3
\$\begingroup\$

Quick question for someone with a bit more experience.

I'm working on a product which will sit in an environment which is likely to experience some static discharge from time to time. Basically, some sensors are connected to an MCU over on average 30 meters of cable, so I guess some static discharge is bound to happen from time to time.

I have been testing the system with the following circuit to protect the MCU pin with no "incidents" so far:

Schematic

As you can see, there are two ESD TVS diodes (D401 and D404), and two Schottky diodes (D402 and D403) protecting the MCU pin "DQ".

For a final production board, I would like to reduce BOM, and hence my question is whether having two sets of diodes (TVS and the Schottky) is completely redundant? If yes, should I remove the Schottky or the TVS diodes to reduce BOM?

\$\endgroup\$
9
  • 1
    \$\begingroup\$ I am not an ESD expert! The MCU pin is an input? What MCU are you using? Can you put a resistor in series between DQ and the pin? If you can keep the current low, then the MCU's built in protection might keep the voltage within range, and then only the TVS diodes are required. \$\endgroup\$
    – gbulmer
    Commented Sep 11, 2014 at 15:37
  • \$\begingroup\$ The MCU pin is "both", it is driving a Maxim/Dallas OneWire bus. The MCU can actively drive the line (and in some transactions it does), then releases the line and senses it's state (slaves can pull the line low actively). Most of the time through, the line is in a passive pull-up state. (1k resistor in the diagram). \$\endgroup\$
    – IgorEE
    Commented Sep 11, 2014 at 15:42
  • \$\begingroup\$ When you say "I would like to reduce BOM" do you mean "reduce BoM cost", and not just the number of different parts? If that is the case, it might be worth adding "cost" to the question. \$\endgroup\$
    – gbulmer
    Commented Sep 11, 2014 at 15:45
  • \$\begingroup\$ Also, can not add a resistor to the line as it will interfere with communication. The MCU is an atmel atmega2560. \$\endgroup\$
    – IgorEE
    Commented Sep 11, 2014 at 15:45
  • \$\begingroup\$ gbulmer: TVS diodes are not that costly, it is not really a matter of cost, it is more a matter of reducing BoM count (number of different parts). This will ultimately make assembly easier and quicker (boards will be manually assembled most likely), and there are 30 sensors * two diodes each = 60 less components to solder. \$\endgroup\$
    – IgorEE
    Commented Sep 11, 2014 at 15:49

1 Answer 1

Reset to default
2
\$\begingroup\$

The number of parts used to protect each signal line might be reduced to 1/2 component, by using an integrated part which handles two signals.

For example this CAN BUS part protects two signals with a single device, replacing a TVS diodes, and the zener diode.

Many companies make these types of products, including NXP, TI, Infineon, On Semi, etc.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.