<p>Yes, this method will work. Some (LED) power supplies have a small potentiometer that you can use to slightly adjust the voltage. On a 12 V printer, adjusting the supply from 12 V to 14 V will give a 36% increase in power.</p>
<p>Obviously there is a limit to how much you can increase the voltage. The voltage itself is unlikely to be the limiting factor, but there is a limit to how much heat the bed can dissipate.</p>
<blockquote>
<p>The PCB heater have 2-part copper and laminate. the reason of failire is the copper go off the board due to high temperature. In this case we can control temperature with firmware.</p>
</blockquote>
<p>This is true, but you have to consider that firmware is not infallible. If you increase the voltage (and thus wattage) of the bed a lot, you risk creating a dangerous situation if something fails. Ideally, you should choose the wattage of the bed such that it does not create a dangerous situation even if it is left on permanently by accident. In extreme cases (e.g. 220V to a nominally 12V heated bed) damage will be almost instantaneous before the firmware can intervene.</p>
<p>You also have to consider that increasing the voltage also increases the current. The wires need to be thick enough to handle the additional current. You also have to pay special attention to the MOSFET that is used to switch the bed; it also needs to be able to handle the current (and voltage, though this is usually less of a problem). Usually the integrated MOSFETs on 3D printer control boards can only handle about ~10 A which is what the bed might normally draw. Also pay attention to fuses and to screw terminal connections. They might not be able to handle the increased current either.</p>

CC BY-SA 4.0
2021-01-10T01:01:32.617Z
What about AC heated bed(Orange silicone pad)? Why it can heat much quicker?  I thought it’s had more power per area so I think add more voltage to PCB bed. And about safety My heated bed got about 100 degree centigrade when permanently connected to supply. If silicone pad got higher wattage what happens when we left them connected?
M lab
CC BY-SA 4.0
2021-01-10T11:31:24.993Z
@Mlab Centigrades is *hundreds of a grad(uation)* and an angular unit. It can also be a reference to any 100 graduation scale, for example Fahrenheit is a centigrade scale. While commonly referred to as a temperature, it is scientifically no longer accepted to speak of °C as centigrad: Temperature is commonly measured in Kelvin, or in the derivate scale K+273.15, aka Degrees Celsius. Do you mean "My heated bed got to about 100 °C / 373.15 K"? That would be beyond the rated temperature range for any heated bed for consumer products. Why do you need a heated bed that operates abve 80 °C?
Trish
CC BY-SA 4.0
2021-01-10T12:36:59.503Z
The reason the AC bed heats faster is indeed because it has more power (per area). If you leave it connected it will reach a higher temperature, roughly proportional to the power (i.e., 2x more power will result in 2x the temperature rise -- a 100 W bed might reach 100 °C (80 °C above ambient) then a 200 W bed will reach 180 °C).
Tom van der Zanden
CC BY-SA 4.0
2021-01-10T12:52:31.933Z
@TomvanderZanden we enforce SI formating on units, so we also should enforce talking about them in terms of SI units or derivate units.
Trish
CC BY-SA 4.0
2021-01-11T08:07:46.507Z
Sorry for making that confusing. I refer to Degree celcius.
@TomvanderZanden I know that for silicone pad the final temperature will higher because of higher power per area. but with your answer " Ideally, you should choose the wattage of the bed such that it does not create a dangerous situation even if it is left on permanently by accident" So  that mean silicone pad can handle higher temperature when silicone pad left connected right?
M lab
CC BY-SA 4.0
2021-01-11T08:49:12.957Z
@Mlab No, not necessarily. A lot of the silicone heaters will create a dangerous situation if left connected. With these, it is a good idea to include a thermal fuse.
Tom van der Zanden
CC BY-SA 4.0
2021-01-09T19:35:26.253Z
2021-01-09T19:35:26.253Z
Tom van der Zanden
<p>You can add a boost converter to the bed's power supply. Assuming that you are using an external MOSFET to control the bed (and you should). Verify what the MOSFET's max voltage rating is, then obtain a boost converter and configure it to give a voltage that is 10% less than the max. I once used this technique when driving a 24V hot end heater using a 12V power supply. Be forewarned, however, the boost converter I was using exploded during use one day.</p>
<p><a href="https://i.sstatic.net/3gAB6.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/3gAB6.png" alt="Commerical, inexpensive boost converter on Amazon" /></a></p>
<p>Or you could get an additional higher voltage PSU and use that as the bed's heater. They make them up to 80V I think now.</p>

CC BY-SA 4.0
2021-01-11T14:27:14.093Z
2021-01-11T14:27:14.093Z
user77232
<h1>Not necessarily</h1>
<p><strong>Potential differential U</strong>, aka <em>Voltage</em> of a part, is not to be ignored: a 24 V part needs only 24 V, not 36 V. A 12 V heartbeat is only safe for 12 V. There is a little tolerance for those measurements, but rule of thumb is about 10-15% of the rated voltage, so a 12 V bed should not be operated at more than 13.8 V for an extended period.</p>
<p>What actually facilitates heating is the <strong>Current I</strong> aka <em>Amperage</em> going through an item, as the formula for the Work P<sub>h</sub> (dissipated in the shape of heat) of the electric resistance R is <span class="math-container">$P_h=R I^2$</span>.</p>
<p>As long as you stay below or at the maximum <strong>rated Power P<sub>r</sub></strong> or <em>Wattage</em> of the heating pad <span class="math-container">$P_r=U I$</span>, you can increase the Current up to the limit of <span class="math-container">$\frac {P_r} U=I$</span>. On the bench, with a regulated power supply, we can use that to get a perfect, maximum output as we want it. But the printer isn't a bench with an expensive PSU, we get only something akin to 12 V out of it, so... what to do?</p>
<h2>DON'T route in extra Power!</h2>
<p>Yea, in DC circuits we can just add batteries behind one another to get twice the Voltage and push a circuit. Or we can put them in parallel, to sum up the current. But that doesn't work just as straightforward in AC circuits (phase shift between parts has to be taken into account). And routing in an extra pair of wires providing 12 V into an already 12 V part would get us something in the order of 24 V and fry the part. You'd accomplish nothing more than turn your heated bed into a fire hazard or a hunk of scrap!</p>
<p>So straight routing in another 12 V on top of what is rated? <strong>NOPE!</strong></p>
<h2>Unhooking from the same PSU?</h2>
<p>Some printers unhook the bed from the board's PSU, running a (differently) regulated power source. In the best case, it's set up to a Voltage/Current pair that maximizes the bed's heating. In such a setup, the whole heating power runs through a MOSFET that acts as a switch: A signal comes from the board to the MOSFET to allow current to flow. No signal on the Gate of the MOSFET leads to no current reaching the bed and no heating.</p>
<p>However, that is a complicated setup - yet one of the only ways how a &quot;mains voltage bed heater&quot; can be done with a board that runs on 5 V. You <em>also</em> will have to route the high power through a properly rated set of wires and connectors. These thicker wires will need proper strain relief as they need larger bending radii than what is installed before. In other words: <strong>you need to know what you do!</strong></p>
<h2>Finetuning the power supply to the bed!</h2>
<p>In many printers, there's also a less invasive method to finetune the power curve of a bed. usually, there is a Potentiometer that is used to tune the output of the heating wires. Altering the potentiometer's setting with a screwdriver results in the output voltage shifting. <span class="math-container">$U=R I$</span> does not change when flipping to AC but U and I became wave functions instead of constants. However, R of a long wire (such as a bed) does depend only on the frequency of the signal and not on either the current nor the driving potential, we can assume R to be constant. So, we know we want to maximize I². So what can we do easily?</p>
<p>A 24 V heatbed has something in the order of 2 Ω and accompanied mainboard in my Ender3 is - according to the labeling - good for about 13 A on the bed output, while a typical 12 V bed runs in the order of 1.2 Ω while such boards typically are limited to 10-11 A. <span class="math-container">$U= RI$</span> to the resuce and... Voila: For the 11 A/1.2 Ω case we can tune the potentiometer to get just a smidge below 13.2 V - just at the 10 % point, for the 13 A/2 Ω case is <em>technically</em> safe at 26 V - and still well within the 10% rule of thumb. However, if you have a 1.2 Ω bed and your borard only allows to draw 10 A, then you are limited to 12 V.</p>

CC BY-SA 4.0
2021-01-09T20:46:50.853Z
(also, wherever in my comment I mentioned voltage or current I meant RMS current/voltage)
Tom van der Zanden
CC BY-SA 4.0
2021-01-09T23:27:30.537Z
Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/118265/discussion-between-tom-van-der-zanden-and-trish).
Tom van der Zanden
CC BY-SA 4.0
2021-01-11T15:35:27.660Z
Typically, going above the %20 power safety margin risks burning out your heater.  For a resistive heater this is about 10% above the voltage rating.  At any rate, increasing the heater voltage shortens the life of your heater.
Perry Webb
CC BY-SA 4.0
2021-01-09T19:07:55.637Z
2021-01-11T01:01:52.020Z
2021-01-11T01:01:52.020Z
Trish
<p>From my understanding, the power of heater must higher than heat dissipate to ambient air so the bed can heat up. The reason why a bed heats up too slowly is due to its heat capacity compared to heater power.</p>
<p>As the heater is a resistive load, I think we can put higher voltage to get more heating power.</p>
<p>The PCB heater has two parts: copper and laminate. The reason of failure is that the copper can come off the board due to high temperatures. In this case we can control temperature with firmware. The questions are:</p>
<ol>
<li>Will this method work?</li>
<li>What can go wrong or what is the risk of this method?</li>
</ol>

<heated-bed><electronics>
Can we apply more voltage to a PCB heated bed to get faster heating?
M lab
# Can we apply more voltage to a PCB heated bed to get faster heating?

(Asked by: username_0 on 2021-01-09)

From my understanding, the power of heater must higher than heat dissipate to ambient air so the bed can heat up. The reason why a bed heats up too slowly is due to its heat capacity compared to heater power.

As the heater is a resistive load, I think we can put higher voltage to get more heating power.

The PCB heater has two parts: copper and laminate. The reason of failure is that the copper can come off the board due to high temperatures. In this case we can control temperature with firmware. The questions are:

1. Will this method work?
2. What can go wrong or what is the risk of this method?

**Tags:** heated-bed, electronics

---

## 3 answers

### Answer by username_1 on 2021-01-09. Score: 3 | Accepted answer

Yes, this method will work. Some (LED) power supplies have a small potentiometer that you can use to slightly adjust the voltage. On a 12 V printer, adjusting the supply from 12 V to 14 V will give a 36% increase in power.

Obviously there is a limit to how much you can increase the voltage. The voltage itself is unlikely to be the limiting factor, but there is a limit to how much heat the bed can dissipate.

> The PCB heater have 2-part copper and laminate. the reason of failire is the copper go off the board due to high temperature. In this case we can control temperature with firmware.

This is true, but you have to consider that firmware is not infallible. If you increase the voltage (and thus wattage) of the bed a lot, you risk creating a dangerous situation if something fails. Ideally, you should choose the wattage of the bed such that it does not create a dangerous situation even if it is left on permanently by accident. In extreme cases (e.g. 220V to a nominally 12V heated bed) damage will be almost instantaneous before the firmware can intervene.

You also have to consider that increasing the voltage also increases the current. The wires need to be thick enough to handle the additional current. You also have to pay special attention to the MOSFET that is used to switch the bed; it also needs to be able to handle the current (and voltage, though this is usually less of a problem). Usually the integrated MOSFETs on 3D printer control boards can only handle about ~10 A which is what the bed might normally draw. Also pay attention to fuses and to screw terminal connections. They might not be able to handle the increased current either.

> **Comments:**
> **username_0**: What about AC heated bed(Orange silicone pad)? Why it can heat much quicker?  I thought it’s had more power per area so I think add more voltage to PCB bed. And about safety My heated bed got about 100 degree centigrade when permanently connected to supply. If silicone pad got higher wattage what happens when we left them connected?  
> **username_2**: @Mlab Centigrades is *hundreds of a grad(uation)* and an angular unit. It can also be a reference to any 100 graduation scale, for example Fahrenheit is a centigrade scale. While commonly referred to as a temperature, it is scientifically no longer accepted to speak of °C as centigrad: Temperature is commonly measured in Kelvin, or in the derivate scale K+273.15, aka Degrees Celsius. Do you mean "My heated bed got to about 100 °C / 373.15 K"? That would be beyond the rated temperature range for any heated bed for consumer products. Why do you need a heated bed that operates abve 80 °C?  
> **username_1**: The reason the AC bed heats faster is indeed because it has more power (per area). If you leave it connected it will reach a higher temperature, roughly proportional to the power (i.e., 2x more power will result in 2x the temperature rise -- a 100 W bed might reach 100 °C (80 °C above ambient) then a 200 W bed will reach 180 °C).  
> **username_2**: @TomvanderZanden we enforce SI formating on units, so we also should enforce talking about them in terms of SI units or derivate units.  
> **username_0**: Sorry for making that confusing. I refer to Degree celcius. @TomvanderZanden I know that for silicone pad the final temperature will higher because of higher power per area. but with your answer " Ideally, you should choose the wattage of the bed such that it does not create a dangerous situation even if it is left on permanently by accident" So  that mean silicone pad can handle higher temperature when silicone pad left connected right?  
> **username_1**: @Mlab No, not necessarily. A lot of the silicone heaters will create a dangerous situation if left connected. With these, it is a good idea to include a thermal fuse.  

---

### Answer by username_3 on 2021-01-11. Score: 0

You can add a boost converter to the bed's power supply. Assuming that you are using an external MOSFET to control the bed (and you should). Verify what the MOSFET's max voltage rating is, then obtain a boost converter and configure it to give a voltage that is 10% less than the max. I once used this technique when driving a 24V hot end heater using a 12V power supply. Be forewarned, however, the boost converter I was using exploded during use one day.

[![Commerical, inexpensive boost converter on Amazon](https://i.sstatic.net/3gAB6.png)](https://i.sstatic.net/3gAB6.png)

Or you could get an additional higher voltage PSU and use that as the bed's heater. They make them up to 80V I think now.


---

### Answer by username_2 on 2021-01-09. Score: 0

# Not necessarily

**Potential differential U**, aka *Voltage* of a part, is not to be ignored: a 24 V part needs only 24 V, not 36 V. A 12 V heartbeat is only safe for 12 V. There is a little tolerance for those measurements, but rule of thumb is about 10-15% of the rated voltage, so a 12 V bed should not be operated at more than 13.8 V for an extended period.

What actually facilitates heating is the **Current I** aka *Amperage* going through an item, as the formula for the Work Ph (dissipated in the shape of heat) of the electric resistance R is $P\_h=R I^2$.

As long as you stay below or at the maximum **rated Power Pr** or *Wattage* of the heating pad $P\_r=U I$, you can increase the Current up to the limit of $\frac {P\_r} U=I$. On the bench, with a regulated power supply, we can use that to get a perfect, maximum output as we want it. But the printer isn't a bench with an expensive PSU, we get only something akin to 12 V out of it, so... what to do?

## DON'T route in extra Power!

Yea, in DC circuits we can just add batteries behind one another to get twice the Voltage and push a circuit. Or we can put them in parallel, to sum up the current. But that doesn't work just as straightforward in AC circuits (phase shift between parts has to be taken into account). And routing in an extra pair of wires providing 12 V into an already 12 V part would get us something in the order of 24 V and fry the part. You'd accomplish nothing more than turn your heated bed into a fire hazard or a hunk of scrap!

So straight routing in another 12 V on top of what is rated? **NOPE!**

## Unhooking from the same PSU?

Some printers unhook the bed from the board's PSU, running a (differently) regulated power source. In the best case, it's set up to a Voltage/Current pair that maximizes the bed's heating. In such a setup, the whole heating power runs through a MOSFET that acts as a switch: A signal comes from the board to the MOSFET to allow current to flow. No signal on the Gate of the MOSFET leads to no current reaching the bed and no heating.

However, that is a complicated setup - yet one of the only ways how a "mains voltage bed heater" can be done with a board that runs on 5 V. You *also* will have to route the high power through a properly rated set of wires and connectors. These thicker wires will need proper strain relief as they need larger bending radii than what is installed before. In other words: **you need to know what you do!**

## Finetuning the power supply to the bed!

In many printers, there's also a less invasive method to finetune the power curve of a bed. usually, there is a Potentiometer that is used to tune the output of the heating wires. Altering the potentiometer's setting with a screwdriver results in the output voltage shifting. $U=R I$ does not change when flipping to AC but U and I became wave functions instead of constants. However, R of a long wire (such as a bed) does depend only on the frequency of the signal and not on either the current nor the driving potential, we can assume R to be constant. So, we know we want to maximize I². So what can we do easily?

A 24 V heatbed has something in the order of 2 Ω and accompanied mainboard in my Ender3 is - according to the labeling - good for about 13 A on the bed output, while a typical 12 V bed runs in the order of 1.2 Ω while such boards typically are limited to 10-11 A. $U= RI$ to the resuce and... Voila: For the 11 A/1.2 Ω case we can tune the potentiometer to get just a smidge below 13.2 V - just at the 10 % point, for the 13 A/2 Ω case is *technically* safe at 26 V - and still well within the 10% rule of thumb. However, if you have a 1.2 Ω bed and your borard only allows to draw 10 A, then you are limited to 12 V.

> **Comments:**
> **username_1**: (also, wherever in my comment I mentioned voltage or current I meant RMS current/voltage)  
> **username_1**: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/118265/discussion-between-tom-van-der-zanden-and-trish).  
> **username_4**: Typically, going above the %20 power safety margin risks burning out your heater.  For a resistive heater this is about 10% above the voltage rating.  At any rate, increasing the heater voltage shortens the life of your heater.  

---

