Long tail pair inverter

Discussion in 'Amps/Cabs Tech Corner: Amplifier, Cab & Speakers' started by aro, Mar 4, 2015.

  1. aro

    aro Member

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    I'm looking at this Aiken explanation of the long tail pair inverter.

    Is the V1B output in phase because the negative feedback fed into it is out of phase?

    Also, he says
    Is that a type, should R1 and R2 be reversed?

    What about a circuit like this one, without negative feedback? Where does the out of phase signal come from into V1B?
     
  2. potatofarmer

    potatofarmer Member

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    V1B is in-phase with the input signal regardless of if any NFB is present.

    The part you quoted is easy to misunderstand without the full context:

    It is not a typo.

    Merlin explains it better than I could:
     
  3. Jeff Gehring

    Jeff Gehring Silver Supporting Member

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    (1) No. You can forget about the NFB. To determine the primary signal driving the V1B triode, consider what a triode does: it amplifies the voltage difference between the grid and the cathode. By virtue of C2, (which for purposes of this discussion appears as a short to audio frequencies) the V1B grid is at the same AC potential as the junction of VR1 and R6. However, due to the signal on its grid being amplified by V1A, the DC cathode current of V1A is being modulated at the audio rate. So at the tied cathodes of V1A and B there will be a signal voltage developed (over R5 and R6) that is larger than the signal voltage developed at the junction of VR1 and R6. V1B will amplify this instantaneous difference in voltage between its grid and its cathode. The signal at the cathodes of V1A/B is of the same polarity as the grid drive to V1A. Since V1B will amplify its grid-to-cathode voltage (not its cathode-to-grid voltage), the drive to V1B is inverted.

    (2) Yes, clearly a typo. R1 develops the "out of phase" output, R2 develops the "in phase" output, as Aiken has correctly identified in the schematic drawing.

    (3) Same as (1), the right-side triode is amplifying the signal voltage as developed over Rk.
     
    Last edited: Mar 4, 2015
  4. Malcolm Irving

    Malcolm Irving Member

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    The way I find it easiest to understand is to put some numbers in. Suppose, at quiescence, the common cathode junction (at the top of the tail) is sitting at 100V. The grid of V1A is (say) at 98V and the grid of V1B is also at 98V.
    Now if a +1V signal swing hits the V1A grid, it goes up to 99V. The other grid (V1B) stays held at 98V by the capacitor.
    The common cathode junction goes up to 100.5V.
    So the grid-to-cathode voltage on V1A has changed from -2V to -1.5V and it conducts more plate current. The grid-to-cathode on V1B has changed from -2V to -2.5V and it conducts less plate current.
    The voltage across the tail has only changed by 0.5%, so it is almost conducting the same current as before.
     
    Last edited: Mar 4, 2015
  5. aro

    aro Member

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    Thanks, gentlemen, great explanations. Once I wrapped my head around it, it was pretty simple.
     
  6. reaiken

    reaiken Member

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    Yes, it is a typo, thanks. I've corrected it. I also realized that the way I worded that sentence is confusing. One of these years, when I get some time, I'll go back and rewrite it.
     

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