One-Sided Junction Capacitance and Depletion Width MCQ Quiz - Objective Question with Answer for One-Sided Junction Capacitance and Depletion Width - Download Free PDF

Concept:

Depletion approximation: It is assumed that the depletion region is devoid of any mobile charge carriers. We assume that the carrier concentration (n and p) is negligible as compared to the doping concentration (Na and Nd) in the depletion/junction region. Outside this region, the net charge density is assumed to be zero.

The depletion approximation simplifies the calculation of electric field, space charge region and electric potential.

But the approximation cannot be applied for diffusion current calculation.

$J_p=-q\times D_p\times \frac{dp}{dx}$

$J_n=q\times D_n\times \frac{dn}{dx}$

With the depletion approximation, the rate of change of charge carriers becomes infinity and hence the diffusion current becomes infinity.

∴ The depletion approximation will not be considered for the calculation of the diffusion current. 

Concept:-

For p⁺ - n diode, the value of N_A is very large $\frac{1}{N_D}+\frac{1}{N_A}\simeq \frac{1}{N_D}$

Write the expression for depletion width,

$w=\sqrt{\frac{2{\epsilon }_0{\epsilon }_r}{q}(\frac{1}{N_D}+\frac{1}{N_A})V_r}$

$=\sqrt{\frac{2{\epsilon }_0{\epsilon }_r}{q}\left(\frac{1}{N_D}\right)V_r}$

Calculations:

Given that,

ε₀ = 8.85 × 10^-14 F/cm.

ε_r = 12

N_D = 10¹⁶/ cm³

V_r = 100 V.

after substituting the value

$w=\sqrt{\frac{2(8.85\times {10}^{-14}(12)}{1.6\times {10}^{-19}}\left[\frac{1}{{10}^{16}}\right](100)}$

w = 364.34 μm

Expression of electric field on n-side

$E=-{\displaystyle \frac{q}{\epsilon }}{X}_{no}{N}_d$

Substitute ${\displaystyle \frac{W}{2}}$ for X_no

$E=\frac{-q}{{\epsilon }_0{\epsilon }_r}\left({\displaystyle \frac{W}{2}}\right)N_d$

$E=\frac{-(1.6\times {10}^{-19})}{(8.85\times {10}^{-14})(12)}\left[\frac{364.349\times {10}^{-6}}{2}\right]{10}^{16}$

E = -274462.5 V/cm

The cut-off frequency is dependent on transmit time of holes.

${\tau }_t=\frac{W_b^2}{2D_p}$

D_p = μ_p V_T = 800 × 0.0259 cm²/s

= 20.72 cm²/s

$\Rightarrow {\tau }_t=\frac{{\left(2\times {10}^{-4}\right)}^2}{2\times 20.72}=9.65\times {10}^{-10}s$

∴ Cut-off frequency $f_c=\frac{1}{2\pi {\tau }_t}$

$=\frac{1}{2\pi 9.65\times {10}^{-10}}$

= 164 MHz

We have, junction capacitance given by, $\mathrm{C}={\left\{\frac{\mathrm{e}{ϵ}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{d}}}{2\left({\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{R}}\right)\left({\mathrm{N}}_{\mathrm{a}}+{\mathrm{N}}_{\mathrm{d}}\right)}\right\}}^{\frac{1}{2}}$ 

Now, since a one-sided junction is assumed and the pn junction is ${\mathrm{p}}^{+}\mathrm{n}$. Thus, ${\mathrm{N}}_{\mathrm{a}}\gg {\mathrm{N}}_{\mathrm{d}}$

$\Rightarrow \mathrm{C}={\left\{\frac{\mathrm{e}{ϵ}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{d}}}{2\left({\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{R}}\right)}\right\}}^{\frac{1}{2}}$

$\Rightarrow \frac{1}{{\mathrm{C}}^2}=\frac{2\left({\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{R}}\right)}{\mathrm{e}{ϵ}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{d}}}$           _____1)

When, ${\mathrm{V}}_{\mathrm{R}}=-{\mathrm{V}}_{\mathrm{b}\mathrm{i}}\Rightarrow \frac{1}{{\mathrm{C}}^2}=0$

Thus, $-{\mathrm{V}}_{\mathrm{b}\mathrm{i}}$ is the intercept on ${\mathrm{V}}_{\mathrm{R}}$ axis.

$\begin{array}{l}\Rightarrow -{\mathrm{V}}_{\mathrm{b}\mathrm{i}}=-0.855\mathrm{V}\\ \Rightarrow {\mathrm{V}}_{\mathrm{b}\mathrm{i}}=0.855\mathrm{V}\end{array}$

Built in potential potential ${\mathrm{V}}_{\mathrm{b}\mathrm{i}}$ is given by

$\begin{array}{l}{\mathrm{V}}_{\mathrm{b}\mathrm{i}}={\mathrm{V}}_{\mathrm{t}}\ln \left(\frac{{\mathrm{N}}_{\mathrm{a}}{\mathrm{N}}_{\mathrm{d}}}{{\mathrm{n}}_{\mathrm{i}}^2}\right)\\ \Rightarrow {\mathrm{N}}_{\mathrm{d}}=\frac{{\mathrm{n}}_{\mathrm{i}}^2}{{\mathrm{N}}_{\mathrm{a}}}\cdot {\mathrm{e}}^{\left(\frac{{\mathrm{V}}_{\mathrm{b}\mathrm{i}}}{{\mathrm{V}}_{\mathrm{T}}}\right)}\\ =\frac{{\left(1.5\times {10}^{10}\right)}^2}{\left(5.34\times {10}^{18}\right)}{\mathrm{e}}^{\left(\frac{0.855}{0.0259}\right)}\\ \Rightarrow {\mathrm{N}}_{\mathrm{d}}=9.15\times {10}^{15}\mathrm{c}{\mathrm{m}}^{-3}\end{array}$

Now, slope of the plot from 1)

$\frac{1}{{\mathrm{C}}^2}=\frac{2{\mathrm{V}}_{\mathrm{b}\mathrm{i}}}{\mathrm{e}{ϵ}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{d}}}+\frac{2}{\mathrm{e}{ϵ}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{d}}}{\mathrm{V}}_{\mathrm{R}}$

Comparing with $\mathrm{y}=\mathrm{c}+\mathrm{m}\mathrm{x}$

We see slope of curve, $\mathrm{m}=\frac{2}{\mathrm{e}{ϵ}_{\mathrm{s}}{\mathrm{N}}_{\mathrm{d}}}$

$\begin{array}{l}=\frac{2}{\left(1.6\times {10}^{-19}\right)\left(11.7\right)\left(8.854\times {10}^{-14}\right)\left(9.15\times {10}^{15}\right)}\\ \Rightarrow \mathrm{m}=1.32\times {10}^{15}{\left(\mathrm{F}/\mathrm{c}{\mathrm{m}}^2\right)}^{-2}{\mathrm{V}}^{-1}\end{array}$.