\begin{array}{l} V_{\text {ox }} \text { , the voltage across a } 2 \mathrm{nm} \text { thin oxide, is }-1 \mathrm{~V} \text { . The } \mathrm{P}^{\text {+ }} \text

{ poly- } \\ \text { gate doping is } N_{\text {poly }}=8 \times 10^{19} \mathrm{~cm}^{-3} \text {and substrate } N_{d} \text { is } 10^{17} \mathrm{~cm}^{-3} \text {. } \\ \text { Find (a) } W_{\text {dpoly }}, \text { b) } \phi_{\text {poly }}, \text { and }(\mathrm{c}) V_{g} \end{array} \begin{aligned} \text { (b) } \quad W_{\text {dpoly }}=\sqrt{\frac{2 \varepsilon_{s} \phi_{\text {poly }}}{q N_{\text {poly }}}} \\ \phi_{\text {dpohy }}=q N_{\text {poly }} W_{\text {dpoly }}^{2} / 2 \varepsilon_{s}=0.11 \mathrm{~V} \end{aligned} \begin{aligned} \text { (c) } V_{g}=& V_{\beta b}+\phi_{s t}+V_{\alpha x}+\phi_{p o b y} \\ & V_{f b}=\frac{E_{g}}{q}-\frac{k T}{q} \ln \left(\frac{N_{c}}{N_{d}}\right)=1.1 \mathrm{~V}-0.15 \mathrm{~V}=0.95 \mathrm{~V} \\ V_{0}=& 0.95 \mathrm{~V}-0.85 \mathrm{~V}-1 \mathrm{~V}-0.11 \mathrm{~V}=-1.01 \mathrm{~V} \end{aligned}