Question
(A) An ideal silicon pn junction diode at $T=300 \mathrm{~K}$ is forward biased at $V_{a}=+20 \mathrm{mV}$. The reverse-saturation current is $I_{x}=10^{-13} \mathrm{~A}$. Calculate the smallsignal diffusion resistance. (b) Repeat part ( $a$ ) for an applied reverse-biased voltage of $V_{a}=-20 \mathrm{mV}$
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However, it is often more convenient to use its reciprocal, which is given by $\frac{1}{R_D} = \frac{dI_D}{dV_a}$. Show more…
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Consider a silicon pn junction diode with an applied reverse-biased voltage of $V_{R}=$ 5V. The doping concentrations are $N_{a}=N_{d}=4 \times 10^{16} \mathrm{~cm}^{-3}$ and the cross-sectional area is $A=10^{-4} \mathrm{~cm}^{2}$. Assume minority carrier lifetimes of $\tau_{0}=\tau_{n 0}=\tau_{p 0}=10^{-7} \mathrm{~s}$. Calculate the ( $a$ ) ideal reverse-saturation current, (b) reverse-biased generation current, and ( $c$ ) the ratio of the generation current to ideal saturation current.
(A) Consider an ideal pn junction diode at $T=300 \mathrm{~K}$ operating in the forward-bias region. Calculate the change in diode voltage that will cause a factor of 10 increase in current. $(b)$ Repeat part $(a)$ for a factor of 100 increase in current.
Calculate the applied reverse-biased voltage at which the ideal reverse current in a pn junction diode at $T=300 \mathrm{~K}$ reaches 90 percent of its reverse-saturation current value.
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