Page 185 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง
P. 185

177
                     4.




                                                   C B
                                          R FLCB             r bb′

                                                                 r b e ′   R ≈  R EP  ≈  R EP 1
                                                                            E
                                            E
                                             g
                                                                      β +
                                                                     ( o  ) 1 R E


                                                 1     4.9 . % ก	   (
	   	 R
                                                                           FLCB

                       !    	ก  R ≈  R EP  ≈ R
                               E
                                           1
                                         EP
                                                                               ) }
                                                               r
                                                                 R FLCB  = R +    R BB     { bb′  +  r b e ′  + ( β + 1 R E  
                                                                           o
                                                     g
                                                                              ) }
                                                         R BB  { bb′  +  r b e ′  + (β + 1 R E
                                                              r
                                                                          o
                                                                 R FLCB  = R +  ) }
                                                     g
                                                         R
                                                                           o
                                                        { BB  +  r bb′  +  r b e ′  + ( β + 1 R E
                                                                    X C  =  R FLCB
                                                 B
                                         1
                       !    	ก  X  =
                               C B
                                     π
                                            C
                                    2 F  ( L C B ) B
                                                         1
                                                                 R  =
                                              FLCB
                                                     π
                                                    2 F  ( L C B ) B
                                                            C
                                                         1                                                                          (4.18)
                     %    &                                  F  =
                                              ( L C  B )  2 Rπ  FLCB B
                                                            C
                     *&          4.14   	ก 1     4.5 .5( 4.9      (
	   	 F
                                                                    ( L C  B )
                                                         1
                       +  ,    	ก 
ก	  (4.18)    F  =
                                              ( L C B  )
                                                     π
                                                            C
                                                    2 R FLCB B
                                                              ) }
                                              r
                                          R BB  { bb′  +  r b e ′  + (β + 1 R E
                                                          o
                     $%         R FLCB  =  R +                 ) }
                                     g
                                          R
                                         { BB  +  r bb′  +  r b e ′  + ( β + 1 R E
                                                           o
                                              ×
                                         8.432 10 3 {2 461.981+  + (101 560×  )}
                                   R FLCB  =  75 +                         = 7.420 kΩ
                                        { 8.432 10 +  2 461.981+ (101 560×  )}
                                                 3
                                                      +
                                              ×
                                   C = 470 pF
                              B
                                                                  1
                                                  =
                                                                  F  ( L C B  )  2 3.14 7.420 10 × 470 10 − 12
                                                                           ×
                                                           ×
                                                     ×
                                                                 ×
                                                                     3
                                            F     =  45.660 kHz
                                              ( L C B  )
                     *     F     = == =  45.660 kHz
                             ( L C B )
                                                                                  ก         	
    
    ก  
  ก
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