<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>

[{'answer_id': 15266, 'author': 'Trish', 'author_id': 8884, 'author_profile': 'https://3dprinting.stackexchange.com/users/8884', 'pm_score': 0, 'selected': False, 'text': '<h1>Not necessarily</h1>\n<p><strong>Potential differential U</strong>, aka <em>Voltage</em> of a part, is not to be ignored: a 24\xa0V part needs only 24\xa0V, not 36\xa0V. A 12\xa0V heartbeat is only safe for 12\xa0V. There is a little tolerance for those measurements, but rule of thumb is about 10-15% of the rated voltage, so a 12\xa0V bed should not be operated at more than 13.8\xa0V for an extended period.</p>\n<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>\n<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\xa0V out of it, so... what to do?</p>\n<h2>DON\'T route in extra Power!</h2>\n<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\xa0V into an already 12\xa0V part would get us something in the order of 24\xa0V and fry the part. You\'d accomplish nothing more than turn your heated bed into a fire hazard or a hunk of scrap!</p>\n<p>So straight routing in another 12\xa0V on top of what is rated? <strong>NOPE!</strong></p>\n<h2>Unhooking from the same PSU?</h2>\n<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>\n<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\xa0V. 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>\n<h2>Finetuning the power supply to the bed!</h2>\n<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>\n<p>A 24\xa0V heatbed has something in the order of 2\xa0Ω and accompanied mainboard in my Ender3 is - according to the labeling - good for about 13\xa0A on the bed output, while a typical 12\xa0V bed runs in the order of 1.2\xa0Ω while such boards typically are limited to 10-11\xa0A. <span class="math-container">$U= RI$</span> to the resuce and... Voila: For the 11\xa0A/1.2\xa0Ω case we can tune the potentiometer to get just a smidge below 13.2\xa0V - just at the 10\xa0% point, for the 13\xa0A/2\xa0Ω case is <em>technically</em> safe at 26\xa0V - and still well within the 10% rule of thumb. However, if you have a 1.2\xa0Ω bed and your borard only allows to draw 10\xa0A, then you are limited to 12\xa0V.</p>\n'}, {'answer_id': 15267, 'author': 'Tom van der Zanden', 'author_id': 26, 'author_profile': 'https://3dprinting.stackexchange.com/users/26', 'pm_score': 3, 'selected': True, 'text': '<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>\n<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>\n<blockquote>\n<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>\n</blockquote>\n<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>\n<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>\n'}, {'answer_id': 15284, 'author': 'user77232', 'author_id': 12857, 'author_profile': 'https://3dprinting.stackexchange.com/users/12857', 'pm_score': 0, 'selected': False, 'text': '<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>\n<p><a href="https://i.stack.imgur.com/3gAB6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3gAB6.png" alt="Commerical, inexpensive boost converter on Amazon" /></a></p>\n<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>\n'}]
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