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.

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    \$\begingroup\$ 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). \$\endgroup\$ – Dor Jul 3 '16 at 22:40
  • \$\begingroup\$ Thank you. What is V in your expression? Also I am confused why the lecture said that Cj and Cox are in parallel here. \$\endgroup\$ – anhnha Jul 4 '16 at 11:48
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    \$\begingroup\$ 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) \$\endgroup\$ – Dor Jul 4 '16 at 12:55
  • \$\begingroup\$ 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? \$\endgroup\$ – anhnha Jul 4 '16 at 14:45
  • \$\begingroup\$ 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. \$\endgroup\$ – Dor Jul 4 '16 at 17:35

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