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Here is my task and solution for a)

enter image description here

Any suggestion for task b) ?

enter image description here Here are my thoughts: When v12(t) reaches in amplitude value of VD+VZ=6.8V, DZ1 will start conducting in forward direction DZ2 will start conducting in reverse direction. While conducting, they will provide constant voltage drop on R, 6.8V.

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    \$\begingroup\$ You are on the right track, what about the parallel R? \$\endgroup\$
    – jippie
    Commented Nov 23, 2015 at 20:19
  • \$\begingroup\$ @jippie, Diodes will provide voltage drop of 6.8V on R, if you mean that :) \$\endgroup\$
    – hari
    Commented Nov 24, 2015 at 17:38
  • \$\begingroup\$ Not what I'm aiming at. What would the feedback parallel branch look like for different voltages? And how would the opamp respond to that? Can you define different modes of operation for different voltages? \$\endgroup\$
    – jippie
    Commented Nov 24, 2015 at 17:42
  • \$\begingroup\$ @jippie, I'm not sure I can answer \$\endgroup\$
    – hari
    Commented Nov 24, 2015 at 17:45
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    \$\begingroup\$ Right, so for low'ish voltages you'll have a regular amplifier with R. Higher voltages will be clamped by the diodes, so the output voltage will not swing beyond 6.2+0.6=6.8V. This is done to prevent the amplifier to saturate. An amplifier in saturation will be considerably slower to recover than an amplifier in its linear region. \$\endgroup\$
    – jippie
    Commented Nov 24, 2015 at 18:14

1 Answer 1

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Hari - regarding the second question (amplitude of the Allpass Oscillator) , I am afraid there is no answer which will be sufficiently correct (unless you are using a circuit simulation program). Let me explain:

As you have shown, the calculation of the output amplitude starts with the limiting action of the Z-diodes. As another input to support your calculation you could assume that at the oscillation frequency the peak voltage at both input terminals of the right opamp is 3 dB lower than the peak voltage Vi1 (This is because the RC lowpass is operated at its 3dB cut-off).

However, this applies for sinusoidal waveforms only. And that is the problem. The Z diodes will cause hard-limiting (clipping) of the signal. Hence, we can expext that the signal is far away from being sinusoidal (can be verified by simulation). As a consequence, the above assumption (voltage at the opamp terminals) is not valid anymore. I think, the error is too large for using this as an input for calculation purposes. Hence, it makes no sense to start amplitude calculations.

In any case, the output amplitudes will be larger than the amplitudes across the diodes.

More than that, the result (distortion) strongly depends on the selected ratio R3/R2 because this ratio determines loop gain and, hence, the required amount of limiting (gain reduction) provided by the diodes.

General comment: For this reason (large signal distortion) , the shown method for limiting and fixing the oscillation amplitude is very poor. There are much better alternatives (soft clipping).

EDIT (UPDATE): Assuming clean sinusoidal signals and R3/R2=1.01~1) we can calculate the output amplitude (peak) to app. Vout,p=Vz/0.6 (with Vz=6.8 volts).

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