Hello again.

The following is food for thought that came to light during a recent project
development meeting.  I invite your comments and anecdotes on the following.

My earlier post regarding test probe 11 of IEC 61032 was related to comparing
force test methods to deflect enclosures to reduce effective clearances that
provide (at least) reinforced insulation.  The standards are IEC 60950-1:2005
and IEC 60065:1998 (I don't have a copy of the 2001 version).

In 60950-1, the method is to apply a force of 250 N using a 30 mm diameter
circular plane surface.  This corresponds to a pressure of 0.35 N/mm².

In 60065, the method is to apply a 30 N force using test probe 11 of IEC 61032
(which bears a significant resemblance to probe in 60950-1, Figure 2A and, for
the purposes of this discussion, treated as dimensionally identical at the
tip).  To calculate a pressure, a little cheating** is needed.  For the
purposes of this discussion, I've assumed that the contact area corresponds
approximately to cross-section B-B in 60950-1, Figure 2A, and further
approximated by a rectangle (not particularly realistic, but close enough for
this discussion)).  This corresponds to a pressure of ~33 N/mm².

60950-1, §2.10.3.1, first dashed paragraph, the minimum clearance is 10 mm.

60065, Figure 9, assuming the peak voltages involved are between 354 V and 900
V, the minimum clearance is 6 mm.

Looking only at the forces involved and a design cramped for space, 6 mm is
far more attractive than the 167% larger 10 mm.

Looking only at the forces involved, it's easy to expect that the 250 N test
is more severe.  But, since materials respond to applied pressures resulting
>from applied forces (an 'other things being equal' statement that ignores
quite a few 'other' things), the 30 N test looks more severe by nearly two
orders of magnitude.

60065, while easier on the volumetric real estate budget, requires some good
basic reinforcement applied to the mechanical design in the relevant areas.

60950-1, while easier on the other mechanical design features needed, requires
a sacrifice of volumetric real estate.



** Approximately calculation of the area of cross-section B-B:

Side A = 2×(2 mm)×[sin(37°/2)] = 0.93 mm

Side B = 2×(4 mm)×[sin(14°/2)] = 0.97 mm

Area = 0.9 mm²


Regards,

Peter L. Tarver, PE
ptar...@ieee.org 

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