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I have a 10k poti. I use it as a variable resistor by soldering two of the three terminals. It is connected to a 1/4 watt 220 ohm resistor in series(this is the load). The voltage across this circuit is 12V DC.

Is the poti safe even it is set to zero ohm? Would 220 ohm prevent it to burn in this case? If so, what should be the quantative method to calculate the minimum value of series resistor which would prevent this poti?

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  • \$\begingroup\$ a variable resistor is a potientiometer potientiometer is the former name for it \$\endgroup\$ – Old_Fossil May 13 '16 at 0:56
  • \$\begingroup\$ "A potentiometer, informally a pot, is a three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. If only two terminals are used, one end and the wiper, it acts as a variable resistor or rheostat." wikipedia \$\endgroup\$ – user16307 May 13 '16 at 0:59
  • \$\begingroup\$ I believe I said that. A rheostat is the wire wound high power version. \$\endgroup\$ – Old_Fossil May 13 '16 at 1:06
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If you know the power rating for your potentiometer, you can start with that as a guide as to how much current you can pass through it. Little trimpots adjusted by screwdriver or mini ones that you turn with your fingers are often 0.5 watt, but may not be so you'd best check.

The power rating for a potentiometer tells you how much power it is rated to dissipate if the current passes along the entire resistance. For example, say the power rating is 0.5 Watts for your 10k pot. Then, the current that would flow to generate that power:

$$ P = I^2R $$ Rearranging... $$ I = \sqrt{P/R} = \sqrt{0.5 / 10000} = 0.0071 A $$

This is the most current that should be allowed to flow through the 10k 0.5 watt potentiometer at any position.

Let's go to ohm's law and see what series resistor can prevent that:

$$ R = V/I = 12 / 0.0071 = 1690 \Omega $$

Substitute the actual power of your potentiometer, and maybe use one a bit bigger to be safe, and note that it could heat up a fair amount if used at the rated power.

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    \$\begingroup\$ But if I use pot with 220 ohm and turn the pot knob to zero. the pot will be a conductor wire. 220 ohm might be in danger but why pot? Is pot in danger around zero but not at zero? \$\endgroup\$ – user16307 May 13 '16 at 0:57
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    \$\begingroup\$ Does 0.5 watt potentiometer mean, it should not dissipate more than 0.5 watt at any resistor value? \$\endgroup\$ – user16307 May 13 '16 at 1:02
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    \$\begingroup\$ Probably in danger a bit away from zero but not at zero, but I've never tried to operate one like that, maybe someone else would know better. Right at zero, the metal of the pin at the end of the resistive track is there to pull heat out of the track, so near zero might very well be ok. There is probably a better way to accomplish what you're trying to do, without having to worry about this however. Can you give a bit more info on what you're trying to achieve, and maybe someone might have a suggestion that avoids the potentiometer or adds something to make sure it doesn't burn up? \$\endgroup\$ – alexwarrior May 13 '16 at 1:08
  • \$\begingroup\$ The power rating is for the full resistor value, so for the full 10 kiloohms. I.e. the current passing along the entire resistive track inside the potentiometer. Like this, there is the most surface area from which heat can escape through. If you were to dissipate the same amount of power through a smaller part of the resistive track, it would heat up more in that section of the track. \$\endgroup\$ – alexwarrior May 13 '16 at 1:11
  • \$\begingroup\$ Im confused so the power rating changes by turning the knob. So you say the current shouldn't exceed in any case 0.0071A? But if the pot is set to 5k it cannot pass 0.0071A safely because it cannot dissipate as in 10k case? \$\endgroup\$ – user16307 May 13 '16 at 1:17
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A good rule of thumb is that the maximum wiper current (if not stated on the datasheet) is the lesser of the current when passed through the entire element would reach the maximum power dissipation for the pot and 100mA.

So if you have a 10K 1W pot, then the maximum current would be 10mA, and the minimum resistor value that would be safe (at 12V) would be 1.2K.

That's a conservative figure suited for carbon and cermet pots- pots that designed for use as rheostats (eg. some wirewound types) can handle more current because they have a thermally conductive core attached to the element to spread the heat.

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  • \$\begingroup\$ thanks for explanation, i thought it is more easier. but sounds like it is not as straightforward as i was thinking. \$\endgroup\$ – user16307 May 13 '16 at 1:21
  • \$\begingroup\$ So the principle is one should not exceed 10mA for ant pot position? \$\endgroup\$ – user16307 May 13 '16 at 1:22
  • \$\begingroup\$ One should not exceed \$\sqrt{\frac{Pd_{MAX}}{R_{ELEMENT}}}\$ or 100mA, whichever is less, for any pot position. For a 1 W 10K pot that works out to 10mA. For a 0.5W 10K pot that works out to 7.1mA as @alexw points out. \$\endgroup\$ – Spehro Pefhany May 13 '16 at 11:38
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If you set it to \$0\Omega\$, then it is no different from a piece of wire.

So to ask your question another way, "is it safe to connect a \$220\Omega\$ resistor to a \$12\mathrm{V}\$ supply?". I'll let you answer that.

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  • \$\begingroup\$ I see, (12*12/220) 0.65 watt. So it is not good because it is greater than quarter watt?? but how about if there would be no series resistor? Around zero the power supply would damage? What is the minimum resistor for a 10k poti under 12V DC? Poti doesnt have any watt ratings on it. \$\endgroup\$ – user16307 May 13 '16 at 0:31
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    \$\begingroup\$ Yes, it is higher than the rated power for the resistor. Further, if there is no series resistor, you have \$0\Omega\$ across the power supply - me thinks that is not a good idea... \$\endgroup\$ – Tom Carpenter May 13 '16 at 0:36
  • \$\begingroup\$ Ok one last question. I have this 10k poti. Do potentiometers have their max watt ratings written on them normally? So I can calculate the min point by using V*V/Rpot \$\endgroup\$ – user16307 May 13 '16 at 0:49
  • \$\begingroup\$ No, they generally don't. If you have a manufacturer and part number you can look it up, otherwise you're stuck with looking at the specification for a similar-looking part. \$\endgroup\$ – nekomatic May 13 '16 at 8:08
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Let's use the power formula.

Power law

So in the circuit: enter image description here

The total current through the circuit is

P = (12^2)/220
P = 0.654545 Watts

Now all you need to do is check whether the 220Ω resistor and the 10KΩ Potentiometer can both handle the wattage (0.654545 Watts). As you said, the 220Ω resistor is 0.25 Watts rated so that will definitely blow up. If the potentiometer is not rated for that Wattage, they both blow up.


Now let's calculate what resistance you need to replace the 220Ω resistor. If the potentiometer has a higher wattage rating than the 220Ω resistor, it makes no difference.

I am going to assume the Potentiometer has a wattage rating of 0.1 W

Since the Potentiometer is 10k, max current is 3mA due to

I = root(P/R)
I = root(0.1/10000)
I = 0.003162A

We are going to replace the 220Ω with something better. Assuming your other resistors are rated 1/4 Watts, let's find the minimum required resistance at 12V

R = (V^2)/P
R = (12^2)/(1/4)
R = 144*4
R = 576Ω

That's the minimum resistance for the resistor. Let's see about the potentiometer.

R = V/I
R = 12/0.003162
R = 3794Ω

Jeez a high value...

Now although the resistor only requires about 600Ω for itself, the potentiometer requires a whopping 4K resistor, so you have to choose the higher one.

Now we don't want our resistor to be on the plain boundary of the maximum Wattage rating, so get like a 4KΩ resistor or more (1/4 watt rated). Also pay attention to the tolerance. For eg if the tolerance was 10%, then the resistor could be 3600Ω which is very dangerous.

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  • \$\begingroup\$ What if the pot was set to 5k? Can u use the same logic for the pot? According to some other answers the wattage of the poti depends on its resistance. They use current. \$\endgroup\$ – user16307 May 13 '16 at 6:31
  • \$\begingroup\$ Well if the pot is 5k, you got a larger region where the 220Ω resistor can blow up. Best solution is to change the 220Ω resistor to a larger value. Let me add to my question, and I'm going to assume the potentiometer max wattage is 1/4 Watts unless you specify otherwise. (It doesn't matter if the potentiometer has a higher wattage rating, cuz the 220Ω resistor will still blow up, but if it has lower wattage rating than 1/4 then specify.) \$\endgroup\$ – Bradman175 May 13 '16 at 6:40
  • \$\begingroup\$ Also 5k set to 0 position is still pretty much equivalent to a piece of wire, like a 10k or a 1M potentiometer set to 0 position. \$\endgroup\$ – Bradman175 May 13 '16 at 6:42
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    \$\begingroup\$ No forget bout the series value Im asking bout the poti. I think pot's wattage depends on its resistance. \$\endgroup\$ – user16307 May 13 '16 at 6:45
  • \$\begingroup\$ @user16307 yes you are right. It is about the current. Well I'm not sure about the current specifications though. \$\endgroup\$ – Bradman175 May 13 '16 at 6:52

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