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LSF0108-Q1: Regarding pull-up resistor values for low frequency applications.

Part Number: LSF0108-Q1

Hello TI Support,

The datasheet shows good depiction of translation with high frequency signal. However, information about low frequency signals & pull-up sizing is missing. Pertaining to that I will appreciate answers to below queries,

1. Can we use 3.3K to 10K pull-up for applications having signal frequency< 100KHz without compromising signal integrity for applications requiring 1.8V--> 3.3V conversion?

2. Is there any effect on Ron of LSF0108-Q1 if higher resistances are used? If yes, can TI provide the curve of pull-up values Vs RON of LSF0108-Q1?

3. Can we reduce pull-up from Vrefb to EN from 200K to 10K or so due RF immunity concerns. Our product undergoes much high levels of RF immunity & hence weak pull-ups can create issues. Please advice.

4. It would be nice to get some data of pull-up resistor values Vs signal frequency it will support. I would recommend to have this data in datasheet. Will resolve most queries from customers.

Regards,

jagdish

  • Ron is independent from the pull-up resistors.

    The 200 kΩ resistor limits the current that can flow into Vref_B and out of Vref_A. A higher current will not affect the operation of the translator. (If you do not need the feature that allows you to omit the pull-up resistor on the low-voltage side when downtranslating from B to A, then you can use a different circuit without the 200 kΩ resistor.)

    The pull-up resistors and the parasitic capcitances form a low-pass R-C filter. The achievable speed depends on the cut-off frequency of that filter. So it is not possible to specify a frequency without knowing the capacitances, which depend mostly on the trace lenghts and the number of other devices connected.

  • Considering node & other parasitic capacitance to 50pF, can we now use 3.3K to 10K pull-up for applications having signal frequency< 100KHz without compromising signal integrity for applications requiring 1.8V--> 3.3V conversion?

  • The cut-off frequency would be 1 / (2 × π × 10 kΩ × 50 pF) ≈ 300 kHz. A 100 kHz sine wave would pass through unmolested, but a square wave needs some overtones and would lose its sharp edges. For 100 kHz, 3.3 kΩ would be better than 10 kΩ.

  • This helped. Thanks for your comments & support.

    Regards,

    jagdish