Creepage distance accross insulation of step up transformer


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I am a little confused about determination of creepage distances. There is switchin PSU in which mains voltage (230) are transformed to 400V and transformed again to 1500 V (output) and then rectified to DC. 60601-1 standard requires to apply creepage distance according working voltage accross that insulation. Working voltage is defined as the maximum voltage accross that insulation. In this case the maximum voltage in transformer (400 V to 1500 V) is 1500 V, so I should apply values for 1500 V?
Somewhere I have found that voltages are summed, but in this case (400 V + 1500 V = 1900 V) the creepage distance would be enourmously large.
Also if my output "-" is earthed. Therefore, in case we intend to make protection from dangerous currents/voltages on the enclosure we should provide protection from positive output to negative output. However, positive and negative outputs are connected by a diode bridge rectifier and some other components (like capacitors etc.). So, these components (diodes, capacitors) are bridging a MEANS OF PROTECTION just the same way as resistors are mentioned in Subclause 4.8. In fact, these compotes are mounted in parallel to feedback resistors. I do doubt that we will be able to find diodes, which can "have a construction that cannot fail short-circuit".
Thnak you for your answers in advance!


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What is this circuit used for / connected to?

Creepage distances are required between Hazardous Voltages and Accessible Parts or Accessible Circuits - they are not required between Harzardous Voltages and Hazardous Voltages


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I attached insulation diagram bellow:
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