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LM26480-Q1: LM26480-Q1 nPOR & pool up resistor

Part Number: LM26480-Q1

Hi TI team,

I have some questions regarding the LM26480-Q1 : 

- What is the maximal current that can support the open drain logic output nPOR when it is pulled to ground? it is not specified in the datasheet.

-I  use this signal to enable other component as serilalizer, is it allowed connecting a capacitor C1 for emc filtering, and what is the maximal value of the capacitor ?

-In the data sheet, the resistor R1 100K is used as pull-up, can I use 10K connected to 1.8V?

Kind regards,

Florent 

  • nPOR has a VOL value specified on page 8 of the LM26480-Q1 datasheet:

    Output level low (with load condition, IOL = 500 µA), VOL = 0.23 V typical, 0.5 V maximum.

    If VIO = 3.3V, 3.3V/100kΩ = 33 µA, which is within the specification.

    If VIO = 3.3V, 3.3V/10kΩ = 330 µA, which is also within the specification. It is OK to use a 10kΩ resistor,

    Smallest resistor allowable (worst-case): 3.3V/500µA = 6.6 kΩ

    Assuming the resistance through the FET is static, then Rds,on = 0.5V/500µA = 1 kΩ (max) then the time constant of discharge will be   = Rdson*C = (1 kΩ)*C and the time constant of charging will be   = Rpu*C where Rpu is the value of the pull-up resistor you have selected.

    You will need to determine what is an acceptable rise time and fall time, which will determine the maximum capacitance you use for your EMI filter.

  • Hi Brian,

    Thanks for your answer make sense to me now ! 

    Regarding the load condition, do you have more info about how it was tested ( the load itself, which permits to have 500uA max ) ? 

    Kind regards,

    Florent

  • Florent,

    For most I/Os, it is common to test the VOL (and VOH for push-pull outputs) by testing at the load current for which the I/O was designed to handle and measure the output voltage.

    For this test, the nPOR signal is forced to pull low (regulated outputs not within regulation), which is the initial condition of this I/O. So during start-up, nPOR pin pulls low and a current source with Test Condition, IL = 500 µA, is applied at this pin. The source-meter will force 500 µA into the pin and measure the voltage. The voltage must be below 0.5V for the test to PASS. This ensures that the Rds,on of the internal open-drain FET is within spec and will pull low (VOL<0.5V) with the recommended pull-up resistor used.

    It is not a measurement of the maximum load current, just a way to verify that VOL will be low enough to meet the target value. This measurement is to ensure Rds,on is low enough, but no one cares about the measured value of Rds,on. The only spec which matters is VOL, which must be lower than VIL for the input pin the nPOR connects to on the processor or other IC.

  • Hey Brian, 

    Thanks for your quick answer.

    Another question : The capacitor C1 will initially be loaded to 1.8V.

    When the transistor is active, the capacitor is initially loaded to 1.8V, and will be discharged through the Rdson (1KOhm) of the transistor.

    The peak current through the transistor is 1.8V/1K=1.8mA during a time constant Rdson x C1.

    Is the peak current 1.8mA is acceptable? , if yes how long time? This will define the maximum value of the capacitor.

    Kind regards & thanks for support,

    Florent

  • I am not concerned with 1.8mA going through the FET. Anything in the single-digit mA range is reasonable for an I/O pin.

    Your only concern should be the timing of the RC constant and how that may impact your rise and fall times for the logic signal: Some logic inputs have a maximum slew rate. You want to make sure you do not violate this condition, if it is present in your system.

  • Hello Brian;

    "I am not concerned with 1.8mA going through the FET. Anything in the single-digit mA range is reasonable for an I/O pin", on what are you based on to affirm this ?

    What commonly happens if the slew rate if violated then ? 

    Thanks for support ! 

    Kind regards,

    Florent

  • Florent,

    Well, I2C is a good example, but not the problem you would face here: For I2C (or any other clocked communication protocol), the data must be valid before the data-valid clock edge (rising or falling), which is called the setup time. It must also be valid for a time period after the data-valid clock edge, which is called the hold time. If the pull-up resistors are too weak, or there is too much capacitance on the bus and a weak open-drain FET (high rds,on) then the rise or fall time for data may be too slow and it will not be ready when the data-valid clock pulse occurs (setup time violation).

    It is difficult to think of another example, except for the most obvious one which is edge-triggered flip flops, a type of logic circuit that is notorious for creating race conditions

    Put simply: unless the nPOR ouput pin of the LM26480-Q1 is tied directly to an nPOR input (or enable, EN) pin of another IC that is physically tied to the POR block in that IC, it would be a best practice to confirm that the other IC does not have any risk related to the slew rate of this signal.

    Please keep in mind that this question is no longer related to the LM26480-Q1 device itself and there is very little TI can do to provide reassurance that the other IC at the other end of nPOR will behave as expected.