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TLC5925: does the TLC5925 have proper power on reset?

Part Number: TLC5925

When the TLC5925 powers up, what state is the output latch? I would like to avoid the need to control the !OE pin with a uC and instead tie it right to ground, but I don't want the outputs to be on unexpectedly when the device powers up initially.

  • Hi, greyltc,

         Sorry for the delay. Your thread could not be opened yesterday. Proper engineer has been assigned to take a look at this now and will reply you soon.

  • Hi greytlc,

         OE pin is recommended to be used to avoid the LED being randomly illuminated immediately after power-up. You can test the device to confirm the result before connecting it to ground directly.

  • Unfortunately, I can't test the device because I don't have one, can you test it for me?

    Unless I've missed it, the datasheet seems to make no mention of what state the output latch resister is when the device is first powered up. Can you tell me what state it takes?

  • Hi Charles,

    Your link leads me to a page that shows

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  • Hi greyltc,

        Sorry. The thread is as following.

        Q:My customer is planning to use TLC5925 by connecting only CLK, SDI and LE signals to host MUC. \OE will be hardwired to GND. Is there any items to be considered when using TLC5925 with \OE hardwired to GND? What is the state of output pins after and during the power-up when \OE is tied to GND?

        Re: 

        Hi Tuomas, 

        If \OE is tied to ground, the TLC5925 will not be able to enter the special mode to update the reference current. This won't be an issue if the LED current is not expected to change.

      

      

  • Thanks Charles,

    That's great to know. I'm not planning on entering the special mode to update the reference current, so that's good.

    This does not seem to answer my question though. I'll re-write it for more clarity:

    What state does the 16-BIT output latch register take on power up?

    Will any of the outputs ever be on if the device is powered on with \OE tied to ground?

    Why doesn't the datasheet tell either: (A) tell me that the output latch will be cleared on power up or (B) warn me that it will be in an unknown state when the device is powered up?

  • Hi greyltc,

        Q1:What state does the 16-BIT output latch register take on power up? Will any of the outputs ever be on if the device is powered on with \OE tied to ground?

        Unfortunately, I don't have the device in my hands right now. However, similar device will have problem of being  randomly illuminated immediately after power-up. So if you want me to test it and confirm the result, please allow me more time to buy the chip and test it.

        Q2:Why doesn't the datasheet tell either: (A) tell me that the output latch will be cleared on power up or (B) warn me that it will be in an unknown state when the device is powered up?

        Normally, the OE/ pin is supposed to be used by customer to avoid those issues. I need to test and confirm the result in your application. If it is urgent, you can test it by yourself since it is simple.  It will take some time for me to buy the chip owing to the purchasing process.

  • Understood. Thanks very much for helping me find an answer to my question Charles. I would really appreciate it if you could physically do the test to resolve my question.

    Alternatively, is there anyone else you might know that's more familiar with the part or already has one that might be able to answer the question without having to actually go through the purchase and testing procedure?

  • Hi greyltc,

        I have applied for the device and I will test it for you. It will take about two or three days to get the result. Since it is an old device, we need to test it and confirm the result.

  • Hi greyltc,

         The test results are available now. I do hope they are still helpful to you.

         The following are the waveforms during power on stage of TLC5925. In the experiments, OE/ is tied to ground. As can be seen in Fig.1 and Fig.2, OE/ pin is always low level after power up, which is the situation in your application. In the Fig.1, LE is always low level and there is no current of each channel after power up. Hence, the LED will not turn on. In the Fig.2, LE pin is used to write 0xFFFF and 0x0000 to device in turn. As shown in Fig.2, the current of channel appears until the LE signal begins to toggle and new channel data is written to device. Fig.3 is the zoomed area of Fig.2.

        Therefore, there will be no current of each channel(the LED will not turn on) after power up until you use LE signal to write new channel data to device according to the experimental results.

        Any more questions ,please let me know.

    Fig.1

    Fig.2

    Fig.3

  • Thanks very much Charles for looking into this. It is still very relevant to me. To be very sure, could you please try the following (all with the ~OE pin tied to ground)?

    1. Show output current waveform as the device powers on (you did this)

    then

    2. Shift 0xffff into the device

    3. Pulse the LE pin ( all outs now on)

    4. Remove power from the device

    5. Wait 30 sec

    6. Power the device up ( I assume the device will show no output current after its powered up )

    7. Pulse the LE pin ( very curious to know what I_out shows here after this in order to know if the registers have been truly set to 0x0000 by the POR )

    It looks like you've shown me 1 - 3, could you please continue by exploring 4-7 for me?

  • Hi greyltc,

         I will test it for you according to your requirements.

  • Hi greyltc,

         The following are the waveforms of device, which are tested according to your requirements. Fig.2 and Fig.3 are the zoomed area of Fig.1.

         Any more questions, please let me know.

    Fig.1

    Fig.2

    Fig.3

        

        

  • Hi greyltc,

         Do you have more questions? I want to confirm whether you have addressed your problem or not.

  • Hi greyltc,

        I'll close this thread due to inactivity and if you have more questions please reply to it directly and it will open again.