Chapter 7, Problem 9E

To determine

Find the capacitance of the diode if $V_A=-1\text{ V}$, if $V_A=-3\text{ V}$ and if $V_A=-10\text{ V}$.

Answer to Problem 9E

The capacitance of the diode if $V_A=-1\text{ V}$ is $\text{10}\text{.1582 fF}$, if $V_A=-3\text{ V}$ is $\text{6}\text{.7952 fF}$ and if $V_A=-10\text{ V}$ is $\text{3}\text{.9674 fF}$.

Explanation of Solution

Given Data:

The given expression for the junction capacitance is $\frac{K_s{\epsilon }_{0}A}{W}$,

The given expression of depletion width of diode is $\sqrt{\frac{2K_s{\epsilon }_0}{qN}\left(V_{bi}-V_A\right)}$.

The cutoff voltage of the diode is $0.62\text{ V}$,

Magnitude of charge of electron is $1.6\times {10}^{-19}\text{ C}$,

Value of $K_s$ is given as $11.8$ for silicon diode,

The cross sectional area is given as $2\mu {\text{m}}^2$ and

Volume density of electrons is given as $5.0\times {10}^{18}{\text{ cm}}^{-3}$.

Formula used:

The expression for junction capacitance is as follows,

$C_j=\frac{K_s{\epsilon }_{0}A}{W}$ (1)

Here,

$A$ is the cross sectional area,

$W$ is the depletion width,

$C_j$ is the junction capacitance,

${\epsilon }_0$ is the permittivity of free space and

$K_s$ is the constant.

The expression for depletion width of the diode is as follows,

$W=\sqrt{\frac{2K_s{\epsilon }_0}{qN}\left(V_{bi}-V_A\right)}$ (2)

Here,

$q$ is the magnitude of charge of electron,

$N$ is the volume density of electrons,

$V_{bi}$ is the cutoff voltage of diode,

$V_A$ is the voltage across the diode and

$W$ is the depletion width.

Calculation:

Substitute $11.8$ for $K_s$, $1.6\times {10}^{-19}\text{ C}$ for $q$, $5.0\times {10}^{18}{\text{ cm}}^{-3}$ for $N$, $0.62\text{ V}$ for $V_{bi}$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_0$ and $-1\text{ V}$ for $V_A$ in equation (2),

$\begin{array}{c}W=\sqrt{\frac{2\times 11.8\times 8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}}{\left(1.6\times {10}^{-19}\text{ C}\right)\times \left(5.0\times {10}^{18}{\text{ cm}}^{-3}\right)}\left(\left(0.62\text{ V}\right)-\left(-1\text{ V}\right)\right)}\\ =\sqrt{\frac{2\times 11.8\times 8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}}{\left(1.6\times {10}^{-19}\text{ C}\right)\times \left(5.0\times {10}^{18}\times {\text{10}}^6{\text{ m}}^{-3}\right)}\left(\left(0.62\text{ V}\right)-\left(-1\text{ V}\right)\right)}\left\{\because {\text{ 1 cm}}^{-3}={10}^6{\text{ m}}^{-3}\right\}\\ =2.057\times {10}^{-8}\text{ m}\end{array}$

Substitute  $11.8$ for $K_s$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_0$, $2.057\times {10}^{-8}\text{ m}$ for $W$ and $2\mu {\text{m}}^2$ for $A$ in equation (1),

$\begin{array}{c}{C}_j=\frac{11.8\times \left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times \left(2\mu {\text{m}}^2\right)}{2.057\times {10}^{-8}\text{ m}}\\ =\frac{11.8\times \left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times \left(2\times {10}^{-12}{\text{ m}}^2\right)}{2.057\times {10}^{-8}\text{ m}}\left\{\because \text{ 1 }\mu {\text{m}}^2={10}^{-12}{\text{ m}}^2\right\}\\ =101.582\times {10}^{-16}\text{ F}\\ \text{=10}\text{.1582 fF }\left\{\because \text{ 1 F}={\text{10}}^{15}\text{ fF}\right\}\end{array}$

Substitute $11.8$ for $K_s$, $1.6\times {10}^{-19}\text{ C}$ for $q$, $5.0\times {10}^{18}{\text{ cm}}^{-3}$ for $N$, $0.62\text{ V}$ for $V_{bi}$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_0$ and $-3\text{ V}$ for $V_A$ in equation (2),

$\begin{array}{c}W=\sqrt{\frac{2\times 11.8\times 8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}}{\left(1.6\times {10}^{-19}\text{ C}\right)\times \left(5.0\times {10}^{18}{\text{ cm}}^{-3}\right)}\left(\left(0.62\text{ V}\right)-\left(-3\text{ V}\right)\right)}\\ =\sqrt{\frac{2\times 11.8\times 8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}}{\left(1.6\times {10}^{-19}\text{ C}\right)\times \left(5.0\times {10}^{18}\times {\text{10}}^6{\text{ m}}^{-3}\right)}\left(\left(0.62\text{ V}\right)-\left(-3\text{ V}\right)\right)}\left\{\because {\text{ 1 cm}}^{-3}={10}^6{\text{ m}}^{-3}\right\}\\ =3.075\times {10}^{-8}\text{ m}\end{array}$

Substitute  $11.8$ for $K_s$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_0$, $3.075\times {10}^{-8}\text{ m}$ for $W$ and $2\mu {\text{m}}^2$ for $A$ in equation (1),

$\begin{array}{c}{C}_j=\frac{11.8\times \left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times \left(2\mu {\text{m}}^2\right)}{3.075\times {10}^{-8}\text{ m}}\\ =\frac{11.8\times \left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times \left(2\times {10}^{-12}{\text{ m}}^2\right)}{3.075\times {10}^{-8}\text{ m}}\left\{\because \text{ 1 }\mu {\text{m}}^2={10}^{-12}{\text{ m}}^2\right\}\\ =67.952\times {10}^{-16}\text{ F}\\ =\text{6}\text{.7952 fF }\left\{\because \text{ 1 F}={\text{10}}^{15}\text{ fF}\right\}\end{array}$

Substitute $11.8$ for $K_s$, $1.6\times {10}^{-19}\text{ C}$ for $q$, $5.0\times {10}^{18}{\text{ cm}}^{-3}$ for $N$, $0.62\text{ V}$ for $V_{bi}$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_0$ and $-10\text{ V}$ for $V_A$ in equation (2),

$\begin{array}{c}W=\sqrt{\frac{2\times 11.8\times 8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}}{\left(1.6\times {10}^{-19}\text{ C}\right)\times \left(5.0\times {10}^{18}{\text{ cm}}^{-3}\right)}\left(\left(0.62\text{ V}\right)-\left(-10\text{ V}\right)\right)}\\ =\sqrt{\frac{2\times 11.8\times 8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}}{\left(1.6\times {10}^{-19}\text{ C}\right)\times \left(5.0\times {10}^{18}\times {\text{10}}^6{\text{ m}}^{-3}\right)}\left(\left(0.62\text{ V}\right)-\left(-10\text{ V}\right)\right)}\left\{\because {\text{ 1 cm}}^{-3}={10}^6{\text{ m}}^{-3}\right\}\\ =5.2667\times {10}^{-8}\text{ m}\end{array}$

Substitute  $11.8$ for $K_s$, $8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}$ for ${\epsilon }_0$, $5.2667\times {10}^{-8}\text{ m}$ for $W$ and $2\mu {\text{m}}^2$ for $A$ in equation (1),

$\begin{array}{c}{C}_j=\frac{11.8\times \left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times \left(2\mu {\text{m}}^2\right)}{5.2667\times {10}^{-8}\text{ m}}\\ =\frac{11.8\times \left(8.854\times {10}^{-12}\frac{\text{F}}{\text{m}}\right)\times \left(2\times {10}^{-12}{\text{ m}}^2\right)}{5.2667\times {10}^{-8}\text{ m}}\left\{\because \text{ 1 }\mu {\text{m}}^2={10}^{-12}{\text{ m}}^2\right\}\\ =39.674\times {10}^{-16}\text{ F}\\ =\text{3}\text{.9674 fF }\left\{\because \text{ 1 F}={\text{10}}^{15}\text{ fF}\right\}\end{array}$

Conclusion:

Thus, the capacitance of the diode if $V_A=-1\text{ V}$ is $\text{10}\text{.1582 fF}$, if $V_A=-3\text{ V}$ is $\text{6}\text{.7952 fF}$ and if $V_A=-10\text{ V}$ is $\text{3}\text{.9674 fF}$.