Surge Protector Ground Indicator Circuit Operation

[Bushougoma]

I came across this ground indicator circuit inside a surge protector I disassembled and I have some questions about how it works. I've cut out much of the circuit not related to the ground indicator (extra MOVs, thermal fuses that disconnect power should the MOVs fail, etc).

It consists of two LEDs the first green LED D2 just indicates power on and will go out if any of the thermal fuses (not shown) blow due to a MOV failure. The second red LED D3 is the ground presence indicator. If your outlet is properly grounded it will illuminate if not it will go out simple enough.

We start out with 120 volts AC which immediately passes through a half wave rectifier D1 lowering the voltage to 60 V with lots of ripple. It then passes through R1 the current limiting resistor for both the green and red LEDs.

Here's how I think the circuit works lets start when a proper ground is NOT present (the ground wire is not connected) and D3 the red LED should be OFF.

A small amount of leakage current through the upper MOV which is connected to the line (forms a voltage divider with the lower MOV connected to neutral) this reduced voltage passes through R3 the base current limiting resistor for the KSP13 NPN darlington transistor. This causes it to conduct and it essentially bypasses the red LED across it's collector and emitter causing it to go out. So current would flow from the cathode of the green LED D2 through the transistor and to neutral. The transistors base is being fed AC so I guess they are using the base emitter junction as a rectifier. It won't conduct on the negative portion of the sine wave but since the LEDs are being fed through a half wave rectifier they don't see the negative portion and would be off anyway.

With a proper ground present (the ground wire is connected) the red LED should be ON.

As before a small amount of leakage current flows through the upper MOV connected to the line but this time a voltage divider isn't formed because the ground wire is connected and essentially both sides of the lower MOV from ground to neutral are at the same potential. The leakage current from the upper MOV connected to the line flows to ground instead of through R3 this means 0 volts is present at the base of the KSP13 transistor and it is essentially open circuit. This puts the green and red LEDs in series so both glow.

Now to my question and hopefully I have the theory of operation right so far . What is the purpose of the parallel 10K resistor? It seems like the circuit would work just fine without it.

Thanks for any help in advance.