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monisha's avatar
monisha
Icon for New Contributor rankNew Contributor
4 years ago

max 10 GPIO VOH and VOL @IOH=8mA

Hi,

I plan to use 10M25DC FPGA for to drive 7mA load @3.3V. As per MAX 10 General purpose I/O user guide, 3.3VLVTTL only have capability to drive 8mA(3.3V LVCMOS is 2mA maximum). Kindly correct me if I am wrong also reply below queries.

1.For IOH=-8mA,What will be the VOH(min) and VOL(Max) values?For 4mA,Voh(min)=2.4V, Vol(max)=0.45V is found from datasheet. whether this value can i use for IOH=-8mA?

2.Also the logic level 3.3V LVTTL and 3.3LVCMOS can i set for induividual pins?Or for each bank do i need to set either LVTTL or LVCMOS?

Kindly consider this as urgent and reply.

Thanks,

Monisha OM

3 Replies

  • Ash_R_Intel's avatar
    Ash_R_Intel
    Icon for Regular Contributor rankRegular Contributor
    4 years ago

    Hi,

    IOH and IOL values that in the datasheet are not max values, but recommended ones and are based on JEDEC standards.

    For your application, you can use 3.3V LVTTL, 8mA setting in Quartus.

    VOHmin and VOLmax remains same for all the IOL and IOH settings.

    3.3V LVTTL and 3.3LVCMOS levels can be set for individual pins in a bank.


    Regards.


  • Ash_R_Intel's avatar
    Ash_R_Intel
    Icon for Regular Contributor rankRegular Contributor
    4 years ago

    We do not receive any response from you to the previous answer that I have provided. This thread will be transitioned to community support. If you have a new question, feel free to open a new thread to get the support from Intel experts. Otherwise, the community users will continue to help you on this thread. Thank you


    • jasmel's avatar
      jasmel
      Icon for New Contributor rankNew Contributor
      2 years ago

      Hello Ash_R_Intel.

      We have a major concern because with the LVTTL for the MAX 10, you DON'T follow the Jedec JESD8-B for the VOL max of 0.40V. Your datasheet claim a higher value of 0.45V at 4mA.

      The Jedec JESD8-B and JESD8-C request a VOL (max) = 0.40 at IOL 2mA.

      Can you explain?

      Regards,