Using resistive foil for heated bed?

Has anyone tried this?
I have found this on my local electronics parts supplier: [www.gme.cz] (sorry, it's in Czech, I was unable to find any equivalent in English) and wondered, if it could be used to create a heatbed very simply...
The parameters for this particular foil are:
Resistance: 60Ohm/cm2,
Working temperature: -60 to 200°C,
Working voltage: 3 to 380V,
Electrical energy to temperature conversion effectivity: >99%,
Width: 50um...

This seems quite ideal for a heatbed, so I wonder - is it worth a try or have I overlooked something important? What would be the downsides?
Even the dimensions are ideal - 26x20cm - with the excess 6cm stripped off, it can be ready-used for a standard Mendel/Průša-sized heatbed!

Edit: Oh and I forgot to say that it costs only 76,- Kč, which is like 3-4€, so...

Edited 1 time(s). Last edit at 07/08/2013 07:25PM by Matouš.

Isn't this Polyimide (Kapton) tape with fine conductive tracks embedded in the surface?

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Waitaki 3D Printer

Yes, as the text says: "Electrically conductive surface is made from a specially created resistive carbon layer with defined surface resistivity, applied on polyimid (PI) foil".

Sounds good, try it!

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Generation 7 Electronics	Teacup Firmware	RepRap DIY

Hi !

That sounds good, especially on the price scale, but is it powerfull enough ? As if i have understood right, the resistive foil resistance is measured by square. So no matter how big the square is, as long as it is square, it has the resistance stated ? In this case it is announced as a 60 ohm/cm2, but i woul say that the resistance is the same 60 ohms for a square that measures 20cm x 20 cm.

So with 12 volts it would be : I=U/R -> 0.2 Amps => 2,4 wats -> a bit too low power

230 volts would make + 800 wats ,Too much So aprox 80 volts would give nice 100w power.

Well I'm not sure, how is the resistance supposed to be calculated, but when I simply connected my multimeter to opposite sides of the foil, the resistance was about 220ohms, if I remember correctly, which is a bit too high

I thought about connecting the wires a bit differently, like one around the edge of the foil and the other in the center, which could lower the total resistance, but would make it heat unevenly (the corners would be colder).
Or maybe connecting the wires next to each other, in parallel - that should lower it even more.
I have added a picture of the connections I have in mind.

I'll have to experiment when I get to it! So far I'm a bit stuck, because I don't have enough uninsulated wire (which is a bit lame... ).

With such resistance, the "center to border" wiring wont work. As the goal for total resistance (if using 12 volts) is in the area of 2 ohms.

The parallel method might work if the wires are close enough. Please do report how the tests end up