I'm looking at this output driver circuit, and two things come to mind:
1) It appears to be more complex than it could be
2) I don't like the 170-volt DC supply (D6-D9 bridge to F2 section), and know it's possible to run this same circuit with 12VDC, which then loops back to observation 1.

The pulse signal is a square wave with variable duty cycle, and SW1 is a representation of the PIA output that enables this circuit.

The 'Drive' output feeds a coupling transformer to the next stage.

I anticipate I would be keeping everything up to C40, it's just C40 and everything to the right I'd like to simplify, mainly to get the output to a safer value and eliminate T1 (custom transformer)...

Could I use an op-amp and a push-pull pair of transistors?

 

It looks like this circuit produces a 160 V peak-to-peak squarewave. Why do you think you can get that from a 12 V circuit with no transformer?

Also, the circuit has the same GND symbol on both sides of the transformer, indicating that there is no isolation between the primary and secondary sides. This means that the microcontroller driving U5B and everything else upstream also are not isolated from the AC mains.

ak

 

The original version of this unit could be powered from AC mains, 12VDC vehicle power, or internal 12VDC battery by way of a power distribution board. For my purposes, and for safety's sake, I want to use 12V for that section and eliminate that 'mains DC' supply and just feed it from 12VDC.

I have seen a schematic of a newer version where the transmit stage is powered at a much lower voltage from a dedicated 35VAC winding on the transformer, so if I adjust the windings on the toroidal transformer (which I need to do anyway), the lower voltage shouldn't be an issue? That newer version had a far simpler output stage, as I said, an op-amp and push-pull transistors.

In the default 'load terminal' arrangement, the T1 primary and secondary circuits are completely isolated, as they should be, but in the 'configuration unit' arrangement, a jumper was connected from the low-voltage DC ground to the negative of C10(!!!).

 

Here's another newer version of this circuit I found in my notes.