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As far as I understood heating elements make use out of Joule heating. Hence, I wonder why one would preferably use materials with a negative temperature coefficient, therefore? Isn't it, that, the higher the resistance, the more warmth will be generated? So why would you want that the resistance decreases with temperature?

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(I am putting this forward as a opinion to be validated by others more knowledgeable than myself).

I am not certain that the resistance will have the effect the original OP expected. i.e. I assume that the author was thinking of the formula:

$$P = I^2 R$$

However, this is a derivation of the more basic $P=VI$, and in an alternative form: $$P=\frac{V^2}{R}$$

In the latter form, the smaller the resistance the higher the power. I expect that the latter form is more applicable because in most cases, the Voltage V is the constant (while the I is calculated from $I=\frac{V}{R}$. So larger resistance, means smaller current, which in turn produces heat. So, I would expect that a Negative temperature coefficient could be beneficial.

Of course there is also a limit imposed by the maximum power theorem, better known as the maximum power transfer theorem which in my simple interpretation states that the maximum power transfer occurs when source impedance is exactly matched to load impedance.

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    $\begingroup$ Opinion validated! $\endgroup$
    – Transistor
    Oct 25, 2021 at 11:37
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Hence, I wonder why one would preferably use materials with a negative temperature coefficient?

Most metals will have a positive temperature coefficient. Platinum, for example, is commonly used as a temperature sensor and the Pt100 has 100 Ω at 0°C and about 138 Ω at 100°C.

Isn't it, that, the higher the resistance, the more warmth will be generated? So why would you want that the resistance decreases with temperature?

No. You're forgetting that for a constant voltage supply that current will decrease with increasing resistance.

$$ P = VI = I^2R = \frac {V^2}R $$

You can see from the last part that for a given supply voltage that power increases as resistance decreases.

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It really depends. If you have multiple heating elements (or areas within a single element), in parallel, they will end up at slightly different temperatures. If the application is temperature limited, then the max total system power is determined by the "weakest-link" -- i.e. the hottest individual element.

If the elements are PTC, then hotter elements' increasing R automatically reduces their share of the overall power. It tends to equalize the per-element power. This is generally good.

If the elements are NTC, then the hotter elements' decreasing R will make them draw relatively more current, making them hotter, and this can make the temperature variation between elements more extreme, thus in effect lowering the overall system power that is allowed.

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  • $\begingroup$ I agree with your reasoning about the greater variation with ntc-- one small note/question though: the variation is greater to heating elements connected in parallel that have different currents. With heating elements connected in series the variation should be less evident.(?) $\endgroup$
    – NMech
    Oct 26, 2021 at 5:58
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    $\begingroup$ @NMech - Yes, I think so. The individual element power would be (Rn)(V^2/Rtotal^2). And if the system was current driven (which I think is less common) and series connection, it would be the other way around. power would be (I^2)(Rn), so NTC would be the self stabilizing case for that, and PTC would hotspot. $\endgroup$
    – Pete W
    Oct 26, 2021 at 6:45

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