Narrow TVS margin between operating and overvoltage

Question

Devices like the PIC16F18323 have an extremely narrow margin on I/O pins. The standard logic-high voltage level is 5V, but the absolute maximum is Vcc+0.3V, i.e. 5.3V.

Reading about TVS devices, in particular those offered on DigiKey, for a diode whose Vwm is 5V, the lowest clamping voltage I can find is 7.5V for the 1N6373. I can assume that by that point my I/O port will be destroyed.

So what gives? Should I be using a TVS diode with a lower Vwm and have it already in avalanche when the device is under normal conditions? Or just cross my fingers with a diode whose avalanche and clamping voltages are significantly above what the pin will tolerate?

Answer 1

You need to choose a TVS that has a working voltage greater or equal to the maximum normal voltage you will see.

This means performing some tolerance analysis with regards to the supply & expected IO (5V could potentially be 5.25).

Take a littelfuse part ( 1.5SMC series) a 5.8V working has a min & max breakdown of 6.45 --> 7.14V & this could RISE to 10.5V (depending on the avalanche current). All this would imply that you are unable to use these devices right on the inputs.

Except you do not need to just use a TVS. Additional steering diodes right at the IC in question will clamp the I/O to Vcc+0.3 (schottky used). Additional input L-C will again reduce what the actual pin will see

^{simulate this circuit – Schematic created using CircuitLab}

A strike (indirect lighting or ESD) is a short duration waveform and thus "waveform shaping" is a viable option to ensure the IC IO does not exceed its maximum rating & other components do not exceed their current or power rating

Answer 2

The reason why the absolute maximum Vcc on the pins is Vcc+0.3V is because above that voltage, the internal clamp diode to Vcc will turn on -- as long as there's a series resistor between the I/O pin and outside world to limit that current to what the clamp diode can handle, it will not permanently harm the IC. (Modern datasheets give an Iik, Iok, or Ik specification for this maximum current.)

So, use the higher clamping voltage TVS to absorb the bulk of the energy and a series resistor to keep the port pin's clamp diode from being burned out.

Answer 3

The usual way of protecting I/O pins with narrow margins (and pin injection currents in the order of 5mA usually) is with a resistor and perhaps an external clamp schottky:

^{simulate this circuit – Schematic created using CircuitLab}

The reason for the 0.3V is usually that there is a schottky internally, but it cannot handle much current; adding one (or two for bidirectional protection) that can handle the current due to a voltage excursion event is a simple and effective means of protecting the pins. The resistor limits the current in the schottky and therefore maintains \$V_f\$ at a nice low voltage.

TVS devices have wide variations of turn-on voltage, full clamping voltage and those parameters are also thermally dependent, making them unsuitable for protecting I/O pins in this particular case.

In the case of I/O pins with schottky devices and a rated pin injetion current (usually the maximum sustained current the internal schottky can handle) a simple resistor can suffice if the maximum voltage can be defined.

Typical arrays for mult-line protection can be found at some manufacturers

Note that if you expect a high voltage (too fast even for a front end TVS as JonRB alludes to - some have slow reaction times) you may want to use pulse-withstanding resistors. These maintain their resistance even after a large transient event; standard resistors may not.