Page 87 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง
P. 87

79
                     2.




                                                                   Q 1  2SC3355 V CBO  =  20 V,V CEO  = 12 V,
                                                                                                Ω
                                                                   V EBO  =  3 V, I C (MAX)  =  100 mA, r bb′ =  2  ,
                                                                   P =  600 mW, β =  β =  120,V BE  =  0.6 V,
                                                                               F
                                                                                   o
                                                                    D
                                                   Q 1             F =           =
                                                               V    T  6.5 GHz, C ob  0.65 pF;
                                                                o
                                  E
                            E       i
                             g

                                           ,%   2.18              	 
    
     %   '"  ก !!

                                     ,C    ก
  ก(

                                                        1
                       ก  ก   (2.17)                    F =
                                                H
                                                     π
                                                    2 R   C
                                                        FH T
                                                        1
                                                                      C =
                                                T
                                                     π
                                                    2 R   F
                                                        FH H
                                                        1
                                C b e ′  +  C b c ′  {1 g+  m  ( R out )} =
                                                     π
                                                          F
                                                    2 R FH H
                                                        1
                                    C b e ′  +  C b c ′  +  C g R  =
                                          ′
                                          b c m out
                                                     π
                                                          F
                                                    2 R FH H
                                                        1
                                                         C g R  =  −  (C b e ′  +  C b c ′  )
                                          ′
                                         b c m out
                                                     π
                                                          F
                                                    2 R FH H
                                            g m
                     (
-     ก (C b e ′  +  C b c ′  ) =    %
   
  ก
                                           π
                                           2 F T
                                                        1       g m
                                                         C g R  =  2 Rπ  FH H  −  2 Fπ  T
                                          ′
                                         b c m out
                                                          F
                                                     π
                                                            π
                                                    2 F −  2 g R    F
                                                         C g R  =  T  m FH H
                                          ′
                                          b c m out
                                                       4π 2 R FH H T
                                                              F F
                                                    2π  F −  g R  F      F −  g R   F  )
                                                                       =
                                                         C g R  =  ( T  m FH H  ) ( T  m FH H
                                          ′
                                          b c m out
                                                                           π
                                                       4π 2 R FH H T      2 R FH H T
                                                                                 F F
                                                              F F
                                                     ( F −  g R  F  )
                                                                      g =  T  m FH H
                                                m
                                                     π
                                                          F F C R
                                                    2 R FH H T  b c out
                                                                 ′
                                    2 g Rπ  m FH H T  b c out  =  ( T  g R  F  )
                                                     F −
                                      F F C R
                                             ′
                                                          m FH H
                                                     π
                                                                      F =  2 g R  F F C R  +  g R  F
                                                                   ′
                                                                   b c out
                                                T
                                                       m FH H T
                                                                           m FH H
                                                                      F =  g R  F  (2 F C Rπ  T b c out  +  ) 1
                                                                     ′
                                                T
                                                     m FH H
                                                              F T
                                                                      g =
                                                m
                                                             π
                                                    R FH  F H  (2 F C R  +  ) 1
                                                                  ′
                                                               T
                                                                  b c out
                                    I
                     (
-     ก  g =  C (dc )
                               m
                                   (k T q )
                                    B
                                                             I C (dc )  =  F T
                                                             π
                                                        F
                                          (k T q )  R FH H  (2 F C R    +  ) 1
                                                                   ′
                                                               T
                                            B
                                                                  b c out
                                                                                  ก         	
    
    ก  
  ก
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