Page 182 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง
P. 182

174
               4.




               .     E r    5    V  (4%
                      i b e ′
                                  b e ′
                     R i (0- )
                        π
                                                   E g r R                                   (4.12)
                                                         ′
                                                    i m b e E
               %    &                                       I  =
                                          d 2                    )
                                              R     ( R +  r  +  R
                                                  π
                                               i (0- )  E  d 2  L
                                     !  .  %  0ก   
 R   ,    (0- )  	
	    (
	 4%  	ก 
ก	  (4.12)

                              V o (0- )               L         π
                                  π
                1 ก - R
                       L
               %    &                                  V o (0- )  = I R                                                                       (4.13)
                                          π
                                                  L
                                               d
                                                2
               *&          4.11      (
	   	 I .5(V o (0- )  
!  ก 	
 % E =  25 mV peak
                                                    π
                                          d
                                                                 i
                                           2
                                                   E g r R
                                                         ′
                                                    i m b e E
                 +  ,    	ก 
ก	  (4.12)         I  =
                                          d 2                    )
                                              R i (0- )  ( R +  r d 2  +  R L
                                                  π
                                                      E
                                  Ω
               $%          r d 2  =  4.762 k , g = 216.459 mS;
                                      m
                                   Ω
                                                     Ω
                             r b e ′  =  461.981  , R i (0- )  =  56.507 k , E =  25 mV peak ;

                                         π
                                                         i
                                              (  25 10 − 3  × 216.459 10 − 3 ×  461.981 560 )
                                                 ×
                                                                ×
                                                                              ×
                                                                I d  =              = 404.025 nA
                                                     ×
                                                                             ×
                                           2
                                                              +
                                                                        3
                                                                     ×
                                               56.507 10 3 ( 560 4.762 10 +  56 10 3 )
                	ก 
ก	  (4.13)              V o (0- )  = I R
                                          π
                                                  L
                                               d
                                                2
                                               Ω
               $%          I d  = 404.025 nA, R = 56 k ;
                                        L
                         2
                                                              ×
                                                     ×
                                                          V o (0- )  =  404.025 10 − 9  × 56 10 3
                                          π
                                      V     =  22.625 mV
                                          π
                                       o (0- )         peak
               *    I d  2  = = = =  404.025 nA peak  ,V o (0- )  = = = =  22.625 mV peak  ;
                                            π ππ
                                            π
                        4.2.3  ก 
    
      	
 !!" ก
 #$$%&    
   ก%       ( - π ππ π
                                                                           π π
                                                                              π
                                                                              π
                                                                             2 )
                              R in ( -2 ) R i ( -2 )
                                 π
                                   π
                                       π
                                         π
                                      r bb′  b′                             D 1
                                                                            r  1 d  I  1 d
                                                 r b e ′  V b e ′  g V
                                                            m b e ′
                     E g                                                     D 2
                   ( -2π π )                        I e
                                               R EP 1  R E                                 V o (0- )
                                                                                               π

                           1     4.8  . %     4==D	ก (.  5 -  	
   ก5	 ,    ( - π ππ π #   1     4.5
                                                                         π π
                                                                            π
                                                                            π
                                                                           2 )
                                                                            ก         	
    
    ก  
  ก
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