I am using Some arduinos (5V Pro-Mini, but planing to move to actual ATMEGA chip) over a distance of few metres.
These are connected using standard Alarm cable (I had problems with Ethernet cable and 12 volts leaking between cores) In mine set-up the cabling caries DC~12V (bit bigger closer to the source and backup car battery)I am protecting mine inputs using resistive divider (all the inputs where designed to be around 12VDC(then dropped to around 4.3VDC when HIGH) ground is connected through diode to protect polarity.
Problem is How to protect the RS-232 from being either shorted to or connected to 12DC (or ground) incorrectly. I am planning the possibility when cabling may extend up-to 60-70 metres.

I have been considering the Zener diodes, but I am not sure how effective would 5.1V + margin be. while the 4.7 model might be too low and short the standard communication Although I do not know what level the internal protection has from over-curent against ground.

Edit: I am using UART on a chip (directly on a logic level) the baud rate is intentionally slow and data send (including self-diagnostic for data corruption)is minimal generally less then 2 bytes per second.

  • 3
    \$\begingroup\$ For clarification are you using an actual RS-232 transceiver? Or are you running UART directly off the MCU pins and incorrectly calling it RS-232? Because actual RS-232 uses +/-12V signals. You talking about 5V signals indicates you are just running a UART straight off the MCU pins at TTL logic levels. Adding a RS-232/422/485 transceiver is what turns it into RS-xxx. Even if you are truly using RS-232, it won't even run 30m. The RS-232 standard only specifies operation out to 15m. Switch to RS-422/485 instead. RS-422/485 transceivers will solve all the issues you're asking about. \$\endgroup\$
    – DKNguyen
    Apr 13, 2019 at 16:40
  • \$\begingroup\$ Agreed. The 15 m limit is at 9600 baud. Halve the baud rate and you can double the distance, etc., (roughly). "These are connected using standard alarm cable ..." RS232 requires screened / shielded cable. "I had problems with Ethernet cable and 12 volts leaking between cores." This is very unlikely unless the cables were obviously damaged. \$\endgroup\$
    – Transistor
    Apr 13, 2019 at 16:53
  • 2
    \$\begingroup\$ @Tomas: Clarence Melvin Zener may be upset at your spelling of his name. \$\endgroup\$
    – Transistor
    Apr 13, 2019 at 16:54
  • \$\begingroup\$ I admit, I am dysgraphic , but I will correct the name \$\endgroup\$
    – Tomas
    Apr 13, 2019 at 17:31
  • \$\begingroup\$ "... the baud rate is intentionally slow ..." Why not give us the baud rate number? \$\endgroup\$
    – Transistor
    Apr 13, 2019 at 17:46

1 Answer 1


RS-232 transceivers such as MAX232 are guaranteed to survive connection (inputs or outputs) to +/-25V without damage. That's what they are designed for.

If you're using CMOS levels direct from the MCU chip, the limits are more like Vss - 0.3 to Vdd + 0.3V. You may be able to protect them by adding some series resistance (a few K) and operating at a relatively low baud rate to compensate for the resulting more sluggish waveforms. The added resistance limits the current, but does not clamp the voltage (except via the on-chip protection networks, so it will exceed the voltages but current is limited), which may be acceptable. You could add unidirectional TVS diodes to the pins and a bit of series resistance for more protection.

  • \$\begingroup\$ For the purpose of added info. Any recommendation for MAX232 alternative that would use Dual supply? Since I use voltage regulator to drop from 12 volts to 5V; it feels wrong to lift it back to 12 and increase load on a voltage regulator \$\endgroup\$
    – Tomas
    Apr 15, 2019 at 0:04
  • \$\begingroup\$ The 232 creates both positive and negative rails from the 5V. There were older chips without the charge pump converters but they needed external + and - supplies. It’s possible you could omit the pump cap for the V+ and connect the the +12V directly, I believe some of the data sheets cover that option explicitly. I don’t see much advantage and possible risks. \$\endgroup\$ Apr 15, 2019 at 0:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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