AC silicone heating mat Red, Yellow, Blue?

Hi all, I am about to set up my A/C silicone heating mat with an SSR, but have been thrown a curve ball with the wiring colors (Red, Yellow, Blue?). Its a generic heating pad, and other than it being labeled for 110/220v and its wattage I can't find a manual to tell me which of these wires are ground ( I purchased it about a year ago and got rid of all the receipts so my wife/accountant wouldn't... i'm sure you get the point). Use a multimeter? Thanks in advance.

I've never seen this before but the fact that it's labeled for two voltages is a clue. Between two wires, say, A and B, you'll get a resistance appropriate for the rated wattage at 110v. Between B and C, you should see about twice that resistance, which would give the appropriate wattage at 220v. This is extremely important because if 220v were applied to the 110v side of the heater, it would likely burn up in short order, heating twice as fast as expected.

In short, none of the wires are ground, they are two separate heater circuits. Devices which do not have a metal chassis typically do not include a ground wire, for future reference.

Thank you for saving me from making a mess.

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Turtle_Cat
I've never seen this before but the fact that it's labeled for two voltages is a clue. Between two wires, say, A and B, you'll get a resistance appropriate for the rated wattage at 110v. Between B and C, you should see about twice that resistance, which would give the appropriate wattage at 220v. This is extremely important because if 220v were applied to the 110v side of the heater, it would likely burn up in short order, heating twice as fast as expected.

In short, none of the wires are ground, they are two separate heater circuits. Devices which do not have a metal chassis typically do not include a ground wire, for future reference.

Let's assume we have a 500W heater.

P=IV > P/V=I
That gives us 2.27A for the 220V, and 5.54A for the 110V.

V=IR > R=V/I
That gives us 96.9Ω for the 220V, and 24.2Ω for the 110V.

That makes sense since "I" is a squared part of the function: P=(I^2)R


EDIT: I initially thought you said that the resistance was reversed.

Edited 2 time(s). Last edit at 06/19/2018 12:16PM by boredom.is.me.

Ok just running my math by you guys before i run the wall pixies into my heater. I used a multi meter to measure the resistance across all the leads
btw its a 450W heater

blue, yellow =53.3 olms
blue, red = 53.3 olms
yellow, red = 106.4 olms

If im calculating P=IV
450 = I* 110
i = 4.09

450 =I * 220
i = 2.045

R = V/I
110/4.09
R = 26.89

R=V/I
220 / 2.04
R= 106.4



So i can connect o either Blue, Yelllow or Blue Red for the 110V circuit? Or am i missing something important? BTW you guys are awesome. The only math i've done in the last 8 years has had to do with drip rates and if the power is out I might have to do some conversions. I seriously appreciate it!

Update I went red+blue the bed heated up and I didn't die

You are running the heater at half power. To get full power with 110V supply, connect red and yellow together, and connect one wire to red+yellow and the other to blue. But you may find half power sufficient.

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Disclosure: I design Duet electronics and work on RepRapFirmware, [duet3d.com].

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dc42
You are running the heater at half power. To get full power with 110V supply, connect red and yellow together, and connect one wire to red+yellow and the other to blue. But you may find half power sufficient.

Thank you dc24 I wasn't entirely sure if the missing half of 53 olms meant the wiring of this board was analogous to the one that came with my di3 or if I forgot to double something, but this clarifies my thoughts. Your advise on these forums is priceless. This is why I'm glad I got the duetwifi.