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SN74LVC2G07: Pull-up resistor calculation for open-drain output

Part Number: SN74LVC2G07

Hi,

Customer plans to use SN74LVC2G07 with output connected to a input of a chip with Vih(min)=2V.  The output of SN74LVC2G07 output will be pulled up through the pull-up resistor to 3.3V.

By checking Choosing an Appropriate Pull-up/Pull-down Resistor (SLVA485), customer is asking the value of Ipull-up, but I can't find the corresponding parameter in SN74LVC2G07 datasheet.  What is your comment here?

In parallel, I'm cross checking SCEA035A, and in page 8 it mentioned that max pull-up resistor should be chosen considering the rise time of the output signal.  Will you also let us know how to decide the range of pull-up resistor value for such application scenario?

Thanks!

Antony

  • Hi Antony,

    I'm going to assume a supply voltage of 3.3V for the SN74LVC2G07 since you didn't give me one.

    In the Electrical Characteristics table, there are two V_OL values that could apply -- one for 16mA and one for 24mA.  I'm going to use 16mA for my example, since I seriously doubt that they will need 24mA to do what they want.

    If you want to know the absolute smallest resistor that can be used, then you just ask our old friend Ohm:  R = V/I = (3.3-0.4)/0.016 = 181.25 Ω

    Unless they just need incredibly fast edges or want to waste some excess power, it would be inadvisable to use a 181.25 Ω ohm resistor as a pull-up.

    To determine the largest possible value, we need to know the capacitance on the line and the worst-case edge rate that can be supported.  I'm going to assume that they are connecting to a modern CMOS device's input and the trace isn't excessively long, so we can use 10pF as the load and 330ns as the worst-case rising edge time (100ns/V).

    The time constant for the capacitive charge curve is R*C, and 3*R*C should be close enough to the 10% to 90% rise time for our purposes.  Putting this together I get the equation:

    t_rise = 3*R*C --> rearrange to solve for R to get: R = t_rise/(3*C) = 11 kΩ

    So there's your range of pull-up resistors that should work -- 181.25 Ω to 11 kΩ.  Personally, I would choose 10 kΩ since it's basically what everyone uses, is cheap, and widely available, but they might want slightly faster edges.

  • Hi Emrys,

    Thanks for the clear explanation. The further customer want to ask is the minimum output voltage when output is at high level. Based on the application scenario you described above, please let us know how to calculate the minimum output voltage level when output is high. Which should be Voh(min), right?

    Thanks!

    Antony
  • Our device does not drive the line high (it's open-drain), the primary concern would be leakage current.

    Let's say we use the 10 kΩ resistor I suggested above, and the leakage is absolutely awful at 20uA (our device is rated to 10uA across process variation, voltage, and temperature, but I don't know what else is on the line).

    In this case, we use Ohm's law again to calculate the voltage drop:

    V_OH = V_CC - 20uA * 10 kΩ = 3.3V - 0.2V = 3.1V

    It's important to note that I took the worst possible case I could think of for this value -- it's more likely that the leakage is going to be on the order of nA.