A one sided abrupt pn junction diode has a depletion capacitance CD of 50 pF at a reverse bias of 0-2 V- The plot of 1-C2D versus the applied voltage V for this diode is a straight line as shown in the figure below- The slope of the plot is - 1020 F-2 V-1

Depletion or transition capacitance is, nbsp; C_D = AϵW For one sided PN junction (Ex: P^+ N junction) nbsp; W= √(2ϵ V_BeN_D) = √(2ϵ (V_bi V)eN_D) where ...

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A one sided abrupt pn junction diode has a depletion capacitance $C_{D}$ of $50 p F$ at a reverse bias of $0.2 V$ . The plot of $1 / C_{D}^{2}$ versus the applied voltage $V$ for this diode is a straight line as shown in the figure below. The slope of the plot is ___ $\times 10^{20} F^{- 2} V^{- 1}$

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Solution

Depletion or transition capacitance is,
$C_{D} = \frac{A ϵ}{W}$
For one-sided PN junction (Ex: $P^{+} N$ junction)
$W = \sqrt{\frac{2 ϵ V_{B}}{e N_{D}}} = \sqrt{\frac{2 ϵ (V_{b i} - V)}{e N_{D}}}$
where $V$ is anode toc athode applied potential
$\Rightarrow C_{D} = \frac{A ϵ}{\sqrt{\frac{2 ϵ (V_{b i} - V)}{e N_{D}}}}$
$\frac{1}{C_{D}^{2}} = \frac{2}{A^{2} ϵ e N_{D}} (v_{b i} - V)$ $\frac{1}{C_{D}^{2}}$
becomes zero at $V = V_{b i}$
From above graph , $y = \frac{1}{C_{D}^{2}} = 0)$ at $x_{1} = V_{b i}$
and $y_{2} = \frac{1}{C_{D}^{2}} = 4 \times 10^{20} a t x_{2} = - 0.2 V$
Slope $= \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{4 \times 10^{20} - 0}{- 0.2 - V_{b i}}$
$∵ V_{b i}$ is not provided, slope cannot be found

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A one sided abrupt pn junction diode has a depletion capacitance CD of 50 pF at a reverse bias of 0.2 V. The plot of 1/C2D versus the applied voltage V for this diode is a straight line as shown in the figure below. The slope of the plot is ___ ×1020F−2V−1