Below circuit seem to use negative feedback in a weird way. There is no input resistor at - terminal, and instead of putting a resistor in the feedback branch, they put two 4.3V zener diodes. I kinda get why the output may be a 5V square wave. But I'm really not so sure about it as two things are throwing me off:

1) Input is a voltage source without a resistor. Since the + input of op amp is at ground, the op amp should try to pull the - input also to ground. Doesn't this actually short the input voltage source to ground and infinite current flow?

2) Is it legal to put zener diodes in feedback branch as shown in the figure? enter image description here

  • 1
    \$\begingroup\$ Almost the purest form of op amp zero-crossing detector. \$\endgroup\$
    – Long Pham
    Jun 7, 2018 at 13:31
  • 1
    \$\begingroup\$ Yeah, I think it is a zero-crossing detector, only the output swing is smaller. Removing those zener diodes, it swings between +15 and -15.. \$\endgroup\$
    – AgentS
    Jun 7, 2018 at 13:37
  • 1
    \$\begingroup\$ Indeed you can use this as a zero crossing detector, limiter (amplitude information is removed) or "squarer". Basically different names for the same circuit. It is more the context in which you use it that determines what we would call it. \$\endgroup\$ Jun 7, 2018 at 13:40
  • 1
    \$\begingroup\$ It is legal, maybe not in Utah though \$\endgroup\$
    – PlasmaHH
    Jun 8, 2018 at 17:23

3 Answers 3


You are right, you have spotted a flaw in this circuit. The circuit does rely on a resistor to be present in series with Vin. If we would apply +25 mV (from an ideal voltage source) at Vin then the opamp would try to pull its - input to 0 Volt (ground level). The opamp will do that by lowering its output voltage to the lowest level it can achieve. That is -15 V, not -5 V. An infinite current will (try to) flow through the 2 zenerdiodes. But in the end the - input of the opamp will never reach 0 V and the ideal voltage source will "win".

As drawn too much current can flow and this could break the opamp and/or the zener diodes.

So you're right, there needs to be some series resistance present to limit the current into Vin. That will then also limit the current flowing through the two zener diodes and the output current of the opamp.

There's no "law" to prevent you from using those two zener diodes in anti series :-) So yes, perfectly legal.

The result is a bi-directional zener diode of in this case 5 V. Meaning that together the diodes will drop 5 V when a current flows through them. Like this, the current can flow in both directions. When using one zener diode you get 4.3 V in reverse and 0.7 V in forward.

  • \$\begingroup\$ Ahh so the input resistor is a must to limit the current. Such a relief! Thank you. I wish they had put that zenar diode structure parallel to the output. (I can't help feeling that the author wants to show off how clever he is by using the virtual ground at the - input. ) Thanks again :) \$\endgroup\$
    – AgentS
    Jun 7, 2018 at 13:35
  • 3
    \$\begingroup\$ Putting the Zener diodes in the feedback loop has the huge advantage of preventing the opamp itself from saturating, which preserves its performance. Many opamps take a long time to come out of saturation, which can lead to very poor performance in the overall circuit. \$\endgroup\$
    – Dave Tweed
    Jun 7, 2018 at 13:41
  • 1
    \$\begingroup\$ indeed and a feedback loop relies on the output being able to influence the input in some way. If that's not possible then it is not feedback. That forgotten resistor makes the influence possible. Without the resistor there can be no "proper" feedback. It is a shame it was not included because it is essential. \$\endgroup\$ Jun 7, 2018 at 13:44

maybe the designer of this circuit is relying in the output impedance of the sinusoidal voltage source circuit (there is no ideal voltage source ,every voltage source has an equivalent output impedance) i thinks thats why the input voltage is very low to get a low current suitable for the zeners becouse output impedance is usually low ,for example if the sinusoidal signal comes from a function generator who has an output impedance equals 50 ohms (common) then the current through the zeners will be just .5mA (piece of cake for the zeners)


The Zeners already have a Zz = 9Ω @58mA and Zzt = 400 Ω @ 1mA or the threshold value.

The sum for Vf=0.7V and Vz=4.3 add up to 5.0V yielding a convenient regulated +/-5V clipped signal.

The Vz tolerance is +/-5% at 58mA but would rise 9mV/mA with more current and drop by 400mV/mA at Vzt. THus for stable operation using more current towards the rated current is preferred.

Also nobody really uses zeners for this due to the soft limiting characteristics using a Zener to change gain and limit the voltage .

Why is this a terrible design example?

  • there are no specs for Voltage tolerance and squareness of peaks.
  • Zeners are poor tolerance to begin with due to the exponential VI curve
  • The Source impedance must be higher than Zeners to reduce the limit the gain and thus the voltage output, thus the current is low and the knee is very rounded
  • the Zener voltage drops well below rated Vz by more than 10% so it will never be 4.3+0.7=5V , rather more like 4.4V with 1K source
  • at this point you ought to be re-examining the real requirements and specs.

    Why do I need this?
    Why do we take for granted examples in books and websites are perfect? Trust but verify. Never assume unless you know for sure.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.