Thus, as the bottom tube's plate voltage pulls down, the cathode follower's output follows, pulling the plate resistor with it and increasing the grounded-cathode amplifier's gain as a result. Had the cathode follower output moved in the opposing direction, the bottom tube's gain would be drastically reduced.
      The effect is similar to power steering in a car. Because the plate resistor is being pulled down in phase with the first stage's movement, the effective load impedance is magnified by the degree of following achieved by the cathode follower. If the cathode follower could deliver true unity gain, the effective load impedance would equal infinity, as every movement in plate voltage would be perfectly matched by the cathode follower. Imagine a power steering so accurate that the wheel felt as if it were mounted on a frictionless bearing and the only resistance to rotation you felt was the steering wheel's inertia. Since no cathode follower delivers true unity gain, the following formula is useful to determine the effective load impedance seen by the grounded-cathode stage:
     r = Ra / (1 - Acf),
where Acf equals the cathode follower's gain.
      For example, if the cathode follower is able to deliver 97% of its input signal at its output, then 1 -  0.97 equals 0.03, which when divided into the 10k plate resistor yields a 33-fold increase in effective load impedance for the first stage. In this example, the 5k plate resistor effectively acts as if it were 165k in value in loading the grounded-cathode amplifier. (To actually run a 165k plate resistor would require a power supply voltage of 1750 volts!)
      Unfortunately, there are no free lunches. The marvelous increase in effective load impedance came at a price, which brings us to the second result of the modification. The low output impedance of a cathode follower relies on its grid seeing a fixed voltage, which the bootstrapping undoes. If the grid voltage moves in phase with the cathode voltage, then the cathode follower's usually-low output impedance falls apart.

    Connecting the cathode follower's input to its output undoes its low output impedance 

    This is a key point, which maybe one out of ten tube-circuit designers catch. In fact, not realizing this point is so widespread that it is difficult to convince otherwise, as so large a body of misunderstanding-filled magazine articles stand in the way.  (Often you will read about how bootstrapping delivers high gain, while retaining a low output impedance; it can't. Once again, there are no free lunches, not even in tube electronics.)
    Here's the secret to making sense of circuits: think in extremes. (Eric Hoffer rightly pointed out that all thought requires exaggeration and that an unwillingness to exaggerate often betrays an unwillingness to think.) Imagine a load impedance of zero and then infinite ohms; imagine an rp of zero and then infinite ohms; and imagine a plate resistor of zero and then infinite ohms. In this case, imagine a cathode follower whose grid was directly connected to its cathode. What would its output impedance be?
    In the bootstrapped circuit, the 5k resistor and the first stage's rp define a voltage divider, whose output feeds the cathode follower's input. If the cathode follower's output is forced +1-volts more positive by a voltage pulse, 40% of the 1-volt pulse will make it to the cathode follower's grid (in phase with the grid's signal), effectively making the pulse 60% as large as the grid-to-cathode voltage relationship would be in a pure cathode follower circuit. If the bottom tube had been a pentode, closer to 100% of the pulse would make it to the cathode follower's grid, increasing the output impedance.

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