Op amp overdrive saturation recovery circuit

Discussion in 'Analog & Mixed-Signal Design' started by sp2821, Oct 4, 2017.

  1. sp2821

    Thread Starter New Member

    Jan 8, 2014
    21
    0
    Hi all,

    I'm trying to build a circuit that can help an op amp to recover from saturation.

    After some online search, I found few clipping circuits where it limits output of an op amp but does nothing to keep a difference between two input terminals of an op amp to a reasonable range.

    So I tried building circuit which clips an output voltage AND keeps Vneg and Vpos virtually shorted.

    My rationales were
    1. If the input voltage reaches 2.8V, output will be at -2.8V and D29~D32 will start to conduct (assuming Vf=0.7V). Output will remain at around -2.8V and Vneg will be around 0V.
    2. For Vin=-2.8V, D25~D28 will conduct and Vout=2.8V and Vneg=0V.
    3. Since Vneg is kept close to 0V even when the circuit is over-driven, the op amp will recover quickly.

    The simulation result shows that the output is well clipped at around -3V (35ns to 80ns) and Vneg is about 0V. But from 80ns to 120ns, it seems the op amp is losing control of Vout. I was expecting Vout to remain at -3V until around 110ns.

    Can anyone tell me why the op amp is behaving weirdly? Is the design fundamentally wrong and won't work?

    P.S The application of this circuit specifically requires 1MOhm input resistance and X1 gain. So I don't have a choice but to use 1MOhm Rf.

    upload_2017-10-4_16-51-1.png
     
  2. #12

    Expert

    Nov 30, 2010
    18,078
    9,619
    The op-amp is fast enough but your diodes have capacitance of 2 pf each. That makes your feedback path equal to 31K ohms of capacitive reactance at 5MHz.

    I don't actually know what's wrong. This is something to think about.
     
  3. crutschow

    Expert

    Mar 14, 2008
    20,042
    5,638
    What is the nature of the input signal?
    Why not just attenuate it so it doesn't clip?
     
  4. wayneh

    Expert

    Sep 9, 2010
    14,854
    5,335
    What is the advantage of clipping the output, which will be clipped at the power rails -2V anyway? Why not just place two zeners at opposite polarity on the inputs, to protect the inputs from exceeding the rail voltages?
     
  5. RichardO

    Late Member

    May 4, 2013
    2,274
    890
    Some general comments...
    What you are doing makes sense to me.

    I get a -3dB point of about 300 KHz. Is this good enough? If not then you should consider a T arrangement for the feedback to lower the feedback resistance.

    There are two sources that cause slow recovery time. The first is the transistors saturating and coming out of saturation slowly. The second is the op-amp's compensation capacitor. I gets "overcharged" during overdrive and takes a while to get back into the linear region. I have speculated that a current feedback will recover faster because it does not have a compensation cap. The problem with a current feedback is that it will not work well with 1 Meg ohm resistors.

    Since this is an inverting amplifier, you can use a couple of diodes back to back on the input to limit the overdrive if the amplifier does saturate. Be warned that these diodes must be very low capacitance so the amplifier does not become unstable.

    You might be able to use fewer diodes by putting a Zener diode inside a bridge circuit of fast diodes in your feedback. That would be 5 diodes no matter what the clamp voltage.
     
    wayneh likes this.
  6. ramancini8

    Active Member

    Jul 18, 2012
    473
    145
    The output circuit of an op amp is not designed for saturation use; thus, keep the op amp from saturating bu limiting the circuit gain with external circuits.
     
    GopherT likes this.
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