Finding formula for this circuit...!!

Discussion in 'Homework Help' started by Himanshoo, Mar 2, 2016.

  1. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    Hey friends please help me on this..

    q.1 what will be the gain expression of this circuit..?
    q.2 what will be the equivalent resistance between point x and ground..with formula..??
    q.3 what effect will be on gain if the value of Cfb1 is increased??

    Screen Shot 2016-03-02 at 5.31.52 pm.png

    Thanks!!
     
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Since this is a 'Homework help' forum, not a 'Homework done for you' one, show us your efforts so far.
     
  3. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    yeah..the gain expression for a common emitter circuit is
    (Rc//RL)/re'
    where re' is the ac resistance of base emitter junction and its value is 26mv/Ie...
    now the thing which I am asking first is that how the involvement of Cfb1 will modify the gain expression...??


    btw nice humour:)
     
  4. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    This circuit/question looks like it may be set up wrong or something is missing.
    For example, with very low input voltage we have essentially a constant current source in series with a capacitor. Eventually the cap charges up and then there's nothing to stop it from charging more. Using pure theory with no component limits, the cap would charge up to an infinite voltage level.

    So something else has to be specified, such as the leakage of the transistor, min input, or something else to make this a real solvable problem.
     
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  5. Jony130

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    Feb 17, 2009
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    Also I don't see any biasing network. So what is the operation point ?
    Also try find a gain expression for this circuit
    Sch.png
     
    Last edited: Mar 2, 2016
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  6. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    The whole circuit is below..including a VAS and output stage...
    I am just asking a vague expression...

    polesplit3.png
     
  7. Jony130

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    I would like to see the original diagram?
     
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  8. Himanshoo

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    Apr 3, 2015
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    Hi jony...
    Actually these days I am focussing on Miller Compensation aka pole splitting..and Ive found a magnificent and simplest explanation ever on it unlike regular textbooks..here the link below...
    http://stephan.win31.de/music.htm#polesplit

    So actually now its not a Homework problem...must be redirected...to other section..
     
  9. Alec_t

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    Sep 17, 2013
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    For very small input signals Q1 won't conduct, so the voltage gain will be zero.
     
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  10. Himanshoo

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    Apr 3, 2015
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    I think so that you are talking about bootstrap action of Cfb1...which ceases the voltage gain of the circuit..
     
  11. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    The gain expression of this circuit should be..

    Av = {-β(Rc2//Rc1)/αRs} / 1+β(Rc2//Rc1)/αRf

    Am I right??
     
  12. Jony130

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    But this time the gain is negative so we do not "bootstrap" anything. And this is why we have a "Miller capacitor" this time.
    The voltage gain without RF resistor is equal to
    Av1 = - Rc2/re' * alpha if we driven our amplifier from the ideal voltage source (Rs = 0Ω).
    And now if we add RF into the circuit the voltage gain will almost remains unchanged if Rc2 << RF.
    And if we include Rs and using a miller effect we have
    Avf ≈ - Rin*(Rs+Rin) * (Rc2||RF)/re' * alpha
    where
    alpha = β/(β+1)
    Rin ≈ RF/(1-Av1)||(β+1)*re'

    Notice the "miller effect" on RF.
     
  13. Himanshoo

    Thread Starter Member

    Apr 3, 2015
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    yeah the feedback resistance appears to be decreased by (1-Av1) times due to miller effect..

    So what effect will it make in the expression if Rf will be replaced by a capacitor..however text says that for a common emitter circuit miller effect acts as a degenerative feedback decreasing the voltage gain...
     
  14. Jony130

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    But for Rs = 0Ω and RF>>Rc2 the gain will not change much. But with Rs > 0Ω and together with RF/(1 - Av) (decreased by Miller effect) form a voltage divider and this will decrease the voltage gain.
    Well, at first glance we can treat Cf as a frequency dependent resistor (Xc). And resistance of this resistor decreases with frequency.
    For DC the gain is Rc/re' but as signal frequency increasing the gain of a amplifier start to drop at rate 20dB per decade (the roll-off will start at F ≈ 0.16/(1+Av*Cf*Rs)), because now this amp start to behave just like a poor's man integrator.
     
  15. Himanshoo

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    Apr 3, 2015
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    Any change in gain expression if Rf is replaced by Cfb...??
     
  16. Jony130

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    But why you want to do such a thing ? But if you want you can replace RF with 1/s*Cfb where s = j*2*pi*f.
     
  17. Himanshoo

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    Apr 3, 2015
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    Because I am thinking that if Cfb1 increases gain decreases because Xc decreases and if Xc decreases total impedance decreases ….

    here what is important for me to know is..in calculating the impedance between point x and ground..does it involves only Rout1 and Xc (reactance of Cfb1)..or any other value like re’ or resistance between collector and emitter terminal…etc.
     
  18. Jony130

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    Cfb capacitor is seen at the output as a capacitor equal to (1 + 1/Av)*Cfb.
    But the author was talking about this circuit
    aa.png

    And, as I said earlier the VAS stage for the DC and low frequency signals behaves like an ordinary common emitter amplifier with gain -Rc/re. But as signal frequency increases the gain of a amplifier start to drop at rate -20dB per decade (the roll-off will start at Fc ≈ 0.16/((1+Av)*Cf*Rs||(β+1)*re) open loop). Because Cfb capacitor controls the high frequency gain of the VAS (Cfb forms a shunt feedback loop around the VAS.) And this means that for F > Fc this stage becomes an Miller integrator where the output voltage is the integral of the input current.
    http://www.ecircuitcenter.com/Circuits_Audio_Amp/Miller_Integrator/Miller_Integrator.htm
     
  19. Himanshoo

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    Apr 3, 2015
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    Sorry for delayed response...

    Ok now talking about the shift of second pole to a higher frequency..the text says..

    Screen Shot 2016-03-11 at 6.12.07 pm.png

    The text means that the second pole occurs due to an RC Filter caused by Rout1 and Cin2 ....
    now what I think is that if
    if Rout1 decreases then Cin2 increases ..if Cin2 increases then its Xc decreases..and if Xc decreases ..more signal from the previous stage will pass on to the ground thus robbing of the second stage with sufficient input signal..which causes the output of second stage to roll down early for low frequencies as well...so where does the second pole seems to shift up in frequency according to text..

    Screen Shot 2016-03-11 at 6.19.05 pm.png

    http://stephan.win31.de/music.htm#polesplit
     
  20. anhnha

    Active Member

    Apr 19, 2012
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    Pole s = 1/(Rout1*Cin2). As Rout1 is decreased, 1/(Rout1*Cin2) will increase or move up in frequency.
     
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