For the particular case at hand, there is no reason to require all the resistors be the same wattage, as you can calculate the wattage needed for each step.
Yes, you can make a crude pot this way, but it's not a very practical solution unless you have a bunch of power resistors and a high-current rotory switch around.
Unless the original poster wants to go this route, I think this is confusing the issue.
Hello again,
I thought by now you would see the 'reason' but you dont seem to be getting the point. Just because you dont see a reason for something does not mean there is no reason
Take another look at what i had posted before, then try to relate that to the topic of this thread. You'll see the line of reasoning.
The resistors MUST be the same, because that is the way i specified them. If a company says they want all the resistors the same, you make them all the same. You might suggest that they dont have to do that, but once you've said it you dont have to imply that the resistors MUST be all different, because they dont want them all different they want them the same. One reason not related to this is that they might want to stock only one part number. One 10 ohm resistor at whatever wattage is required. But in any case, for my purpose and the purpose of this thread, the resistors must be all the same. Same value, same wattage.
If you followed my lead instead of shooting off on a tangent, you would have noticed that when we follow those specifications we end up with a 'pot' that has 10 indents, and this roughly approximates a regular pot with no indents, so what we find out about the power rating we can later apply to a regular continuous pot which was part of the subject of this thread. What is interesting is the pot rating seems to have to be higher than expected for a job like this one.
See next, we figure out the wattage rating of the 'pot' which would be the required rating of a pot *IF* we were to follow this rough procedure. Since a pot is made of a linear resistive element, the element is like that of 10 resistors in series, which is in the style of a finite element model of the resistance of a linear pot with 10 elements. We do this just to get the idea how the wattage of one section of the resistive element (one 10 ohm resistor in our case) affects the rating of the whole pot. We do this because it's harder to see the required rating of the pot when we look at the whole thing, but looking at one small section reveals that the wattage must be higher than expected.
For example, around the web some have suggested using a pot that is rated the same as the resistor. This would mean that we would go out and buy a 100 ohm pot rated for 10 watts (because the fixed resistor is rated for 10 watts assuming good cooling conditions). However, it appears that the 10 watt rating would not be enough, which is part of the point of this analysis. Using a finite element model of the pot helps to see the reason behind a higher rated pot.
To save time, i'll take it to the next step so as to illustrate the way this works...
The next step is to note that when we buy a pot, the linear element has a certain width and length. When we turn the pot shaft, we make the resistance smaller and the length shorter but the width stays the same. So what we do is we make the resistance less but we also make the surface area of the resistive element smaller because we are only using part of the element now. This smaller area has to be able to dissipate enough power to keep the pot from overheating.
This is approximated with 10 resistors, each 10 ohms. Since we found out that using just one of those resistors requires a power rating of 2.5 watts, that means all the resistors must be rated for 2.5 watts because the resistive element inside common pots does not vary in wattage along the length. Since we have 10 resistors, this adds up to 25 watts. Thus for that one small detent section to be able to handle the 2.5 watts, the pot must be rated for 25 watts. If we buy a 10 watt pot, that one small section will only be able to handle 1 watt so when we turn it to 10 percent of the travel it burns up that section of the pot element and renders the pot useless.
So you see how this works now? We dont really intend to build a pot with 10 resistors, we just want a model we can work with that reveals the reason for the higher pot rating requirement.
The next step after that would be to improve the finite element model by lowering the value of the first resistor down from 10 ohms to just 1 ohm. We then construct the 'pot' from 100 resistors each 1 ohm , then calculate the power requirement of some of the resistors and find the one with the max power. We then know that we need 100 times the power rating of that one max powered resistor.
Continuing with this idea, we let the value of the incremental resistance drop to zero as the number of resistors goes to infinity, and what we end up with is a more normal continuous pot and a calculated power rating.
But it's interesting enough just to look at the 10 ohm increment and the 1 ohm increment and see what becomes of the power rating of a pot that was made with those resistances in series. It comes out higher than most people seem to expect. When i look around the web i see lots of lower estimates which cant be correct, and the finite element model helps to explain this.
Try it and see what you get, it's very interesting to do it this way.
Of course it is usually more practical to use a PWM or even a transistor in the linear mode. Either way we loose power though so the efficiency isnt that great. I have several high power rheostats around but only use them for loads for testing different things.