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ISO7742: Reverse current at high output

user5397266
Prodigy 230 points
Part Number: ISO7742

Hi team,

I would like to realize a circuit to connect the OUTA pin of ISO7742 and turn on/off the Pch FET as shown below.
The maximum rating of the OUTA pin (Vccx+0.5V) is not exceeded because of R1 and R2.
If I increase the value of R1 and R2 and suppress the reverse current when OUTA is outputting High, will this configuration work?
If so, what is the maximum rating of the reverse current?

Supplementation:
I am using a Pch FET because I do not want to reverse the logic.
Also, we do not want to add FETs to reduce the number of components.

Thanks.

  • 0
    Prodigy 230 points

    Sorry. The circuit was wrong.
    I would like to realize the following connection.
    I think it is the same that reverse current flows when OUTA is High.

  • 0 in reply to user5397266
    TI__Genius 15625 points

    Hello,

    Thank you for your inquiry.

    I have several points to make in regards to the information you have provided:

    1. For the operation of the p-channel FET, the R1 and R3 are not necessary. Since R1 would be removed, there would be no reverse current into the OUTA pin of the ISO7742 device.
    2. It seems like the VCC2 of the ISO7742 is 3.3V while the supply to the p-channel FET is 5V. Since the OUTA is 3.3V while the p-channel FET is 5V, OUTA might not be able to completely turn off the device. it is possible that the p-channel FET could be continuously on due to the OUTA being lower voltage than the source pin.
      1. Changing VCC2 to 5V would make OUTA turn on and off the p-channel FET properly.
    3. Since the p-channel FET is used on the high side, the output of the p-channel FET will be in reverse logic in comparison to the OUTA. So when the OUTA is HIGH, the led will be off. When the OUTA is low, LED will be on.

    Regards,
    Aaditya Vittal

  • 0 in reply to Aaditya V.
    Prodigy 230 points

    Hi Aaditya,

    For 1 and 2, R1 and R3 are needed to adjust the Vgs of the FET.
    For example, if the resistance ratio is R1:R2:R3=3:2:2, Vgs=0.44V when Vsw is High and Vgs=2.5V when Vsw is Low, which is the desired value.
    Also, 3.3V for Vcc and +5V for LEDs are values that cannot be changed.

    On top of that, I am asking about the maximum rating of the reverse current.
    Please answer.

    Best regards,

  • 0 in reply to user5397266
    TI__Genius 15625 points

    Hello,

    Thank you for your response.

    user5397266 said:
    For 1 and 2, R1 and R3 are needed to adjust the Vgs of the FET.

    Since the input to the FET is digital, I do not believe VGS needs to be adjusted. VGS can be 0V or -3.3V/-5V to turn ON or OFF and it doesn't need to be any intermediate value, assuming VGS(th) of FET is >-3.3V/-5V.

    When R1 and R3 are removed, the following two states occur. VGS(th) of -2V is assumed for the example below.

    State 0:
    OUTA = 0V
    VGS = -5V which is lower than VGS(th) and hence, Q1 = ON
    LED1 = ON

    State 1:
    OUTA = 3.3V
    VGS = -1.7V which is higher than VGS(th) and hence, Q1 = OFF
    LED1 = OFF

    Let me know if the above stated example is not clear to you.

    user5397266 said:
    On top of that, I am asking about the maximum rating of the reverse current.

    Assuming you are referring to the current back into OUTA pin when it is HIGH (3.3V), we recommend the current is <1mA to avoid any damage to device internal ESD bypass diode.

    Regards,
    Aaditya Vittal

  • 0 in reply to Aaditya V.
    Prodigy 230 points

    Hi Aaditya,

    +5V is an example.

    Also, there is a limit to the FETs that can be used, so the calculation must be based on the threshold I wrote.

    >we recommend the current is <1mA to avoid any damage to device internal ESD bypass diode.

    →　In other words, am I correct in assuming that as long as R1+R2 > 1.7kΩ = (5V-3.3V)/1mA, there is no problem?

    Best regards,

  • 0 in reply to user5397266
    TI__Genius 15625 points

    Hello,

    Thank you for your response.

    Yes, your assumption is correct. However, the ESD bypass diode is designed to protect against ESD transient current and not continuous current. Therefore, we do not recommend passing the 1mA current continuously through the ESD bypass diode.

    Regards,
    Aaditya Vittal

  • 0 in reply to Aaditya V.
    Prodigy 230 points

    Hi Aaditya,

    What is the value of the current that can flow continuously?

    Best regards,

  • 0 in reply to user5397266
    TI__Genius 15625 points

    Hello,

    The ESD bypass diode is not designed for continuous current use, so we do not have a continuous current value to share. If you would still like a value to be shared, then that would be 1mA. However, since we do not characterize the device for continuous current, we do not guarantee reliable operation of the device.

    Regards,
    Aaditya Vittal