In my work I want to use the ADL5380. On the website analog.com I found CN-0374,where it is used ADL5380. I have a few questions on CN-0374. 1) Why on page 1 for ADA4940-2 Zin=500 Ω, if Rg=249 Ω and Rf=499 Ω, then Zin= 750 Ohm, isn't it? 2) On page 3 stated that power gain = -4.643 dB for the ADL5380. Where is the number? What is the difference between conversion gain and power gain?

1. The amplifier is driven with balanced differential input signals. For this case, the input impedance is 2 * Rg which is 500Ohms in our case. For more information, please look at the datasheet for ADA4940 Page 24, section "Calulcating the input impedance of an application circuit".

2. On page 1 of the datasheet, it states that the voltage gain for the ADL5380 is 5.36dB. This is equivalent to -4.64dBm of power gain. To convert voltage gain to power gain, you would have to add the correction factor to compensate for impedance mismatch between the input and output of the ADL5380 as the ADL5380 has a 9:1 impedance transformer on each of the differential baseband ports. This presents a 450 Ohm differential load to each baseband port when interfaced with 50 Ohm equipment. There is more on this in the ADL5380 datasheet on page 38, section "characterization setups".

Pgain = Vgain + 10*log(Rin/Rout)

= 5.36 + 10*log(50/(450+50))

= -4.64dBm

Hope this helps!