Match 50 ohm impedance to LNA input impedance

Question

I have following PCB version of circuit: on SMA input we have antenna strictly matched to 50 ohm, then we have regular RF switch IC, after RF switch impedance matching network and final stage is transceiver(transceiver LNA input -- this from datasheet diagram).

If complex impedance of transceiver is unknown but it is for sure not 50 ohm, how correctly measure impedance to which 50 ohm should be converted for best match transceiver LNA. Connect VNA to transceiver itself before any matching network(test point 1 on image) or VNA should be connected to full path of signal propagation(test point 2), or 3rd variant most logical as for me before any impedance transformation happen(test point 3).

So i sticked with 3rd variant, i desolder matching components (L1, C1, C2) and solder 0 Ohm resistor jumpers to L1 and C2, attach coax cable to C2 jumper(test point 3 on image), compensate this small coax cable in VNA settings and got complex impedance for my test frequency(750Mhz) as 7.8 -j16.9, so my initial 50 ohm impedance that arrived from antenna should be transformed into this complex impedance, i was using smith chart tool to cast 50 ohm to complex impedance like following:

Looks like for my case if i use 10 pf capacitor i will get 7.63 -j18.0 that nearly my initial measured impedance, so i solder 10 pf capacitor to C1(L1 & C2 - 0 ohm jumpers) and result was totally disappointed, and not really what i was waiting for.

I choose wrong place to plug my VNA probe and should test from different one or test point was valid and i totally cracked logic for matching impedance ?

Answer 1

To measure the LNA input impedance you can use testpoint 1 or alternatively, as you already did, testpoint 3 with port extension to move the reference plane to the input of the LNA.

and got complex impedance for my test frequency(750Mhz) as 7.8 -j16.9

7.8 -j16.9 is equivalent to 44.4 Ohm || 10.35pF @ 750MHz. The resistive part is already close enough to 50 Ohm, which means you would have to only compensate the capacitive component with a small shunt inductor.

If you add 150pF as a DC block in front of the LNA, the resistive part of the input impedance increases to about 50 Ohm. The complete matching network looks like this:

Another possibility would be to compensate the -j16.9 Ohm reactance with a series inductor and then transform with an additional LC lowpass matching network from 7.8 up to 50 Ohm. The 1nF capacitor works as a DC block:

The series inductor can be replaced with a short microstrip transmission line (e.g. 50 Ohm line with vf = 0.667):