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MSP430F67651IPZR ADC maximum injection current

subramani pasupathy
Intellectual 575 points

Hi All,

I am using MSP430F67651IPZR. I have 19V DC input to my system that I need to read through ADC. I used resistor divider to scale it down to 3V

If the bottom resistor fails and becomes open, then we will get 19V to ADC pin. How do I protect it. What is the maximum injection current that an ADC input pin can take

Thanks for your reply in advance

regards

subbu

 

 

  • 0
    TI__Guru**** 202256 points

    Hello subbu,

    Any pin can only handle Vcc+0.3 V before damage can occur.  The MSP430 includes some internal clamp diodes on each pin that can sink up to 2 mA of current in an overvoltage situation, but 19 V to an ADC pin will most definitely destroy that pin and possibly the MCU as well.  So let's try our best to avoid that!  Here is a great blog that covers some microcontroller input protection techniques, definitely worth a read.

    Regards,

    Ryan

  • 0 in reply to Ryan Brown1
    Intellectual 575 points

     Hello Ryan

    Thanks and appreciated

    regards

    subbu

  • 0 in reply to subramani pasupathy
    Guru 227245 points

    Ryan has given the solution without noticing :)

    Just make sure that any excess voltage drops on a series resistor on no more than 2mA.
    This means, adding 1k series resistance for each 1V above VCC will limit the injection current to 1mA (assume VCC=0V in case the MSP Is unpowered, to be extra safe).

    In case of 19V, a series resistor of 15-20k will do. Increase the sampling time of the ADC accordingly, so the sampling capacitor has enough time to charge through the additional resistance.

  • 0 in reply to Jens-Michael Gross
    Intellectual 575 points

    Thanks Micheal. I have provided 18K series resistor to all ADC that can be read slowly. 

  • 0 in reply to subramani pasupathy
    Guru 46710 points
    Adding series resistor to adc input is not good solution in your case. Better just use proper resistor values of divider. Protection against divider failure would be series dode between adc input and vcc.
  • 0 in reply to Ilmars
    Intellectual 575 points

    Hi IIImars, 

    Thanks,

    I have sent the gerber with 18K series resistor. Let me check with that by removing the bottom resistor divider in one board once I receive PCBA. If it works without damaging the IC, then we can leave as it is.

    I think it will work as our ADC acquisition is about 100ms.  

    What do you think.

    regards

    subbu

  • 0 in reply to subramani pasupathy
    Expert 2740 points

    subbu,

    I would like you to check the VCC voltage before and after removing the bottom resistor divider . 

    The concern is that the VCC may then exceed the safe voltage limit. ( > ~4V)

    Peter

  • 0 in reply to Peter Dvorak
    Guru 227245 points

    Ilmars, a diode isn’t required if the resistor divider’s top resistor is large enough to act as proper series resistor in case of bottom resistor failure.

    On 19V, you want to have a voltage drop of >16V for a current of <2mA. Which is true for any resistor >8k. So an 18k/3k3 resistor divider will reduce the 19V to 2.95V, and in case of the 3k3 failing will limit the current to 1mA.
    However, as peter mentioned, these 1mA will go into VCC, and when the total load on VCC is <1mA, this will raise VCC.

    However, some MSPs have dedicated analog pins (mainly those with multiple SD16/SD24 converter modules with differential inputs and +-0.6V input range). Those pins are not multiplixed with port pins and therefore have no clamp diodes. Here, external protection diodes are a good idea.

  • 0 in reply to Jens-Michael Gross
    Guru 46710 points

    Jens-Michael Gross said:
    Ilmars, a diode isn’t required if the resistor divider’s top resistor is large enough to act as proper series resistor in case of bottom resistor failure.

    Proper protection shall protect input pin, not bottom resistor :D I just suggested universal solution that includes both - unlikely bottom resistor failure and much more real surge/overvoltage into top resistor.

    [edit] Thou to protect against both positive and negative surges, two diodes needed - one connected to VCC another to GND

  • 0 in reply to Ilmars
    Guru 227245 points

    Ilmars said:
    Proper protection shall protect input pin, not bottom resistor :D

    Yes, that’s what I wrote and meant. Because in case the bottom resistor fails (bad soldering, broken etc.), the full 19V input will go to the pin, with only the top resistor in series.
    With a top resistor of 18k, in case of bottom resistor failure and 100% permanent overvoltage, this would result in 1.94mA into the pin, which is still acceptable. Moreover, the 2mA are rated current, not peak current, so spikes or surges even beyound 100% overvoltage will be handled without permanent damage.

    But of course, thsi does not apply to the SD16/SD24 inputs with +-0.6V input range, as these have no internal  clamp diodes (and not figital I/O logic). Here, external diodes are probably needed to protect them.

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