do you know the leakage inductance of your transformer?
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Let us assume the low figure of 18pF per diode, at 200kHz this is 44k-ohm, lets say we have 900 volt reverse and 1V forward across the diode at the 200kHz, this is approx 315Vrms ( AC ) giving a capacitive current of 7mA (much higher peak current for a square wave applied voltage), if the true capacitance is a bit higher say 36pF the current is 14mA rms + peaks, this would be 900 x 0.14 = 12.6 VA out of the transformer - just for this one 1kV stage ( x100 for 100kV = 1260VA), before any real power is being delivered. This also ignores the losses of the snubbers you will find necessary to use to stop the diodes going pop. Say we use 33pF(3kV) and 1k-ohm, losses due to snubber will be 2.916W at 200kHz ( x 100 = 291.6 watts), so each 1k-ohm resistor needs to be 5W. Also let us assume the reverse recovery of the diodes is a bit worse than the data sheet suggests - if we drive the diodes with a square wave rising from zero to 900V in 250nS say the dv/dt in the diode capacitance alone will be: [i/C = dv/dt] = 64.8mA for 18pF - it is likely the reverse rec current will be a few times higher than this for square wave drive - all this current builds up in the Tx leakage inductance - then the diodes try to turn off with the current still trying to flow backwards thru them - if they go overvolts then they go bang, without snubbers there will be a LOT of HF ringing - radiating RFI every where - including into your 200kHz control ckt - with snubbers the ringing and overvolt spike may be suppressed to low enough levels to stop the diodes going pop - but they will absorb a lot of heat at 200kHz if they are big enough...
So you see - a few issues...
One way to compensate for the above is to have a sine wave voltage drive out of the Tx as this limits the dv/dt at the zero xing - peak dv/dt lower at 20kHz than at 200kHz - but perhaps you know better than me ...