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Feedback Loop Broken question : Stability part 3

Navroop Singh47
Prodigy 30 points

Hello,

I am going through Stability part 3 and needed more clarification on below:

It is mentioned on slide 5 ( stability Part 3) that  Aol_loaded = Vo / Vfb ( What does loaded mean here ? )

 1/ Beta = 1 / Vfb  ( Can you explain how did we get this relationship ? we know Beta = Vfb / Vout in closed loop)

Thanks in advance for your help.

-Navroop-

  • 0
    TI__Guru 64485 points

    Hello Navroop,

    Slide 5 describes the standard op amp circuit configuration to break the feedback loop and generate the open-loop curves.  We've labeled it "Aol_Loaded" because it represents the open-loop gain of the op amp (Aol) loaded with the output load presented to it in the circuit.  The resistive and/or capacitive load that is placed on the op amp output must remain on the output while generating the open-loop gain curves or effects of the loading will not appear in the simulation results. This is shown in the examples on Slide 6 where the output load always remains directly connected to the op amp output.

    Beta (Vfb/Vo) is the feedback factor of the op amp circuit configuration and represents the ratio of the voltage presented to the feedback node (Vfb) to the output voltage (Vo). 1/Beta is the inverse of this relationship and is therefore Vo/Vfb.  In this circuit configuration we've broken the loop between the output of the op amp and connection to the feedback network.  The node where the small-signal voltage source (Vin) is injected into the circuit is where the output (Vo) was connected before the loop was broken.  Therefore the inverse feedback factor, 1/Beta, is Vin/Vfb.  Because the Vfb node gain/phase results already represents Vfb/Vin, 1/Beta equals 1/Vfb. 

    More information behind where and why the loop is broken can be found in Stability 6 where we analyze a circuit with multiple feedback loops.