Re: NanoVNA #calibration


Radu Bogdan Dicher
 

Larry - I re-executed the calibration over 10.4 through 11MHz, saved to
"1," and I always recall it before measuring. However, the letter changes
every time I start the scan (from "C1" to "c1") - I press sweep, flop!, it
changes. Not sure why, as far as I can tell I have the same settings on
device and computer application (center: 10.7MHz, span: 600kHz), and I
RECALL 1 it every time I prepare for scanning.

David - yes, I am using 56.2ohm and 301ohms, 1% tolerance resistors. It's
the closest I could come up with by calculating for minimum loss. The
filters are 150kHz Muratas (JAs), but I also measured about four different
bandwidths, and from multiple sources. I have at least 7 or 8 different
batches and they all have the 5dB difference L/R. I think my calibration is
not getting applied, based upon Larry's feedback above.

Still trying to figure out how to maintain calibration settings while
executing the scan.

Thank you both,
Radu.

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On Sun, May 24, 2020 at 6:19 AM <david.hostetler@ieee.org> wrote:

What I see in the plot that you posted is a filter that is about 170 kHz
wide at the -3 dB points and centered at 10.715 MHz. This is consistent
with a Murata SFELF10M7HA00-B0 that has a spec of 180 +/-40 kHz and a
center frequency of 10.7 MHz +/-30 kHz. The fact that all the filters
exhibit the same response wouldn't surprise me if you purchased them at the
same time, they probably came from the same batch, and the tolerance is
likely based on batch to batch differences caused by process variations
rather than random variations within a batch.

It is a little harder to understand/explain the fact that the low
frequency side runs out of dynamic range at -60 dB and the high side gets
there at 5 dB higher at -55 dB. Since these filters are symmetrical, have
you tried swapping the input and output by turning the filter around at the
fixture?

Also, you haven't really detailed your circuit for matching the 50 ohm of
the VNA to the 330 ohm of the filter. If I were doing it, I would use a
minimum loss matching pad. This would consist of a 54.281014769949586 ohm
resistor (it is the value that the calculator gave - LOL) across the input
and output coax to the VNA and a series resistor of 304.01167774007837 ohms
up to the filter input/output. This would give a loss of 13.867585162369918
dB on each side. 5% values: 56 and 300. 1% values: 54.9 and 301. I am sure
the 5% values would be just fine. You could bridge the filter location with
a short on your test fixture to determine the exact loss through the pads,
then place the filter in the fixture and easily find the filter insertion
loss.



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