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Heated bed power calculation

Posted by zipmsp 
Heated bed power calculation
April 02, 2014 07:27PM
I'm looking at the MK1 heated bed and I'm trying to figure out what Prusa meant when he said it is easily converted to 24v.

Current = Voltage / Resistance
Power = Voltage * Current

With a 12v power supply and 1.2 ohm heated bed, current = 10A and power = 120W.
With a 24v power supply and 1.2 ohm heated bed, current = 20A and power = 480W.

So I need a much larger power supply for 24v, and it seems like I might start tripping breakers!

1. Is this the right application of Ohm's law/the power law?
2. What is physically different about the MK2b heated bed that allows it to run 24v? It is spec'd for 3.4 ohm, so current=7A and power=169watts.
3. Is there a way to modify the MK1 to increase its resistance so I can run 24v at a lower power level?
Re: Heated bed power calculation
April 02, 2014 09:28PM
My guess is that you'd need to cut a track or two to get it to run 24V without issue.

The Mk2b and Mk3 have three terminals. It's effectively 2 resistors with one end on each joined to the other (Point B below). You run them either in series (24V - Point A to Point C) or parallel (12V - Point A+Point C [shorted] to Point cool smiley.

Text pic for reference:
  +––––+     +––––+  
+–+ R1 +––+––+ R2 +–+
| +––––+  |  +––––+ |
+         +         +
A         B         C

Note: I know that the board uses 1, 2, 3 but not sure if the numbering for the Mk2b and Mk3 are the same, or if the clones get the numbering right, so I used letters to avoid any possible confusion.

Edited 1 time(s). Last edit at 04/02/2014 09:29PM by Cefiar.
Re: Heated bed power calculation
April 02, 2014 10:16PM
I forgot about how resistance adds in parallel versus series. I can convert the parallel board into series by making three cuts and moving one of the power connections.

The total resistance of the parallel circuit is 1.2 ohms and there are four resistors in parallel. Solving for the resistance of one resistor:

1.2 ohms = 1 / (1/x + 1/x + 1/x + 1/x)
x = 3.33 ohms

Resistance adds together in series, so the resulting series circuit has 13.32 ohms of resistance. I=V/R and P=IV, so at 12v this is a 0.9A 10.8W circuit, and at 24v this is a 1.8A 43.2W circuit. The original 12v circuit was 120W, so both are worse than just using 12v in parallel.

What about making an MK2b board out of an MK1. That is, two pairs of series resistors. Resistance in each circuit is 6.66 ohms. At 12v this is two 1.8A 21.6W circuits = 3.6A 43.2W. At 24v this is two 3.6A 86.5W circuits = 7.2A 173W.

Getting 173 Watts out of 7.2 Amps at 24 Volts seems better than 120 Watts from 10 Amps at 12 Volts.
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