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Let us take the case of the circuit in the image, while splitting the capacitor connected from base to collector into separate input and output capacitors as shown in the second figure. We use Miller's theorem.

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C1= C ( 1- v2/v1) and C2= C ( 1- v1/v2)

Here for calculating the value v2/v1 we must use the closed loop values or the circuit where the BJT has a feedback path. Most of the text books recommend using the v2/v1 value when there is no feedback (capacitor 'c') present or simply to use the open loop voltage gain. My questions are:

  1. Wouldn't it be wrong to use the open loop voltage gain as it might give wrong input and output capacitor values?

  2. Moreover, we use Miller's theorem to simplify any complex circuits that has parallel connections, but to use Miller's theorem it seems we must first find the voltages across the particular impedance in parrallel connection. To find that voltage drop we must first solve the circuit containing that particular impedance. Wouldn't it make the theorem useless?

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To use Miller's Theorem, we must use the actual V2/V1 ratio. If there is local feedback around the transistor, then this must be taken into account. However, CB feedback is in parallel to the capacitor. Emitter degeneration needs its own careful handling, simple application of Miller's needs that emitter grounded.

Most of the text books recommend using the v2/v1 value when there is no feedback.

Context, context! If the stage is part of a larger circuit, and there is feedback around that, the feedback does not affect this one particular stage. Generally when we are worrying about such a capacitance multiplication, there is no local feedback, as in the diagram you have drawn.

An important case for using the actual V2/V1 ratio is when the collector is driving a zero or low impedance, like the emitter of a cascode stage. Here the first stage voltage gain is radically reduced, and the effect of Miller with it.

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  • \$\begingroup\$ So does it mean that we will not be able to apply miller's theorem in the above circuit? If so could you kindly give methods on how to split the capacitance? \$\endgroup\$
    – Moulee
    Commented Feb 27, 2023 at 9:25
  • \$\begingroup\$ @Moulee It's common emitter, no prblems to apply Miller. \$\endgroup\$
    – Neil_UK
    Commented Feb 27, 2023 at 9:56
  • \$\begingroup\$ I mean can we apply the miller's theorem to the circuit using the open loop voltage gain? \$\endgroup\$
    – Moulee
    Commented Feb 27, 2023 at 11:02

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