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SN6501: how many current the device can handle with the transformer for 3.3v input 5v output

Part Number: SN6501
Other Parts Discussed in Thread: SN6505B

Hi team,

my customer wants to use the SN6501 as the isolation power supply. We want to check how many current the device can handle with the transformer(PM2180.012NLT)  for 3.3v input 5v output. i have checked it and the transformerP759.pdf can be used. Do we have some excel or formula to do the calculation?

Thanks.

  • Hi Frank,

    On the primary side of the transformer, SN6501 can sink up to 350mA at Vin = 5V. This is reflected on the secondary side as an inverse relationship with turns ratio, so maximum load current, Iload max, is calculated as Iload max = Iprimary / N.

    At Vin = 3.3V, the max current SN6501 can sink on the primary side is 150mA. PM2180.012NLT has a turns ratio N = 2/1, so Iload max = 150mA / 2 = 75mA. This equation is not included in the SN6501 datasheet since manufacturing tolerances of components in an isolated power supply combined may result in a lower number than the equation yields, and the greater concern is accounting for sufficient turns ratio as shown in SN6501 datasheet equation 10:



    If our customer's expected load current is very close to this 75mA limit or higher, please encourage them to use an SN6505 device like the SN6505B.


    Please let me know if there is more information I can assist with.


    Thank you for posting to E2E!
    Manuel Chavez

  • Hi Frank, all,

    I updated the post above to include that SN6501 can sink 150mA of current on the isolation transformer's primary side at Vin = 3.3V and 350mA of current at Vin = 5V operation.


    Respectfully,
    Manuel Chavez

  • Hi Manuel,

    Thanks for your reply. Why the 3.3v input current is 150mA, 5v is 350mA. I used to think it means output load current. And the datasheet list the 100mA(3.3v to 5v),

    If the device can only handle 75mA load current, why datasheet give 100ma curve? 

    Thanks.

  • Hi Frank,

    You're welcome! SN6501's datasheet shows 150mA for 3.3V operation and 350mA for 5V operation limits as current maxima for the device to sink on the transformer's primary side. A section of page 1 is shown below:



    How much of the primary-side current is available for the load depends on the transformer turns ratio. For an ideal transformer, Power in = Power out, meaning there is a tradeoff where voltage and current have an inversely proportional relationship. It is not possible to sink 150mA on the primary side of a transformer at 3.3V and at the same time draw 150mA from the secondary side at 5V.

    The load current curves on page 1 of the datasheet are for Würth Elektronik's 760390013 transformer, which has a turns ratio of N = 1.7. With this turns ratio, current available to the load at 3.3V operation is Iload = 150mA / 1.7 = 88mA. This curve extends up to 100mA because the SN6501 has an absolute capability to sink 500mA on the primary side, although operating the device under this condition is not recommend.

    I apologize since this graph is misleading. I will work with our team to update this datasheet.

    Will higher currents be necessary for your system?


    Thank you,
    Manuel Chavez