> I don't believe this is correct. Warm-up or down time is like an RC
> time constant curve. When charging a capacitor, the last few percent
> takes much longer than the first few percent. The reverse holds true
> for discharging. When power is removed, the initial drop rate is much
> faster than the later drop rate. Kind of like your bank account - it
> goes out a lot faster than it comes in.
>
>
> 73, Bill W7TI
Hi Bill,
While this is something I rarely consider, I'm not aware of the
temperature having a different slopes for increase and decrease
anywhere on the curve, unless the filament were removed and replaced
with an ice cube. In that case you would be correct, the slope would
be like a capacitor R/C circuit in both directions.
Are you sure the heater or filament "sinks" heat energy
(like a resistor would "sink" charge from a capacitor)? If the
filament does this, what did it do with it's own thermal
energy?
I'd bet the slope is pretty even up near the top, and near the bottom
heating occurs faster than cooling.
Before publishing a circuit or recommending a solution, it is
always a good idea to understand what is really needed. Even the
Edisonian method requires verification before selling the bulbs.
73, Tom W8JI
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