be used by wiring the + and - together and putting the ~ leads in series. This is actually 2 sets of paired diodes in opposite directions so a bridge wired in this manner would drop AC by about a volt and a half.

You can easily make voltage droppers by soldering or mounting them in Euro style barrier strips. Bend the leads u shaped and scew them into the terminals. Thes are handy inserts to reduce voltage to individual accesories instead of wiring a lot of transformer taps.

Dale Hz

[jkerklo]

Dave, don't worry about the power for the resistor yet. Go ahead and construct the series circuit and use it to measure the hot resistance of the light bulb. Most toy train bulbs are between 1 and 4 watts. For example, a #57 bulb (at 14 volts) is .25 amperes and 3.36 watts. An effective hot resistance of 56 ohms. (See the light bulb chart in "LIGHTING" on www.Three-Rail.com )

When you are done, I expect the final power for the resistor will be less than 1/2 watt.

Dale's suggestion for using a diode for the battery pack is a good one. Since it is DC, only one diode for each .7 volts is practical. (The light bulbs would require two diodes per .7 volts)

Using a resistor for reducing the voltage to a motor has another problem. Motors draw more current as load increases. More current through a resistor means a higher voltage drop across the resistor and, therefore, less voltage to the motor. Backwards from what would be wanted.

John Kerklo
TCA 94-38455
www.Three-Rail.com

[FJ and G]

Old,

Thanks, Didn't see your post as I was typing and yes. There are those handy dandy Internet math calculators for those of us who never liked numbers crunching and wi***o compute milliwatts time something to the something power.