Hi Glen -

Is multiplying I with Q enougth to demodulate FM as in a quadrature
demodulator?
The IQ signals multiplied do not make an FM quadrature
detector as the phase does not really shift much over the NBFM
bandwidth. There may be a clever way to extract FM more direcly from
the IQ channels without first detecting a sideband, but I don't know
it.
direct way:
I read about something like:
phase = (I * diff Q + Q * diff I)/(I*I + Q*Q)
but I cannot remember it exactly.


Thank you for given the link to Peregrine. I read the datasheet. How do
you
mix it?

Maybe the wrong part? I'm talking about the PE4140 FET ring mixer.
You use two, driven with LOs ninety degress out of phase with respect
to each other.
Yes, the correct part!
But I cannot see a great benefit using your concept. What is better than
others?

How much dynamic range do I need? I thought about 100dB?
If I recall theory I loss 2dB if limiting the signal to remove AM
sensitivy.techniques,

You need good dynamic range up to the limiter. This can be done with
analog circuits as discussed. After the limiter an A/D converter is
not even needed in theory--a fast-running timer hooked to a digital
port could do the trick. I don't remember any loss by removing AM in
the limiter (since all the the information is contained in the
frequency of the signal), but my theory is in the distant past, back
when FM meant 'funny math'.
Everywhere I read about the necessity to remove AM. If the band is used for
FM only, why then remove the not found AM in it?
I ran a simulation with Spice doing FM demodulator concepts comparision. The
difference between limited and not limited FM product detector was not of
significance. Doing the correct low-pass filtering after the detector was of
much higher importance.

- Henry