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Bootstrapping and miller's theorem

  1. Nov 26, 2013 #1
    I am trying to understand how the input impedance of a bootstrapped darlington emitter follower is analysed. They make use of Miller's theorem. But honestly, I have not understood this theorem as well... All I know is bootstapping increases input impedance but i have no clue as to how? And why is a capacitor connected between the input and output terminals? Can someone link me to some place where a bootstrapped darlington circuit is analysed? I googled it but did not find any helpful links...
     
  2. jcsd
  3. Nov 26, 2013 #2

    berkeman

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    Staff: Mentor

    IIRC, the Miller capacitance is between base and collector, and provides negative feedback (slowing down frequency response). Bootstrapping is positive unity feedback from output to input (for any amplifier) to increase input impedance, so it would be between emitter and base in this case.

    The most common bootstrapping I'm familiar with is bootstrapping the shield capacitance on an input to an amplifier. If the input coax cable (or triax cable) has enough capacitance to cause a LPF rolloff of the input signal, you can drive the shield with a buffered version of the signal to eliminate the capacitance to the adjacent shield.

    http://www.ti.com/general/docs/lit/getliterature.tsp?literatureNumber=snoa664&fileType=pdf

    .
     
  4. Nov 26, 2013 #3
    Hi, Miller's theorem is better know as a Miller effect

    http://en.wikipedia.org/wiki/Miller_effect
    http://web.mit.edu/klund/www/papers/jmiller.pdf

    We simply have a ideal voltage amplifier with gain equal to A = -10V/V

    Next we connect a resistor R = 10Ω between the input and the output of the amplifier.

    attachment.php?attachmentid=64300&stc=1&d=1385502637.png

    Now let as try to find a input resistance.

    Rin = Vin/Iin

    In = (Vin - Vout)/R = (Vin - A*Vin)/R = Vin * (1 - A)/R

    Rin = Vin/Iin = R/(1 - A)

    Iin = (1V - (-10V))/10Ω = 1.1A

    So Rin = 1V/1.1A = 0.909Ω

    So as you can see our Rin resistance is (1 - A) smaller then R if we have inverting amplifier .
    And this is what we call a Miller effect

    Now let us consider different situation. We replace our amplifier with "voltage follower" amplifier.
    But now the gain is equal to A = 0.5V/V

    So if Vin = 1V we get 0.5V at the output. So the input current is equal to:

    Iin = (Vin - Vout)/R = 0.5V/10Ω = 50mA and therefore

    Rin = 1V/50mA = 20Ω

    If we increase the gain to 0.9V/V we have

    Iin = (Vin - Vout)/R = 0.1V/10Ω = 10mA

    So the input resistance is equal to

    Rin = 1V/10mA = 100Ω


    attachment.php?attachmentid=64301&stc=1&d=1385504371.png

    Also we can use Miller's theorem to find Rin.

    Rin = R/(1 - A) = 10Ω/(1 - 0.5) = 20Ω

    Rin = R/(1 - A) = 10Ω/(1 - 0.9) = 100Ω


    But this time we call this bootstrap
    http://electronics.stackexchange.co...analysis-of-a-emitter-follower-with-bootstrap

    Any questions ?
     

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