[PSES] FCC market surveillance

[Amund Westin]

Do FCC carry out any market surveillance to check out FCC approved
equipment?
If yes, do they publish their findings?

Best regards
Amund

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Re: [PSES] Optos for PELV isolation

[McDiarmid, Ralph]

Like very low-drift analog circuits, high impedance circuits require 
attention to detail normally dismissed as negligible in 'everyday' circuit 
design.  
___ 


Ralph McDiarmid  |   Schneider Electric   |  Solar Business  |   CANADA  | 
  Regulatory Compliance Engineering 




From:
Ted Eckert 
To:
EMC-PSTC@LISTSERV.IEEE.ORG, 
Date:
06/29/2015 12:55 PM
Subject:
Re: [PSES] Optos for PELV isolation



I’ve designed systems where the optocouplers were required to have an 
unusually high insulation resistance. I worked for a company that designed 
industrial water treatment equipment. The pH probes we used act like a 
battery. The more current you draw out of them, the sooner they need to be 
replaced. If you wanted to get at least 6 months operation out of a pH 
probe, the current draw had to be limited to no more than a few 
picoamperes. 
 
DC-DC converters with a high level of insulation (and lousy efficiencies) 
were used to get power to the pH probe’s analog amplifier circuitry and 
optocouplers were used to get the data back to the digital electronics. 
The high insulation values were not for safety; they were just to limit 
current from the pH probe finding a lower impedance leakage path. I 
remember a case where a mishandled board failed because a fingerprint was 
left on the solder side of the board under the optocoupler. It got 
encapsulated by the conformal coating and the oil from the fingerprint 
provided enough of a current path that the system was depleting pH probes 
about once every two weeks.
 
Ted Eckert
Compliance Engineer
Microsoft Corporation
ted.eck...@microsoft.com
 
The opinions expressed are my own and do not necessarily reflect those of 
my employer.
 
From: Nyffenegger, Dave [mailto:dave.nyffeneg...@bhemail.com] 
Sent: Monday, June 29, 2015 12:07 PM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Optos for PELV isolation
 
HI Rich,
 
The standard also says 5 second interval between discharges.  And the test 
must be run on 20 samples.  We know UL likes to test.  First  edition of 
the standard is 1984, the test has probably been in there since the 
beginning.
 
-Dave
 
From: Richard Nute [mailto:ri...@ieee.org] 
Sent: Monday, June 29, 2015 1:51 PM
To: Nyffenegger, Dave; EMC-PSTC@LISTSERV.IEEE.ORG
Subject: RE: [PSES] Optos for PELV isolation
 
 
Hi Dave:
 
 
“subjected to 50 discharges from a 0.0005 microfarad capacitor that has 
been charged to a potential of 20 kV between the short-circuited input and 
short circuited output terminals.” 
 
Okay, 500 pF charged to 20 kV then applied to the opto isolation.  The 
input-to-output capacitance is on the order of 1 pF.  That means nearly 
the full 20 kV is applied across the isolation barrier, and exponentially 
decays by the insulation resistance.  One opto specs the insulation 
resistance at 100 Gohms minimum!  Time constant is 500e-9 x 100e+9 or 
5 seconds or 833 minutes or 13 hours!  Repeated 50 times!  50 days to 
do the test, at 5 days per week, 10 weeks, 2-1/2 months!
 
I must be missing something.  Test does not make much sense. 
 
 
Best regards,
Rich
 
 
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Re: [PSES] rapid shut down requirement for solar inverter

[McDiarmid, Ralph]

And there are going to be more stringent requirements in the 2017 NEC, 
related to rapid shutdown. 
___ 


Ralph McDiarmid  |   Schneider Electric   |  Solar Business  |   CANADA  | 
  Regulatory Compliance Engineering 




From:
Scott Aldous <0220f70c299a-dmarc-requ...@ieee.org>
To:
EMC-PSTC@LISTSERV.IEEE.ORG, 
Date:
08/05/2015 07:35 AM
Subject:
Re: [PSES] rapid shut down requirement for solar inverter



Hi Bostjan,

With regard to the US, the rapid shutdown requirement is in the 2014 NEC, 
see 690.12. The NEC is not automatically normative throughout the USA. 
What is required is controlled at the state and sometimes more local 
level. More information here. It is already mandatory in quite a few 
states.

On Wed, Aug 5, 2015 at 1:02 AM, Boštjan Glavič  
wrote:
Dear Experts,
 
Do you know when this requirement will become mandatory in US/CAN? It is 
intended to protect firemen.
 
Thank you for your support.
Best regards,
 
Boštjan Glavič
Laboratory of Electronic Engineering
Head of Laboratory
SIQ Ljubljana, Tržaška cesta 2
SI-1000 Ljubljana, Slovenia
T: + 386 (0)1 47 78 265
F: + 386 (0)1 47 78 444
E-Mail: bostjan.gla...@siq.si
 
 
 
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-- 
Scott Aldous
Compliance Engineer
Google
650-253-1994
scottald...@google.com
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[PSES] IEC 61010-1 Table K.17

[Doug Powell]

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
http://www.linkedin.com/in/dougp01




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Re: [PSES] Calculating Reflection Angles on OATS/SAC

[Brent DeWitt]

Hi James,

 

The image concept again is useful.  By definition, the  ground reference
plane is at zero potential.  For that to be true, charges on the real
antenna and its image must be equal and opposite.  Put a plus on one end of
a dipole and a minus on the other and look at them.  If they are vertical,
and the bottom of the real dipole has the minus sign, the top of the image
must be plus for the charges to cancel.  For the horizontal example, if the
left end is plus the same end of the image must be minus for the same
reason.

 

In the extreme thought experiment, if you lowered the vertical dipole so its
center point were at the ground plane (now a monopole), its image would
complete the dipole.  The same extreme applied to the horizontal dipole
would have the two cancelling each other out entirely.  We can see this in
reality, since the vertical polarization with the antenna at one meter
height is usually the strongest emission at low frequencies where the path
length in wavelengths is small.  The first maximum from the horizontal
dipole occurs when there is a 180 degree path length difference between the
real antenna and its image.

 

Does that help any?

Brent

 

From: Pawson, James [mailto:james.paw...@echostar.com] 
Sent: Thursday, August 06, 2015 5:13 AM
To: EMC-PSTC@LISTSERV.IEEE.ORG
Subject: Re: [PSES] Calculating Reflection Angles on OATS/SAC

 

Many thanks for all of the replies on this topic. The conceptual key I
lacked was the “image” of the receiver below the ground plane which made the
calculations a lot simpler and I’ve now got an up and running spreadsheet.
I’ve also been introduced to things like cotangents and arctangents which
are new to me.

 

The only thing I still remain confused about is the phase of the reflection
from the ground plane.

 

 Gert wrote: “Note that vertical waves invert in polarity on reflection
with the ground plane, where horizontal polarized waves do not.”

 

 Brent wrote: “…and take the difference for phase, remembering that the
horizontally polarized image is 180 degrees out of phase to start with while
the vertical image is in phase.”

 

I might be misunderstanding but these statements seem to contradict each
other. I can kind of see how a vertically polarised wave would be reflected
inverted. If this was the case, could this be compensated for by subtracting
180° from the reflected ground ray to ensure the phases added/subtracted
correctly at the RX antenna?

 

Thanks again

James

 

 

 

_
From: Pawson, James 
Sent: 31 July 2015 15:59
To: EMC-PSTC@LISTSERV.IEEE.ORG  
Subject: Calculating Reflection Angles on OATS/SAC

 

 

Hi,

 

I’m trying to calculate the distances/angles at which a maximum (in phase)
or minimum (anti-phase) signal would occur on an OATS/SAC.

 

I can do this simply when the TX and RX antennae are the same height above
the reflecting surface as the point of reflection lies halfway between the
two antennae, Distance_tx = Distance_rx. The direct and reflected paths can
be calculated using simple geometry and the wavelength is given by lambda =
c / f.

 

However when the height of the RX antenna is different to the height of the
TX antenna then the horizontal distance to the reflection point is no longer
equidistant. I can see that the ratio Height_tx / Distance_tx = Height_rx /
Distance_rx remains the same because the angle of reflection is the same.
But I’m left with two unknown Distance terms in the equation.

 

Is there a standard equation for calculating the reflection angle on an
OATS/SAC with a varying height antenna? Or can someone give me some pointers
to help me figure it out myself? I was so distracted thinking about this
that I missed my turnoff whilst cycling home the other day.

 

I’ve tried Googling but maybe I’m not putting in the right search term.

 

Any assistance gratefully received.

Thanks and regards,

James

 

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