So, I have created my own heater block for my 3D printer and mistakenly was using a 12 V, 40 W heater cartridge thinking that was the standard for my Ender 3, when in fact it is supplied with a 24 V, 40 W.

You may have seen my earlier post about when using autotune the temperature would overshoot by a large margin and returned the following error.

PID Autotune failed! Temperature too high

Graph of Overshoot:

PID Autotune fail graph

I was unable to remove this overshoot even through manual tuning.

Now I have switched from the 12 V cartridge to the 24 V this problem is resolved, and I can now run the autotune. It follows a much more gradual curve when heating up so doesn't trigger the same error.

However, I am unsure why this is the case? Can anyone explain why the 12 V heater cartridge results in too high of a temperature increase?

It seems counterintuitive to me as I would have thought the higher voltage cartridge would heat up faster as opposed to vice versa?


It is all about resistance.

This requires some formulae:

$U = Voltage$
$I = Current$
$R = Resistance$
$P = Power$

$U = I \times R$

$P = U \times R$

$ R = \dfrac{P}{I^2} = \dfrac{U^2}{P} $

The 12 V, 40 W cartridge has a resistance of about 3.6 Ω.

If you use this cartridge at 24 V, this caculates to a power of 160 W!

This means that there is an enormous influx of heat that is hard to control, hence the overshoot.

| improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.