Asking about Agilent ATF-35143

Discussion in 'Homework Help' started by jjohn, Apr 15, 2009.

  1. jjohn

    Thread Starter Member

    Feb 18, 2009
    13
    0
    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.
     
  2. jjohn

    Thread Starter Member

    Feb 18, 2009
    13
    0
    I forget to say Agilent ATF-35143 is low noise transistor.(LNA)which is square inside the diagram . thanks.
     
  3. jjohn

    Thread Starter Member

    Feb 18, 2009
    13
    0
    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.
     
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