<p>It is all about resistance. </p>

<p>This requires some formulae:</p>

<p><span class="math-container">$U = Voltage$</span><br>
<span class="math-container">$I = Current$</span><br>
<span class="math-container">$R = Resistance$</span><br>
<span class="math-container">$P = Power$</span><br></p>

<p><span class="math-container">$U = I \times R$</span><br></p>

<p><span class="math-container">$P = U \times R$</span><br></p>

<p><span class="math-container">$ R = \dfrac{P}{I^2} = \dfrac{U^2}{P} $</span><br></p>

<p>The 12&nbsp;V, 40&nbsp;W cartridge has a resistance of about 3.6&nbsp;&ohm;.</p>

<p>If you use this cartridge at 24&nbsp;V, this caculates to a power of 160&nbsp;W!</p>

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

CC BY-SA 4.0
2020-04-02T22:27:33.937Z
2020-04-03T10:07:28.093Z
2020-04-03T10:07:28.093Z
0scar
<p>So, I have created my own heater block for my 3D printer and mistakenly was using a 12&nbsp;V, 40&nbsp;W heater cartridge thinking that was the standard for my Ender 3, when in fact it is supplied with a 24&nbsp;V, 40&nbsp;W.</p>

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

<pre><code>PID Autotune failed! Temperature too high
</code></pre>

<p>Graph of Overshoot:</p>

<p><a href="https://i.sstatic.net/NmPFA.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/NmPFA.jpg" alt="PID Autotune fail graph"></a></p>

<p>I was unable to remove this overshoot even through manual tuning.</p>

<p>Now I have switched from the 12&nbsp;V cartridge to the 24&nbsp;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.</p>

<p>However, I am unsure why this is the case? Can anyone explain why the 12&nbsp;V heater cartridge results in too high of a temperature increase?</p>

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

<creality-ender-3><hotend><pid>
PID autotune fails 'Temp too high' with 12 V heater cartridge but works with 24 V?
FEA42
# PID autotune fails 'Temp too high' with 12 V heater cartridge but works with 24 V?

(Asked by: username_0 on 2020-04-02)

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](https://i.sstatic.net/NmPFA.jpg)](https://i.sstatic.net/NmPFA.jpg)

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?

**Tags:** creality-ender-3, hotend, pid


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## 1 answers

### Answer by username_1 on 2020-04-02. Score: 1 | Accepted answer

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 &ohm;.

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.


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