Chris' "Maxwell" equations look correct. But the point at which ferrites are
placed will not always have a common mode impedance of 50 ohms. Here's An
example: a large DUT has a 1 meter long cable that connects to the ground
plane. At 75 MHz the common mode impedance of the cable, at the DUT, is
about 3k ohms. Adding a 1k +j1k ferrite at the DUT knocks down the radiation
by 10 dB. But the radiation is not reduced mainly by losses but by
"detuning" the "antenna". The resonant frequency has shifted from 76 MHz to
60 MHz. Now 60 MHz could be a problem. I noticed this type of thing while
trying ferrites to reduce emissions from a digital device that had a DUT
cable (not grounded at the end). I could fix one frequency with ferrites but
it would just tune the cable/system to resonance at another frequency. The
person I was working with didn't believe this theory so I ADDED wire to the
end of the DUT cable to make it 1/2 wavelength, rather than 1/4 wavelength
at the offending frequency and dropped the signal by 20 dB.
If anyone has any changes to the model or a what-if, I can simulate it and
send you the simulated data.
Dave Cuthbert
Micron Technology
From: Pettit, Ghery [mailto:ghery.pet...@intel.com]
Sent: Wednesday, January 29, 2003 12:09 PM
To: 'Chris Maxwell'; Pettit, Ghery; neve...@attbi.com;
emc-p...@majordomo.ieee.org
Subject: RE: EN55022:1998 + A1:2000
Chris,
You can indeed make your own, but my bet is that A2LA or NIST NVLAP
inspectors will want to see calibration data, not calculations.
Now, if we just had a published calibration technique...
Ghery
From: Chris Maxwell [mailto:chris.maxw...@nettest.com]
Sent: Wednesday, January 29, 2003 10:57 AM
To: Pettit, Ghery; neve...@attbi.com; emc-p...@majordomo.ieee.org
Subject: RE: EN55022:1998 + A1:2000
Ghery,
If the standard is assuming a 50 Ohm system, doesn't this breakdown to a
simple calculation?
Insertion Loss = IL = 20 x log((50 + Zf) / 50)----where Zf is the ferrite
impedance
This could easily be solved for Zf if you assume IL to be 15dB (in this
case the dB are truely dimensionless; as you are calculating a pure loss
>from an arbitrary level).
I would think that you would just have to:
1. Solve the above for Zf. By the way, I get 231.2 Ohms. Can someone
check this?
2. Gather up a box of "doughnuts" such that the total Zf is above the
answer for step 1 at all frequencies from 30Mhz to 1Ghz
3. Color code the doughnuts (or whatever) and write a procedure that says
something like "clamp three blue doughnuts and two red doughnuts over each
cable ..."
Have I over simplified this??? Wouldn't this be proof enough for any
accreditation body? I know that $300 may not seem like alot to some; but it
adds up after a few cables. Besides; we went to college to learn all of
that math; why not use some of it? I don't mind paying for stuff that I
can't make; but this one seems possible to me.
Chris Maxwell | Design Engineer - Optical Division
email chris.maxw...@nettest.com | dir +1 315 266 5128 | fax +1 315 797 8024
NetTest | 6 Rhoads Drive, Utica, NY 13502 | USA
web www.nettest.com | tel +1 315 797 4449 |
> -----Original Message-----
> From: Pettit, Ghery [SMTP:ghery.pet...@intel.com]
> Sent: Wednesday, January 29, 2003 11:59 AM
> To: 'neve...@attbi.com'; emc-p...@majordomo.ieee.org
> Subject: RE: EN55022:1998 + A1:2000
>
>
> Amendment 1 to CISPR 22:1997 (Amendment A1:2000 to EN 55022:1998) requires
> that the clamps provide at least 15 dB of loss in a 50 ohm system over the
> frequency range of 30 MHz to 1000 MHz. The use of extension cords is
> prohibited. Can you guarantee that your bucket of doughnuts will meet
this
> requirement? How will you demonstrate that to your lab's accrediting
body?
>
>
> Fischer's clamps are around $300 each. Compared with what we had to
choose
> from prior to their product, these are not big bucks.
>
> Ghery Pettit
> Intel
>
>
>
>
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