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Bootstrapping Amplifiers, and Signal Bypasses

  1. Oct 17, 2014 #1

    I am reading The Art of Electronics, and I have two specific questions regarding examples it has introduced. I am dealing with bipolar transistors the current chapter. I think my questions mostly stem from not understanding concepts.

    The text has introduced the technique of 'bootstrapping'. The circuit is the first image attached to this post. While I understand that bootstrapping is meant to increase the input impedence seen by the signal source, I do not understand how it works.
    In particular, R3 is very strange to me, because both its ends are connected to the same node. Whether I approach this circuit from a DC mindset (Capacitors acting as open circuits, or having an effectively infinite impedence), or an AC mindset, the inclusion of R3 seems irrelevant to the circuit. I see R3 as irrelevant because it seperates no nodes from one another in the circuit, and neither does R3 connect anything which wasn't already directly connected.
    The text tries to illustrate the operation of the circuit by showing that the current through R3 is zero. This much seemed obvious to me. Why the input impedence would be approximately infinite (the text implies this is the case because of R3's current being zero)? If a time-varying signal is applied, again, why would R3 be relevant when it has been shorted like in the image?

    My second question deals with a common-emmitter amplifier. I understand how this circuit works. Additionally, I can monkey my way through the process of selecting the values for resistors and caps and understand the rationale behind it... except for one thing. When alternative paths are included at the emmitter (one path with a cap for the signal, the other lone resistor being for the DC bias 'signal'), I do not understand why the capacitor's value is determined the way it is. Referring to the image included, I would like to tell you all that there is very often a pair of resistors R1 and R2 which form a voltage divider. The base connects between the two resistors.
    The capacitor is calculated to place the 3db point at a given frequency on a Bode (gain vs. frequency) plot, but the question is what is the R value we include when calculating 1/(2*pi*f*R)? To me, this would the impedence from the capacitors perspective, looking into the circuit.
    To me, this would be ((R3//Re)+re)//(R1//R2)/beta. In words, I think I would reflect the resistors in the voltage divider (R1//R2) across the transistor (remembering to divide by beta), add the internal 'resistance' re because it is in series (here re is the reciprocal of the transconductance, I think), then because Re is grounded put it in parallel too, then FINALLY, add R3 because it is in series to the capacitor.

    The text determines that the correct value of R should be R3 + re. Nothing else.

    Even if the voltage divider network is omitted due to becoming smaller when it is brought across the transistor (Rdivider/Beta)... should Re still not be included in the calculation by putting it in parallel?

    Please keep in mind that I am an absolute beginner and examples like this make me think I do not know how to Thevenin properly yet, which is exactly what I do when determining Zin or Zout (are low/high pass filters special and subtlely different? Why is a simple Thevenin not sufficient if they really are different?)

    Thank you for reading my novel, I have done my best to explain my interpretations and questions clearly.


    If my images have not been included properly, the links are as follows:
  2. jcsd
  3. Oct 18, 2014 #2


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    R3 provides base current to the transistor from the R1 : R2 voltage divider.
    C1 is connected to R3 and the Q1 base. It is not connected to the junction of R1 and R2.
    There is no black dot where those wires cross.

    The output impedance of an emitter follower is not Re, but is closer to Re divided by the transistor current gain. With a beta of 100, Re = 100 ohms, looks like 10 ohms.
    The RC circuit then becomes 180 + 10 = 190 ohm in series with 68uF.
    Last edited: Oct 18, 2014
  4. Oct 18, 2014 #3
    Thank you, it makes so much more sense now knowing that those dots mean a connection! Newbie mistake I hope.

    I completely understand the math you are doing here, but I still do not understand why it is done this way. Since the resistor Re is on the very same side of the transistor as our reference point, why does the value beta of the transistor get involved at all?
    I understood that when looking through a transistor from the base side, impedences on the emitter side are made to look larger by a factor of beta. When looking into a transistor from the emitter side, impedences on the base side are divided by a factor of beta. It seems to me that from the reference point of the low-pass RC circuit, we don't have to look 'through' the transistor at all when considering resistor Re.

    Can you tell what I have misunderstood? I am trying my best to digest all this stuff conceptually.

    Thank you very much for the help!
    Last edited: Oct 18, 2014
  5. Oct 19, 2014 #4
    You are right, the impedance from the capacitors perspective is equal to
    [tex](\frac{R1||R2}{Hfe +1 } + re)||RE + R3[/tex]
    But in real life we don't use this "precise/pedantic" approach because we pick the capacitors from E6 series (10 ; 22 ; 33; 47; 68).
    We use a simplified approach C = 0.16/(R * F) = 0.16/( 180 + re * F) = 0.16/(180 + 26 * 10Hz) = 77μF use 100μF or 68μF
    Last edited: Oct 19, 2014
  6. Oct 19, 2014 #5
    Thank you so much for your help! It really clarified things for me.
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