For Vsb=0, Vth=2*phi+Qdep/Cox

The energy bands Ei and Ef have a difference of 2*q*phi representing that the surface of the silicon has the same concentration as the bulk, but a different polarity.

The depletion charge is given by: sqrt(2*e*Nsub*q*V), where e is the dielectric constant and V is the voltage across the negative charge half of the junction, i.e. from bulk to surface...V=2*phi.

Now, when Vsb!=0, Vth changes to:
Vth,bias=2*phi + Qdep,bias/Cox

where Qdep,bias=sqrt(2*e*Nsub*q*V) where now V=2*phi-Vbs

I am confused, because I would think Vth,bias=2*phi-Vbs + Qdep,bias/Cox.  That is to say, Qdep changes because of the applied voltage, so it would seem like the difference between bulk and surface would also change to 2*phi-Vbs.  But, this isn't correct...I looked it up in my Streetman devices text.

So, why does the difference between Ei and Ef still have to equal q*2*phi?  I would say because the surface still needs to have the same concentration of negative charge for strong inversion.  I think what's confusing me is that the voltage from bulk to surface remains at 2*phi, but the voltage that defines the depletion layer charge has changed to 2*phi-Vbs.  

If the voltage difference between bulk and surface is 2*phi, how can the voltage defining the depletion layer simultaneously be 2*phi-Vbs???  My explanation is that the charge concentration at the surface remains the same as it was when Vbs=0, but now there is less depletion region charge and more carrier charge.  Another way of saying this is the voltage drop across the bound charge has changed, which changes the width of the depletion layer (as evidenced by Qdep,bias), but the total voltage drop between bulk and surface must remain at 2*phi, to maintain the right concentration for strong inversion; the remaining voltage (2*phi-(2*phi-Vbs)=Vbs) is dropped across the mobile inversion charge.

Does this make sense and is it correct?

Do the energy band diagrams for Ei and Ef look the same for both cases where Vbs=0 and Vbs!=0?  I am basing my thoughts on the fact that they remain the same, and the difference is still q*2*phi (which is correct).

Thanks,
Marc

Here is a page with some better versions of the equations:
http://ece-www.colorado.edu/~bart/book/book/chapter7/ch7_4.htm

Did anyone make sense of my long-winded post?

My real question is, how come the depletion charge is determined by 2*phi-Vbs but the voltage from bulk to surface remains steady at 2*phi?

Thanks,
Marc