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Re: Substrate heating cables
On Wed, 30 Dec 1998, Mortimer Snerd wrote:
> Finally, I need to get a transformer for the cables, and am a little
> confused. I was under the impression that the cables would run hotter
> if you put more amps through them. I am now wondering if this is the
> case. Do the cables develop only so much heat, and the rest of the
> power put out by the transformer is unused, or will the transformer only
> put out as much power as the cables need to develop a particular
> wattage? Or (as seems to happen with frightening frequency :-) am I
> completely wrong?
I'm sure you'll get a handful of responses to this one.
Just like a fuse, the current rating on a transformer designates the
maximum current it can deliver before it fizzles out. The amount of
current it actually supplies is determined by the transformer voltage and
the resistance of the load, in this case the heating cables:
The cables have a fixed Resistance R, which means for a given Voltage V
applied across them you will have a current I = V/R (This is Ohm's law of
V = I*R). The amount of heat delivered (Power P) = I * V = V^2/R. In
practical terms, the wattage goes as the applied voltage squared. To get
more power, you need a higher voltage transformer.
BUT there is a caveat: it's probably not a good idea to go above the rated
voltage for a given cable; there's no telling what it will do (melt the
insulation? burn out? etc), so a Dupla 150 watt cable shouldn't be used to
give 200 watts by hooking it up to a 30-volt transformer. I have,
however, used lower voltages in order to reduce the wattage on a 100 watt
cable to about 50 watts.
So let me do a practical example which might help illustrate.
You have a 150 watt Dupla heater designed to work at 24 volts.
First, we find out how much current that will draw:
P = I * V -> I = P / V = 150 watts / 24 volts = about 6 amps.
So a 7 amp transformer would work just fine here. It won't fizzle.
A 4 amp transformer cannot handle this load, and may burn out.
Additional fun: Since we know the power goes as the voltage squared for a
given heater, it follows that if you hook the same heater up to a 12-volt
transformer, you will get 1/4 power or 37 watts. At 18-volts, you get a
bit more than half the power. The corresponding current draws will be
proportional to the voltages, and will be 3 amps for the 12 volt
transformer and 4.5 amps for the 18-volt transformer. This has been left
as an exercise to the reader. :)
erik at thekrib dot com