Junction Potential MCQ Quiz in தமிழ் - Objective Question with Answer for Junction Potential - இலவச PDF ஐப் பதிவிறக்கவும்

Concept:

The width of the depletion region for a reversed bias pn junction is given by:

$W={\left[\frac{2ϵ(V_o+V_R)}{q}\left(\frac{N_a+N_d}{N_a{N}_d}\right)\right]}^{\frac{1}{2}}$

where V_o = Built-in voltage

V_R = Applied Reverse biased voltage

|V_o + V_R| = voltage across the depletion region.

W ∝ $\sqrt{V_o+V_R}$, for the same doping profile.

Calculation:

Given |Vo + V_R|₁ = 0.75

W₁ ∝ ${\left(\sqrt{V_o+V_R}\right)}_1$, i.e.

W1 ∝ (0.75)½

1 μm ∝ (0.75)^½    ---(1)

Similarly W₂ ∝ ${\left(\sqrt{V_o+V_R}\right)}_2$

Given |Vo + VR|₂ = 3 V

W₂ ∝ (3)½      ---(2)

Dividing (2) by (1), we get:

$\frac{W_2}{1}={\left(\frac{3}{0.75}\right)}^{\frac{1}{2}}$

W₂ = (4)^½ μm

W2 = (4)^½ μm

W2 = 2 μm

Inside the depletion region

$\frac{dE}{dx}=\frac{\rho }{{ϵ}_s}$

$\rho ={ϵ}_s\frac{dE}{dx}=-1.05\times {10}^{-12}\left(\frac{F}{cm}\right)\times \frac{{10}^5\left(N/cm\right)}{50\times {10}^{-7}cm}$

= -2.1 × 10^-2 c/cm³

The doping level is then

ρ = -q Na

$N_a=\frac{-\rho }{q}=\frac{2.1\times {10}^{-2}c/c{m}^3}{1.6\times {10}^{-19}C}$

= 1.3 × 10¹⁷/cm³

The electric field is related to the potential by:

$E(x)=-\frac{dV}{dx}$

The built-in potential is, therefore, the integral of the electric field at zero bais, i.e.

${\varphi }_B={10}^5\frac{v}{cm}\left[\left(50+100\right){10}^{-7}\right]\times \frac{1}{2}$

$\Rightarrow \frac{{10}^5\times 150\times {10}^{-7}}{2}$

⇒ 75 × 10^-2 V

$E=-\overline{\mathrm{\nabla }}.V$

$V=-\int E.dx$

= Area under E - x

$=\frac{1}{2}\left(4\times {10}^{-6}\right)\left(15\times {10}^4\right)$

The potential across junction $V=-\int Edx$

$\Rightarrow E_m=\frac{2V}{W}$

$=\frac{2\left(V_O+V_r\right)}{W}$

$=\frac{2\left(0.75+5.25\right)}{5\mu m}$

Given V_f = 0.5 V = V_bi 0.7 V

In depletion region electric field is maximum at centre, its value is

$E_{max}=\frac{2V_{junction}}{w}$

$V_{junction}=V_{bi}-V_f$ (Or V_bi + V_r; V_r is reverse voltage)

$4\times {10}^6=\frac{2\left(0.7-0.5\right)}{W}$

$\Rightarrow W=\frac{2\times 0.2}{4\times {10}^6}$

$=0.1\times {10}^{-6}m$

The p⁺n diode has a heavily doped p- region.

Under forward bias, the electrons are injected from the n region to the p region and holes are injected form p region into the n region.

These injected carries become minority carriers in the other region and recombine with the majority carriers there and decay exponentially with distance

The switching speed of p⁺n (heavily doped p – region) junction depends on the lifetime (τ) of the majority carriers (i.e. electrons) in the n – region which is a lightly doped region.

Depletion width on base side ${\mathrm{x}}_{\mathrm{d}\mathrm{B}}$ is

$\begin{array}{l}{\mathrm{x}}_{\mathrm{d}\mathrm{B}}={\left\{\frac{2}{\mathrm{q}}\left(\frac{{\mathrm{N}}_{\mathrm{c}}}{{\mathrm{N}}_{\mathrm{B}}}\cdot \frac{1}{\left({\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}\right)}\right)\left({\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{C}\mathrm{B}}\right)\right\}}^{\frac{1}{2}}\\ \Rightarrow {\mathrm{x}}_{\mathrm{d}\mathrm{B}}={\left\{\frac{2\times 11.7\times 8.85\times {10}^{-14}}{\left(1.6\times {10}^{-19}\right)}\left(\frac{2\times {10}^{15}}{5\times {10}^{16}}\cdot \frac{1}{5.2\times {10}^{16}}\right)\left({\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{R}}\right)\right\}}^{\frac{1}{2}}\\ \Rightarrow {\mathrm{x}}_{\mathrm{d}\mathrm{B}}={\left\{\left(9.96\times {10}^{-12}\right)\left({\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{C}\mathrm{B}}\right)\right\}}^{\frac{1}{2}}\end{array}$

Built-in potential, ${\mathrm{V}}_{\mathrm{b}\mathrm{i}}={\mathrm{V}}_{\mathrm{T}}\ln \left(\frac{{\mathrm{N}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{n}}_{\mathrm{i}}^2}\right)$

$\begin{array}{l}\Rightarrow {\mathrm{V}}_{\mathrm{b}\mathrm{i}}=\left(0.0259\right)\ln \left(\frac{5\times {10}^{16}\times 2\times {10}^{15}}{{\left(1.5\times {10}^{10}\right)}^2}\right)\\ \Rightarrow {\mathrm{V}}_{\mathrm{b}\mathrm{i}}=0.695\mathrm{V}\end{array}$

Now, $\frac{\mathrm{d}{\mathrm{x}}_{\mathrm{d}\mathrm{B}}}{\mathrm{d}{\mathrm{V}}_{\mathrm{C}\mathrm{B}}}=\frac{3.156\times {10}^{-6}}{2\sqrt{{\mathrm{V}}_{\mathrm{b}\mathrm{i}}+{\mathrm{V}}_{\mathrm{C}\mathrm{B}}}}$

$\begin{array}{l}\Rightarrow {\frac{\mathrm{d}{\mathrm{x}}_{\mathrm{d}\mathrm{B}}}{\mathrm{d}{\mathrm{v}}_{\mathrm{C}\mathrm{B}}}|}_{{\mathrm{V}}_{\mathrm{C}\mathrm{B}}=1.11}=\frac{3.156\times {10}^{-6}}{2\sqrt{0.695+1.11}}\\ \Rightarrow {\frac{\mathrm{d}{\mathrm{x}}_{\mathrm{d}\mathrm{B}}}{\mathrm{d}{\mathrm{v}}_{\mathrm{C}\mathrm{B}}}|}_{{\mathrm{V}}_{\mathrm{C}\mathrm{B}}=1.11}=1.18\times {10}^{-8}\mathrm{m}/\mathrm{V}\end{array}$