Thermal Protector

[hbiss]

Here's something that I can't remember ever having to do. Customer calls and says that the recessed lighting fixture in their 2nd floor closet won't work. It's the usual square incandescent, probably Atlite, from about 30 years ago. I find out that the thermal protector is open. Since everything is LED these days I'm tempted to just eliminate the thing, but being the conscientious guy I am, I decide to replace it. So the question is with what. No markings on it other than the GTE (telephone company!) logo. So now I do a little research.

The thermal protector is supposed to protect against not only overlamping but covering the fixture with insulation or otherwise so as to trap heat. It contains a bi-metal switch and a resistor. The resistor is across the line and serves as a heat source. This is supposed to provide enough heat so that the switch can sense if it's covered in insulation. But what I can't see is how it senses overlamping. It's located at the far end of the wiring compartment so I would think that the fixture would have to be almost on fire for the heat to get to it.

Anyway, I do a search and see that there are many choices here. 120, 277, 480V/ 3 wire, 4 wire, value of the resistor and 60 deg c or 110 deg c.

So I order obviously 120V, 3 wire (resistor is across black & white, blue goes to lamp). Resistor measures 6.7k so I get one with a 7k. Then I chose 110 deg c only because I'm thinking that this fixture (and thermal switch) is exposed on top to an uninsulated attic that can easily get to 120 deg F. Maybe that's the wrong approach but then maybe that's why it failed in the first place. But then again I'm not privy to how that 110 deg c is actually arrived at.

Anybody have any experience with this?

-Hal

 

[Coppersmith]

Since everything is LED these days I'm tempted to just eliminate the thing

I would have stopped here and added a label to the fixture that said "LED Bulbs Only".

 

[gar]

190427-1236 EDT

hbiss:

My guesses are:

1. Because you indicate 3 wires I suspect lamp current thru the sensor is part of the thermal equation.

2. C to F is done with F = 32 + C*9/5 . 60 C is thus 32 + 9*12 = 140 F. I would expect this to be too low for a hot incandescent in an insulated box in an ambient of 120 F.

LEDs produce a lot of internal heat because they radiate little of their input power thru any window. Compared to an incandescent of equal visible light output I would expect less internal heat buildup from the LED, but not as much of a difference as you might think.

I ran an experiment on this some time back, but I don't remember the details.

.

 

[hbiss]

190427-1236 EDT

hbiss:

My guesses are:

1. Because you indicate 3 wires I suspect lamp current thru the sensor is part of the thermal equation.

2. C to F is done with F = 32 + C*9/5 . 60 C is thus 32 + 9*12 = 140 F. I would expect this to be too low for a hot incandescent in an insulated box in an ambient of 120 F.

LEDs produce a lot of internal heat because they radiate little of their input power thru any window. Compared to an incandescent of equal visible light output I would expect less internal heat buildup from the LED, but not as much of a difference as you might think.

I ran an experiment on this some time back, but I don't remember the details.

.

1. You could be right. Current flows from black to blue. Black is hot, blue to the lampholder. Resistor is from black to white.

2. That's what I thought too. 140 F is too close to the ambient. BUT we don't know how that spec is obtained.

I find that the driver creates heat but not nearly as much as an incandescent of an equal lumen output.

-Hal