Choosing the diode for protecting an ADCs inputs

Question

I am designing a a board based on a microcontroller that have differential analog pins. The data sheet says that although the MCU's ADCs are internally protected against ESD strikes I must still take care to prevent the the input does not exceed the voltage and current limits of the pins which is Vcc (3.3 V) and +- 10mA. I am also planning to tie the Vref to 1.5 V. The analog pins will be both used in single ended and differential mode depending on the situation.

Doing a bit of reading, the conventional way seems to be placing diodes.

^{simulate this circuit – Schematic created using CircuitLab}

Since my Vref is set to 1.5 volts anything beyond that and cutting it beyond 3.3 V would destroy the pin. So anything in between should where diode cut-off the voltage. Would say at 2.5 V would be good point? But looking for parts I'm greeted by three identical sounding specifications Reverse Standoff, Breakdown, Clamping. Which specification is the one I should look at?

For current, since assuming we successfully find a part that will keep the input voltage at 2.5 V I should also adjust R1 and R2 so that it won't exceed 10mA.

R = V/I
R = 2.5/0.010
R = 250 Ohms 

It should not be less than 250 ohms. Am I approaching this correctly?

Answer 1

The practical ESD protection design. We know that for 40 yrs, all CMOS has a latchup effect. That is when the input exceeds either rail by 0.4~0.5V enough to trigger a high impedance SCR substrate and latch the power supply thru RdsOn in the substrate Vdd to Vss.

The solution has always been to use cascade bulk resistors around 10k and Schottky diodes after each of 2 series current limiting resistors to meet the 3kV minimum specs for the HBM.

There are models for human fingers and carts, but not the capacitance of cables with 30pF/ft. So the discharge of a floating cable connection can be far more SERIOUS than a finger.

Thus to add this type of additional protection I assert all you need is to cascade another 10k series R and diode pair or TVS diodes if you prefer.

Proof of concept

I included a logical ideal switch with 0pF and 0 Ohms to allow you to see the difference in signal voltages and component voltages (differential) and current. When you point at the plot the corresponding test point lights up in aqua colour.

If you don't understand, please ask a good question.

We used to blow up SCSI ports in production final test with long cables until I implemented ESD safe practices in '85 due to this problem.

I used 10kV 10% with an RC coupling impedance and CMOS input circuit with 200kHz rep rate just for convenience to display it.

Slow down the interactive simulation with the slider to suit your preference.

Answer 2

So most of the terms are mostly self-explanatory, they are just sorta hidden in the context of how these TVS diodes work. Use of reverse in only one term also confuses the issue. All of these terms are in the "reverse" direction. That is to say a more positive voltage on the cathode than on the anode.

Reverse standoff Voltage

This is the maximum voltage that you can apply to the diode in the "reverse direction" before any significant amount of current will flow through the diode. Basically anything less than this, and the diode isn't there.

Breakdown Voltage

Is higher than Reverse standoff Voltage and is the voltage at which some significant amount of current begins to flow. The current that flows will be specified in the datasheet. Make sure that this current is not so high that it will significantly load the signal that you are trying to read with your ADC.

Clamping Voltage

Higher still than the breakdown voltage. Now the TVS diode is in full swing, lots-o-current will flow after you reach this threshold. So much so that the voltage on your input lines will find it difficult to go above this voltage. Hence clamping voltage. This is likely the voltage at which the diode is rated.

BUT NOTE: Even though the diode is rated at the clamping voltage and sold as a "2.5v" diode, current starts flowing all the way down at the breakdown voltage. Be mindful of this.

Answer 3

I must still take care to prevent the the input does not exceed the voltage and current limits of the pins which is Vcc (3.3 V) and +- 10mA.

Here's where you may have missed a trick.

You have series protection resistors R1 and R2 at 50 ohms - they will limit the current into the MCU analogue pins on both positive and negative excursions. If the current limit is 10 mA and the resistor is 50 ohms, the maximum positive safe voltage that can be applied to the left of those resistors is defined by ohms law: -

$$I_{IN_{MAX}} = \dfrac{V_{IN_{MAX}}- \text{3.3 volts}}{R_{IN}}$$

So this becomes: -

$$V_{IN_{MAX}}=I_{IN_{MAX}}\cdot R_{IN} + \text{3.3 volts}$$

With 10 mA and 50 ohm, \$V_{IN_{MAX}}=\$ 3.8 volts.

This doesn't leave much headroom to apply a TVS protection diode at the input so, if the series resistor was 500 ohm instead of 50 ohm, you'd have a maximum input voltage of 8.3 volts and it becomes much easier to find a TVS that will do the job.

This allows you to pick a TVS with a stand-off voltage of around 3 volts. The stand-off voltage will mean that it draws no appreciable current at normal signal levels. The breakdown and clamping specifications now can be up to 8.3 volts (higher if R1 and R2 are greater than 500 ohms).

So, if you opt for a TVS diode that you think might do the job, let me know and I'll walk you through the data sheet so see what sort of surge current it can handle before the input voltages to the MCU might become excessive.

It does eventually come down to handling surge currents though - you have to decide what standard of ESD/surge protection your circuit is intending to comply with because that surge spec will define the peak pulse voltage and the duration and the surge impedance. Your TVS has to meet those requirements.