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OPA377: OPA377 Power OFF PIN internal Status

Vincent Lin
Prodigy 240 points
Part Number: OPA377
Other Parts Discussed in Thread: TINA-TI

Dear Sir:

Regarding the OPA377 Power OFF PIN  internal Status

When the OPA377 Power OFF(no power supply).

*Which the following the Pin internal status of +IN(Pin1) at OPA377 Power OFF?

1.Pull Up Resistor to VCC Power.

2.Pull Down Resistor to GND.

3.High impedance: Open.

* How many lekage current of +IN(Pin1) at OPA377 Power OFF?

 

Please check to reply it asap.

Thanks

B.R

Vincent

 

  • +1
    TI__Guru 64290 points

    Since there is no enable pin in OPA377, I presume you refer to the supply pin being turned off.  If there is no power supply provided, ALL pins (including input and output pins) will collapse on its negative supply (presumably GND).  Once this happens, all pins will show high impedance when left floating.  However, the impedance will decrease and result in forward-biasing ESD protection diode, when any pin is driven above or below the voltage at the negative supply (GND) - see an equivalent circuit for OPA377 below.

    All in all, the leakage current (Id) on any of the pins including IN+ will be a function of forward-bias voltage (Vd) applied to the pin above or below negative supply:  

    Assuming Is ~1fA, driving any pin above of below GND would result in the following approximate leakage current:

    Notice that for Vd above 0.7V, the current would exceed the absolute maximum rated value of 10mA, which could damage the part.

  • +1 in reply to Marek Lis
    Prodigy 240 points

    Dear Sir:

    When the Pin internal status of +IN(Pin1) is GND at OPA377 Power OFF.

    U301_PWM7(High): 3.3V

    RLIMIT(R751) : 1Kohm.

    ILIMIT = 3.3V / 1K ohm = 3.3mA.

    The ILIMIT current doesn’t exceed the absolute maximum rated value of 10mA.

    It wouldn’t damage the part.

    Is it correct?

    Please check to reply us asap.

    Thanks

    B.R

    Vincent

  • +1 in reply to Vincent Lin
    TI__Guru 64290 points

    Vincent,

    Originally you did not mentioned applying 3.3V on positive input. In this case the input current will be around 2.7mA - see below.

    Vincent OPA377.TSC

  • 0 in reply to Marek Lis
    Prodigy 240 points

    Dear Sir:

    1.How to get the Vin 601.1mV?

    2.The blue circle the diode which means the OPA377 IN+ internal protection diode.

    Is it correct?

    3.In this case the input current (IN+)will be around 2.7mA ( at OPA377 Power OFF)

    It doesn’t exceed the absolute maximum rated value of 10mA.

    It wouldn’t damage the part.

    Is it correct?

    Please check to reply it asap.

    Thanks a lot.

    B,R

    Vincent

  • +1 in reply to Vincent Lin
    TI__Guru 64290 points

    Please see inline answers below:

    1. How to get the Vin 601.1mV?  
    You may simulate it as I did using attached Tina-TI schematic. However, in my previous post instead of Vcc I floated negative supply - below is a correction.  Simulating two cases with floated vs grounded Vcc, respectively, I get input current equal to quiescent current of 681uA in Fig 1 vs 2.8mA shown in Fig 2.

    2.The blue circle the diode which means the OPA377 IN+ internal protection diode.

    Is it correct?  YES

    3.In this case the input current (IN+)will be around 2.7mA ( at OPA377 Power OFF)

    It doesn’t exceed the absolute maximum rated value of 10mA.

    It wouldn’t damage the part.

    Is it correct?  YES

    6153.Vincent OPA377.TSC

  • 0 in reply to Marek Lis
    Prodigy 240 points

    Dear Sir:

    When the OPA377 Power OFF

    The U301_PWM7 is still around 0.9V.

    The 0.9V is leakage current flows the 1k and the internal ESD diode(around 0.6V) to GND.

    Is it correct?

    Please check to reply us asap.

    Thanks

    B.R

    Vincent

  • 0 in reply to Vincent Lin
    TI__Guru 64290 points

    Vincent,

    It is an external source that drive the PWM7 input of the op amp and NOT all the way around (op amp does not drive its own input). Thus, either the PWM7 has a wrong input setting before turning on of 3.3V supply or circuit is in a lockup state from a previous cycle.  If you need further assistance, you need to show us full multiple cycles what you witness - giving us pieces and bits at the time, does not allow for the full picture of the problem to be formulated.

  • 0 in reply to Marek Lis
    Prodigy 240 points

    Dear Sir:

    Please check to reply us asap.

    Thanks a lot.

    B.R

    Vincent

  • 0 in reply to Vincent Lin
    Prodigy 240 points

    When the OPA377 Power OFF(P3V3_CODEC).

    The SOC internal Pull up RES. 90Kohm to 3.3V

    The voltage of U301_PWM7 is around 0.9V. [ the 1k and the internal ESD diode to GND.]

    Is it correct?

    Please check to reply us asap.

    Thanks a lot.

    B.R

    Vincent

  • +1 in reply to Vincent Lin
    TI__Guru 64290 points

    Yes, with 90k pullup resistor to 3.3V, it all makes much more sense.  But for PWM7 to be at 900mV could only be possible by the bypass cap holding Vcc voltage around 600mV - see below.

  • 0 in reply to Marek Lis
    Prodigy 240 points

    Dear Sir:

    The SOC internal Pull up RES.to 3.3V.

    The voltage of U301_PWM7 : The voltage of (R751+ U710 OPA377 internal ESD diode+P3V3_CODEC)

    Is the calculation correct?

    Please check to reply us asap.

    Thanks a lot.

    B.R

    Vincent

  • 0 in reply to Vincent Lin
    TI__Guru 64290 points

    Yes. that's correct but only when P3V3_CODEC << 3.3V.

  • 0 in reply to Marek Lis
    Prodigy 240 points

    Dear Sir:

    According to the measurement.

    U301_PWM7 : 0.9V

    P3V3_CODEC : 0.24V

    The SOC internal Pull up RES.(90Kohm) to 3.3V.

    ILIMIT = (3.3V-0.9V) / 90Kohm = 26.7uA.

    The Voltage of U301_PWM7 = P3V3_CODEC + ILIMIT*R751 + VD (U710 OPA377 internal ESD diode)

    0.9V = 0.24V + 26.7uA*1Kohm+ VD (U710 OPA377 internal ESD diode)

    VD (U710 OPA377 internal ESD diode) = (0.9V-0.24V- 0.0267V)= 0.6333V

    Is the calculation correct?

    Please check to reply us asap.

    Thanks a lot.

    B.R

    Vincent

  • +1 in reply to Vincent Lin
    TI__Guru 64290 points

    Yes, your calculation must be correct. Based on your direct measurements the internal ESD diode forward bias voltage under 26.7uA current must be 633.3 mV.  That’s very close to my initial estimate - see below.