HW6 - EE2 Problem Set#6 Due June 2 in class Question#1(10 points A pn junction with Vbi = 0.7 V and a minority carrier lifetime of 100 ns conducts

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EE2 Problem Set#6: Due June 2 in class Question#1 (10 points): A pn junction with V bi = 0.7 V and a minority carrier lifetime of 100 ns conducts a forward bias current of 4.3 mA. At time t = 0, the circuit conditions change so that a reverse bias voltage of 20 V is applied in series with the diode and a 1-kΩ resistor. (a) If the contribution of recombination to the stored charge removal can be neglected, due to the large value of the discharging current, how long will it take for the stored charge to be removed? (b) If the reverse bias voltage is instead 0 V, it will take a much longer time to remove the excess charge, meaning that the role of recombination cannot be neglected. If it is found that it takes 200 ns to remove the stored charge then how much charge is removed by recombination? (a) From Subject 14, Eq. 14.9 we saw that the stored charge is given by Q s = . For the forward- bias condition, the stored charge is therefore 4.3 mA × 100 ns = 4.3 × 10 -10 C. To remove this charge under reverse bias take an amount of time Q s / I , where I is the reverse-bias current. Immediately after switching, the value of this current is approximately 20.7 V/1 kΩ = 20.7 mA (note how we take the reverse-bias voltage plus the built-in voltage), which then yields a time of around 21 ns . Since this time is much less than the minority carrier lifetime, this justifies neglecting recombination as a route to restore the charge density. (b) If the external reverse bias voltage is set to zero, then in the initial period where the stored
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This note was uploaded on 02/28/2011 for the course EE 2 taught by Professor Vis during the Winter '07 term at UCLA.

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HW6 - EE2 Problem Set#6 Due June 2 in class Question#1(10 points A pn junction with Vbi = 0.7 V and a minority carrier lifetime of 100 ns conducts

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