Page 160 - 30105-2003 การวิเคราะห์วงจรอิเล็กทรอนิกส์ความถี่สูง
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152

               3.


                        3.7.4  ก
  ก     	
  
 	     @A*      

  กCK  "



                                                                                Ω
                                                                      Ω
                               * ก,
.  @ ก
  ก    V     = 12 V, R =  75  , R =  75  ,T = 25 C;
                                                      CC         g        L         A


                                    C  F  1
                                                                Q 1
                                                                                V o
                        E         E 1 i  E 2 i
                         g


                                                                                           Ω
                      C    F  = 5.5 MHz, B  =  ±  60 kHz,V  =  0.6 V, β =  β =  85, C  = 1 pF, F =  2 GHz, r bb′ =  2  ;
                       F 1  R        W           BE        F   o     ob      T
                             	+    3.21              
,  "- /(,. .  !(ก01  &  "-   ก
                                                                             '

                              3.7.4.1  ก
ก,
.    
 I
                                                  C
                                                        k T
                                                     F T  ( B  ) q
                 ก !ก    (2.20)               I C (dc )  =
                                                  F
                                              R FH H  ( 2 F C Rπ  T  b c out  +  ) 1
                                                             ′
               ก  
 "
   F =  5.5 MHz, R FH  =  R CF 2 , R out  =  R L  , C b c ′  =  C ob ;
                          H
                 ก !ก   (3.30)                 Z  =  R  +  R     (r  +  r  )}
                                         incf  CF 2  { BB   bb′  b e ′
                
      ก R !  9  	   !  8 % ($
                         BB


                                                             Z incf  =  ( R CF 2  +  r bb′  +  r b e ′  )


                
      ก C &   ก   9    !&  
  
   
  ก+ %!   600 Ω     
 $
 Z incf   %&   !  9  600 Ω
                          1
                         F
                                              β
                                                                r b e ′  =  o
                                              g m
                             I
                
      ก  g =  C (dc )   %
                         m
                              V T
                                              β V
                                                                r b e ′  =  o T
                                              I
                                               C (dc )
                            β V
                 
 9   r b e ′  =  o T  "
 !ก   Z incf   %
                            I
                            C (dc )
                                                          β V
                                        Z   =  R   +  r  +  o T
                                         incf  CF 2  bb′
                                                            I C (dc )
                                       β V
                                        o T  =  Z  −  ( R  +  r  )
                                       I C (dc )  incf  CF 2  bb′


                                                                         ก         	
    
    ก  
  ก
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