Numerator: kT

The passage around this formula

The reason that fourth side matters can be written down directly. A transistor’s subthreshold swing — how many millivolts of gate voltage it takes to change the drain current by a factor of ten — is well approximated by $S = \frac{kT}{q}\ln(10)\left(1 + \frac{C_{\mathrm{dep}}}{C_{\mathrm{ox}}}\right)$. where $C_{\mathrm{dep}}$ is the capacitance of the depletion region the gate does not fully control and $C_{\mathrm{ox}}$ is the gate’s own oxide capacitance. As $C_{\mathrm{dep}}$ falls toward zero, S falls toward its physical floor of about 60 millivolts per decade at room temperature. Wrapping the gate fully around the channel is, in this equation, a direct attack on $C_{\mathrm{dep}}$ : there is no substrate face left for the depletion region to leak through,…

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