B
billcalley
- Jan 1, 1970
- 0
Hi All,
I have been trying to figure out this basic concept for quite some
time now, but without much success: When purposefully mis-matching an
RF stage (such as a PA) in a 50 ohm system, I sometimes see references
alluding to the fact that the next stage will *not* be able to see 50
ohms due to this mismatched stage. The part that really confuses me
is: Why not? Why can't the PA's output matching network not only
purposefully be designed to conjugately mis-match the PA's output (for
max Pout) AND also present exactly 50 ohms at the other end of this
same matching network? And if, for some reason, this is not
possible, then why not just add another pi (or L) network so as to
give the next stage exactly 50 ohms? Am I understanding this
correctly, or do I have it completely FUBAR'ed up?
Many thanks!
-Bill
Hi Guys,
From what I have been able to absorb now from all of your really
terrific responses is that it is totally and completely impossible to
LC match, even theoretically, a purposefully mis-matched active
device*, and then look back into that mis-matched network and see a
perfect 50 ohms for the next stage. However, the purposefully mis-
matched active device WILL get to see the impedance it wants to see.
And all this is due to the reciprocal nature of LC matching networks.
Thus, it would be wise, as you have all mentioned, to design any
non-50 ohm LNA or PA to be as close to 50 ohms as possible, or to
design the next connecting stage so that it properly works with
something other than 50 ohms -- 'cause there is no way to "fix" this
mis-match issue with an LC matching network. Is that correct, or have
I misunderstood something? (I can't believe I didn't know -- or
didn't understand -- this stuff from the get-go! A major glitch in my
knowledge-base, that's for sure.)
Many Thanks,
-Bill
*The mis-match created so as to optimize an LNA transistor's input for
NF, or a PA transistor's output for P1dB, to name two common reasons.