Variable resistor to change 12 volts from 1 to 12

This should work fine except for the 3A requirement. To be on the safe side, just get a potentiometer that has the highest power rating you can get.

Thank you, Vulcan,

According to my High School class formulas. V xA =watts, so at 3 amps, this resistor should be rated at least 36 watts. And, even if it were rated at 100 watts, the resistor would still vary the voltage correctly? I believe so. In any case, I will get a pot so designated, and test it under load.

As I am not an electron traveling on the surface of a copper wire, I have no first hand experience with the dynamics of electricity flow. We have only formulas and metaphors to describe these anomalies. Eventually, I suppose, with everyday immersion in these principles, I will have an epiphany, and all concepts will jell. In the mean time, I must rely on these formulas to practically solve the immediate problem at hand. May I ask the formula you used to arrive at these resistance values? Ohms law incorporates amps- that I do not see in this conclusion. Or, I can just go back to school. In any case- thank you.

That is a brute force method.

I would get hold of a voltage regulator, buffer it with an Darlington or FET output transistor (if you really need three amps) and use a small pot just to control the regulator chip. Plenty of heat sinking!

Use something like an LM317 (that was from memory, but I think its correct!) or something more modern....

...Or use a variable switched mode power supply, which are very efficient!

May I ask the formula you used to arrive at these resistance values?

Sure!

First of all, I used a commercially available potentiometer value (1000Ω). Designate this as R1. Then I calculate what value R2 should be. According to Ohm's law, the current going through the two series-connected resistors is

I = 12V÷(R1+R2) = 12V÷(1000+R2)

The required voltage across R2 is 1V. To achieve that R2 needs to be:

R2 = 1V/I = 1V/(12V÷(1000+R2))

Moving things around we get (I hope you remember your algebra !):

R2 = 1000/11 = 90.91Ω

If you can get a 91Ω resistor, you'd be more closer to 1V than with a 100Ω resistor but the difference is very small. The current going through both resistors is approximately 0.011A.

Thanks again for taking the time to apply Ohms law to my problem. I appreciate you answering my basic logic questions with math and figures. Your choice of 1000 ohms to plug into the equation was good, and easy to figure. AND most appropriate- because the amperage range that results is likely to replicate the temperature sensors that I intend to replace with this variable resistor. Next, I thought that I would plug into the equation some " what ifs". That is, What if I really was encountering a load of 3 amps. I know that this is a lot of power, and actually equivalent to pedaling a bicycle at 8 mph on level ground (about). However, in automotive circuits, hi amperage draw at 12 volts is not unusual. If my math is correct, to balance E=IxR, then to vary 12 volts from 12 V , to 1volt at a 3 amp draw, I would need a resistor that ranged from 4.0 Ohms to .33 Ohms. Is that correct? I'm sure that it is not efficient, as I remember that once when reducing voltage to a 6 volt VW wiper motor converted to a new 12 volt system, the motor got so hot ( as did the ceramic resistor) that the motor overran into its own park circuit. You had to physically grab the wipers to switch them off. In any case, if that math is correct, the I can see more clearly how load relates to the choice of resistance, and gives me a more relational feeling regarding keeping the smoke contained.

from 12 V , to 1volt at a 3 amp draw, I would need a resistor that ranged from 4.0 Ohms to .33 Ohms

What makes you think that the current draw at 12V will be the same as when it's 1V? You're load resistance is constant (I'm assuming) and your voltage is variable. If you drop the voltage the current should drop as well.

However, your pot still needs to handle the power being dropped across it. If you can't find a 12V x 3A = 36W potentiometer (at these sizes, they're called rheostats), better forget the pot and use linear regulators instead. RS Components have units from 25W to 500W.

As for your wiper motor, your old motor was 6V and you changed the power supply to 12V? Since you increased the voltage, the current would certainly increase also (resulting in your experience).

You tried putting a series resistor to drop the voltage down to 6V but this, too, became hot. My guess is your resistor was too small (in resistance and power rating) and you were just using one resistor. In the example I gave, I used two resistors.

Vulcan is correct in what he says and the reason he mentions to use more resistors is to "spread the load" so to say, so that the heat generated is not concentrated in one resistor.

A good rule of thumb for such circuits is to at least double (even more could be better) the needed wattage that you have calculated, this keeps things cool-ER. I also feel that a small fan to cool things further might not be a bad idea!

What I mean is, if you have calculated that your maximum dissapation is say 40 watts, design a resistor network that will handle 80 watts....at the same resistance of course, but using more resistors with a higher ohmic values, but the same wattage.

Going for a regulator design would be much, much better!

RS Components have units from 25W to 500W.

A little clarification. I'm referring to rheostats here. RS Components has rheostats from 25W to 500W.

Ok, Obviously, I have a number of miss understandings. First, I have redrawn your schematic in a way that I am familiar with, That is, where positive is at the top of the page, and negative is at the bottom- similar to many automotive wiring diagram formats. The only "loads" in my diagram are the resistors. And, when I plug into my formula at the top of the page the source voltage, 12 volts, and then the resistances supplied, I can calculate amperage loads on those resistors. I italicize, because, at this point, I am confused as to how the resistor changes the voltage at the two (?)s, knowing that amperage draw must decline with decreased voltage. And, I cannot plug into the formula any figures that make sense to me regarding the voltage reduction. At this point, you will may say- "look at my schematic- it is simply stated there" . I am sorry that I do not understand your schematic, yet have tried to plug in your supplied resistance figures to make sense to me.. My first assumption is that you used two resistors to describe the high and low parameters of the Single potentiometer that I initially asked for. Then, as I saw that the bottom of you schematic is labeled "0 v" , a diagram that I am not able to decipher. Another assumption is that the arrow is the wiper contact in a pot where the accumulated resistance is measured. (standard) However, as you are more educated in this area than I, could you look at the suppositions that I have outlined in my schematic, and re-arrange- or describe what I have left out- or added incorrectly?

If you have given me up as a "lost cause" just say OPT OUT- and I will go back to school, Thanks for your help so far- ODD Parts

I have a calculations and equations for your problem but I don't know how to attach MsExcel table. Can some one show me.

This is what it looks like if cut&pase is done. Messy! sorry to show you

# V R1 R2 I =V/(R1+r2) P1=I*I*R1 P2=I*I*R2 V1=I*R1 V2=I*R2 Chk V=V1+V2 1 12 100 1000 0.01090909 0.011901 0.119008 1.090909 10.90909 12 2 12 100 500 0.02 0.04 0.2 2 10 12 3 12 100 250 0.03428571 0.117551 0.293878 3.428571 8.571429 12 4 12 100 0 0.12 1.44 0 12 0 12 5 12 4 0 3 36 0 12 0 12

There's something wrong with your drawing there. One end of your potentiometer is hanging or unconnected. It's supposed to be connected to ground. In my drawing, the 0V corresponds to ground.

I've been going back over the posts and I've come to the realization that this scheme might not work as it is now. I need to ask a few questions:

What, exactly, are you planning to do with this? If you are connecting the output to a motor, we need to do it differently. The motor also represents a resistance and it's in parallel with the lower resistor. We'll have to recalculate if this is what you intend to do.

If you're going to use this circuit for sending a signal, it might work as I drew it. You'll have to give more info. If it's a secret, you can send me a personal message by clicking on my name and then click on "Send Vulcan a message".

Vulcan is correct to caution about the current draw. For a hobby project I have a large "power" potentiometer I bought cheap on Ebay. It's nearly 6" in diameter and weighs about three pounds. I later learned to use voltage regulator IC's as someone else recommended. But the pots are more intuitive as a learning tool. It's a good lesson when you let the smoke out of them.

Go to the auto parts store and get a dash light dimmer pot. It is often built into the headlight switch but is available as a separate control. This will handle a fairly large load. The actual output voltage for any given position of the pot will vary depending on the load, so you may want to incorporate a volt meter in the circuit to know what voltage you actually have. -- JHF