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Altera_Forum's avatar
Altera_Forum
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17 years ago

Ris/fall time?

Hello,

I am developing a simple test board for a cyclone FPGA (EP2C8T144). It is my very first FPGA project, and I want to build my own FPGA board.

Now, my question is:

Which rise and fall times can I expect on any I/O pin of the FPGA? I am not using LVDS. I just use simple 3.3V LVCMOS.

The problem is, that there will be reflections on longer tracks on the PCB, if the rise time is too short. How can I calculate the maximum allowed track length if I know the driver strength of the I/O pin (4 mA)?

Hope, somebody can help me! This would be very great.

Thanks a lot in advance and excuse my bad english (I am a student in Switzerland) :D

11 Replies

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    I suggest you to avoid connecting the FPGA output pins to input pins of the device that you are driving. Put a small series resistor (22ohm or 33ohm) in every track and you will avoid this kind of reflection problems independently of the track length.

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Hello OrchestraDirector,

    thanks for your answer. The device which I am driving with the FPGA is an ARM7 microcontroller (the FPGA is attached to the micro's memory bus).

    How long can I make the trace length, if I don't put any resistors in the tracks? Longest track is ~25 mm. Will this work if I program driver strength to minimum? I don't need very fast I/O (I'm working with low frequencies).
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Hi, if your frequencies are low enough (let's say under 50MHz) you can go on without series resistors also with 25mm, with higher frequencies a resistor array of 22ohm is recommendable.

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Dear OrchestraDirector,

    thank you a lot! My ARM works with 72 MHz, but the external data bus is slower. So, I think this could work ;) Is it maybe possible to use longer tracks than 25 mm?

    And is it correct that only the Pins, which are outputs on the FPGA, need a series resistor?

    Thank you for your help!
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Yes, correct, but be careful to place the resistor array next to the destination pins, which are in your case the microcontroller input pins.

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Okay. What if some pins are bidirectional (like the data bus from the micro to the FPGA) ? place the resistors anyway near the microcontroller?

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    If you can, it is advisable to use a separate resistor array for the unidirectional signals (near the destination pins) and a resistor array for the bidirectional signals, which can be placed wherever you like between the pins.

  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    Some suggestions haven't been clear to me, so I like to mention, what I would do.

    Basically I use a series resistor a the driver, that means at both end for a bidrectional signal. On chip series termination respectively selecting a low current strength (another name for the same thing) with FPGA can save external resistors in most cases.

    A typical single ended trace impedance on a multilayer PCB is in the 50 to 90 ohms range. You can use calculation tools to determine the nominal impedance for your board. So if you have a digital output with high current strength (e. g. a logic device with 10 to 20 ohm output impedance) a series resistor of 33 to 68 ohm would be a suitable choice.
  • Altera_Forum's avatar
    Altera_Forum
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    17 years ago

    @FvM and vernmid:

    Thank you very much!

    That's exactly what I wanted to know :-)

    And this pdf article looks interesting.

    Hmm, does Cyclone II support on-chip termination? I found this feature only in Cyclone III devices.