Page 151 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง
P. 151

143

                     3.



                                                                    g m
                       ก !ก   (1.16)                  C b c ′  =  C ob  =  C re ,C b e ′  =  −  C ob ;
                                                                   π
                                                                  2 F T
                                               2    C N
                                                        2
                       ก !ก   (3.11)             C N  S  =  P  P
                                             S
                                                        2
                                                    C N
                                                                      C =  S  S
                                                P      2
                                                     N  P
                     ,           C =  C  ,  N =  N  , C =  C  , N =  N  ;
                                    ′
                              P    b ep  P    11  S   b e ′  S  21
                                                         2
                                                    C N   21
                                                     b e ′
                                                                  C b ep  =                                          (3.25)
                                                ′
                                                      N 2 11
                                                       1
                       ก !ก   (3.1b)                      F =
                                                R
                                                    2π  LC
                     ,           F =  F  ( R L 11 ) , L =  L 11 , C =  C BT  +  C b ep ;
                                                           ′
                              R
                                                             1
                                                                F R (L 11 )  =
                                                    2π  L 11  ( C BT  +  C b ep )
                                                                   ′
                                                           1
                                                    ( C BT  +  C b ep ) =  2
                                               ′
                                                             ( )
                                                         π
                                                    L 11 ( 2 F R L 11 )
                                                           1
                        
 $
                                    C BT  =  2  −  C b ep                                        (3.26)
                                                                      ′
                                                             ( )
                                                         π
                                                    L 11 ( 2 F R L 11 )
                                                       1
                       ก !ก   (3.1b)                      F =
                                                R
                                                    2π  LC
                     ,           F =  F  ( R L 22 )        !8    ,. 

.'    L
                              R
                                                              22
                                     L =  L 22 , C =  C CT ;


                                                           1
                        
 $
                                    C  =                                                               (3.27)
                                               CT                 2
                                                    L  ( 2 Fπ    )
                                                     22     R (L 22 )
                     "#    
     3.19    + %!   9  C   % C
                                                 BT      CT
                                                                    g m      ;
                       -  ,
    ก !ก   (1.16)       C b c ′  =  C ob  =  C re ,C b e ′  =  −  C ob
                                                                   π
                                                                  2 F T
                                                             ×
                                                      397.937 10 − 3        − 12  209.919 pF
                                                                         ×
                                                                     C b e ′  =  6  − 1.3 10  =
                                                     ×
                                                           ×
                                                                ×
                                                    2 3.14 300 10
                                                                       ×
                                                    C N  2 21  =  209.919 10 − 12  ( ) 1  2  209.919 pF
                                                     b e ′
                       ก !ก   (3.25)                 C  =   =                    =
                                               ′
                                               b ep    2                2
                                                      N  11          ( ) 1
                                                                               ก         	
    
    ก  
  ก
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