Question 19.3: An A/D converter incorporates a resistor ladder consisting o...
An A/D converter incorporates a resistor ladder consisting of 128 units made of n-well to generate equally-spaced reference voltages (Fig. 19.33). If the two ends of the ladder are connected to V_1=+1 V \text { and } V_2=+2 V , calculate the ratio R_{128} / R_1 .
The width of the depletion region inside the n-well is given by x_d=\sqrt{2 \epsilon_{s i}\left(\phi_B+V_R\right) /\left(q N_{w e l l}\right)}, where N_{w e l l} denotes the n-well doping level and V_R the reverse bias voltage. Assuming that the zero-bias depth of the n-well is equal to t_0, we have
\begin{aligned} \frac{R_{128}}{R_1} &=\frac{t_0-\sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}}\left(\phi_B+V_1\right)}+\sqrt{\frac{2 \epsilon_{s i}}{q N_{w e l l}} \phi_B}}{t_0-\sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}}\left(\phi_B+V_2\right)}+\sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}} \phi_B}} &(19.10)\\ &=\frac{t_0+\sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}} \phi_B}\left(1-\sqrt{1+\frac{V_1}{\phi_B}}\right)}{t_0+\sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}} \phi_B}\left(1-\sqrt{1+\frac{V_2}{\phi_B}}\right)}&(19.11) \end{aligned}If the difference between R_1 \text { and } R_{128} is small, we can divide the numerator and denominator of (19.11) by t0 and approximate the result as
\begin{aligned} \frac{R_{128}}{R_1} & \approx\left[1+\frac{1}{t_0} \sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}} \phi_B}\left(1-\sqrt{1+\frac{V_1}{\phi_B}}\right)\right]\left[1-\frac{1}{t_0} \sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}} \phi_B}\left(1-\sqrt{1+\frac{V_2}{\phi_B}}\right)\right] &(19.12)\\ & \approx 1+\frac{1}{t_0} \sqrt{\frac{2 \epsilon_{s i}}{q N_{\text {well }}} \phi_B}\left(\sqrt{1+\frac{V_2}{\phi_B}}-\sqrt{1+\frac{V_1}{\phi_B}}\right)&(19.13) \end{aligned}For example, if t_0=2 \mu m , N_{\text {well }}=10^{16} cm ^{-1} \text {, and } \phi_B=0.7 V , the mismatch between R_{128} \text { and } R_1 is nearly 60%.