Heated Bed Conversion to 24V Issues - Resistence is wrong


#1

So I have the heated bed that came with the kit: https://www.tindie.com/products/mjrice/pcb-heatbed-for-3d-printers-200-x-300mm/

I am in the middle of converting it over to 24V, when I notice the resistance is wrong.

According to the docs (on that page) the resistance at 12V should be ~1.4 Ohms, and at 24V it should be ~2.8 Ohms.

When I change from A-B and C-D connections to a B-C connection the resistance is actually 6 Ohms. This is slightly less power than at 12V. I did try putting a wire from only A-B or C-D (not both) and the resistance was where it should be at 2.8 according to the documentation.

Does anyone know if the documentation might be wrong or if this is an issue? Can I just hook it up at A-B or C-D?


#2

My take on the instructions is that to convert from 12 V to 24 V, you will need to remove the jumper wires that connect A to B and C to D, and connect a jumper between B and C. The power leads should remain as originally connected, to W1 and W2.
Heatbedwires
Is this what you’ve done?
Someone please confirm or correct my schematic…?


#3

Yes, the wiring is as you describe.

The main issue is that the resistance is twice what it should be, and ultimately the power is less by 5 to 10%.

I have traced the patterns on the board, they are in a zig zag pattern going down the board like a ladder.

When A-B and C-D are connected, every trace is heating, like 1111

When B-C are connected, every other trace is heating, like 1010

When A-B are connected (C-D open), every other trace is heating in groups of two like 1100.

The whole point of 24V is to make heating faster, and A-B only is the only way that can be done here.


#5

I am thinking that it is worth trying out if the A-B works because the power will be greater, the current will be the same, and the power would be absorbed by the glass plate anyways.


#6

Hmm… Respectfully, I think that is a bad idea zzing. Maybe you did this math already, I don’t know, but here it is:
At V= 12V and R=1.4 ohms, I=V/R=8.57 A. P=VI=102 Watts
At V= 24V and R=2.8 ohms, I=V/R=8.57 (the same, as you say.)
BUT now P=V
I=206 Watts.
The problem is, You are now putting twice the power through half the board traces, a factor of -->4<-- times the power per active trace. A good recipe for magic smoke at some point, I think.

I’d suggest waiting until mjrice (hopefully) weighs in here to clear up what may be happening.