I haven't personally used these yet, but I came across the three legged LITTELFUSE "iTMOV" a while back. These thermally protected varistors allow their condition to be monitored. See page 5 of the datasheet below for some examples. Maybe this type of MOV can be of use to you.
Regards,
Frank
Littelfuse_Varistors_iTMOV_Datasheet.pdf
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Thanks Frank. I wasn't aware of these devices; very neat. MOV and fuse in one package with extra lead connecting to the junction of the two. LittleFuse's Application Example 1 is exactly the circuit about which I speculated. However, I do wonder why they use a current-limiting resistor with mains voltage. At 120-240 V, the resistor will consume a lot more power than the usual types of LED's which are 20-30 mA. At 240 VAC RMS, the peak voltage is 339 V. For a 30 mA LED with a 2 V drop, the resistor would need to be (339 - 2)/0.03 = 11K ohms. Assuming a 50% duty cycle based on the circuit in their example, the LED would consume 0.03 watts and the resistor 2.5 watts! A transformer plus resistor or a capacitor or a mini switch-mode PS would be a better choice for current-limiting at higher voltages.
Their example also shows a diode in series with the LED, which I don't understand. Unless that diode has a really whopping voltage drop (like 220 V for a 240 V mains voltage!), it's not going to significantly lower resistor power consumption nor protect the LED from high reverse voltage during the negative 1/2 of the AC cycle. Typically in an AC application, there is a diode in parallel, but opposite polarity, with the LED to protect it during the "reverse" half of the cycle. This configuration will give the resistor a 100% duty cycle though, so in the example above, the LED would consume 0.03 W and the resistor 5 W! That's a lot of power to throw away, and even a 10 W rated resistor is going to get more than a little warm at 5 W.
If anyone knows of a diode that can be placed in series with an LED and resistor on mains voltage that will do the job of protecting the LED from high reverse voltage and allow low power consumption by the resistor, I'd like to know about it.
Mike