MOSFET gate capacitance in strong inversion

This lecture (page 4 and 5) said about gate capacitance of a MOSFET in various operating condition.

From the table below, as you can see in strong inversion $C_{ox}$ and junction capacitance $C_{j}$ are considered to be in parallel.

How is it possible for them to be in parallel here? Could anyone explain that?

Thank you.

• I think that it relates to the ("new") definition of capacitance: $C = \frac{d Q}{d V}$. Let's define $Q_d \triangleq Q_{depletion}$. In Strong Inversion, for low frequencies, $Q_d$ does NOT change - it's constant. That means $\frac{d Q_d}{d V} = 0$ . If it were in series then that would imply that the charge of $C_{ox}$ shouldn't have changed (but it does change). – Dor Jul 3 '16 at 22:40
• Thank you. What is V in your expression? Also I am confused why the lecture said that Cj and Cox are in parallel here. – anhnha Jul 4 '16 at 11:48
• V is the voltage between Bulk and Gate. Cj and Cox are in parallel for the reason that I already wrote: $C = \frac{dQ}{dV} = \frac{dQ_{ox}}{dV} + \frac{dQ_{dep}}{dV} = \frac{dQ_{ox}}{dV}$ (it works out, because that $Q_{dep}$ does not change with voltage when in Strong Inversion) – Dor Jul 4 '16 at 12:55
• Thank you. However, from your formula above C = Cox + Cdep. And the two are always in parallel. But as from the table, it is only parallel in strong inversion not in the four remaining regions. Could you explain? – anhnha Jul 4 '16 at 14:45
• That is because of two things: (1) The new definition of capacitance: $C \triangleq \frac{dQ}{dV}$, which implies that $C \ne 0$ only when the charge varies with voltage. (2) The fact that only at Strong Inversion the charge $Q_{dep}$ does not change. In all other cases - $Q_{dep}$ changes with voltage. By this behavior we change the equivalent electrical schematic of the MOS structure. If $Q_{dep}$ were to change with voltage at Strong Inversion, then $C_{dep}$ and $C_{ox}$ were considered to be connected in series. – Dor Jul 4 '16 at 17:35