# How do I use a clipping Zener at the input of an Opamp inverting amplifier?

I have a $V_{in}$ signal for which I cannot guaranty that it would have an upper limit. I want to clip this input signal to a limit (e.g.; 8.2V) with a circuit like above.

Is this circuit alright? Are there better Opamp circuit for this purpose?

This modified circuit is not a standard inverting amplifier. What is the input-output relation/formulation in this circuit when $V_{in} < 8.2V$?

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## 2 Answers

That would work, but you don't need RL. Also (of course), Vout be negative for positive Vin.

Your two 10 kohm resistors cause the zener to see half of $V_{in}$. So, the trip point for $V_{in}$ is not the zener voltage itself, but twice of it (16.4 V). The equation for $V_{in}$ < 16.4 V is exactly the same as if the zener wasn't there. So, you would have two 10 kohm resistors in series.

$V_{out}=-V_{in}\dfrac{100,000}{20,000}=-5V_{in}$ , for $V_{in}$ < 16.4 V.

If $V_{in}$ >= 16.4 V, then you just have a constant voltage source, of 8.2 V, applied to an amplifier with G=-10.

$V_{out}=-82$ V, for $V_{in}$ >= 16.4 V (assuming the opamp were able to deliver that output voltage, which is not the usual case).

(Note that the two equations converge, for $V_{in}$=16.4 V.)

Also, if $V_{in}$ can be high, but not very very high, you may not need a diode, because the input resistors (20 kohm in your case) limit the input current, and therefore the damage that you could do to the circuit. What is the maximum Vin you would like to work with?

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Very good explanation, thank you very much. "What is the maximum Vin you would like to work with?" > I'm trying to sense current on a 33.33mOhm resistor (3 x 0.1 Ohm connected in parallel) which will be connected on a 220V AC power line in series. Hmm, while writing this comment, I realized that I really don't need a voltage clipping part at the input stage; for reaching 8.2V on the resistor, it must flow 250A through, which is not practically possible; the circuit breaker must have broken the current far before reaching that current level. Again, thank you for your answer. – hkBattousai May 2 '12 at 20:04
@hkBattousai Thank you. You may have problems if the current-sense resistors have power ratings that can be exceeded in your practical case. If you blow them, your circuit will see a high input voltage. Either have enough margin for those power ratings, or plan a voltage protection mechanism. – Telaclavo May 2 '12 at 20:44
@hkBattousai if you wanted to be on the safe side you can use your current topology, but you would need to insert a buffering filter at the node where the zener diode exists. This will give you that clipping characteristic without you worrying about the current draw. – CyberMen May 3 '12 at 15:05

Your clipping circuit topology looks fine, but the 8.2V value should maybe be adjusted. Most opamps don't want to see inputs outside their power range. This means at the clipping voltage should be less than the positive supply voltage. Note also that the zener diode will conduct forwards like a normal diode, so you are effectively clipping to about the range -0.5 to +8.2 Volts.

You don't need to clip any higher than what it would take to cause the largest negative output. This depends on the negative supply voltage and the fact that there is a gain of -10 from the zener voltage to the output. For example, if the negative supply is only -20 V, then you could clip at 2 V and have the same output signal. Of course there is no need to clip that low, and it's a good idea to stay a bit away from the zener knee so that it doesn't load the signals you do want. You should have a range of possible clip voltages, from the minimum to drive the amp to full negative output to the maximum allowed input voltage after attenuation by the R2-R3 divider. Perhaps 8.2V is within this range and then everything is fine. Without knowing the power supply voltages we can't tell.

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