2 Questions regarding Resistors for toy train applications

Classic Toy Trains Forum
1

Still in “Dummies” chapter 1 and studying Ohm’s law. I have a question of 2 applications of resistors:

  1. Electronics for Dummies, page 27 (let’s keep LEDs out of this discussion to simplify it):

“What if you find your light [could be a locomotive light] is too bright? A lower current reduces the brightness of the light, so just add a resistor to lower the current.”

The book then cites an example:

I (current expressed in amps)=V (voltage) divided by R (resistance expressed in ohms)

Using a multimeter, you find your light circuit is powered by 12 volts and has 9 ohms of resistence.

Thus, 12 volts divided by 9 ohms equals 1.3 amps.

If you add a 5 ohms resistor to the circuit, you will increase the original 9 ohms by 5 to equal a total of 14 ohms resistance.

Thus, 12 volts divided by 14 ohms equals 0.9 amps.

As you can see, we’ve lowered the current to the bulb and the bulb will be dimmer and also will last much longer.

Question: Is there any disadvantage to using this method for a locomotive lightbulb; e.g., heat buildup in the resistor adding to the heat buildup in the motor?

  1. For my second example, I’ll use my remotely controlled toy train, which uses a battery pack. Say I’ve hooked up 2 battery packs in a series, which produces a total of 14.4 Volts. However, I’d like to decrease the output a couple of volts.

Question: Would I simply put 1 or more resistors in-line on either the positive or negative wires leading out from the battery pack (using Ohms law, of course to figure out what’s needed)?

2

Dave, your technique for the lightbulb is fine, but there is more going on. The resistance of most things, and light bulbs in particular, goes UP as temperature increases. So, 9 ohms is the cold resistance of the bulb, but it will change.

Go ahead and construct the circuit with the 5 ohm resistor, then measure the voltage across the resistor with power applied. Use ohms law to compute the current through the resistor.

Note that the SAME current must be flowing through the light bulb and resistor and the SUM of the voltages across the light bulb and the resistor will be the applied voltage (12 volts). By measuring the voltage across the resistor you can use ohms law to compute the hot resistance of the bulb.

Adjust the resistor value, if you are not satisfied with the brightness of the bulb.

The total heat generated by the bulb and resistor will be less than that of the bulb alone, since current draw will be less. You can compute power consumed using P (watts) = IV across the bulb, resistor, or both. Ohms law can be substituted for I or V to get various forms of the power equation.

The same can be applied to the motor (you only need one resistor, either wire leg), but again, measure the current via a small resistor in series with the motor. Motors don’t get that hot, but the actual resistance depends on the position of the commutator, so a direct (motor off) ohms meter reading will be deceptive.

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

3

The problem with resistors is that it becomes a series circuit that as pointed out can change with components heating up. In example 2 since you have 14.4 volts DC you can simply drop the voltage by stringing diodes in the direction of the flow. Each diode will drop voltage by a steady .7 volts regardless of the load. So 3 diodes in a string would drop the 14 volts to about 12 volts. The diodes also do not consume power as much as resistors. Just get diodes 3 or 6 amp that will handle the load. If you drop the voltage using a resistor the sharing would vary with each individual engine ran on the track. Lighted cars would also change things.
As far as the locomotive bulb I would just get a higher voltage bulb that fits in the socket,or I would convert to constant voltage lighting if space allowed. You could also install a single diode in the circuit and half wave it.

Dale H

4

Hello Dale:

You seem to be a big fan of diodes! Could you explain why a diode reduces the voltage .7 volts? Is that the general characteristic of a diode? Do different diodes have different voltage consumptions? My (limited) understanding of diodes is that they restrict current flow in one direction (useful for AC circuits). In DC, I would assume current would just flow through the diode normally, but would use up some voltage just like a resistor. Therefore, why would a diode be preferable in a DC circuit over a resistor?

Thank you for your help. I find these topics fascinating and appreciate the knowledge you and others are willing to share.

Regards,

JO

5

900ma flowing through a 5 ohm resistor will dissapate 4.05 watts of power. That’s a BIG resistor. You will not find a good selection at radio shack where the majority of the resistors are 1/8w, 1/4x and 1/2w (I think they have 8 ohm 10 watters for dummy speaker loads.)

And It will probably get hot enough to melt plastic, burn fingers, etc. They make nice power resistors that have alumninum cases and mount to heat sinks. Most folks would use a 10 watt resistor in this application. (Try Mouser.)

Check out this online calculator:

http://www.the12volt.com/ohm/page2.asp

Think carefully about using simple resistor here.

Old 2037

6

John,

Ah, yes, the power equation: power (watts) equals volts time current (amps)

Quote: “The total heat generated by the bulb and resister will be less than that of the bulb alone, since current draw will be less.”

Quote: “The same can be applied to the motor (you only need one resistor, either wire leg), but again, measure the current via a small resistor in series with the motor.”

Quote: “The resistance of most things, and light bulbs in particular, goes UP as temperature increases. So, 9 ohms is the cold resistance of the bulb, but it will change.”

GREAT STUFF@!

BTW, Dale’s solution sounds ideal, but I too had the same question that JO asks.

Thanks for this valuable (invaluable?) feedback.

I have a hunch I’m not the only one learning from this.

7

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.

8

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

9

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 c

10

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

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.

If I remember correctly, Yes.

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]

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 dr

11

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~

12

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)…|
…|…|
…|…

13

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)

14

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!!!

15

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 conv

16

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

17

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.

18

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

19

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!

20

You got it!!!

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!

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]