1
\$\begingroup\$

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?

\$\endgroup\$
3
  • \$\begingroup\$ What!? I remember reading exactly this question a while ago. Why is it reposted? \$\endgroup\$
    – KyranF
    Commented Oct 24, 2014 at 20:33
  • \$\begingroup\$ AH, it could be this one: electronics.stackexchange.com/questions/132591/… \$\endgroup\$
    – KyranF
    Commented Oct 24, 2014 at 20:34
  • \$\begingroup\$ 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. \$\endgroup\$
    – xapo
    Commented Oct 25, 2014 at 14:22

3 Answers 3

2
\$\begingroup\$

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.

\$\endgroup\$
1
  • \$\begingroup\$ 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). \$\endgroup\$
    – KyranF
    Commented Oct 24, 2014 at 20:47
0
\$\begingroup\$

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).

\$\endgroup\$
-1
\$\begingroup\$

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

\$\endgroup\$
7
  • 1
    \$\begingroup\$ I don't think this will achieve the desired result. You need a high side switch. \$\endgroup\$ Commented Oct 24, 2014 at 20:58
  • \$\begingroup\$ @SpehroPefhany why wouldnt this work ? \$\endgroup\$
    – efox29
    Commented Oct 24, 2014 at 20:59
  • \$\begingroup\$ 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. \$\endgroup\$ Commented Oct 24, 2014 at 21:01
  • \$\begingroup\$ @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 ? \$\endgroup\$
    – efox29
    Commented Oct 24, 2014 at 21:04
  • 1
    \$\begingroup\$ 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. \$\endgroup\$ Commented Oct 24, 2014 at 21:07

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.