A p-n junction is doped with donor concentration $N_D$ and accepter concentator $N_A$, having depletion width of $x_n$ and $x_p$ respectively. Choose the correct option.
$N_D \propto \frac{1}{x_n}$
Understanding the relationship between doping concentrations ($N_D$, $N_A$) and depletion widths ($x_n$, $x_p$) is crucial for semiconductor device physics.
Charge neutrality requires the total charge ionized on both sides of the depletion region to be equal:
$q N_D x_n = q N_A x_p$
This simplifies to the charge balance equation:
$N_D x_n = N_A x_p$
The depletion width on the n-side ($x_n$) is approximately given by:
$x_n \approx \sqrt{\frac{2 \epsilon_s V_{bi}}{q} \frac{N_A}{N_D(N_D+N_A)}}$
This formula shows $x_n$ depends inversely on factors related to $N_D$ and $N_A$.
When the donor concentration vastly exceeds the acceptor concentration ($N_D \gg N_A$), we approximate ($N_D + N_A \approx N_D$). The expression for $x_n$ simplifies significantly:
$x_n \approx \sqrt{\frac{2 \epsilon_s V_{bi}}{q} \frac{N_A}{N_D^2}} = \frac{1}{N_D} \sqrt{\frac{2 \epsilon_s V_{bi} N_A}{q}}$
This demonstrates the inverse proportionality between $x_n$ and $N_D$ under this common condition:
$x_n \propto \frac{1}{N_D}$
Rearranging the inverse proportionality $x_n \propto \frac{1}{N_D}$ gives the relationship:
$N_D \propto \frac{1}{x_n}$
This relationship correctly describes the behavior under the approximation $N_D \gg N_A$.
In a semiconductor diode, the cut-in voltage is the voltage:
The leakage current in a pn junction is of the order of:
Diode junction breakdowns above 5 V are caused by:
The width of the depletion layer in a P-N junction diode
ln a p-n junction diode the forward current