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1M2 = ( lcn740/ Icn38)A,A,/A,  = 19A, = 19(0.01 3) = 0.25 = 25 percent = 25%
                                                                                                     (1 4-1 5)
            Note:  (I 740/I 38) = (IcQ/I     cQ ) (740/38) = 1(740/38) = 19
                              cQ
                     cQ
            But  what  does  this  25  percent  of 1M2 distortion  mean?  This  means  that the  1M2
            distortion  components  have  an  equivalent  transconductance  of  25  percent  of the
            small-signal  transconductance.  So,  if the  transistor  is  biased  at  1  mA,  the
            transconductance for generating the sum- or difference-frequency signal  is 25%(gm
            =  25%(0.0384) A/V, or 9.5  mA/V.

            So,  for  example,  for an  oscillator  signal  of 0.013  volt  peak  sinusoidal  waveform
            added  or  combined  with  a  small-signal  RF  signal  V RF  into  the  base  and  emitter
            junction  of a  bipolar  transistor  amplifier  operating  at  1 mA,  the  IF  signal  current

            IS;9-IF with the difference frequency is


                       ISI~I F =  V Rf  X  25%(glll)  =  V                  X  9.5 rnA/ V
                                                                        Rf
            Note  that the conversion  transconductance  is  IS;9_IF/VRF.  And  for a O.013-volt  peak
            sinusoidal  oscillator  signal  into  the  mixer,  the  conversion  transconductance  is
            25%(gm).
            Equation  (14-15)  shows  that the  conversion  gain  or IF  signal  transconductance  is

            controlled by the oscillator's signal amplitude.  But it should  be  noted that increasing
            the  oscillator's  peak  amplitude  A,  beyond  O.013-volt  will  raise  that  25  percent
            number to something  higher that is no  longer linear or proportional.  By using  other
            mathematical  techniques  that are  more  accurate  than  the  power-series  expansion

            [e.g.,  Equation  (14-8)  or  (14-9)],  it is  found  that if the  oscillator signal  is  raised  to
            26  mV  peak,  or  2  3  13  mV  peak,  we  get 45%(gm)  instead  of  50%(gm)  for  the
            conversion gain.
            In  previous  chapters  concerning  the superheterodyne  radio,  it was  mentioned  that
            the oscillator voltage into the  mixer or converter is in  the range of about 200  mV to
            300  mV  peak  to  peak  or  100  mV  to  150  mV  peak.  The  next section  will  explore  a

            more accurate model for bipolar mixer circuits with  large oscillator voltages that are
            beyond  13 mV peak.
            Another  technique  to  more  accurately  characterize  the  behavior  of  a  simple
            transistor mixer is  to view the transistor as having  a time-varying transconductance

            that depends on  the oscillator signal.  Recall  that the small-signal  transconductance
            gm =  I /(0.026  volt).  Usually  we  define  a small-signal  transconductance  as  gmQ  =
                    cQ
            I /(0.026  volt),  where  ICQ is  a  constant  DC  current,  which  then  makes  gmQ  a
             cQ
            constant small-signal transconductance.
            But when  an  oscillator voltage is added  with the  bias voltage,  the  collector current

            is  actually  time-varying,  and  thus  the  transconductance  is  changing  with  time  in
            relation  to  the  amplitude  and  frequency  of the  oscillator  signal.  Therefore,  the
            output collector current as a function of time can  be thought of as
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