Timeline for Protect MCU pin from ESD
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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Sep 11, 2014 at 16:25 | comment | added | IgorEE | Thanks Dave, i will do some tests with the ferrite as well, replace the TVS and schottky diodes with an integrated part, and if tests go well I will adopt that for production board. | |
Sep 11, 2014 at 16:22 | comment | added | Dave Tweed | You should at least consider putting a ferrite bead in series with the line, which will present a very high impedance to the fast edges associated with ESD events, while having negligible effect on your 1-wire signals. | |
Sep 11, 2014 at 16:14 | answer | added | gbulmer | timeline score: 2 | |
Sep 11, 2014 at 16:08 | comment | added | IgorEE | The TVS is SP1003-01DTG (Digikey PartNo: F3377CT-ND). It is indeed a unidirectional TVS, thank you for pointing the vulnerability to negative spikes. | |
Sep 11, 2014 at 16:00 | comment | added | George Herold | Part number for TVS? If that is a unidirectional TVS then I think it is only the D403 Schottky that is protecting you from large negative voltages on the input. (Sparks can be of either polarity.) | |
Sep 11, 2014 at 15:49 | comment | added | IgorEE | 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. | |
Sep 11, 2014 at 15:45 | comment | added | IgorEE | Also, can not add a resistor to the line as it will interfere with communication. The MCU is an atmel atmega2560. | |
Sep 11, 2014 at 15:45 | comment | added | gbulmer | 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. | |
Sep 11, 2014 at 15:42 | comment | added | IgorEE | 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). | |
Sep 11, 2014 at 15:37 | comment | added | gbulmer | 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. | |
Sep 11, 2014 at 15:34 | history | edited | JYelton | CC BY-SA 3.0 |
Removed unnecessary thanks/signature
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Sep 11, 2014 at 15:31 | history | asked | IgorEE | CC BY-SA 3.0 |