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ISOW7742-Q1: Supply current calculation

Part Number: ISOW7742-Q1

We are using ISOW7742FQDFMRQ1 in one of our designs with the following configuration:

VDD=3.3V, VIO=3.3V, VISO=3.3V. INB & IND grounded, OUTB & OUTD not connected. We are using 2 channels to provide isolated data at a rate < 1Mbps and other 2 channels are static. The isolated power supply provides 3.3V 60mA load.

For this configuration, how do we calculate max 3.3V supply current for this device?

Power converter (Section 7.10 of datasheet):   VDD = 3.3 V, VSEL = GND2 ILOAD=60mA estimates max IDD=216mA

For VIO, VISOIN = 3.3V Section 7.14 of datasheet states max channel supply current - DC signal for VI=VCCI is 7.6mA for IDD_IO and 8.5mA for IISOIN. DC signal for VI=0 is 4.6mA for IDD_IO and 5.5mA for IISOIN

I assume this is per channel, so if 2 channels have VI=VCCI and 2 channels have VI=0, supply current for DC signal is (7.6+8.5)*2 + (4.6+5.5)*2 = 52.4mA. Please confirm.

Since 2 channels are switching AC the supply current for 2 channels will need to include supply current for AC signal i.e. (6.3 + 7.2) = 13.5mA for all channels switching at 1Mbps. So I assume it will be half this for our case with only 2 channels switching i.e. 6.75mA. Please confirm.

Total current drawn from 3.3V supply will be summation of power converter + DC Signal + AC Signal i.e. 275mA. Please confirm.

Also, is the channel supply current IDD_IO (assumed to be per channel) drawn from VDD as well as VIO? If answer is yes... We have a separate configuration where VDD is 5V and VIO is 3.3V. How will the stated IDD_IO be split between the two supplies?

  • Hi Kiran,

    Thank you for reaching out, please see my inputs below. Thanks.

    Power converter (Section 7.10 of datasheet):   VDD = 3.3 V, VSEL = GND2 ILOAD=60mA estimates max IDD=216mA

    That's right, VISOOUT supports a maximum of 60mA of load current in this voltage configuration. The associated max IDD current for this test condition is 216mA as you indicated. If your load is <60mA then the IDD current will reduce accordingly.

    I assume this is per channel, so if 2 channels have VI=VCCI and 2 channels have VI=0, supply current for DC signal is (7.6+8.5)*2 + (4.6+5.5)*2 = 52.4mA. Please confirm.

    The current into VIO and VISOIN specified in datasheet is the total current going into that pin under the specified test condition. Thus, each of this is a total current for 4 channels on their respective side. i.e., I(IO) (referred as IDD_IO in datasheet) is the total current consumption by VIO pin for all 4 channels on the left side of device while I(ISOIN) is the total current consumption by VISOIN pin for all 4 channels on the right side of device.

    Since we only have current consumption data provided for test conditions where all 4 channels are in the same state, you could take the average of DC and AC currents. But since these currents are much smaller thanpower converter current consumption, you might as well consider the larger one of the DC and AC values to accommodate for margin.

    In summary, you don't have to multiple the values by 2 and don't have to take two sets of values. Just one values for I(IO) and one value for I(ISOIN) should give you total current consumption by all 4 data channels.

    Total current drawn from 3.3V supply will be summation of power converter + DC Signal + AC Signal i.e. 275mA. Please confirm.

    Total current = power converter + [average of DC and AC I(IO) currents] + [average of DC and AC I(ISOIN) currents] = 216mA + 7mA + 7.9mA = ~231mA.

    Also, is the channel supply current IDD_IO (assumed to be per channel) drawn from VDD as well as VIO? If answer is yes... We have a separate configuration where VDD is 5V and VIO is 3.3V. How will the stated IDD_IO be split between the two supplies?

    IDD_IO listed in datasheet is only the current consumed by VIO pin and doesn't include current into VDD. IDD listed for power converter is the only current going into VDD. This should enable you to consider IDD for 5V and I(IO) for 3.3V.

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


    Regards,
    Koteshwar Rao