The excerpt below is from Multi-level conversion: High voltage choppers and voltage-source inverters by T.A. MEYNARD, H. FOCH.


From the figure, how do you know that the voltage drop across each switch is E/2? For realistic model, every switch has parasitic capacitance between its terminal (and maybe finite open switch resistance). So the voltage drop across each switch is E/2.
Do you agree with this explanation?

Now let's assume that the switch is ideal (no parasitic capacitance, infinite open-switch resistance), what is the voltage drop across each switch? Is it still E/2?

In the field of High Voltage Power Conversion, the circuit designer is often confronted to a serious problem: there are no semiconductors capable of sustaining the desired voltage (traction application for example). For this reason, circuit designers proposed several converter topologies in which only a fraction of the voltage is applied to each switch.


The first solution involves series connection of several switches controlled in synchronism, thus obtaining the equivalent of a high voltage switch (Fig. 1).

enter image description here


In case of ideal model, in principle you could have any combination of voltage, provided that the sum is E. However, if the devices are exaclty equal, for symmetry and "continuity" (i.e. limit R-->inf) reasons, the voltage on each device would be still E/2.

Instead, I have doubt on the real devices, as, for sure, they are polarized in slightly different conditions, hence their off-resistance might not be exactly equal (not no mention device-to-device parameter dispersion).


You don't. The node between the two open switches has no well-defined voltage. If you try to measure the voltage with a meter (effectively a 10-megohm resistor), it will change the voltage on that node to be the same as the other node you're measuring against, producing a voltage drop of zero.

You need additional components across the switches if you want the open-circuit voltage to divide evenly.


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