• State TI Thinks Resolved
  • Locked Locked
  • Replies 2 replies
  • Answers 1 answer
  • Subscribers 21 subscribers
  • Views 832 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.

LMX2541 - sine wave input to generate 100M sine wave output

Elad Segalis1
Intellectual 760 points
Other Parts Discussed in Thread: LMX2541

Hi team ,

There is something which is not clear from the d/s - on page 19 OScin and Fin sensitivity chart

Does the section in grey is the recomended or all the area between the black line

My cutomer wishes to operate the LMX2541 with 10M sine wave (5-11dBm )  input to generate 100M sine wave output

Is this can be done ?

Thanks

 Elad 

  • 0
    TI__Guru** 110530 points
    Hi Elad,

    10MHz sine wave OSCin is okay as long as its amplitude is high enough, 5 to 11dBm is fine.
    as for the output, as it is divided down from the internal VCO, the output format is a square wave.
  • 0
    TI__Mastermind 28021 points

    Elad,

    The grey area is what you should go off, not the black lines.

    Take the typical performance plots as just typical performance.  On the black line is where the device typically fails, so you certainly want to stay away from that.   We do try to show how a device typically varies, but it does not account for all possible process variations. 

    On the other hand, the electrical specifications are guaranteed.  We guarantee a minimum slew rate of 150 v/us and a maximum Fosc voltage of 2V.

    Now 2V is about 10 dBm and this is why the grey area tops out around 10 dBm.   The graph is intended as a courtesy, but the guarantee comes from the electrical specs.

    If we consider a sine wave of the form A*sin(2pi*f*t), then the slew rate at the zero crossing can be calculated from the derivative of A*f*2pi*cos(2*pi*f*t) (t=0) = A*2pi*f.

    So if f=10 MHz and the minimum slew rate os 150 V/us, this works out to A=(150 V/us)/(10MHz*2*3.1459) = 4.76 V pp, which violates the max value of 2.

    If we start with V=2Vpp (A=1V) and a slew rate of 150 V/us, I calculate:

    f = (150 V/us)/(2*3.1415*1V) = 23.9 MHz, which looks like the minimum frequency for the grey area.

    Regards,

    Dean

    If we consider f=25 MHz and a 2V p-p imput (1V amplitude), this works out to a slew rate of 1V*(25 MHz)