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From the Beaglebone Black SRM Rev C.1: "DO NOT APPLY VOLTAGE TO ANY I/O PIN WHEN POWER IS NOT SUPPLIED TO THE BOARD. IT WILL DAMAGE THE PROCESSOR AND VOID THE WARRANTY. NO PINS ARE TO BE DRIVEN UNTIL AFTER THE SYS_RESET LINE GOES HIGH."

I want to connect a sensor to ADC input. When I apply the power to BBB and to the sensor in the same moment, there is a danger that the signal from sensor comes earlier as BBB is ready. What is the best way how to prevent from this?

I can imagine following:

a) Control a transistor with SYS_RESET signal to switch on the power for the sensor? b) put "something" between the pin and the sensor?

xapo
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  • What!? I remember reading exactly this question a while ago. Why is it reposted? – KyranF Oct 24 '14 at 20:33
  • AH, it could be this one: http://electronics.stackexchange.com/questions/132591/how-to-disconnect-voltage-from-beaglebone-black-expansion-pins – KyranF Oct 24 '14 at 20:34
  • Thanks for notifying me. I'm sorry, I didn't find that and topic is slightyly differnt. Surprisingly when I search for "SYS_RESET" I get only my question. – xapo Oct 25 '14 at 14:22

3 Answers3

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If the sensor is supplied from the same voltage source as the microcontroller, generally there is no problem. It's not a matter of the board being "ready" so much as never applying voltage to an input that exceeds Vdd of the microcontroller. The actual specification is:

-0.5V to IO supply voltage + 0.3 V

So if the IO supply voltage is 0V the input voltage should not be more than 0.3V or less than -0.5V.

If such a voltage is applied without limiting the current, it's possible to damage the microcontroller.

Alternatives include adding some series resistance to limit the current or buffering the input with "something" tolerant of voltage applied when power is off. There are digital buffers that are designed for this purpose (they also translate voltage levels), and for analog circuits you can use an op-amp buffer with some input resistance to protect the op-amp.

Spehro Pefhany
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  • Yes I wonder if the documentation says that line to protect against this case, for people who do not think about this (applying a voltage to the IC without the IC being powered first). It's possible that the power management on the PCB of the BBB has some sequence that makes this a real possible threat, and only when the main core's voltage is 'safe' does it set the SYS_RESET pin high, which basically allows external circuits to then power-up/turn on (an external power sequence perhaps). – KyranF Oct 24 '14 at 20:47
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As suggested in the question, and reinforced in the comments of this question on EE.SE a reasonable solution would be to use the SYS_RESET logic HIGH signal to turn on/chip-enable a tri-state or similar buffer/line driver IC.

A buffer will let most reasonable analog signals through, but you may be able to find a more suitable IC like an analog multiplexer which is designed for ADC style input/output conditioning, with a CHIP ENABLE pin or tri-state mode to prevent it from driving the connection to the BeagleBone Black until the SYS_RESET pin has indeed gone high as required.

If the CHIP ENABLE/Tri state mode pin is ACTIVE LOW, then you may easily invert the logic output of it using a simple NPN transistor inverter circuit, or an actual inverted buffer IC (basically, a NOT gate).

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KyranF
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This circuit may not be suitable for this particular application(see comments). I'll leave this here regardless.

If the BBB has floating outputs, R2 just ensures that the transistor is off (which ensures that your sensor is depowered.

When the BBB is good and ready, it needs to drive the input of R1, with a logic 1, which would then turn the transistor on, and bring V- of your sensor to ground and completing the circuit.

If you want to turn your sensor off, drive R1 with logic 0, and the transistor will turn off.

schematic

simulate this circuit – Schematic created using CircuitLab

efox29
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    I don't think this will achieve the desired result. You need a high side switch. – Spehro Pefhany Oct 24 '14 at 20:58
  • @SpehroPefhany why wouldnt this work ? – efox29 Oct 24 '14 at 20:59
  • Because when the sensor ground is opened up, the output of the sensor will generally be higher than zero, so you're disabling the function but not really preventing it from driving the BB pins to some unhealthy voltage- maybe even more voltage than normal. – Spehro Pefhany Oct 24 '14 at 21:01
  • @SpehroPefhany interesting. What would "generally higher" mean ? If the BB and sensor run off 5V, the sensor output would be more than say..2V when the sensor ground is removed ? – efox29 Oct 24 '14 at 21:04
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    Depends on the sensor, but if it was just a voltage divider, and you remove the ground on the bottom resistor, the open-circuit output voltage goes right to +5V. If the divider was intended to supply 3.3V maximum, you've now exceeded the maximum 'normal' range. – Spehro Pefhany Oct 24 '14 at 21:07
  • But thats because there is no current flow so every part of the resistor is at 5V. You thinking that the sensor might find its "ground" through the BB instead ? I mean this is all hypthetical, it could work or it could not work, but its interesting :) – efox29 Oct 24 '14 at 21:10
  • Yes, the current would definitely flow through the BB. Not hypothetical- this is a plausible connection for (say) a thermistor sensor (resistor to +5 and NTC thermistor to ground, junction to ADC input). – Spehro Pefhany Oct 24 '14 at 21:12