We can determine the distance that the space charge region extends into the p and n regions from the metallurgical junction. $x_p = \frac{N_d x_n}{N}$ $W = x_n + x_p$ Prove that: $W = \left\{\frac{2 \epsilon_s V_{bi}}{e} \left[\frac{N_a + N_d}{N_a N_d}\right] \right\}^{1/2}$
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Start with the equation Naxn = ax W = xn + xp. This equation relates the width of the space charge region (W) to the concentrations of acceptor impurities (Na) in the p-region and donor impurities (N) in the n-region. Show more…
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