Divide by 2, H-Bridge, step-down, switched-capacitor, charge-pump - Operation principle of is shown the attached jpg.
Suppose we look at node “bottom”.
In steady-state it toggles between voltages (Vin+ε)/2 and 0V, while Vout is at ~(Vin-ε)/2.
i.e. the bottom right transistor experiences OFF and ON states in opposite polarities !
The involved voltages are >>5V so NMOS and PMOS are asymmetrical, due to the multiple wells that enable the high VDS operation.
For example for PMOS:
VSD>0 and VSG>Vth => Isd>0 (ON state)
VSD>0 and VSG<Vth => Isd~0 (OFF state)
But when VSD<0 regardless of VSG, the diodes between Drain and Bulk will conduct.
I.e. reverse SD polarity enable current through diodes.
Should the Bottom Right switch be realized by a PMOS?
And if yes, do you agree it will run negative direction current through the its diodes?
Is it OK?
Should it work like that?
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Following FvM replies @ #2 & #4 (please confirm/reject) my summary is:
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in circuits like switched-capacitors charge-pumps, which are based on
high-voltage,asymmetrical-MOSFETs,with reverse polarity body-diodes:
*Devices OFTEN experience straight and reverse polarity (Vsd>0 and Vsd<0) - depend on topology.
*For OFF state - they MUST be ONLY in straight polarity (otherwise the body diodes will open).
*For ON state - they are allowed to experience reverse polarity. In this condition, the channel will run current in the reverse direction, and if Ion*Ron>Vdiode, then, in addition, the diode will open and run forward current in the reverse polarity. For Ion*Ron>>Vdiode, most of the current will run through the body diode.////
- Was it an accurate statement?
- Is such current through a body diode typically safe for asymmetrical MOSFET?
- Is it typically modeled in MOSFETs? By discrete devices vendors and BCD IC process fabs?
Thanks