OK, I think I see what you're getting at. Basically, you want the grid to be 0.7 volts negative with respect to the filament's center. So, dump the diode completely. Connect the battery "+" to ground. Across the battery, from + to -, put a voltage divider set to trickle current- like two 1M resistors in series. Connect the bottom of the grid leak resistor to the junction of the two 1M resistors. That will give you -0.75V.
Another way to do it is to run the filament from a floating CT transformer, then connect the diode from the CT to ground. From the B+, run a resistor to the diode/CT junction to put some extra current through the diode to help linearize it. In this circuit, 47K would be a good value, getting you an extra 2 milliamps.
Bypass the diode with a capacitor.