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OPA192: Active High pass Filter opamp input voltage protection with clamp diodes sinking current problem

Part Number: OPA192
Other Parts Discussed in Thread: XTR111,

I am designing an iepe signal conditioner and as seen in the introduction I have a 2nd order butterworth high pass filter..
When the sensor (load) is not connected, the current source with 24 V supply with 4mA constant current will completely turn on the MOSFET and the output will be 24 volts (xtr111 output will not give exactly 24 volts, it will give a value of approximately 21.6V), but let's assume that there is a 24V input.


If we look at the OPA192 datasheet, the entry values in the Absolute Maximum Ratings section are limited as follows. Since we are supplying (V–) – 0.5 and (V+) + 0.5 :(+15V )(-15V ), -15.5 and 15.5 are the maximum limits we can apply to the inputs


If we focus on transient analysis, when the 24V input signal comes, the current will pass through the clamp diode (BAT64) through the 2k limiting resistor and go to the +15 line. The simulation is here.
Accordingly, the 15V power supply will need to sink a value such as 3.315mA. I use a UWE1215S-3WR3 supply that can output +-15 V to feed the opamp. However, I am not sure how much source sinking it can do, maybe it does not do it at all.

graph of the +15 volt source 

u can see the +15v source is sinking  You can see that it is sinking for up to  10ms.

Briefly, how can I protect the inputs of the opamp when we apply 24v? Do I need to add a 15V zener diode reverse connected to the power rails for sinking?
I don't want the opamp to burn out when there is no load.

If you notice any other errors in the circuit, please let me know.

  • The BAT64-4 shows <0.5V forward drop under 10mA current, will that not stay in spec for max input Vcm voltage for the OPA192?

  • Hey Michael Thanks for your feedback!

    First of all, yes, you are right, the Vin pin seems to remain at a value of 15.37V. However, what I am wondering is that the 3.315 current coming from here goes to the +15V supply point. We can think of it as if we are pumping current here, in which case the voltage of this point should be higher, but the task of our +15v supply is to keep the output pin at +15 relative to Gnd, in this case it needs to absorb (sink) this 3.315mA current. However, not all power supplies have the ability to sink. .I think I'm having a problem understanding whether voltage sources are ideal or not.

  • If it is a transient, will not the supply decoupling caps absorb it?

  • Electronix,

    You are correct that many LDO cannot sink current.  You are also correctly limiting the input current to less than 10mA.  In general, we recommend adding a transient voltage suppressor (unidirectional TVS) to the power supply to help with this issue.  Any transient signal that is channeled through the ESD structure of the amplifier input is channeled to the supply and the TVS diode absorbs (sinks) that signal and limits the supply voltage to a safe level.  It may seem counterintuitive to ad a protection circuit on the power supply when the EOS is on the amplifier input, but I have seen this method work successfully many times. The TVS standoff voltage should be equal to the nominal supply voltage and ideally the breakdown should be below the power supply absolute maximum voltage.  In general, having the TVS on the supply is a good precaution as it helps for the issue you are discussing as well as other power supply transients. The Amplifier Precision Labs Video Series section on electrical overstress covers this topic in detail.

    Best regards, Art

  • Hey Art, thank you for your answer. I watched the videos in the EOS section
    I want to protect the power supply pin, but I also want to protect the vin input pin. How do I determine the value of the current limiting resistor here?
    That is, the more current passes, the greater the voltage drop on the bat64 diode will be. And the voltage drop on the Vin pin (Vsup+Vbat64(will be equal to the voltage drop on bat64))
    In this case, if I increase the 2k resistor value, the current passing through the diode will decrease and thus the voltage drop on the diode will decrease and less voltage will occur on the Vin pin.

    I can understand the situation regarding the TVS diode. The standoff voltage will be +15 voltage (it is an advantage if there is a low leakage current). The reverse breakdown minimum voltage should also be below the absolute max rating for the power pin of the opamp. For this reason, I think the breakdown voltage should be below 15.5. Please correct me if I'm wrong. 

    However, most clampimg voltages are above this value. You can see the list here

    digikey tvs diode list

    It will flow significant current after the breakdown voltage.

    By the way, does the clamping voltage matter for tvs?

    Considering that the Vin pin can have a maximum of 15.5 Volts, the TVS breakdown voltage being 16 volts will not help the Vin pin to remain below 15.5 volts. In this case, would it be a good idea to place a 15V zener diode?

  • For example, if I place a partially imitated TVS diode at the specified intervals, this is what happens. The nonivering input sees a value such as 16.5V.

  • Electronix,

    1. The absolute maximum supply voltage for OPA192 is +/-20V.  So, if you are using a TVS with a standoff voltage of 15V, you need to find a breakdown less than 20V.  Here is a list of devices from digikey with a standoff of 15V and a breakdown between 15.6V and 17V.  These devices should work well for protection so you can decide which one based on your leakage requirements or other criteria.
    2. The nice thing about your circuit from an overstress perspective is that it is AC coupled.  Thus you don't have to worry about a continuous DC fault.  Continuous DC faults are more difficult to protect against as you need to consider the DC power dissipation of all the devices. 
    3. To choose current limiting resistors, find the voltage drop on the resistor and choose the resistor to limit the Schottky input current to 1mA (1mA is will make the maximum drop on the BAT64 350mV).  So, if the fault voltage is 20V and the supply is 15V the voltage drop on the current limiting resistor is approximately 20V - 15V = 5V.  To choose the resistor the use this equation:  Rlimit = (20V - 15V)/1mA = 5k ohm.  In your case, the 1kohm resistor will limit the Schottky current to approximately Idiode = (20V - 15V)/1kohm = 5mA.  This is probably ok also.  Keep in mind that this current is going into the Schottky diode and not into the amp.  If some of the transient input overstress signal "gets through" the Schottky diode the voltage of the transient will be pretty low and likely have a minimal impact on the op amp (i.e. close to zero current into the op amp ESD structure).  Sometimes a second input protection resistor is used between the Schottky and the op amp input to limit any overstress signal that passes through the Schottky.  This second protection resistor can be fairly low (between 100 and 1k).
    4. The disadvantages of using a large input resistor:  increased noise, bias current error (minimal in this case), potential bandwidth limitations from input capacitance x Rin.  In your case, I think the main limitation would be the noise.
    5. Of course, this all depends on the level of the transient.  If this system is exposed to 1000V transients we need to consider other options.  In your case I think you are looking at lower voltage transients.
    6. I think your circuit is pretty robust, and doubt you will see overstress issues.  You could increase the series resistor and add a second series resistor after teh Schottky for improved robustness.
    7. Below is the TINA simulation for a 20V overstress input signal transient.

    opa192 overstress.TSC

    I hope this helps.

    Best regards, Art