Let us assume that an NMOSFET has its source, drain and substrate grounded therefore VDS and VSB=0 V . We apply a gate voltage VGS=2 V and I want to know how this potential is distributed as we move from gate towards substrate.
Let us assume the threshold voltage to be VT=0.7 V. Now I know that since the MOSFET is in inversion, a charge density Qinv= Cox*(VGS - VT) will appear in the inversion layer. The threshold voltage will be divided into two parts:
- Change in surface potential from φS to -2φf where φf is the bulk fermi potential.
- To support the depletion region charge due to increased depletion region width at inversion.
Let us assume that VT is equally divided between these two components and assign 0.35 V to both. Now I believe that potential at the surface under the gate must be atleast at 0.35 V because we still need to account for bending of band and change of surface potential but since the MOSFET is under inversion, an inversion layer links the surface, drain and source and therefore they must be at 0 V (since we assumed they are grounded). So where am I going wrong?
This demonstration simulates the effect of an external bias on a MOS capacitor. I checked the graph of electrostatic potential as I increased the gate voltage and I found that VSurface increases indicating that some part of applied voltage is used to increase the surface potential to cause inversion. There is a potential difference between surface and substrate. But since in MOSFET we have grounded source, drain and substrate to 0 V and source, drain and surface are connected through an inversion layer in inversion mode, we have set VSurface to be at 0 V. So there is no potential difference between surface and substrate. Then how can we achieve inversion?