Base Width Modulation MCQ Quiz - Objective Question with Answer for Base Width Modulation - Download Free PDF

• As the reverse-biased base-collector voltage is increased, the transition (or) the depletion region of the base-collector junction is increased resulting in the extension of this region more towards P-side (or base side), thereby reducing the effective base width.
• This reduction in the effective base-width results in the increase in current gain as well because of the increase in the slope of concentration gradient variation from the emitter side to the collector side.

• As the reverse-biased base-collector voltage is increased, the transition (or) the depletion region of the base-collector junction is increased resulting in the extension of this region more towards P-side (or base side), thereby reducing the effective base width.
• This reduction in the effective base-width results in the increase in current gain as well because of the increase in the slope of concentration gradient variation from the emitter side to the collector side.

Under punch through condition the depletion region covers the entire base region. Hence the base metallurgical width is equal to depletion width in base

$W_B=\frac{\sqrt{2{\epsilon }_{si}{V}_{pt}}}{e{N}_B}$

Here given punch through voltage bias of base and collector V_BC = 10 + V_Bi

V_bi = built in voltage 

$=\frac{KT}{q}\ln \left(\frac{N_C{N}_B}{n_i^2}\right)$

$=25mV\ln \left(\frac{5\times {10}^{18}\times 5\times {10}^{15}}{1.5\times 1.5\times {10}^{20}}\right)$

= 0.808 V

∴ V_BC­ = 10.808 V

$W_B=\sqrt{\frac{2\times {10}^{12}\times 10.808}{1.6\times {10}^{-19}\times 5\times {10}^{15}}}$

= 1.64 × 10^-4 cm

= 1.64 μm

• As the reverse-biased base-collector voltage is increased, the transition (or) the depletion region of the base-collector junction is increased resulting in the extension of this region more towards P-side (or base side), thereby reducing the effective base width.
• This reduction in the effective base-width results in the increase in current gain as well because of the increase in the slope of concentration gradient variation from the emitter side to the collector side.

Concept:

The depletion region extending to the base is defined as:

$x_b={\left(\frac{2ϵ{}_s\left(V_{bi}+V_R\right)}{q}\frac{N_c}{N_B}\frac{1}{\left(N_C+N_B\right)}\right)}^{1/2}$

N_C, N_B = Carrier and base doping concentrations

V_R = Applied reverse bias voltage

Since punch thorough voltage (V_pt) >> V_bi, we can neglect V_bi , the above equation can be rearranged as:

$V_{pt}=\frac{e{x}_{B_o}^2}{2ϵ{}_s}\frac{N_B\left(N_C+N_B\right)}{N_C}$

X_B0 = Metallurgical base width

Calculation:

Putting on the respective values, we get:

$25=\frac{\left(1.6\times {10}^{-19}\right){\left(0.5\times {10}^{-4}\right)}^2\left({10}^{16}\right)\left(N_C+{10}^{16}\right)}{2\left(11.7\right)\left(8.85\times {10}^{-14}\right)N_C}$

$12.94=1+\frac{{10}^{16}}{N_C}$

N_C = 8.38 × 10¹⁴ cm^-3

We have ${\mathrm{W}}_{\mathrm{B}}\cdot {\mathrm{N}}_{\mathrm{B}}={\mathrm{W}}_{\mathrm{C}}\cdot {\mathrm{N}}_{\mathrm{C}}\Rightarrow \frac{{\mathrm{W}}_{\mathrm{C}}}{{\mathrm{W}}_{\mathrm{B}}}=\frac{{\mathrm{N}}_{\mathrm{B}}}{{\mathrm{N}}_{\mathrm{C}}}$

Now, total depletion width of the collector base junction be $\mathrm{W}={\mathrm{W}}_{\mathrm{B}}+{\mathrm{W}}_{\mathrm{C}}=\sqrt{\frac{2ϵ}{\mathrm{e}}\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{B}}\cdot {\mathrm{N}}_{\mathrm{C}}}\right){\mathrm{V}}_{\mathrm{R}}}$

${\mathrm{W}}_{\mathrm{B}}(1+\frac{{\mathrm{W}}_{\mathrm{C}}}{{\mathrm{W}}_{\mathrm{B}}})=\sqrt{\frac{2ϵ}{\mathrm{e}}\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{B}}\cdot {\mathrm{N}}_{\mathrm{C}}}\right){\mathrm{V}}_{\mathrm{R}}}$

${\mathrm{W}}_{\mathrm{B}}(1+\frac{{\mathrm{N}}_{\mathrm{B}}}{{\mathrm{N}}_{\mathrm{C}}})=\sqrt{\frac{2ϵ}{\mathrm{e}}\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{B}}\cdot {\mathrm{N}}_{\mathrm{C}}}\right){\mathrm{V}}_{\mathrm{R}}}(\because \frac{{\mathrm{W}}_{\mathrm{C}}}{{\mathrm{W}}_{\mathrm{B}}}=\frac{{\mathrm{N}}_{\mathrm{B}}}{{\mathrm{N}}_{\mathrm{C}}})$

${\mathrm{W}}_{\mathrm{B}}\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{C}}}\right)=\sqrt{\frac{2ϵ}{\mathrm{e}}\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{B}}\cdot {\mathrm{N}}_{\mathrm{C}}}\right){\mathrm{V}}_{\mathrm{R}}}$

Now, for a reverse bias ${\mathrm{V}}_{\mathrm{R}}={\mathrm{V}}_{\mathrm{p}\mathrm{t}}$, we have depletion width in base region attain its maximum value ${\mathrm{W}}_{\mathrm{B}}=0.5\mu \mathrm{m}$. Thus, we have 

${\mathrm{V}}_{\mathrm{p}\mathrm{t}}={\mathrm{W}}_{\mathrm{B}}^2{\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{C}}}\right)}^2\left(\frac{\mathrm{e}}{2ϵ}\right)\left(\frac{{\mathrm{N}}_{\mathrm{B}}\cdot {\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}\right)$

We have

$\begin{array}{l}{\mathrm{V}}_{\mathrm{p}\mathrm{t}}=\frac{{\mathrm{e}\mathrm{W}}_{\mathrm{B}}^2}{2{ϵ}_{}}\cdot {\mathrm{N}}_{\mathrm{B}}\left(\frac{{\mathrm{N}}_{\mathrm{B}}+{\mathrm{N}}_{\mathrm{C}}}{{\mathrm{N}}_{\mathrm{C}}}\right)\\ \Rightarrow 25=\frac{\left(1.6\times {10}^{-19}\right){\left(0.5\times {10}^{-4}\right)}^2}{2\times 11.7\times 8.85\times {10}^{-14}}\cdot {10}^{16}\left(1+\frac{{10}^{16}}{{\mathrm{N}}_{\mathrm{C}}}\right)\\ \Rightarrow 1+\frac{{10}^{16}}{{\mathrm{N}}_{\mathrm{C}}}=12.94\\ \Rightarrow {\mathrm{N}}_{\mathrm{C}}=8.38\times {10}^{14}\mathrm{c}{\mathrm{m}}^{-3}\end{array}$

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

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