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CD4059A: clock divider

Pankaj Jha81
Intellectual 320 points
Part Number: CD4059A
Other Parts Discussed in Thread: SN74HC4060, SN74LVC1G80

 I have a 16kHz clock. I want to divide it by 2^15 or 32768. The high voltage is 1.8V and low voltage is 0V. Do u have a IC that can do this ???

  • 0
    TI__Mastermind 24665 points
    Hello,

    It looks like the best solution available would be to use the SN74HC4060 paired with a single SN74LVC1G80.
    The SN74HC4060 will give a 2^14 division and then the SN74LVC1G80 will divide the clock one more time to get 2^15 division. The application diagram in the datasheet shows how to do the divide by 2 with the SN74LVC1G80.

    www.ti.com/.../sn74hc4060
    www.ti.com/.../sn74lvc1g80.pdf

    I hope that this helps, please let me know if you have additional questions.

    Best,
    Michael
  • 0 in reply to Michael J Schultis
    Intellectual 320 points

    Dear  Michael,

    Thanks for your reply.

    Do the ICs you've mentioned also meet my criteria for high and low voltage, i.e. , if I input a 16kHz clock with a high voltage of 1.8V and a low voltage of 0V, will the ICs output a clock pulse of high voltage of 1.8V and a low voltage of 0V.

  • 0 in reply to Pankaj Jha81
    TI__Mastermind 24665 points

    Hello,

    You are absolutely correct, the device I mentioned previously actually will not work at the 1.8V spec you need.

    1.8V is a little bit more difficult to be able to acheive. While these solutions are not optimized, they seem to be the best options we have.

    1. Use 4x http://www.ti.com/lit/ds/symlink/cd74ac161.pdf

    This will do the frequency division in stages. 

    2. If you can take the original solution that I provided, but supply the HC4060 device with 2V, then it will actually be able to "understand" an input voltage of 1.8, and output a 2V signal. I understand this is kind of an odd solution, but it is a possibility. You could then power the second stage SN74LVC1G80 with a 1.8V supply and the inputs would be overvoltage tolerant to 5.5V, and would then down translate the output voltage to the 1.8V level that you need.

    Best,
    Michael