two stage common emitter amplifier

Discussion in 'General Electronics Chat' started by mrk, Jan 8, 2010.

  1. mrk

    Thread Starter New Member

    Jan 7, 2010
    3
    0
    Hi , please can somebody help me to solve my assignment . the question circuit is attached and the question is ;

    1) for the curcuit design a two stage common emitter amplifier for for the following specifications ;

    RE1=REA1+REA2 , RE2=REB1+REB2 , IE1=1.8mA , IE2=4mA
    REA1=0.22RE1 , REB2=0.4RE2
    FOR BOTH STAGES ;
    β=200 , R2=0.1βRE1 , R4=0.1βRE1 , VA=100V , VE=0.1VCC , VC=0.5Vcc , VBE=0.65V

    2) PERFORM AC ANALYSIS ( from 10 Hz to 100 MHz ) and print output for V of the both stages separately .

    thanks a lot in advance for your support and so operation .

    best regards
     
  2. peranders

    Well-Known Member

    May 21, 2007
    87
    0
    What kind of help do you ask for?
     
  3. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    This belongs in the homework section, and keep in mind that normally people wont solve problems for you. They will however gladly help you if you show some initiative and show us what you have done so far and where you are stuck.
    So where are you stuck with this assignment ?
     
  4. mrk

    Thread Starter New Member

    Jan 7, 2010
    3
    0
    sorry for the mistake. i have solve the circuit and find all the values , bat i dont know where i do mistake . the values taht i find is ;

    Re1= 1000ohm , Re2=450ohm , Rea1= 220ohm , Rea2=780ohm , Reb1=270ohm , Reb2=180ohm , R4=10kΩ, R2=10kΩ , Rc1=5kΩ , R1=10kΩ , Ic1=1.791mA , Vb1=1.15V , Rc2=2.2kΩ , Bb2=1.15V , R3=10kΩ

    bat my instructor is said this values incorrect . so kindly can you help me and inform me the right values .

    thanks
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    Hmm, for example
    VE=0.1VCC and IE1=1.8mA , IE2=4mA
    So
    Re1=1.6V/1.8mA=888=820

    Re2=1.6V/4mA=400=390


    And Rc1=6.4V/1.79mA=3.5733=3.6K
    and soon


     
  6. mrk

    Thread Starter New Member

    Jan 7, 2010
    3
    0
    VE=0.1VCC so Ve= 0.1 X 18 = 1.8 V am i right ? why 1.6 V
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,957
    1,097
    If so I'm sorry because I assume 16V.
    So for 18V
    Re1=1.8V/1.8mA=1K; Rc1=9V/1.79mA=5K=5.1K;

    Re2=1.8V/4mA=450=470; Rc2=9V/3.98mA=2.2K.

    R2=0.1*200*1K=20K -->
    R1=( Vcc - (Vbe+Ve) )/(IR2+Ib)

    IR2=(Vbe+Ve)/R2=2.45V/20K=122.5uA

    Ib≈1.8mA/β=9uA

    R1=( 18V - (1.8V+0.65V) )/(122.5uA+9uA)=15.55V/131.5uA=118K=120K
     
    Last edited: Jan 9, 2010
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