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LMX2491: Reference schematic diagram

user4779049
Expert 2435 points

Part Number: LMX2491

Hi Team, 

Can you advise me on my question, related to the LMX2491 reference schematic diagram.

Referring it on the data sheet, the output from the external VCO is connected to the FIN pin,

through a 50 ohm power divider and a -6dB attenuator.

The 50 ohm power divider consists of three  18 ohm resisters connected as star topology,  

the -6dB attenuator composed with two 18 ohm resisters and one 58 ohm resister as T shape 

topology.

I think that a 50 ohm resister is necessary on the FIN  pin, to make them work as intended.

If it is true, the 51 ohm resister on *FIN may be replaced with a 51/2 ohm resister to make 

the circuit symmetry. 

 

What do you think of it?

The second question:

If power divider and attenuator make the VCO output totally -12 dB, the power delivered to the

PIN pin is -11 to -8 dBm, because the VCO output power is 1 to 4 dBm.

In contrast, the expected power level of FIN input is -6 to 6 dBm.

How do you deal with this mismatch?

 Mita

 

  • 0
    TI__Guru** 121230 points

    Hi Mita-san,

    Can you post the schematic you are referring to? The application diagram in the datasheet has a different configuration than what you have described.

  • 0 in reply to Noel Fung
    Expert 2435 points

    Noel-san,

    Here is the schematic diagram.

    It is on the first page of the datasheet.

    Please focus on the  part surrounded with orange line.

    I suspect that circuit in green box is power divider and 

    circuit in the blue box is attenuator.

    It needs 50 ohm termination on its output. 

    Mita

  • 0 in reply to user4779049
    TI__Guru** 121230 points

    Hi Mita,

    The resistors are used to split the VCO signal and at the same time, ensure the VCO and Fin can see a 50Ω load.

    The VCO will see a load of 18 + (18 + 50)//(36 + 68) = 59Ω

    Assuming the VCO output impedance is 2Ω, FIN will see a load of {[(18 + 50)//(2 + 18) + 36]//68} + 18 = 47Ω.

    Assuming VCO output power is 0dBm, then Pa will be approx. -3.5dBm. Pb will be approx. -7dBm.

  • 0 in reply to Noel Fung
    Expert 2435 points

    Noel-san

    Thank you for your feedback.

    The Output requires 50 ohm termination.

    Likewise, FIN needs 50 ohm termination resister as well, don't you?

    Mita

  • +1 in reply to user4779049
    TI__Guru** 121230 points

    Hi Mita-san,

    A 50Ω load at FIN shall return better spurs, if possible, add a 50Ω load at FIN and FIN*. The resistor network can be simplified as follows:

  • 0 in reply to Noel Fung
    Expert 2435 points

    Noel-san,

    Thank you for your feedback.

    I think that the power divider circuit you suggested is fine.

    However, the reference schematic on the 1st page of the datasheet is

    slightly different from your circuit.

    I think -6 dB attenuator is placed on the output of the power divider.

    Please take look at the reference schematic on the 1st page of the

    datasheet.  The attenuator needs 50 ohm termination resister.

    Please measure the I/O voltage of the circuit using TINA TI.

     

    Mita

  • 0 in reply to user4779049
    TI__Guru** 121230 points

    Hi Mita-san,

    Pa is approx. -3.5dBm, the 18Ω resistor will create another 3dB loss, so the 50Ω connector will measure approx. -6.5dBm.