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BJT amplifier - calculate resistances

  1. May 30, 2014 #1

    etf

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    Hi!
    Here is my task:
    Calculate RC1, RC2, RE2 for maximum undistorted symmetrical signal at output if DC collector currents are IC1=IC2=1mA. Calculate voltage gain Av.
    Vcc=9V, RE1=1.8kΩ , VBE=0.7V, VCES=0.2V, VT=26mV, beta=100, VA→∞.
    Solution: RE2=700Ω, RC1=5.724kΩ, RC2=3.141kΩ, Av=gm1*gm2*rpi2*RC1*RC2/(rpi2+RC1).


    nik.JPG

    After assuming that BJT's operate in forward active mode and solving circuit bellow I got RE2=700Ω, RC1=5.782kΩ. Is it ok? How to find RC2? I have no problems calculating Av.

    31052014894.jpg
     
  2. jcsd
  3. May 31, 2014 #2
    Re1 looks good
    But you are slightly off the true Rc1 value.
    What is the current that is flow Rc1 ? What is the voltage across Rc1?

    As for the Rc2 maybe this will give you some hint
    https://www.physicsforums.com/showthread.php?t=633117&highlight=swing
    But for your circuit Vc is not equal to 4.5V. You must include DC emitter voltage and Vce(sat).
    9V - (2.518 + 0.2) = 6.282V.
    So to get maximum undistorted signal Vc2 must be equal to ??

    P.S

    What it is ??
    https://www.physicsforums.com/attachment.php?attachmentid=70181&d=1401493487
     
    Last edited: May 31, 2014
  4. May 31, 2014 #3

    etf

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    Thanks, I will take a look on post... My second scheme is scheme for DC analysis, I just used models for BJT's in forward active mode (http://people.senecac.on.ca/john.kawenka/EDV255/images/bjt05.png [Broken]) instead of usual BJT symbol.
     
    Last edited by a moderator: May 6, 2017
  5. May 31, 2014 #4
    Your DC model is incorrect. Try to draw it again.
     
  6. May 31, 2014 #5

    etf

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    Whats wrong with model?
     
  7. May 31, 2014 #6
    The model is ok. The overall diagram is wrong. Why T1 base is short to ground?
    Why Vcc is connect is not connected to ground ?
     
  8. May 31, 2014 #7

    etf

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    E1 should be connected to ground, instead of B1. I don't know why I connected B1 to ground :)
     
  9. May 31, 2014 #8
    Do you managed to find Vc2 voltage ?
     
  10. May 31, 2014 #9

    etf

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    I can't figure it out :(
     
  11. May 31, 2014 #10
    What is the voltage at Q2 emitter ??
    And notice that voltage at collector can swing from Ve2 + Vce(sat) up to Vcc.
    So to get maximum undistorted symmetrical signal at collector we must chose Vc at the middle of this swing.
     
  12. May 31, 2014 #11

    etf

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    VE2=VC1E1-VB2E2=2.518V so voltage at collector (VC2) can swing from 2.518V+0.2V=2.718V up to Vcc=9V (6.282V swing). Now I should find Vc and after that RC2?
     
  13. May 31, 2014 #12
    Exactly, and you should pus Vc at the middle of the swing.
     
  14. May 31, 2014 #13

    etf

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    2*(vc2e2-0.2)=6.282 → vc2e2=3.341v.
    Vc2e2+rc*ic2-rc1*ic1-vb2e2=0 → rc2=(rc1*ic1+vb2e2-vc2e2)/ic2
    rc2=3141Ω
    Thanks a lot!
     
  15. May 31, 2014 #14
    The answer is good, but you over complicate this.
    All you need to find Rc2 is to use a II Kirchhoff law for the output loop.

    attachment.php?attachmentid=70196&stc=1&d=1401570058.png

    Vcc = VRc2 + Vce2 + Ve2

    Ve2 = Ie2*Re1 + Ic2*Re2 = 1.01mA*1.8kΩ + 1mA*700Ω = 2.518V

    The swing

    Vcc - (Ve2 + Vce(sat)) = 9V - (2.518V + 0.2V) = (9V - 2.718V) = 6.282V

    From this VRc2 = Vce2 = 6.282V/2 = 3.141V

    And

    Rc2 = VRc2/Ic2 = 3.141/1mA = 3.141kΩ

    And Rc1 = (Vcc - (Ve2 + Vbe2))/( Ic1 + Ib2 ) = 5.782V/(1mA + 10μA) = 5.724kΩ
     

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  16. Jun 1, 2014 #15

    etf

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    I have another example.
    Calculate resistances R1, R2, RC, RE for maximal symmetrical undistorted voltage on output, if total maximal collector current is Icmax=4mA. Vcc=10V, load resistance RL=10kΩ, β=100, VBE=0.7V, VCES=0.2V, VT=26mV, VA=100.

    prvi.JPG

    In DC analysis, I calculated VCE=VCC-(beta*RC/(RE*(beta+1))+1)*((VCC/R1-VBE/R1-VBE/R2)/(1/R1+1/R2+(1-beta/(beta+1))/RE)). For maximal swing Vce shoud be on middle of load line. I'm not sure I understand what total maximal collector current means :confused:
    Here is solution:
    RE=2*beta*(Vcc-VCES)/(Icmax*(1+beta))-(beta/(1+beta))*(RC*RL/(RC+RL)+RC)
    R2=R1*(2*beta*VBE+(1+beta)*RE*Icmax)/(2*beta*(Vcc-VBE)-[R1+(1+beta)*RE]*Icmax)
     
    Last edited: Jun 1, 2014
  17. Jun 2, 2014 #16
    OMG, how do you manage to derive the formula for RE and Vce ??
     
  18. Jun 3, 2014 #17
    I'm also not sure what total maximal collector current is.
    I see two options.
    First:
    Ic_max = Icq - quiescent DC current or maybe Ic_tot = Icq + Ic_ac ?

    And to find components values we can use ohms law.
    For example assume Veq = 1V and Icq = 1mA and from there we have
    Re = 1V/1.01mA = 990Ω = 1K.
    Rc = 4.5V/1mA = 4.4KΩ ≈ 4.3KΩ and voltage at base Vb = Veq + Vbe.
    So you design voltage divider to get Vb at output. And the current that is flow through divider should be much larger than the Ib

    http://eelinux.ee.usm.maine.edu/cou....BJT Amps. for Undistorted VoltageSwing-X.pdf
     
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