<p>Which of these will heat a bed fastest?</p>

<ul>
<li><p>A. 12&nbsp;V, 10&nbsp;A power supply</p></li>
<li><p>B. 24&nbsp;V, 5&nbsp;A power supply</p></li>
<li><p>C. Both A and B will be the same (only total watts matter)</p></li>
<li><p>D. Depends on the situation</p></li>
</ul>

<p>I originally thought Amperage was what mattered until I realized I needed a 24&nbsp;V power supply to even heat my Lulzbot mini bed by one degree.</p>

<p>I know voltage is used to determine insulation thickness on wires. But thin wires with high current in them also get hot. Is insulation thickness on wires only to prevent you from accidentally cutting through them and shocking yourself, or is it for heat reasons?</p>

<p>I'd like to power my heated bed with a 19.5&nbsp;V, 5&nbsp;A power supply. It's just an old laptop charger - I want to reduce strain on my circuit. It's a big bed and I have a few other laptop chargers lying around so I'd prefer to choose the best one.</p>

[{'answer_id': 11072, 'author': '0scar', 'author_id': 5740, 'author_profile': 'https://3dprinting.stackexchange.com/users/5740', 'pm_score': 3, 'selected': True, 'text': '<p>It depends on whether you are re-using the bed or not, it is actually the resistance of the bed that determines this in conjunction with the voltage (the current you get for free).</p>\n\n<p>Let\'s say that the heatbed resistance is 1.2&nbsp;&ohm; (depending on the heated bed make and model the resistance is typically in between 0.9&nbsp;-&nbsp;1.5&nbsp;&ohm;), this means that the power can be calculated using:\n<span class="math-container">$$P = U \\times I$$</span>\n<span class="math-container">$$U = I \\times R$$</span>\ncombining gives:\n<span class="math-container">$$ P = I^2\\times R = \\frac{U^2}{R} $$</span></p>\n\n<p>For 12&nbsp;V (assumed default printer voltage) this means that the heatbed power equals about 120&nbsp;Watt (at a current of 10&nbsp;A). Running that same bed at 24&nbsp;V means that the power is 480&nbsp;Watt (at a current of 20&nbsp;A). So yes, that will heat up fast, at the expense of an increased current, which is pretty high, and should not be attempted without extra resistance in the loop.</p>\n\n<p>If you\'re using the laptop charger, the current draw equals about 16&nbsp;A, which the adapter cannot deliver.</p>\n\n<p>This means that you need to acquire a new heatbed that is able to handle a higher voltage out of the box (more resistance), or you need to put additional resistors in the loop, but beware of the currents. Note that heated beds for 12&nbsp;V/24&nbsp;V exist, the wiring is different depending on the voltage. Note that such beds heat up faster, it all depends on the resistance and the voltage, but running the 24&nbsp;V circuit on 19.5&nbsp;V (160&nbsp;Watt bed) is definitely an improvement over the 120&nbsp;Watt bed at 12&nbsp;V but still requires about 8&nbsp;A (only applicable to heatbed that can run 12&nbsp;V/24&nbsp;V through extra resistance connections).</p>\n\n<p>Be careful with this and be sure what you are doing!</p>\n'}, {'answer_id': 11077, 'author': 'CrossRoads', 'author_id': 16523, 'author_profile': 'https://3dprinting.stackexchange.com/users/16523', 'pm_score': 0, 'selected': False, 'text': '<p>Bed heaters look like this  <a href="https://i.stack.imgur.com/A9Gqn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/A9Gqn.jpg" alt="enter image description here"></a></p>\n\n<p>They are rated for use with 12V or 24V supplies. 12V supply would take longer to warm it up, as P = V^2/R.  Say it was a 2 ohm total resistance bed, then 12*12/2 = 72 watts, vs 24*24/2 = 288 watts. And 19V*19V/2 = 180W. Then you work backwards, P=IV or P/V = I to determine current draw: 72/12 = 6 amp, 288/24 = 12 amp, 180/19 = 9/5A.</p>\n'}, {'answer_id': 11078, 'author': 'Sean Houlihane', 'author_id': 4927, 'author_profile': 'https://3dprinting.stackexchange.com/users/4927', 'pm_score': 0, 'selected': False, 'text': "<p>Power is the thing that determines how quickly the bed heats up, nothing else.</p>\n\n<p>A specific bed will be defined by it's <em>resistance</em>, this is the only relevant factor which is a constant. Of course, you shouldn't run it at a much higher power than it was designed for, or at a significantly higher temperature (these are related, since the bed looses much more energy to the room as it gets hotter).</p>\n\n<p>All 'low voltage' beds will have the same sort of insulation requirements (effectively none, less than 36V is regarded as safe to touch unless the skin is also penetrated).</p>\n\n<p>The wires used to connect to the heated bed must have a significantly lower resistance than the bed itself (similarly the connectors). Otherwise the wiring overheats and the bed has to work harder (making the overheating worse). Using a heat bed designed for a higher voltage (and a matching higher voltage supply) puts less strain on the wiring because the current is reduced for the same power output.</p>\n\n<p>Since the resistance is constant, increasing a 12V power supply to 13V gives a ~17% power increase (with the same bed) because the current also increases.</p>\n"}]
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