<p>Bed heaters look like this  <a href="https://i.sstatic.net/A9Gqn.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/A9Gqn.jpg" alt="enter image description here"></a></p>

<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>

CC BY-SA 4.0
2019-09-26T12:17:13.147Z
2019-09-26T12:17:13.147Z
CrossRoads
<p>Power is the thing that determines how quickly the bed heats up, nothing else.</p>

<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>

<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>

<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>

<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>

CC BY-SA 4.0
2019-09-27T04:15:57.627Z
I would also consider the voltage to be a constant, in the case of a DC power supply. It's the current draw that changes
K Mmmm
CC BY-SA 4.0
2019-09-27T08:04:10.837Z
@steveantwan absolutely not. You can adjust most modular switching supplies maybe 15%, and a DC-DC switching converter can be had for very little $. If you have a specific bed, you have a resistor. I will not have non-scientific concepts introduced into my posts.
Sean Houlihane
CC BY-SA 4.0
2019-09-27T14:15:29.497Z
In the case of a laptop charger or the big silver boxes most people use, the voltage is constant during operation. The resistance is also an independent variable. Current is the dependent variable.
K Mmmm
CC BY-SA 4.0
2019-09-27T14:23:12.237Z
R u saying in the 12V silver boxes the voltage might change during normal operation? Or only if u modify it?
K Mmmm
CC BY-SA 4.0
2019-09-26T12:36:07.153Z
2019-09-26T12:36:07.153Z
Sean Houlihane
<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>

<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:
<span class="math-container">$$P = U \times I$$</span>
<span class="math-container">$$U = I \times R$$</span>
combining gives:
<span class="math-container">$$ P = I^2\times R = \frac{U^2}{R} $$</span></p>

<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>

<p>If you're using the laptop charger, the current draw equals about 16&nbsp;A, which the adapter cannot deliver.</p>

<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>

<p>Be careful with this and be sure what you are doing!</p>

CC BY-SA 4.0
2019-09-25T19:35:27.953Z
I see, the current here is a dependent variable (as opposed to voltage and resistance which are independent variables). The current draw varies with the resistance and voltage. If i could make a suggestion, change the variable for voltage from U to V (more intuitive for people without an ECE background or from a physics background), and also specify answer D in the first line of the answer.
K Mmmm
CC BY-SA 4.0
2019-09-25T20:14:49.830Z
I calculated most of this and figured out that i could add an extra 1.2 ohms if i used 10 resistors of 10 Ohms each, connected in parallel, and two laptop chargers connected in parallel at 18.5V and 4.3A each. The total watts would only add by about 10% though. So, I will implement this later on if i want to
K Mmmm
CC BY-SA 4.0
2019-09-25T20:39:33.380Z
I just want to add that could still be a huge improvement because there's almost no chance of the circuit overheating if all the heat that normally goes into the MCU instead goes to a separated power supply and relay
K Mmmm
CC BY-SA 4.0
2019-09-27T14:17:21.400Z
^ assuming the MCU is next to the heated bed terminals on the board you're using
K Mmmm
CC BY-SA 4.0
2019-09-25T18:43:38.033Z
2019-09-26T10:12:55.757Z
2019-09-26T10:12:55.757Z
0scar
<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>

<heated-bed>
Does voltage or current affect the time to heat a bed more?
K Mmmm
# Does voltage or current affect the time to heat a bed more?

(Asked by: username_0 on 2019-09-25)

Which of these will heat a bed fastest?

* A. 12 V, 10 A power supply
* B. 24 V, 5 A power supply
* C. Both A and B will be the same (only total watts matter)
* D. Depends on the situation

I originally thought Amperage was what mattered until I realized I needed a 24 V power supply to even heat my Lulzbot mini bed by one degree.

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?

I'd like to power my heated bed with a 19.5 V, 5 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.

**Tags:** heated-bed


---

## 3 answers

### Answer by username_3 on 2019-09-25. Score: 5 | Accepted answer

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).

Let's say that the heatbed resistance is 1.2 &ohm; (depending on the heated bed make and model the resistance is typically in between 0.9 - 1.5 &ohm;), this means that the power can be calculated using:
$$P = U \times I$$
$$U = I \times R$$
combining gives:
$$ P = I^2\times R = \frac{U^2}{R} $$

For 12 V (assumed default printer voltage) this means that the heatbed power equals about 120 Watt (at a current of 10 A). Running that same bed at 24 V means that the power is 480 Watt (at a current of 20 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.

If you're using the laptop charger, the current draw equals about 16 A, which the adapter cannot deliver.

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 V/24 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 V circuit on 19.5 V (160 Watt bed) is definitely an improvement over the 120 Watt bed at 12 V but still requires about 8 A (only applicable to heatbed that can run 12 V/24 V through extra resistance connections).

Be careful with this and be sure what you are doing!

> **Comments:**
> **username_0**: I see, the current here is a dependent variable (as opposed to voltage and resistance which are independent variables). The current draw varies with the resistance and voltage. If i could make a suggestion, change the variable for voltage from U to V (more intuitive for people without an ECE background or from a physics background), and also specify answer D in the first line of the answer.  
> **username_0**: I calculated most of this and figured out that i could add an extra 1.2 ohms if i used 10 resistors of 10 Ohms each, connected in parallel, and two laptop chargers connected in parallel at 18.5V and 4.3A each. The total watts would only add by about 10% though. So, I will implement this later on if i want to  
> **username_0**: I just want to add that could still be a huge improvement because there's almost no chance of the circuit overheating if all the heat that normally goes into the MCU instead goes to a separated power supply and relay  
> **username_0**: ^ assuming the MCU is next to the heated bed terminals on the board you're using  

---

### Answer by username_1 on 2019-09-26. Score: 0

Bed heaters look like this [![enter image description here](https://i.sstatic.net/A9Gqn.jpg)](https://i.sstatic.net/A9Gqn.jpg)

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.


---

### Answer by username_2 on 2019-09-26. Score: 0

Power is the thing that determines how quickly the bed heats up, nothing else.

A specific bed will be defined by it's *resistance*, 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).

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).

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.

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.

> **Comments:**
> **username_0**: I would also consider the voltage to be a constant, in the case of a DC power supply. It's the current draw that changes  
> **username_2**: @steveantwan absolutely not. You can adjust most modular switching supplies maybe 15%, and a DC-DC switching converter can be had for very little $. If you have a specific bed, you have a resistor. I will not have non-scientific concepts introduced into my posts.  
> **username_0**: In the case of a laptop charger or the big silver boxes most people use, the voltage is constant during operation. The resistance is also an independent variable. Current is the dependent variable.  
> **username_0**: R u saying in the 12V silver boxes the voltage might change during normal operation? Or only if u modify it?  

---

