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ISO1042: Output leakage current in high Z

Part Number: ISO1042

Hi Experts,

Good day

The power-off bus input leakage current Ioff(lkg) is given at 4.8µA max. The test condition is not cleared: CANH = CANL = 5 V, VCC2 to GND via 0Ω and 47 kΩ resistor.

Is it really with 0V across CANH to CANL, or CANH = 0 and CANL = 5 V for instance?

About the leakage current test, it is still not clear. Which voltage do you apply to measure 4.8µA of leakage current for the Ioff(lkg) parameter across CAN output? Could not be 0V.

Zout = Vout / Ioff(lkg) = (CANH-CANL)/Ioff(lkg) . In the case of CANH=CANL=5V, vout=0. I will expect to apply 5V and to measure current for instance.



Regards,

Josel

  • Hi Joselito,

    Thanks for reaching out.

    Please allow us to review your question, consult the appropriate team members and come back to you. Thanks.


    Regards,
    Koteshwar Rao

  • Hi Josel,

    Apologies for the delay in response.

    The test condition is not cleared: CANH = CANL = 5 V, VCC2 to GND via 0Ω and 47 kΩ resistor.

    This states that CANH and CANL are shorted together and then applied with 5V through 0Ω resistor and checked for current into the CAN pins. This is repeated again with 47kΩ resistor in place of 0Ω and checked for current into the CAN pins. The maximum of these two currents is the leakage current into CAN bus pins. During this test, VCC2 is made 0V or shorted to GND2.

    Is it really with 0V across CANH to CANL, or CANH = 0 and CANL = 5 V for instance?

    Having CANH = 0V and CANL = 5V would result into valid operation mode, the current during this operation can't be referred as leakage. Leakage current is the current that can pass into device when device is unpowered. For these kinds of test, typically all pins of relevance are shorted and measured for current.

    Let me know if you have any further questions, thanks.


    Regards,
    Koteshwar Rao