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Common Emitter Amplifier Analysis

  1. Apr 20, 2012 #1
    This is my analysis of common emitter amplifier.

    Ic = βIb
    Ic ≈ Ie (because Ib is very small) => Ie = βIb
    Vcc = IcRc + Vce + IeRe
    Vcc = IeRc + Vce + IeRe
    Vce = Vcc - βIb(Rc + Re)

    Vb = IbRb + Vbe + IeRe
    Vb = Ib(Rb + βRe) + Vbe

    Gain = Vce/Vb
    Gain = (Vcc - βIb(Rc + Re))/(Ib(Rb + βRe) + Vbe)
    Gain = - βIb(Rc + Re)/Ib(Rb + βRe)
    Gain = - β(Rc + Re)/(Rb + βRe)

    but everywhere I found only Gain = -Rc/Re

    Please if someone can tell me what's wrong,I've tryed to understand this amplifier
    for some time,and this is the analysis that i've came up with,and sorry for bad english
     
  2. jcsd
  3. Apr 20, 2012 #2
    You are doing DC analysis in all the equations. The Gain =-Rc/Re is AC analysis. You cannot use

    Gain = Vce/Vb

    You cannot use DC parameter for this. Vce as you wrote depend on Vcc, Rc etc. You have to do the AC analysis for this.
     
  4. Apr 20, 2012 #3

    vk6kro

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    Science Advisor

    Just to cut down on the typing, assume
    beta= 300 (small signal current gain or hfe)
    intrinsic base resistance =200 ohms,
    emitter resistor = 350 ohms,
    collector resistor = 2000 ohms
    ib = small signal base current

    Input voltage = ib * 200 + ib * 300 * 350 ie base voltage plus emitter voltage
    Output voltage = ib * 300 * 2000 ie base current times hfe times collector resistor.

    Gain = output voltage / input voltage = ib * 300 * 2000 / ib * 200 + ib * 300 * 350

    Gain = (ib * 600000 / ib *200 + ib * 105000) or (ib * 600000 / + ib * 105200)

    ib cancels, so the gain = 600000 / 105200 = 5.7034

    Notice that the ratio of collector resistance to emitter resistance = 2000 / 350 = 5.7143, so the simple formula is an excellent shortcut if the value of the emitter resistor is fairly large and hfe is large.
     
    Last edited: Apr 20, 2012
  5. Apr 21, 2012 #4
    Thanks for the replies,I understand now why I was wrong,and sorry for posting in the wrong category,I'm new to this forum.
     
  6. Apr 21, 2012 #5

    psparky

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    Gold Member

    I found your thread to be most educational.
     
  7. Apr 21, 2012 #6
    But beside the wrong calculated Gain,the rest of the analysis is right for q active point calculation?
    And I found on the internet that Rb < βRe so that Re feedback to work wich means that
    VRb < VRe
    assuming that Ic ≈ Ie,could anyone please tell me why?
     
  8. Apr 21, 2012 #7
    I'm happy to know that this thread is useful for you :)
     
  9. Apr 21, 2012 #8
    From years of design transistor circuits including IC designs, I never use this way of analysis. As long as β>100 or so, you can in the first past assume Ib≈0, Ic≈Ie, Rb≈0 and really simplify the thinking and calculations. After all, all your parameters are only approximations anyway. There are temperature dependence, process dependence that likely blow out all your effort to predict the result. I suggest you to get a used book by Malvino. It is a very easy book to read, it is cheap. It show you how to look at transistor in DC and AC and super impose them together.

    It is my opinion that you should not get stuck on this way to work with BJT.

    1) Ib is important if you DC input impedance is very high that the little current drawn is causing voltage drop at the input circuit.

    2)Rb is important mainly in noise calculation where the thermal noise of the Rb become important. Also when in some power application where Ib can get high and need a pre driver to even drive the transistor. Other than this, mostly, we don't even think of the Ib.

    How close is Ic≈Ie very seldom even come to mind in circuit design.

    Transistor circuit by default is not very accurate, if you need more accurate result, you need negative feedback to stabilize the circuit. With negative feedback, this will make a lot of the concern irrelevant.

    One thing that is important you never mention. The approximated input impedance of the common emitter stage. Input impedance equal β(re+Re). Where re is 1/Vt which is about 25Ω for Ie=1mA. 250Ω if Ie=0.1mA etc. You have not even include this in and this turn out to be important for low current design like what I am doing in music electronics at the moment.

    Get the Malvino book for cheap, then learn what is important in design. I am still using the materials for designing very advanced transistor circuits.
     
  10. Apr 21, 2012 #9
    Thank you for this ,i will follow your advice,and I'm very glad that I joined this forum,people are very nice here :)
     
  11. Apr 21, 2012 #10
    I forgot to mention:

    Gain ≈ Rc/(re+Re).

    re=1/gm=Vt/Ic

    Vt≈25mV at 25 deg C. It is only approximation. In common design, this is plenty good.

    And this is more important than the other parameters. I am designing circuit this very moment that run 50uA collector current. re≈500Ω, this turn out to be a big part of the gain calculations.
     
    Last edited: Apr 21, 2012
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