Dear All ,
In order to operate the Agilent ATF-35143 , the bias circuit is needed .
They chose Vds=2V , Vgs = -0.65V , Ids= 10 mA.The bias circuit diagram is attached but how they get that value of L and C .???
I calculate total inductance L = 103 nH and total capacitance C = 100.1 nF
Can anyone explain or I appreciate any hint for that .thanks.

 

I forget to say Agilent ATF-35143 is low noise transistor.(LNA)which is square inside the diagram . thanks.

 

For bias point(0.6V),
At 1.42 GHz operating freq, 103nH will give 918.5128 ohm
Z= 1/2*pi*f*C
[ Z = 1/(2*pi*1.42*10^9*103*10^9) ]
C=100pf will give Zc= 1.121 ohm at 1.42 GHz.( It is shorted at 1.42 GHz )
C= 27pf infront of ATF amplifier will block DC from Input RF,
will give 4.15 ohm at 1.42 GHz ,
How about another C( also block DC ) behind ATF amplifier , C= 1pf then Zc = 112 ohm ???
Is it Correct ?
Thanks.