MOSFET gate driver trasformer isolated

Hi
I was working on my old LCD TV ccfl power supply. Incidentally the TV was repaired by a non authorized TV repaier technician and now was problem associted with CCFL gate driver, I traced the circuit diagram for isolated gate driver for the MOSFET to be similar to the image attached. however I am unable to read the components value from the board as they were damaged. Can anyonw help me analyse this circuit from texas instrument website so a to calculate rough or approximate values for this gate driver. FYI the TV was a toshiba 40BP20 serries and I don't have service manual or schematic of the exact circuit. The power supply has a part number PSIV201141A V71A00025300


URL: https://e2e.ti.com/blogs_/b/powerho...n?hootPostID=44c899267481bcf404f3b1e6581e830f

The value of CB will depend on the primary inductance. You will need to tell the primary and secondary inductances, but we presume its 1:1.
The Q1, Q2 can be any 40V rated PNP of the “high gain at high IC “ type …eg the ones from diodes.com
It is unfortunate that the secondary damping resistor has been placed so it can give a high VEB voltage on the PNP, and therefore damage it……it should not be placed in this way.
Also, it really needs a primary side damping resistor to damp out LC ringing. But again we need the primary inductance to tell this….the damping resistor is usually about SQRT(Lp/CB).

Looking at the circuit , i suspect damage due to undamped LC ringing, or the too-high VEB on the PNP as discussed.

The zeners should be just say a volt or two below the VGS max of the FETs

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The gate-source resistor... say 10k or so...after all, you have the PNP turn-off circuit.
The diodes, makes them 1N4148's or similar....fast .....
The base resistor.....depends on the hfe of your chosen PNP....just make it low enough to give enough IC for quick-enough turn off......say 510R or so to start.

The red capacitors...these are y caps...just leave them off altogether, they are for EMC.