How to calculate voltage drop?

[eZAK]

How does one calculate voltage drop?

I have 18vac and I want to drop it down to 14vac, 12vac, & 10 vac.

I will be using a rheostat.
I'm wondering what size I need.
The two I have in mind are;
11 ohm/ 25w/ rated at 1.25a,
OR a 100 ohm/ 25w/ rated at .5a.

I'm unsure of the amperage draw at this point.

Thank you

[Anonymous]

First you **need** to know the current drawn by the device(s) you wi***o power. If you do not know that, you can't calculate what you want to know. You can't simply put in a rheostat and set it and hope all the accessores now work with a new voltage. It does not work that way. And as the load changes, like multiple accessories not turned off in unison on the aft end of the rheostat, the voltage drop will *change*.

The voltage drop depends on the CURRENT and the RESISTANCE involved. E = I*R. So say you have 50ma of current being drawn, and 45 ohms of series resistance.

This yields .05A * 45 = 2.25 volts dropped...

Next you need to be sure the power is OK for the series resistor. P = I*E or in this example .05A * 2.25 V = .1125 Watts.

Sure you don't want to try a Mini Commander ;) No calculations needed... Just kidding, Note however that units that supply fixed volatges by a setting, like a power master or TPC work a lot different - that is why they cost more.

jon

[webenda]

If you wire the rheostate as an adjustable resistor, it is as jon says.

If you wire your rheostat as a potentiometer (see image above), then the no load
voltage is proportional to the slider position. At 50% from reference (0V) it is 18 x .50 = 9V.

[Dave45681]

The key in the above example is that it is a no load voltage. Once you draw some non-zero current, I am pretty sure you have a problem again.



In the above sketch, the blue outline represents the rheostat. The load is your engine/accessory/etc.

Call R1 the half of the rheostat resistance between the set point and the 18V side, R2 the half between the set point and the 0v side.

Lets use the 9V example, just since the math is easier:

The current through R2 will still be 9V/50 ohms = .18 amps.

For no load, the same current will be going through R1, since there is an infinite resistance and no current draw if there is no load.

Once the load draws some larger current, the current going through R1 will be the current going through R2 plus the current drawn through the load.

So if you have a 1 amp draw from your load, you will draw 1.18 amps through R1. This just exceeded the current rating specified in the example above.

Hope this helps too.

[webenda]

Dave,

Nice picture. Illustrates your point well.

The load could not draw 1 amp through 50 ohms with a source voltage of 18V.
If the load were a short, eliminating R2 from the problem: I = V / R1 = 18 / 50 = 0.36 Amp. Now the voltage at the center of the rheostat is zero, the current through R2 is zero amps, the current through the load is 0.36 amp and the voltage across the load is zero volts. As you said, "Once you draw some non-zero current, I am pretty sure you have a problem again."

You could calculate the voltage across the series - parallel combination, or (no math method) just start with the rheostat slider at zero and adjust until you obtain the desired voltage (keeping the constraints of 25W, .85amp in mind.)

One nice thing about our hobby is that we have variable voltage transformers. eZAK, adjust your transformer to the voltage you want, apply the load with your ammeter in series to determine the current. Then use mousecat’s instruction to determine which rheostat to use.

You already posted your voltage requirements. If we knew the load current (or maybe more information about the load) someone would do the calculation and post the answer. Some train nuts are math nuts also.

[lionelsoni]

Unfortunately, the only way to get good load regulation with a potentiometer is to size the potentiometer so that it draws much more current than the load.

[eZAK]

Thanks guys.

To further clarify this situation;

I have edited my orignal message to correct the errors made in the amperage ratings. The ones now posted reflect the actual amperage ratings.

Also let's assume for a minute that the amp draw will be 1a at full load.
This will most likely be less. Maybe 1/2 to 3/4 of an amp.

The voltages stated in the org. post will be for various Acc's from PW to present.

Now for the therory!
Using ohm's law and the 11ohm rheostat at 50%,
RxI=V, 5.5x1=5.5v. 18v minus 5.5v = 12.5v Is this correct?
At 100% it would be 11x1=11v 18v minus 11v = 7v ??

Now using the 100ohm rheostat . [%-)]
I seem to wind up with a negative voltage[banghead]

Math always gives me a head ache.

[lionelsoni]

Start with the voltage and current at the load, for example, 1 ampere and 12 volts. The voltage across the rheostat is then 18 - 12 = 6 volts. The resistance of the rheostat therefore needs to be 6 volts divided by 1 ampere, or 6 ohms.

Here is the problem with the rheostat: Suppose the current drops to .5 ampere. Then the resistance needs to be 6 volts divided by .5 ampere, or 12 ohms. In other words, as the load changes, you need to adjust the rheostat to keep the voltage where you want it. This is what is meant by poor load regulation.

[Dave45681]

Thanks webenda and others for correcting my somewhat faulty explanation.

I hope I didn't create more confusion than I helped by posting the pic breaking down the circuit elements.

I guess I will have to chalk it up to "hadn't had my wheaties yet", since I can't claim a Monday moment for yesterday . [B)]