Analog input over-voltage protection using Schottky and Zener Diodes. Can it work?

Question

I'm developing an analog input module based on TI's ADS8668 and want to protect the input channels from over-voltage. According to the datasheet, each analog input channel withstands voltages from -20V up to +20V.

Although my analog signals are expected to be in the range of 0-10V, wrongly connected inputs (e.g. miswiring) could lead to voltages up to +24V on the ADC input channel.

To protect my analog inputs, I tend to use Schottky diodes, as recommended by TI (Engineer It: How to Design Protection Circuits for Analog I/O Modules) and Analog Devices (Protecting ADC Inputs).

I have no supply rail in 10-20V range on my PCB, that I can clamp my diodes to. Therefore I would just place a Zener diode (D3) as shown below, to protect my analog inputs from voltages above Zener voltage.

Simulating this schematic in LT Spice using a 12V Zener results in a system behaviour according to my requirements (see below).

Is it sufficient (and good practice) to clamp my Schottky diode (D2) to the Zener diode without having any positive supply rail (unlike this solution, which uses a shunt regulator)?

Is there anything I need to beware of, especially the power dissipation of my components in an over-voltage case?

Thanks in advance!

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

Yes, that would work fine, but you can make this even more simple:

^{simulate this circuit – Schematic created using CircuitLab}

Zener diodes behave more or less as a "normal" diode when in forward mode so we can use that to clamp negative voltages.

When using a 12V, 400 mW zener diode the maximum current through the zener diode can be 400mW / 12 V = 33 mA. There will then be 12 V across R1, R1 needs to limit the current to 33 mA so it needs to be: 12 V / 33 mA = 364 ohm or higher. I added some margin and suggest 470 ohms then the actual current will be 12 V / 470 ohm = 26 mA and R1 will dissipate 12 * 26 mA = 0.312 W so you might want to use a 0.5 W resistor (or use two 1 kohm 0.25 W resistors in parallel).

I would make the capacitor C1 10 nF, that should suffice to remove most pulses. Only if you need to measure high frequency signals C1 might need to get a smaller value or must be removed. Then the zener diode might be an issue as well as zeners aren't very fast. But if you stay below 100 kHz or so there should be no issue with this circuit.