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For t_{ox} = 90 A° , we have C_{o x}=3.84 fF / \mu m ^2 Using Eq. (7.42), we write

f_C=\frac{K}{\gamma C_{o x} W L} g_m \frac{1}{4 k T} (7.42)500\; kHz =\frac{K}{3.84 \times 100 \times 0.5 \times 10^{-15}} \cdot \frac{1}{100} \cdot \frac{3}{8 \times 1.38 \times 10^{-23} \times 300} (7.43)

That is, K = 1.06 × 10^{−25} V^2F.