[SOLVED] Leakage current in a driver IC; how to compensate?

Risen:
Your divider solution that sounds like it has some promise. Would you mind explaining it a bit further with how you calculated it?

If R1 is the resistor to +170v and R2 is the resistor to ground, then the voltage divider is equivalent to a supply of 170 * R1/(R1 + R2) volts with a series resistor of value R1 * R2/(R1 + R2). So if you can decide what is the ideal voltage supply for the neon lamp supply (I suggest you aim for 120v) and what series resistor you would want to use with that supply, then you can work out the values of R1 and R2 to use by solving those as simultaneous equations.

I have to admit that I didn't do that, instead I reduced R1 a bit (probably not enough) from the value of 200K to allow for some current through R2, and then chose R2 to give a voltage in the range 100 to 120v at the junction.