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I'm building an industrial circuit, and want to use 0-10V and 4-20mA sensors in this circuit.
I want to protect my ESP32 MCU(NodeMcu board) from transients and high voltages.
I design this circuit and have questions: enter image description here

  • Is this circuit works good for my purpose?

  • which component is necessary and Which one is optional(schottky diodes,100nf capacitor,polyfuse)?

  • NodeMcu does not have any Analog ground ,does it matter?

  • whats your opinion on this circuit?

It's not a precision system, but I want to protect it in a good way.

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    \$\begingroup\$ You tend to want an impedance buffer between your the point of ESD entry and the supression device, like a resistor or ferrite bead. \$\endgroup\$ – DKNguyen Dec 10 '19 at 17:04
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    \$\begingroup\$ A ferrite bead and RF help reduce current and voltage. if dV/dt= 10kV/1us and I=1A then C= I*dt/dV = 1e-10 F= 100 pf then repeat for L lossy bead \$\endgroup\$ – Tony Stewart EE75 Dec 10 '19 at 17:36
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    \$\begingroup\$ @Danesh_sa this is a bit tricky. You have many things you should know about your circuit and application. The frequency of you r input signal the cutoff of your filter. The sample and hold time of your ADC analog front end. And the sampling frequency on your software \$\endgroup\$ – Ashraf Almubarak Dec 10 '19 at 18:33
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    \$\begingroup\$ @Danesh_sa It limits the current through the clamp which keeps the voltage drop across the clamp low, which keeps the voltage that the protected input is exposed to low. It also lets your clamp handle larger spikes due to limiting that current. Combined with parasitic capacitances (or added capacitance), it also slows down the rise time which buys time for the clamp to activate. \$\endgroup\$ – DKNguyen Dec 10 '19 at 18:50
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    \$\begingroup\$ @Danesh_sa Unidirectional ones are better for unipolar signals because they do not let the voltage go negative and it does not need to go negative because it is a unipolar signal. No, do not eliminate D3. Eliminate D2 and replace D3 with a undirectional TVS. It will clamp to the rails like before when there is power, but also protect the line when there is no power by breaking down in reverse. \$\endgroup\$ – DKNguyen Dec 10 '19 at 21:55
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assuming that you modified the D3

-Is this circuit works good for my purpose?

it should be working. However you should not that in the ADC analog front end there is an internal holding capacitor. You potential divider has an output resistor of R2 approx. This may delay the charging of the hold capacitor if you have high sampling rate or high frequency signal. Attached is the ADC front end inside a microchip. It should be the same. enter image description here

-which component is necessary and Which one is optional(schottky diodes,100nf capacitor,polyfuse)?

schottky diodes important. 100nF don't know, you should take care they attenuate your signal and add a delay(phase shift) in to your signal so you may get the wrong value of the sensor in that time. The corner frequency of this circuit is about 400 Hz. If your signal frequency is more than that the capacitor will filter out the high frequency and you will get small readings.

-NodeMcu does not have any Analog ground ,does it matter?

It may be directly connected to the Digital one. check datasheet and technical references.

-whats your opinion on this circuit

You will have negative voting by this kind of questions.

-its not a precision system ,but i want to protect it in a good way.

I will add another resistor (1k) before connecting this circuit to the ADC. so the schottky diode will conduct most of the currents if a surge occurred in the analog side. This resistor should be as close as possible to the ADC pin in the board. enter image description here

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  • \$\begingroup\$ ,thank you for your detailed answer.I edit the schottky diode above:) .as you explain ,i think its better to eliminate C1 (because it may filter some signals ).could you please explain more about 1k resistor,(why 1k?),it may affect on ADC measurments. \$\endgroup\$ – Danesh_sa Dec 10 '19 at 18:34
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    \$\begingroup\$ 1k resistor is a suggestion. Inside your microcontroller, any input will probably have the TVS diodes same as D3 and D1. However, they are smaller in size and operate for a limited number. if you didn't add the resistor the surge current will go both in your protection diode (either D1 or D3) and the diodes inside the microcontroller die. They have higher leakage current which may give false reading. When you add a resistor you will force all surge to go only through D1 & D3 and the inner diodes will not conduct any current. \$\endgroup\$ – Ashraf Almubarak Dec 10 '19 at 19:57
  • \$\begingroup\$ aha,BAT54 has low leakage current.sorry i cant understand " and the diodes inside the microcontroller die", so you mean the clamping diode is just for cut the voltage and it can not protect from surge current.because of this we use 1k resistor in series ,but there are two things!1-i use TVS diode do to absorb surge spikes .2-both clamping diodes has leakage current and i have a series resistor so there is voltage on 1k resistor , so it affect ADC measurements,Right?(or its not big concern?) \$\endgroup\$ – Danesh_sa Dec 10 '19 at 21:11

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