Hi Doug:
A simple (and therefore incomplete) explanation:
Breakdown in air requires the ions to travel from
one pole to the other before the polarity
reverses. At high frequencies, the polarity
reverses before the ions can travel the distance
between the poles.
At high frequencies (compared to mains
frequencies) the clearance can be less. See:
http://lss.fnal.gov/archive/other/ssc/sscl-539.pdf
http://ewh.ieee.org/r10/taiwan/pses/archive/2012_0
4_27/IEEE%20PSES%20April/TC108%20hf_FC_v.1.2_IEEE%
20PSES%20TAI.pdf
Best regards,
Rich
---------- Original Message ----------
From: Doug Powell <doug...@gmail.com
<mailto:doug...@gmail.com> >
To: EMC-PSTC@LISTSERV.IEEE.ORG
<mailto:EMC-PSTC@LISTSERV.IEEE.ORG>
Subject: [PSES] IEC 61010-1 Table K.17
Date: Fri, 7 Aug 2015 16:47:28 -0600
All,
I am evaluating spacings for an RF product that
operates at 13.56 MHz and can produce maximum
voltages of 5,000 Vrms (7,070 Vpk). If I do an
interpolation of using Table 6 (Mains 230 V, OV
Cat II, indexing on 5,000 Vrms), I get a minimum
clearance requirement of 14.9 mm. When I do the
same calculation on Table K.17 (column 3, indexing
on 7,070 Vpk), I get 12.7 mm.
Now I understand the effects of high frequency
voltage stress causing air molecules to become
more energetic and therefore more likely to break
down at lower voltages. So why in this case does
IEC 61010-1 Table K.17 result in lower clearance
values than Table 6? Somehow, this just seems
wrong.
Please note that in the case of high frequencies
paragraph K.3.1 indicates I am to skip over
section K.3.2 with the D1 + F × (D2 D1)
calculations.
Thanks a bunch!
Doug
Douglas E Powell
doug...@gmail.com <mailto:doug...@gmail.com>
http://www.linkedin.com/in/dougp01
-
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