Question

Derive (23) from (20) and (22) $|Q_n(y)| = [V_G - V_{FB} - \Delta\psi(y) - 2\psi_B]C_{ox} - \sqrt{2\epsilon_s q N_A} [\Delta\psi(y) + 2\psi_B]. (20) I_D = \frac{Z}{L}\int_0^L |Q_n(y)|\psi(y)dy. (22) I_D = \frac{Z\mu_n}{L}\int_0^L |Q_n(y)|\mathcal{E}(y)dy = \frac{Z\mu_n}{L}\int_0^L |Q_n(\Delta\psi)|\frac{d\Delta\psi(y)}{dy}dy = \frac{Z\mu_n}{L}\int_0^{V_D} |Q_n(\Delta\psi)|d\Delta\psi, = \frac{Z\mu_n}{L}C_{ox}\left\{\left(V_G - V_{FB} - 2\psi_B - \frac{V_D}{2}\right)V_D - \frac{2}{3}\frac{\sqrt{2\epsilon_s q N_A}}{C_{ox}}[(V_D + 2\psi_B)^{3/2} - (2\psi_B)^{3/2}\right\}. (23)

          Derive (23) from (20) and (22)
$|Q_n(y)| = [V_G - V_{FB} - \Delta\psi(y) - 2\psi_B]C_{ox} - \sqrt{2\epsilon_s q N_A} [\Delta\psi(y) + 2\psi_B]. (20)
I_D = \frac{Z}{L}\int_0^L |Q_n(y)|\psi(y)dy. (22)
I_D = \frac{Z\mu_n}{L}\int_0^L |Q_n(y)|\mathcal{E}(y)dy = \frac{Z\mu_n}{L}\int_0^L |Q_n(\Delta\psi)|\frac{d\Delta\psi(y)}{dy}dy = \frac{Z\mu_n}{L}\int_0^{V_D} |Q_n(\Delta\psi)|d\Delta\psi,
= \frac{Z\mu_n}{L}C_{ox}\left\{\left(V_G - V_{FB} - 2\psi_B - \frac{V_D}{2}\right)V_D - \frac{2}{3}\frac{\sqrt{2\epsilon_s q N_A}}{C_{ox}}[(V_D + 2\psi_B)^{3/2} - (2\psi_B)^{3/2}\right\}. (23)

$Derive (23) from (20) and (22) |Qn(y)| = [VG - VFB - Δψ(y) - 2]Cox - √(2 q NA) [Δψ(y) + 2]. (20) ID = (Z)/(L)∫0^L |Qn(y)|ψ(y)dy. (22) ID = (Z)/(L)∫0^L |Qn(y)|ℰ(y)dy = (Z)/(L)∫0^L |Qn(Δψ)|(dΔψ(y))/(dy)dy = (Z)/(L)∫0^VD |Qn(Δψ)|dΔψ, = (Z)/(L)Cox{(VG - VFB - 2 - (VD)/(2))VD - (2)/(3)(√(2 q NA))/(Cox)[(VD + 2)^3/2 - (2)^3/2}. (23)$

Added by Sharon C.

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