Chapter 3 · Visualization 6 of 6 · Section 3.2
C–V With a Contacted Inversion Layer
With an n⁺ region next to the gate, electrons enter and leave the inversion layer in nanoseconds, so the capacitance returns to C′ox in inversion at any frequency. VCB moves the rise.
Small-signal gate capacitance C′gb = dQ′G /dVGB at fixed VCB
Where the small-signal charge goes at your VGB
What to look for 1/C′gb = 1/C′ox + 1/(C′b + C′i ) (all charges follow the signal)
Below VTB all curves are the same: accumulation and depletion do not know about terminal C. Above VTB the inversion electrons, supplied by the n⁺ region, answer the signal at the surface: C′ returns to C′ox . Larger VCB : the dip is wider and deeper, because the depletion region grows further before inversion begins. Try this Measure VTB from the rise of the curve for VCB = 0, 1, 2 V and compute VT = VTB − VCB . Fit γ. Set VGB just above VTB . How is the signal charge split between electrons and the depletion edge? In depletion the isolated and contacted curves coincide, whatever VCB . Why? Thin the oxide to 2 nm. Why does the dip become shallower?
Model: exact charge-sheet equations of Chapter 3 (3.2.8), (3.2.21), with hole, acceptor and electron terms, Boltzmann statistics, ni = 10¹⁰ cm⁻³, φ0 = 2ΦF + 6φt , Q′o /q = 10¹⁰ cm⁻², T = 300 K. The same equations are in the Colab notebook. Contacted curve: numerical derivative of Q′G = −Q′C − Q′o from (3.2.8) at fixed VCB . Isolated curve: Chapter 2 high-frequency model. Section 3.2 and Fig. 3.6 of Tsividis and McAndrew, Operation and Modeling of the MOS Transistor .