agilent lna

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
Agilent ATF-35143 is low noise transistor.(LNA)which is square inside the diagram .

Can anyone explain or I appreciate any hint for that .thanks.

The reason of the two bias inductors in series is to get higher Series Resonant Frequency of the resultant inductor (smaller inductor have greater SRF), and the reason of the two capacitors in parallel is to get wideband frequency decoupling (due to different reactances at different frequencies).

Dear vfone ,
thanks.
You means.

For 47nH and 1uf , its SRF =2.3 MHz
For 47nH and 100pf , its SRF = 73 MHz

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.

I guess not.
if you see the specifications of inductors alone, let say for 103 or near 100 nH , 47nH and 56nH, The SRF of 100nH is less than 56nH.

my misunderstanding . SRF is based on its internal stray capacitance.