Question 5.5: Assume a p^+-n diode is biased in the forward direction, wit...
Assume a p^+-n diode is biased in the forward direction, with a current I_f. At time t = 0 the current is switched to –I_r. Use the appropriate boundary conditions to solve Eq. (5–47) for Q_p(t). Apply the quasi-steady state approximation to find the storage delay time t_{sd}.
(5-47): i(t)=\frac{Q_p(t)}{\tau_p}+\frac{dQ_p(t)}{dt}
From Eq. (5–47),
i(t)=\frac{Q_p(t)}{\tau_p}+\frac{dQ_p(t)}{dt} \quad for \space t<0,Q_p=I_f\tau_p
Using Laplace transforms,
-\frac{I_r}{s}=\frac{Q_p(s)}{\tau_p}+sQ_p(s)-I_f\tau_p \\ \space \\ Q_p(s)=\frac{I_f\tau_p}{s+1/\tau_p}-\frac{I_r}{s(s+1/\tau_p)}\\ \space \\ Q_p(t)=I_f\tau_pe^{-t/\tau_p}+I_r\tau_p(e^{-t/\tau_p}-1)=\tau_p[-I_r+(I_f+I_r)e^{-t/\tau_p}]
Assuming that Q_p(t)=qAL_p\Delta p_n(t) as in Eq. (5–52),
\Delta p_n(t)=\frac{\tau_p}{qAL_p}-[I_r+(I_f+I_r)e^{-t/\tau_p}]
This is set to equal zero when t = t_{sd}, and we obtain:
t_{sd}=-\tau_p\ln{\Big[\frac{I_r}{I_f+I_r}\Big]}=\tau_p\ln{\Big(1+\frac{I_f}{I_r}\Big)}
(5-52): \Delta p_n(t)=p_n(e^{qv(t)/kT}-1)=\frac{Q_p(t)}{qAL_p}