jason_class wrote on Mar 14th, 2006, 4:32pm:Hi Murphy
Thank you for spending time on my enquiry.
Vb is the separation between the Ei and Ef at the substrate side (The potential)
As for Vfb flatband voltage but in the equation, the Qi/Ci is assumed to be zero ,
making Vfb= Vms
Eg is the bandgap energy
I saw in the book for the below 3 cases;
1) n+ poly gate on p silicon, Vms= - (Eg/2q)- Vb
2)p+ poly gate on n silicon , Vms = +(Eg/2q) +Vb
3)p+ poly gate on p silicon , Vms = +(Eg/2q) -Vb
Do you know how to get the equation for the 3) case?
That is my exact question.
Kindly enlighten
Thank you Murphy
best regards
jason
Hi Jason,
You can derive all three equations very easily in the following manner:
1. Draw the band diagram of the bulk semiconductor, and of the gate polysilicon separately.
Both have the same Ec (Conduction band), Ev (Valence band) and Ei (Intrinsic level). Only the Ef (Equilibrium Fermi level) is different.
2. The bulk is lightly doped and so (Ei-Ef) is finite. For the gate, assume that Ef is either at Ec (n+ gate) or Ev (p+ gate).
3. Now, the Vfb can be easily found by taking the difference of potentials that would be required to align the 2 Fermi levels on either side. The (Eg/2q) is just half the bandgap.
For your first question, (p+gate on p bulk), Ef(gate) = Ev.
Thus, aligning the 2 sides needs the gate to be raised up by Ef(bulk)-Ev which is equal to
(Ei-Ev)-(Ei-Ef(bulk)) = (Eg/2) - q*Vb
You can derive the other 2 equations in a similar manner.
Regards
Vivek