Suppose the pn junction is reverse biased, so the width of the depletion region is large as shown. Now I disconnect the battery. What happens to the ions in the depletion region ? I think the depletion region tries to get back to the width corresponding to its barrier potential. But I have difficulty in visualizing how exactly this happens. Any help is appreciated. Thanks!
Electrical – How to discharge the pn junction capacitance
diodes
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Best Answer
It self-discharges in much less than a second usually.
Ir at Vr rises with many factors usually power rating. Similarity Cr rises with Pd rating but may drop with high Vr ratings for diodes and rises in Schottky diodes over Silicon diodes.
Consider: https://www.fairchildsemi.com/datasheets/FJ/FJH1100.pdf
"Ultra low,leakage signal diode"
It is 2pF at Vr = 0V and Ir is 3pA at Vr=5V.
We expect C at Vr=5V to be much less , let's say <1pF. thus Rr=5V/3pA and C<1pF so RC=T < 5/3 seconds
Now let's look at the body diode in a 500A IGBT.
http://www.pwrx.com/pwrx/docs/cm450dx-24s1_e.pdf. (see note 1)