How is a diode damaged by exceeding its Vrrm (maximum peak repetitive reverse voltage)? At this point it conducts more current than usual, so is it the current flow that damages it? Why would that be any different from being forward biased when it conducts almost unlimited current? Or is there some other factor?
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From looking at a diode curve you can see that its reverse current sharply rises at Vr much like the forward current does. From this I would imagine that repetitively hitting the diode with reverse voltage in excess of Vrrm would cause the diode to conduct significant current and the resultant power dissipation to either melt the silicon and fail open, or, if the supply is limited, fail shorted.
A diode will certainly not "conduct almost unlimited current" -- the device has a designed power dissipation and will not conduct more than this amount for very much time. When reverse-biased you are essentially damaging the thin junction and it is for this reason that it's not the same as simply forward-biasing the diode. Avalanche and Zener diodes are designed to operate reverse-biased and have differently-formed junctions which are designed to withstand this mode of operation. Zener diodes especially have a much more heavily doped junction than normal diodes.
answered Oct 26, 2010 at 15:47
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\$\begingroup\$ sure, write what i do in more detail. +1 as your reward. \$\endgroup\$– KortukCommented Oct 26, 2010 at 15:50
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3\$\begingroup\$ My answer was posted 2 minutes before yours, I think it's more accurate to say that you summarized my answer, not that I elaborated on yours. :-) \$\endgroup\$ Commented Oct 26, 2010 at 16:14
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\$\begingroup\$ An ideal diode will conduct almost unlimited current. Of course one would be damaged by excessive current. But how would being forward biased be different from being reverse biased - one drops 0.7V and one drops 0V (at the point of breakdown), right? \$\endgroup\$– Thomas OCommented Oct 26, 2010 at 16:21
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\$\begingroup\$ Take a look at the characteristic curves. Diodes do not conduct zero current when reverse-biased. \$\endgroup\$ Commented Oct 26, 2010 at 17:11
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\$\begingroup\$ Andrew, I did not mean to say that you expanded on mine. I wrote it in class and as I hit post the message popped up saying yours had been posted. It told me a minute before, but the concept of peak reverse voltage has one clear answer. \$\endgroup\$– KortukCommented Oct 26, 2010 at 20:02