...am I overstaying my welcome?

 

There is no such thing as overstaying your welcome here. At least, I have never seen the least indication of that with any poster or question.

This thread began with a question about low-voltage relays and getting your PS to output useful current. Those questions seem to have been solved, right?

Now, the next step is a question that involves switching mains power with devices that may or may not be suitable for that. I would not want to give yes/no type of answers to what could present significant risk to you without seeing your proposed circuit diagram and getting a little more information about the thermal switches you are planning to use.

And, as I suggested in a previous post, the answer may be as simple as finding the standby/shutdown pin for the power supply, if it has one. Have you searched for it? Can you provide more information about the PS, so we might look for product information on it ourselves?

John

 

Perhaps a clearer way to explain the divider issue:

The output voltage is directly related to the proportion of the resistors as you have read in all the theory. The part you are missing is that your load is in parallel with one of the resistors; thus the load itself is part of the divider circuit(by changing the effective operating resistance of that branch of the circuit).

Thus, you measure your 12 volts with a tiny load of a voltmeter which has an insignificantly high resistance and get very close to the theoretical voltage. When you connect an ammeter in the same manner(which has near zero resistance) it so significantly alters the branch that you detect little or no current because the new circuit actually has near zero volts. When you put a fan motor in parallel with this same resistor, the motor winding being very low in resistance, results in a much different resistance value for that branch; and significantly changes your output voltage to a value to low to operate the device.

In other words, when calculating the resistance values for the divider, you have to include the load resistance in parallel with the resistor you are paralleling.