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TINA/Spice: stability analysis of sallen-key high pass filter

Cera Wei
Intellectual 2010 points
Other Parts Discussed in Thread: TINA-TI, OPA191, LM2904

Tool/software: TINA-TI or Spice Models

hi,

I'm designing a sallen-key high pass filter as below, the output is instable.

So I analyze the stability of this circuit using the following circuit.

The bode plot of Aol*beta is:

 

At frequency of 12Hz, there is a pole, why the frequency shift is up to 300 degree? My basic knowledge is that a pole may result in 90 degree phase shift, is that wrong? Besides, the slope at 0dB is -20dB/dec, why the circuit is instable? 

Thank you！

Cera

  • 0
    TI__Prodigy 20 points

    Hi Cera,

    Circuit with OPA191: R1, C1, R2, C2 add 2 poles at ~12Hz without much change in gain during that phase. Hence a big dip in phase is observed.

    Circuit with LM2904: Please try keeping the Opamp in voltage follower mode for Sallen-Key HPF. Later R1, C1, R2 and C2 can be tuned to get the required cut off frequency.

    I hope this helps.

    Thanks

    Pranesh

  • 0 in reply to Pranesh M
    TI__Intellectual 2010 points
    hi, Pranesh,

    Thank you for your reply. I still have a question, why the Sallen-key HPF should be follower mode? What's the largest gain of Sallen-key filter?

    Thanks
    Cera
  • 0 in reply to Cera Wei
    TI__Prodigy 20 points

    Hi Cera,


    I am no expert in Sallen-key HPF. But, as i understand, the opamp is used in voltage follower mode as a buffer. Meaning, the load should not affect my HPF characteristics(pole-zero locations). And, since it is a filter, 0dB gain (unity gain) is preferred.

    I hope, this helps.


    Thanks

    Pranesh

  • 0 in reply to Pranesh M
    Intellectual 405 points

    Hi Cera

    By my calculation the Q of your circuit is negative. This is due to the gain of 5 in your circuit. Looking at the component values in your circuit (equal C, one resistor twice the other), are you trying to design a Butterworth filter with Q=0.707? If you make the gain = 1 by removing R4 then by my calculations Q=0.707.

    regards

    Steve

  • 0 in reply to Stephen Power
    TI__Intellectual 2010 points

    Hi, Steve,

    Thank you for your reply. Your analysis is right!

    By the way, I still can't understand why the phase shift is so large.

    Thanks,

    Cera