Miller Effect on a BJT

Discussion in 'General Electronics Chat' started by nylonman, Jul 28, 2011.

  1. nylonman

    Thread Starter New Member

    Jul 28, 2011
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    Hi, I'm readting The Art Of Electronics and I have a question about the Miller Effect.

    The theorem says that the capacitance between the base and collector looks like a bigger one between base and ground due to the voltage amplification at the collector, so the equivalent capacitance is Ccb(1+G).

    I understand the principle but my question is: how can the gain be the same than without the effect? The gain without the miller effect depends on the collector current, but wouldn't this gain be reduced because of the current passing through the "capacitor" betwen base and collector?

    [​IMG]
     
    Last edited: Jul 28, 2011
  2. Ron H

    AAC Fanatic!

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    That's a good question. The the multiplier is the amplifier's (gain +1) without the Miller cap.
     
  3. ErnieM

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    Apr 24, 2011
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    Yes the calculated gain G is reduced when the miller effect is taken into account.

    That's the beauty of algebra: it takes all these terms into account simultaneously.
     
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  4. nylonman

    Thread Starter New Member

    Jul 28, 2011
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    So, is the formula an approximation?
     
  5. Adjuster

    Well-Known Member

    Dec 26, 2010
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    An approximation, yes, but usually not a very gross one for frequencies within the circuit passband, or not too far beyond it.

    The point is that provided that the base circuit impedance is not too low in comparison with the collector impedance, and also that Av is reasonably large, then Rs*Ccb*(1+ Av) >> Rl*Ccb.

    This means that the loss of gain due to the Miller effect in the input circuit will often be significant at frequencies well below the point where the direct loading of Ccb on the collector will make much difference.
     
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  6. ErnieM

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    Apr 24, 2011
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    Ccb(1+G) is an exact representation of the equivalent capacitance of Ccb when it is reflected to be in parallel with Cbe, assuming gain is calculated correctly, here that means G = Vc/Vb.
     
    Last edited: Jul 29, 2011
  7. Ron H

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    For an amplifier with finite gain G, the Miller capacitance will be Ccb*(1+G). The reduced high frequency gain of the closed loop amplifier does not cause the Miller capacitance to drop as gain drops.
     
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