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ISO1228: When an external load resistor is connected to the field side

TAGA KOUKI
Prodigy 30 points
Part Number: ISO1228

Tool/software:

I hope this message finds you well.

We are currently considering the adoption of the ISO1228. In the circuit we plan to use, an input current of over 10mA is required.

However, according to the datasheet, the maximum input current for the ISO1228 is specified as 3.5mA. This presents a challenge in using the ISO1228 as intended.

To address this issue, we are contemplating the addition of a load resistor on the input side to divert the current accordingly.

Questions:

Is it acceptable to attach a load resistor to the ISO1228? The circuit in question is similar to the one depicted in "Digital Input Module, Digital Output Module, Figure 6."

If it is acceptable, can the VIH and VIL calculations be performed using the method described in the datasheet?

We would greatly appreciate your prompt response to these inquiries.
Best regards,

  • +1
    TI__Expert 7325 points

    Hi Taga,

    Thanks for reaching out and detailed inputs.

    The above Solution will work absolutely fine to leak the extra current.

    The VIH/VIL Calculation will also remain same as mentioned in ISO1228 Datasheet

    Regards
    Varun

  • 0 in reply to Varun Kumar
    Prodigy 30 points

    I hope this message finds you well. Thank you for your prompt response to our previous inquiry.

    The information you provided has been very helpful in our consideration of adopting the ISO1228.

    However, we have one additional question that we would like to clarify: Regarding the input of DC24V, if an external load resistor (Rload) of 1kΩ is attached to the ISO1228, is the following calculation for the current flow correct?

    The current through Rload: I(Rload) = DC24V / 1kΩ = 24mA The current flowing into the ISO1228 (Rthr and RPAR) is set to IL = 2.5mA as specified in the datasheet, meaning 2.5mA flows into Rthr.

    Therefore, the total current flowing from the DC24V source would be 24mA + 2.5mA = 26.5mA.

    We would greatly appreciate your clarification on this matter.

    Thank you for your continued assistance.

    Best regards,

  • +1 in reply to TAGA KOUKI
    TI__Expert 7325 points

    Hi Taga,

    TAGA KOUKI said:
    Therefore, the total current flowing from the DC24V source would be 24mA + 2.5mA = 26.5mA.

    Yes, This is right.

    The 1kOhm Load Resistor and (RTHR+RPAR+ISO1228) can be considered as Parallel Networks.

    Regards
    Varun