To: | peter.chadwick@Zarlink.Com |
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Subject: | Re: [Amps] Equalising resistors with HV diodes |
From: | R.Measures <r@somis.org> |
Date: | Tue, 21 Sep 2004 01:23:53 -0700 |
List-post: | <mailto:amps@contesting.com> |
On Sep 21, 2004, at 12:30 AM, peter.chadwick@Zarlink.Com wrote:
- Only if the total PIV of the diode string exceeds the actual voltage encountered on the job . Now that leads to an interesting question: suppose you have a 1 kV diode in series with a 200v diode, and that the capacitances are equal. - If the junction capacitance at 200v on one diode equals the junction C of another diode at 1000v, the die of the 200v diode must be much larger than that of the other diode. Then as the reverse volts increase, the 200 volt diode will get half - However, if the capacitances of the diodes are small, the allowable avalanche current will be capable of charging the unequal capacitances (the voltage across 2 series capacitors is inversely proportional to their capacity, even on DC) but unable to break down and avalanche because no leakage current can flow - the 1 kV diode is reverse biased. When testing the 56 diodes for the current project's FWB/FWD anode PS, they exhibited less piv-dependency on temperature than I have seen with older diodes. ... ... ... ...Richard L. Measures, AG6K, 805.386.3734. www.somis.org _______________________________________________ Amps mailing list Amps@contesting.com http://lists.contesting.com/mailman/listinfo/amps |
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