Question 10.1: Maximum Electric Field Sketch the maximum lateral field ξm ...
Maximum Electric Field
Sketch the maximum lateral field ξ_m as a function of V_D .
ξ_m has different functional forms depending on whether V_D is larger or smaller than V_{Dsat}. \text{For } V_D\lt V_{Dsat} , the device is in the linear region. From Equation 9.2.8, we have
I_D=C_{ox}W[V_G-V_T-V(y)]\frac{\mu _{eff}\xi (y)}{1+[\xi (y)/\xi _{sat}]} (9.2.8)
Expressing the field as the negative derivative of voltage and integrating to an arbitrary location y, we have
I_Dy=\mu WC_{ox}\left(V_G-V_T-\frac{V(y)}{2} \right) V(y)\frac{1}{1+(V_D/\xi _{sat}y)}By rearranging terms, we can solve for V(y):
V(y)=(V_G-V_T)-\sqrt{(V_G-V_T)^2-\frac{2[1+(V_D/\xi _{sat}y)]I_Dy}{\mu WC_{ox}} }The field ξ(y) is
\xi (y)=-\frac{dV(y)}{dy} =\frac{I_D}{\mu WC_{ox}\sqrt{(V_G-V_T)^2-\frac{2[1+(V_D/\xi _{sat}y)]I_Dy}{\mu WC_{ox}} }}ξ _m at the drain (y = L) is
\xi _m(V_D)=\frac{\frac{\mu WC_{ox}}{L}\left(V_G-V_T-\frac{V_D}{2} \right)V_D\frac{1}{1+(V_D/\xi _{sat}L)} }{\sqrt{[\mu WC_{ox}(V_G-V_T)]^2-2(\mu WC_{ox})^2\left(V_G-V_T-\frac{V_D}{2} \right)V_D } }For V_D > V_{Dsat}, \ ξ_m is given by Equation 10.1.14 and can be approximated by Equation 10.1.15.
\xi _m=\left[\frac{(V_D-V_{Dsat})^2}{\ell ^2}+\xi _{sat}^2 \right] ^{1/2} (10.1.14)
\xi _m\approx \frac{(V_D-V_{Dsat})}{\ell } \ \ \ \text{ for } \ \ \ (V_D-V_{Dsat})\gt 1 \ V (10.1.15)
Hence, a sketch of the field can be drawn.
