<p>A more powerful PSU only would solve the problem in two cases: Either your PSU is anemic and underpowered in the first place, or you want you'd separate the bed's power supply from the rest of the machine - by using a higher <strong>Voltage</strong> for the bed. This would however need you to regulate the heating by having the board control not the bed directly but, control a (Normally Open for safety!) MOSFET, which in turn throttles the power to the bed.</p>
<p>In that case, you can use the resistance R of the bed with whatever voltage your alternate PSU provides to get the power that is turned into heat from the bed using <span class="math-container">$P_\text{bed}=\frac{U_\text{bed}^2}{R_\text{bed}}$</span>. Our MOSFET can regulate the power that is turned into heat in the bed as it is a Variable Resistance: The total potential differential stays the same, but the voltage available to the heated bed is governed by the resistance of the bed and the MOSFET's resistance. Since the two are in line, they have the same Current flowing through them:</p>
<p><span class="math-container">$$U_\text{supply}=U_\text{bed}+U_\text{MOSFET}=I_\text{total}\times(R_\text{bed}+R_\text{MOSFET})$$</span></p>
<p>That results in what is commonly called a Voltage Divider: the voltage that is available for the bed comes from a derivate of that: <span class="math-container">$$U_\text{bed}=U_\text{supply}(\frac{R_\text{bed}}{R_\text{MOSFET}+R_\text{bed}})$$</span></p>
<h2>Why the hazzle?</h2>
<p>Often, a board also might have a potentiometer for each power exit, and these are generally nothing else but variable resistances - and give us the same effect as a MOSFET for controlling the voltage available to a bed. If available, turning the Bed-Potentiometer a tiny bit will provide just a little higher voltage to the bed and allow slightly faster heating.</p>

CC BY-SA 4.0
2021-04-13T18:46:26.470Z
2021-04-13T20:13:47.500Z
2021-04-13T20:13:47.500Z
Trish
<p>May I recommend an alternative approach, which does not require any change of hardware?  The time required to heat the bed is not huge, so either via USB from your computer or from the front panel,  instruct your printer to heat the bed first, while you're setting everything else up (loading gcode files, changing filaments, or whatever).  This way tasks are completed in parallel.</p>

CC BY-SA 4.0
2018-05-03T14:39:06.843Z
I am using a preheat custom function in marlin which does the job +1
profesor79
CC BY-SA 4.0
2018-05-03T19:08:53.913Z
Insulation at the bottom works also, also putting a piece of cork on top during pre-heat and remove just before printing
0scar
CC BY-SA 4.0
2018-05-04T06:10:37.160Z
The question at hand is whether getting a higher-rated PSU would speed up heating of the heated bed. This - while helpful - doesn't answer that.
Tom van der Zanden
CC BY-SA 4.0
2018-05-03T14:35:53.120Z
2018-05-03T14:35:53.120Z
Carl Witthoft
<p>I glued (high temperature silicone) an isolation (cork 5-8mm) on the bottom side of my heatbed. It avoids loosing heat thru the bottom side. Effect: minimal faster heatup and less energy consuming over the time of use.</p>

CC BY-SA 4.0
2018-05-04T06:10:41.493Z
The question at hand is whether getting a higher-rated PSU would speed up heating of the heated bed. This - while helpful - doesn't answer that.
Tom van der Zanden
CC BY-SA 4.0
2018-05-04T08:06:05.800Z
this is a correct answer @TomvanderZanden as it focuses not only on the tiny area given by OP, but also provides an alternative way to the solution.
profesor79
CC BY-SA 4.0
2018-05-04T08:49:36.897Z
@profesor79 The question is "Would using a higher amperage power supply result in faster heat up times?". This answer doesn't *also* provide an alternative solution, it provides *only* an alternative solution. The question is very specific - the person asking the question wants to know whether upgrading the power supply makes sense - and this does not address this at all.
Tom van der Zanden
CC BY-SA 4.0
2018-05-04T09:33:43.897Z
@TomvanderZanden the thing is that we can be closed to what it says, or be open to what it means.
profesor79
CC BY-SA 4.0
2018-08-01T22:27:56.337Z
i did the same think with kapton tape on the bottom.  it cut about 15 seconds off the heat up time.
EvilTeach
CC BY-SA 4.0
2018-05-04T05:08:13.737Z
2018-05-04T05:08:13.737Z
Horitsu
<p>Also a thing that helped to preheat faster: make sure no draft is cooling down bed. It sounds obvious but cold-end fan and window draft, even psu fan draft can contribute significantly to preheat time. Eliminating draft source changed bed preheat time from 10 down to 5 minutes in my case...</p>

CC BY-SA 4.0
2018-05-04T06:10:48.080Z
The question at hand is whether getting a higher-rated PSU would speed up heating of the heated bed. This - while helpful - doesn't answer that.
Tom van der Zanden
CC BY-SA 4.0
2018-05-03T18:31:45.123Z
2018-05-03T18:31:45.123Z
Const
<p>I hate to sound like the Toolman Tayler from that old TV show. &quot;But, it comes down to More Power!&quot;
Power is the ability to do work (move a mass certain distance) within certain amount of time.
Power =  mass x distance x Time.</p>
<p>It can also be expressed in electrical terms, as the ability to heat something within certain amount of time.
Power = Voltage x Current</p>
<p>Since most systems have a fixed voltage, it is still possible to increase power by increasing the current, since,</p>
<p>Voltage = Resistance x Current,</p>
<p>And Power = Voltage x Current,</p>
<p>So, Power =  (Resistance x Current) x Current.</p>
<p>So by switching a power supply with the same voltage, but higher current, it will provide additional power to the system. The larger current available would be able to flow through the heating elements, heating them up faster.</p>
<p>However the caveat will be the amount of heat dissipation in the system, the large surface of the bed, will carry away enough of the heat due to air convection, that it may not make much of a difference. Or perhaps the heating element may not handle the larger amount of current and burn out.</p>
<p>It would be worth testing it out, in my humble opinion. Hopefully without causing a fire somewhere. :-)</p>

CC BY-SA 4.0
2021-04-13T17:25:09.053Z
I'm sorry but this completely wrong, the current is the result of the voltage over a resistance, you cannot choose the current by choosing a power supply that has a higher output current... So, if you don't change the voltage (increase it) or change the resistance (decrease it by replacing the bed with a lower resistor heating element), the current will not change.
0scar
CC BY-SA 4.0
2021-04-13T17:36:48.607Z
I am not quite sure, what you mean by that, if what you wrote was correct, there would be no need to provide optional power supplies with higher wattage. And yet, there they are available many 24 VDC power supplies with 300 Watts, 500, 750, 1000 Watts, etc... Their specific purpose is to provide higher current, and satisfy higher loads.
Gustavo Vargas
CC BY-SA 4.0
2021-04-13T17:42:53.520Z
Indeed, there are different PSU's with different output powers, the higher the power the higher the current the PSU **can** deliver, **but**, if you *do not change something about the bed* (either the voltage or the resistance) a higher rated PSU will *not give the bed more current*. Current is *not something you choose*, it is *something you get* from applying a voltage over a resistor, it is basic electronics.
0scar
CC BY-SA 4.0
2021-04-13T17:46:49.990Z
Actually... the main problem is not the PSU... it's **the board**. The board acts as a throttle to any electricity flowing through it. It can only take a very distinct combination or risk burning out. You can only circumvent that by using a separate switch.
Trish
CC BY-SA 4.0
2021-04-13T17:56:27.240Z
as for the physics: calculate the other side please: The bed has a Resistance R and is rated for Voltage U. $U/R=I$, It's rated power is given for a standard $P=U I$ setup, but actually calculated using $P=\frac U R U = \frac {U^2} R$. The Resistance of the bed is a fixed value but Voltage can be easily modified - as in adding an extra resistance or removing it. THAT is what a potentiometer is: a *throttle* for the Voltage.
Trish
CC BY-SA 4.0
2021-04-13T18:00:27.563Z
That's true! The current is and can be throttle using a potentiometer. The Power (Wattage) available is the limiting factor.
Gustavo Vargas
CC BY-SA 4.0
2021-04-13T18:11:55.230Z
NO! Potentiometers are nothing but a variable value resistor. The result is a Voltage divider: You do change, without adding any load to the potentiometer, **U**, not **I**. If your PSU is anemic, then yes, a higher rated PSU works, but if you only provide a constant 12 Volt, then 144 divided by *the resistance of the whole setup* is what the PSU's Current supply is.
Trish
CC BY-SA 4.0
2021-04-13T18:20:15.967Z
By reducing the resistance (using a higher power heating element), it will increase the current consumption. If the power supply can't supply enough current, you can't increase the temperature. That's why you need a higher wattage PSU.
Gustavo Vargas
CC BY-SA 4.0
2021-04-13T18:31:33.553Z
But ONLY if that is your limiting factor (underpowered PSU). Usually, it's the heating element itself.
Davo
CC BY-SA 4.0
2021-04-13T18:51:17.933Z
Using a Ender 6 as example, they come with a 24 VDC board/system. So the voltage is fixed, the PSU is a chinese brand, 15 A, 300 Watt PSU.
Gustavo Vargas
CC BY-SA 4.0
2021-04-13T14:57:04.320Z
2021-04-13T16:39:23.543Z
2021-04-13T16:39:23.543Z
Gustavo Vargas
<p>Changing the PSU with one with a higher amperage will <em><strong>not</strong></em> make the bed heat up any faster unless the PSU is underrated for the amperage required and the voltage is dropping as a result of the load. This can be checked by measuring the output voltage with a multimeter (when the PSU is loaded e.g. by a heating heat bed). In this case, the PSU has a marginal higher Amperage than the printer consumes (even has some room for the over-voltage; <em><strong>under the assumption that it is a good working PSU</strong></em>). Increasing the voltage will decrease the heat up time. There is a screw next to the 12 V connectors that can change the output voltage of the PSU. Usually, it is safe to increase the voltage up to 14 V, but that depends on your whole setup (and 14 V is applied to the whole setup, increasing the current for <strong>all</strong> parts, including your printer controller board, this board must be rated for the 14 V).  <strong>Please do check the stability of the voltage during load.</strong></p>
<p>Although it can be done, it is not something I used. What is an extra minute on a print of several hours?</p>
<p>You can do the math: say the heat bed has a resistance of 1.2 Ω. We only need two formulas:</p>
<ul>
<li><span class="math-container">$U=R\times I$</span> - potential Difference <em><strong>U</strong></em> is Resistance <em><strong>R</strong></em> times Current <em><strong>I</strong></em></li>
<li><span class="math-container">$P=U\times I=I^2\times R=\frac {U^2} R$</span>. The power <em><strong>P</strong></em> of an item the potential difference times the current through the item.</li>
</ul>
<ul>
<li>at <strong>12 V</strong> that will draw 10 Amps (12 V / 1.2 Ω) resulting in a 120 Watt bed: <span class="math-container">$P= 12^2 \text V \times 10^2 \text A= {10^2 \text A}\times {1.2\ \Omega}=\frac{12^2 \text V} {1.2\ \Omega} $</span>),</li>
<li>at <strong>14 V</strong> that same bed will draw 11.7 Amps (14 V / 1.2 Ω) resulting in a 163.3 Watt bed.</li>
</ul>
<p><em>Use at your own risk!</em></p>
<p>What you could do to decrease time to heat the bed without changing the PSU or the voltage is to insulate the bottom of the heat bed with heat bed cotton sheets or cork (placemats from IKEA ;) ), put a sheet of cork onto the heat bed before printing and start heating the bed through the LCD panel of the printer or any attached printer controller programs over USB prior to printing.</p>

CC BY-SA 4.0
2018-05-03T07:29:37.910Z
`Changing the PSU with one with a higher amperage will not make the bed heat up any faster. ` - unless the PSU is underrated for the amperage required and the voltage is dropping as a result of that. This can be easily verified with a multimeter (measure the voltage coming out of the PSU with the bed on and with the bed off). Increasing the voltage has a number of other benefits (the hotend heats up faster too, and the stepper will have a higher top speed).
Tom van der Zanden
CC BY-SA 4.0
2018-05-03T08:38:05.027Z
Unfortunately, I had to lower back the voltage on that kind of psu, as the voltage was not stable and temperature readings on thermistors were jumping up and down by 4 to 6 degrees. So this is something that you need to take into account.
profesor79
CC BY-SA 4.0
2018-05-03T06:38:48.260Z
2021-04-13T18:09:56.650Z
2021-04-13T18:09:56.650Z
0scar
<p>Currently I am using a 12 volts, 20 amperes power supply (Model S-240-12) </p>

<p><a href="https://i.sstatic.net/fraEVm.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fraEVm.jpg" alt="enter image description here"></a></p>

<p>The stepper motors and the extruder need 5 amp, and the heated bed build plate needs 11 amp. </p>

<p><a href="https://i.sstatic.net/fXymv.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/fXymv.jpg" alt="enter image description here"></a></p>

<p>Technically you only need to use a 12 Volts, 16 amperes power supply, but I understand that you use the one with 20 amp because pulse currents from extruders and stepper motors can be stressful to supplies loaded to the max, so for reliability and performance, it is better to use a supply rated for 25% more than you need </p>

<p>In the place where I buy the spare parts for my 3d printer they also sell 12 V power supplies capable of delivering 25 amp and 30 amp and they told me that if I use those ones you are going to be able to speed up the heating of the heated bed. Is that true? I understand that the heated bed is only going to take the 11 amp that it needs so is not going to make any difference to use power supplies capable of delivering more current</p>

CC BY-SA 4.0
2018-05-03T08:52:26.993Z
are you trying to print ABS materials? What temperature is required?
profesor79
CC BY-SA 4.0
2018-05-06T00:58:16.850Z
you could add a 2nd power supply that outputs a higher voltage, and use the old 12v to power a relay or switch a mosfet which controls the higher voltage.
dandavis
<heated-bed><ramps-1.4><switching-power-supply>
Speeding up the heating of the heated bed
Nau
# Speeding up the heating of the heated bed

(Asked by: username_0 on 2018-05-02)

Currently I am using a 12 volts, 20 amperes power supply (Model S-240-12)

[![enter image description here](https://i.sstatic.net/fraEVm.jpg)](https://i.sstatic.net/fraEVm.jpg)

The stepper motors and the extruder need 5 amp, and the heated bed build plate needs 11 amp.

[![enter image description here](https://i.sstatic.net/fXymv.jpg)](https://i.sstatic.net/fXymv.jpg)

Technically you only need to use a 12 Volts, 16 amperes power supply, but I understand that you use the one with 20 amp because pulse currents from extruders and stepper motors can be stressful to supplies loaded to the max, so for reliability and performance, it is better to use a supply rated for 25% more than you need

In the place where I buy the spare parts for my 3d printer they also sell 12 V power supplies capable of delivering 25 amp and 30 amp and they told me that if I use those ones you are going to be able to speed up the heating of the heated bed. Is that true? I understand that the heated bed is only going to take the 11 amp that it needs so is not going to make any difference to use power supplies capable of delivering more current

**Tags:** heated-bed, ramps-1.4, switching-power-supply
> **Comments:**
> **username_1**: are you trying to print ABS materials? What temperature is required?  
> **username_2**: you could add a 2nd power supply that outputs a higher voltage, and use the old 12v to power a relay or switch a mosfet which controls the higher voltage.  

---

## 6 answers

### Answer by username_5 on 2018-05-03. Score: 7 | Accepted answer

Changing the PSU with one with a higher amperage will ***not*** make the bed heat up any faster unless the PSU is underrated for the amperage required and the voltage is dropping as a result of the load. This can be checked by measuring the output voltage with a multimeter (when the PSU is loaded e.g. by a heating heat bed). In this case, the PSU has a marginal higher Amperage than the printer consumes (even has some room for the over-voltage; ***under the assumption that it is a good working PSU***). Increasing the voltage will decrease the heat up time. There is a screw next to the 12 V connectors that can change the output voltage of the PSU. Usually, it is safe to increase the voltage up to 14 V, but that depends on your whole setup (and 14 V is applied to the whole setup, increasing the current for **all** parts, including your printer controller board, this board must be rated for the 14 V). **Please do check the stability of the voltage during load.**

Although it can be done, it is not something I used. What is an extra minute on a print of several hours?

You can do the math: say the heat bed has a resistance of 1.2 Ω. We only need two formulas:

* $U=R\times I$ - potential Difference ***U*** is Resistance ***R*** times Current ***I***
* $P=U\times I=I^2\times R=\frac {U^2} R$. The power ***P*** of an item the potential difference times the current through the item.

* at **12 V** that will draw 10 Amps (12 V / 1.2 Ω) resulting in a 120 Watt bed: $P= 12^2 \text V \times 10^2 \text A= {10^2 \text A}\times {1.2\ \Omega}=\frac{12^2 \text V} {1.2\ \Omega} $),
* at **14 V** that same bed will draw 11.7 Amps (14 V / 1.2 Ω) resulting in a 163.3 Watt bed.

*Use at your own risk!*

What you could do to decrease time to heat the bed without changing the PSU or the voltage is to insulate the bottom of the heat bed with heat bed cotton sheets or cork (placemats from IKEA ;) ), put a sheet of cork onto the heat bed before printing and start heating the bed through the LCD panel of the printer or any attached printer controller programs over USB prior to printing.

> **Comments:**
> **username_6**: `Changing the PSU with one with a higher amperage will not make the bed heat up any faster. ` - unless the PSU is underrated for the amperage required and the voltage is dropping as a result of that. This can be easily verified with a multimeter (measure the voltage coming out of the PSU with the bed on and with the bed off). Increasing the voltage has a number of other benefits (the hotend heats up faster too, and the stepper will have a higher top speed).  
> **username_1**: Unfortunately, I had to lower back the voltage on that kind of psu, as the voltage was not stable and temperature readings on thermistors were jumping up and down by 4 to 6 degrees. So this is something that you need to take into account.  

---

### Answer by username_3 on 2021-04-13. Score: 1

A more powerful PSU only would solve the problem in two cases: Either your PSU is anemic and underpowered in the first place, or you want you'd separate the bed's power supply from the rest of the machine - by using a higher **Voltage** for the bed. This would however need you to regulate the heating by having the board control not the bed directly but, control a (Normally Open for safety!) MOSFET, which in turn throttles the power to the bed.

In that case, you can use the resistance R of the bed with whatever voltage your alternate PSU provides to get the power that is turned into heat from the bed using $P\_\text{bed}=\frac{U\_\text{bed}^2}{R\_\text{bed}}$. Our MOSFET can regulate the power that is turned into heat in the bed as it is a Variable Resistance: The total potential differential stays the same, but the voltage available to the heated bed is governed by the resistance of the bed and the MOSFET's resistance. Since the two are in line, they have the same Current flowing through them:

$$U\_\text{supply}=U\_\text{bed}+U\_\text{MOSFET}=I\_\text{total}\times(R\_\text{bed}+R\_\text{MOSFET})$$

That results in what is commonly called a Voltage Divider: the voltage that is available for the bed comes from a derivate of that: $$U\_\text{bed}=U\_\text{supply}(\frac{R\_\text{bed}}{R\_\text{MOSFET}+R\_\text{bed}})$$

## Why the hazzle?

Often, a board also might have a potentiometer for each power exit, and these are generally nothing else but variable resistances - and give us the same effect as a MOSFET for controlling the voltage available to a bed. If available, turning the Bed-Potentiometer a tiny bit will provide just a little higher voltage to the bed and allow slightly faster heating.


---

### Answer by username_4 on 2018-05-03. Score: -1

May I recommend an alternative approach, which does not require any change of hardware? The time required to heat the bed is not huge, so either via USB from your computer or from the front panel, instruct your printer to heat the bed first, while you're setting everything else up (loading gcode files, changing filaments, or whatever). This way tasks are completed in parallel.

> **Comments:**
> **username_1**: I am using a preheat custom function in marlin which does the job +1  
> **username_5**: Insulation at the bottom works also, also putting a piece of cork on top during pre-heat and remove just before printing  
> **username_6**: The question at hand is whether getting a higher-rated PSU would speed up heating of the heated bed. This - while helpful - doesn't answer that.  

---

### Answer by username_7 on 2018-05-04. Score: -1

I glued (high temperature silicone) an isolation (cork 5-8mm) on the bottom side of my heatbed. It avoids loosing heat thru the bottom side. Effect: minimal faster heatup and less energy consuming over the time of use.

> **Comments:**
> **username_6**: The question at hand is whether getting a higher-rated PSU would speed up heating of the heated bed. This - while helpful - doesn't answer that.  
> **username_1**: this is a correct answer @TomvanderZanden as it focuses not only on the tiny area given by OP, but also provides an alternative way to the solution.  
> **username_6**: @profesor79 The question is "Would using a higher amperage power supply result in faster heat up times?". This answer doesn't *also* provide an alternative solution, it provides *only* an alternative solution. The question is very specific - the person asking the question wants to know whether upgrading the power supply makes sense - and this does not address this at all.  
> **username_1**: @TomvanderZanden the thing is that we can be closed to what it says, or be open to what it means.  
> **username_8**: i did the same think with kapton tape on the bottom.  it cut about 15 seconds off the heat up time.  

---

### Answer by username_9 on 2018-05-03. Score: -3

Also a thing that helped to preheat faster: make sure no draft is cooling down bed. It sounds obvious but cold-end fan and window draft, even psu fan draft can contribute significantly to preheat time. Eliminating draft source changed bed preheat time from 10 down to 5 minutes in my case...

> **Comments:**
> **username_6**: The question at hand is whether getting a higher-rated PSU would speed up heating of the heated bed. This - while helpful - doesn't answer that.  

---

### Answer by username_10 on 2021-04-13. Score: -4

I hate to sound like the Toolman Tayler from that old TV show. "But, it comes down to More Power!"
Power is the ability to do work (move a mass certain distance) within certain amount of time.
Power = mass x distance x Time.

It can also be expressed in electrical terms, as the ability to heat something within certain amount of time.
Power = Voltage x Current

Since most systems have a fixed voltage, it is still possible to increase power by increasing the current, since,

Voltage = Resistance x Current,

And Power = Voltage x Current,

So, Power = (Resistance x Current) x Current.

So by switching a power supply with the same voltage, but higher current, it will provide additional power to the system. The larger current available would be able to flow through the heating elements, heating them up faster.

However the caveat will be the amount of heat dissipation in the system, the large surface of the bed, will carry away enough of the heat due to air convection, that it may not make much of a difference. Or perhaps the heating element may not handle the larger amount of current and burn out.

It would be worth testing it out, in my humble opinion. Hopefully without causing a fire somewhere. :-)

> **Comments:**
> **username_5**: I'm sorry but this completely wrong, the current is the result of the voltage over a resistance, you cannot choose the current by choosing a power supply that has a higher output current... So, if you don't change the voltage (increase it) or change the resistance (decrease it by replacing the bed with a lower resistor heating element), the current will not change.  
> **username_10**: I am not quite sure, what you mean by that, if what you wrote was correct, there would be no need to provide optional power supplies with higher wattage. And yet, there they are available many 24 VDC power supplies with 300 Watts, 500, 750, 1000 Watts, etc... Their specific purpose is to provide higher current, and satisfy higher loads.  
> **username_5**: Indeed, there are different PSU's with different output powers, the higher the power the higher the current the PSU **can** deliver, **but**, if you *do not change something about the bed* (either the voltage or the resistance) a higher rated PSU will *not give the bed more current*. Current is *not something you choose*, it is *something you get* from applying a voltage over a resistor, it is basic electronics.  
> **username_3**: Actually... the main problem is not the PSU... it's **the board**. The board acts as a throttle to any electricity flowing through it. It can only take a very distinct combination or risk burning out. You can only circumvent that by using a separate switch.  
> **username_3**: as for the physics: calculate the other side please: The bed has a Resistance R and is rated for Voltage U. $U/R=I$, It's rated power is given for a standard $P=U I$ setup, but actually calculated using $P=\frac U R U = \frac {U^2} R$. The Resistance of the bed is a fixed value but Voltage can be easily modified - as in adding an extra resistance or removing it. THAT is what a potentiometer is: a *throttle* for the Voltage.  
> **username_10**: That's true! The current is and can be throttle using a potentiometer. The Power (Wattage) available is the limiting factor.  
> **username_3**: NO! Potentiometers are nothing but a variable value resistor. The result is a Voltage divider: You do change, without adding any load to the potentiometer, **U**, not **I**. If your PSU is anemic, then yes, a higher rated PSU works, but if you only provide a constant 12 Volt, then 144 divided by *the resistance of the whole setup* is what the PSU's Current supply is.  
> **username_10**: By reducing the resistance (using a higher power heating element), it will increase the current consumption. If the power supply can't supply enough current, you can't increase the temperature. That's why you need a higher wattage PSU.  
> **username_11**: But ONLY if that is your limiting factor (underpowered PSU). Usually, it's the heating element itself.  
> **username_10**: Using a Ender 6 as example, they come with a 24 VDC board/system. So the voltage is fixed, the PSU is a chinese brand, 15 A, 300 Watt PSU.  

---

