MOS capacitance when Vgs is negative

[iVenky]

I have a question on what happens to MOS capacitance Cgs+Cgd, when Vgs goes negative in the NMOS shown. I see that when Vgs goes negative, the channel is full of holes because of p- substrate, which means we see the Cox capacitance (without any other capacitance in series), but I am not sure if this is between Cgs (or Cgd) or Cgb. My intuition says it's between Cgb, but if that's the case, then there is no capacitance between Cgs (or Cgd), ignoring the overlap capacitance.
Attached the figure here

https://imgur.com/a/LPlIcKX

 

[trurle]

I have a question on what happens to MOS capacitance Cgs+Cgd, when Vgs goes negative in the NMOS shown. I see that when Vgs goes negative, the channel is full of holes because of p- substrate, which means we see the Cox capacitance (without any other capacitance in series), but I am not sure if this is between Cgs (or Cgd) or Cgb. My intuition says it's between Cgb, but if that's the case, then there is no capacitance between Cgs (or Cgd), ignoring the overlap capacitance.
Attached the figure here

https://imgur.com/a/LPlIcKX

The capacitance (between gate and source/drain) of NMOS in depletion is reduced because no channel mean Cgb is in series with Csb+Cdb.
with negative voltage, the capacitance is limited by Csb+Cdb, and reverse proportional to square root of voltage between gate and drain/source - because of depletion regions forming around source/drain

 

[evi7538]

Actually when you apply a small negative voltage to the gate the device enters a depletion mode and in that mode what Trurle said is true. But then if you decease the Vgs even lower the device enter the accumulation mode and the gate-bulk capacitance increases again back to the same value that it had in the inversion mode (when Vgs>Vth). In the accumulation mode in the first approximation (neglecting the channel resistance) the drain and source are shorted through the low-resistance of the channel so all capacitors Cgb, Cgd and Cgs are connected in parallel. In a more accurate approximation you have a distributed RC-network across the channel.