• State Resolved
  • Locked Locked
  • Replies 4 replies
  • Subscribers 22 subscribers
  • Views 512 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

LMK04828: LMK Loop Filter Components

LMK04832: Dual loop mode and OSCin_FREQ register programming problem.

Anon Nymous
Prodigy 30 points
Part Number: LMK04832


Hello.

I'm working on a clock generator based on LMK04832. Device spec is:

- dual loop mode with holdover enabled

- PLL1: CLKin1 with 10MHz input and 10MHz output

- PLL2: OSCin 10MHz input, using internal VCO1 with 3200MHz, and 400MHz output

When I'm trying to program required frrequency into register OSCin_FREQ nr 0x162 (value 0 for 0 to 63MHz freq. as stated in the datasheet) , the PLL2 is not locking. If i set default value - it's locking.

Is there any reason for such behavior?

https://e2e.ti.com/cfs-file/__key/communityserver-discussions-components-files/48/TICSPro_5F00_PLLatinumSim_5F00_cfg.7z

  • 0
    TI__Mastermind 28560 points

    Hello,

    Is the 10 MHz a square wave?  Is the slew rate meeting the OSCin requirements of min 0.15 V/ns?

    73,
    Timothy

  • 0 in reply to Timothy T
    Prodigy 30 points

    Hi Timothy,

    10 MHz is a sine wave (from IQD IQO-164-4 1833RC).  We are using single-ended source as in Figure 24. in datasheet.

  • 0 in reply to Anon Nymous
    TI__Mastermind 33840 points

    Hello,

    The slew rate of a 10 MHz sine wave is given by 2*pi*F*Vpk. In order to satisfy the minimum slew rate requirement of 0.15 V/ns, your signal amplitude must be 0.15 * 10^9 / (2 * pi * 10^7) = 2.3 Vpk, which is 4.6 Vpp and which exceeds the input voltage absolute maximum ratings for the OSCin pin. It is not possible to meet the 0.15 V/ns requirement with a single-ended 10 MHz sine wave. Note that with a differential input, each pin can see half the amplitude (2.3 Vpp) while achieving the same slew rate. Therefore, TI recommends differential input with sine wave signals at low frequencies.

    Although there may be some side effect of the default OSCin frequency range selection which allows lock to a 10 MHz sine wave signal, this is not characterized behavior. There should be no expectation that it will always work.

    Regards,

  • 0 in reply to Derek Payne
    Prodigy 30 points

    Thanks Timothy for suggesting direction of problem source and thanks Derek for clarification.