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CD4541B: Why do I need twice the frequency as the data sheet say ?

Dodonny
Prodigy 20 points
Part Number: CD4541B

Hell, Everything is in the title.

I'm working for very sensitive industry. I need to argue why in product I designed I have to introduce an x2 factor in the value of Rtc to get the good timing value.

For instance to get 25s with 270nF capacitor and 1024bit counter my resistor need to be 78K instead of 40k as the formula say. All my test on bench results in a doubled value of the resistor. When I plug my oscilloscope I need to set it to twice the frequency to get the correct timing.

Thanks in advance for your help.

  • 0
    TI__Guru* 87860 points

    Hi Donald,

    To get 25s from a 10-bit counter means the clock frequency is 40.96Hz, which is beyond the accuracy of the equation.

    Somebody in this forum said there is no lower bound limit to the oscillator frequency, so I assume this device can support 40Hz. To determine the appropriate RC becomes kind of trial and error game now. 

    FYI, below link is an example design using the equation.

    https://e2e.ti.com/support/logic-group/logic/f/logic-forum/871042/cd4541b-example-design

  • 0 in reply to Noel Fung
    Prodigy 20 points

    Thanks for your reply. But event if I set the system to 65536 count (16bits counter) with frequency above 1Khz the result is exactly the same. I still have to add a x2 factor to my equation. Plus, all my test are linear. They don't have a magic point where the equation is bad and until I reach a particular frequency. No matter the frequency the result is the same.

  • 0 in reply to Dodonny
    TI__Guru* 87860 points

    Hi Donald,

    I am afraid I don't have any idea why would that be, given the application circuit is super simple. 

  • +1
    Prodigy 75 points

    I think we need clarity with regards to what you mean by "to get 25s".  Are you looking for a single pulse (low->high, then high->low) that is 25 seconds between each edge, or are you looking for 25s between rising edges?  These two situations would differ in timing by a factor of 2.  In the first case, you actually want the output to be half the (1/25s) frequency as you are only looking for half of a cycle at the output - the actual frequency of the output would be 1/50s not 1/25s.  This would explain why you have to double the resistance.

    Also, 78K would set half the frequency not double.  Increasing R reduces F.

  • 0 in reply to Joseph Haas
    Prodigy 20 points

    Thanks for your reply.
    My issue root are because I'm using this Timer in MODE = 0. This mean I'll not output a full square wave divided by 2^n. But only the first part of the period (LOW or HIGH depending of Q/nQ pin). So yes this seems to be the cause of my misunderstanding. Thanks for your help.