2 Questions regarding Resistors for toy train applications

[FJ and G]

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."

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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.

[Anonymous]

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

[johnandjulie13]

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

[Anonymous]

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

[jkerklo]

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

[FJ and G]

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?

2. 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)?