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PCA9306: Question about PCA9306.

Pokkun
Genius 4695 points
Part Number: PCA9306

Hi, team.

I have any questions about PCA9306 from customer.

＜Question＞
１.Is the description of Figure 10 on Datasheet P13 correct?

Because, the explanation of the red line is "VDPUX≤2V", but the waveform line in the range of 2.0 to 5.5 V.
(Black line is the reverse)

２.To calculate the bus capacity, measure the rising waveform of SCL and SDA, is it calculating to use the formula "tr = t2 - t1 = 0.8473 × Rp × Cb" in the slva689 document ?

Example：When the rise time is 1us, the pull-up resistor is 10 kΩ.
"Cb = 1*10^-6 / (0.8473*10000) ≈ 118pF"

Best regards,

Masumi Sekiguchi

  • 0
    TI__Prodigy 280 points

    Hi Pokkun, you may relay the following information to our customer.

    1. The graph is incorrect here. The red line should be valid for VDPU >=2V. 

    2. In order to best help, we may need some clarifying information such as:

    - Are the measurements of the waveform taken from 0.3*VCC to 0.7*VCC or somewhere else?

    - Was the PCA9306 enabled or disabled during the measurement?

    - What sort of equipment was used to perform the measurement?

  • 0 in reply to Paul Gomez - TRX
    Genius 4695 points
    Thank you for the response.

    I heard below to customer..

    > • Is the measurement of the waveform "0.3 * Vcc ~ 0.7 * Vcc range" or "other range"?

    →The rising waveform is acquired, at the condition of "0.3 * Vcc ~ 0.7 * Vcc".

    > • Whether the status of PCA 9306 under measurement is "enable" or "disable"?

    → It's "enable" state.

    > · What kind of equipment was the measurement carried out?

    →Measured with an oscilloscope.
    ※ Do you have necessary information such as measuring equipment information?

    Best regards,
    Masumi Sekiguchi
  • 0 in reply to Pokkun
    TI__Prodigy 280 points

    Taking the measurement at 0.3*Vcc and 0.7*Vcc is the correct method. However, note that the equation tr = t2 - t1 = 0.8473 × Rp × Cb assumes using the correct levels and also that you decouple the two sides of the PCA9306. In other words, the equation assumes just the Rpullup and Cbus capacitance in series. With the PCA9306 enabled, you have a variable rsdon across the device, which in turn realizes the Cbus capacitance on side 1 and 2 in parallel. Please try to disable the device and then take the rise time measurements again so that the direct equation can be used. Also note that the scope probe will add some capacitance to the measurement as well.

    Regards,

    Paul

  • 0 in reply to Paul Gomez - TRX
    Genius 4695 points
    Thank you for contacting.

    Just to be sure, confirm me.

    ＜Confirm＞
    Is it okay for them to understand that the capacity of Cb increases?

    That is,Is it OK with understanding below?
    ・Measure with "enable state + probe capacity" included.
    ・Read the graph of Cb value calculated from tr.
    ・There is able to consider "no problem" if the currently used Rp is less than the maximum resistance on the graph.
    (disable state,Considering the probe capacity, if Cb becomes smaller, the allowable resistance becomes larger)

    Best regards,
    Masumi Sekiguchi
  • 0 in reply to Pokkun
    TI__Prodigy 280 points
    -Please measure with the PCA9306 disabled and calculate each Cbus independently.
    - using an Rp< Rpmax is ok and based on the equation, and yes once you know your Cb, you can play with the Rp value. Note that a larger Rp will increase your rise time, but a lower Cb will decrease your VOL.
  • 0 in reply to Paul Gomez - TRX
    Genius 4695 points

    Thank you for your reply.

    Measured in the disabled state, I understood from the result that Cbus calculated is a correct numerical value.

    However, it seems that it is difficult for customers' design circuits to disable the IC and verify it.

    So, is it OK that like following recognition ?

    ＜Recognition＞
    (From slva 689 Figure 3)
    Using the premise that there is no problem if it is using at the maximum Rp or less, If Cbus measured with enable is used, it becomes larger than the proper value, it can be said that it is a worst condition.

    So that, if it culculate about Cbus with considering "Rp <Rp (max)" on the premise of that IC's enable state, No probrem.

  • 0 in reply to Pokkun
    TI__Prodigy 280 points
    If possible, is the customer able to unsolder the device?

    Otherwise, the equation will be a similar approximation of the Rp, Cb. To simplify, say that when Vgs >Vth, Rdson~0ohm. Essentially that means that Rp1//Rp2 and Cb1//Cb2. But yes, Rp and Cb are still inversely proportional, so having your "worst case" Cb will mean your Rpmax simply decreases, satisfying Rp<Rpmax.
  • 0 in reply to Paul Gomez - TRX
    Genius 4695 points
    Thank you for the answer.

    >If possible, is the customer able to unsolder the device?
    ⇒I told you in response to this customer, but it may not be able to be verified due to the schedule of the development schedule.

    Also, we gather the contents that we have responded so far, and the customer seems to proceed with the design with the contents below.
    (This issue is "CLOSE", at below contents.)

    <Customer opinion>
    It's no probrem to proceed with the design with Rp (max) which is measured with "enable state + probe capacitance included" and assumed Cb larger than the ideal state.