in/output protection

On a sunny day (12 Dec 2006 06:55:05 -0800) it happened "CeeRox" wrote in :

1)

--- R -----------> MCU |k zener 5.1V | ///

2) +5V |k diode |a

--- R ---------> MCU

2) will make the +5V higher if R is low enough, or there is little load.

If it's an input to a CMOS chip (TTL levels), usually a series resistor of a few K will suffice.

If it's an output you may need a small value resistor plus some kind of clamping device such as a TVS or Zener diode.

Best regards, Spehro Pefhany

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Jan Panteltje skrev:

R would be at least 50 Ohm if the zener can handle 1 Watt continous (7V).... There would be problems if a CMOS-output should have an input protection for eg. 75V. But then maybe a VDR could be useful, if the zener-diode would survive the time before the VDR gets "high-ohmic".

Don't do this. If the zener or diode clamps just a wee bit beyond the substrate diode then the substrate diode will "win", take the brunt of the current and the MCU goes up in smoke. They can only take so many mA on the substrate diodes and after that it can latch up. If you are really lucky it'll take the regulator and everything else with it to the grave.

Use several 10K or provide Schottky diodes to VCC and GND. Or use a regular double diode such as the BAV 99 (for low current cases) but a second resistor between that and the MCU pin.

Another often unwanted effect is that a current into the upper substrate diode can actually power your MCU side. If it doesn't present enough of a load, maybe because it has entered a low power mode, that voltage might exceed abs max and then bzzzt ... poof.

--
Regards, Joerg

http://www.analogconsultants.com

Joerg skrev:

If we skip the 75V case....what would be the difference with a schottky diode..isn't there a risk that the substrate diode would win in that case aswell..?

If we consider the case with a buffer curcuit between the MCU and the connector, e.g. if the MCU has LV-TTL outputs and there is a need for an 5V output. A 74LVC2G07 (powered with 3V3) and the outputs pulled up to 5V could be used. Then a zener might be suitable...?

Buffer--- R -----------> connector |k zener 5.1V | ///

CeeRox skrev:

Skip the buffer disscussion... =P I wrote to quickly there. The main problem is that the output should survive 12V continously, not just a quick spike...

The Schottkys win provide they are large enough for the expected current. Schottkys begin to conduct around 400mV while the substrate diodes on all but the most esoteric chips start at 600mV. But again you can also use regular silicon diodes here if you spring for another resistor between diode protector and MCU pin. This resistor can be much smaller. Reason is that there will be very little voltage differential between that point and the MCU pin.

Then we'd need to know: How much current must the output be capable to drive and what's the type of load it will see?

Protecting power outputs that must supply lots of current is a lot more tricky than to protect weak outputs or inputs.

--
Regards, Joerg

http://www.analogconsultants.com

...

The 'easy' character may be disputed, but usually one would want to feed the 'dirty' signal into a chip intended for such signals. The old MC1489A receiver for RS-232 will take +/- 20V without complaint, and convert it to a TTL output. It's an inverter, of course.

What is the likely frequency range of the signal?. What is the allowable input voltage range on the MCU?. What is the driver feeding the signal (CMOS, or TTL, or something else)?. Could differential signalling be used?. What is the likely noise enviroment?. How is the return path handled?. Different answers depending on all these factors, and a few others. Ranging from Pi networks, an input resistor with a MOV, through various types of ESD protected buffer, to opto-isolation.

Best Wishes