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LMH6518 Aux O/P

Supratim Majumdar
Intellectual 270 points
Other Parts Discussed in Thread: LMH6518

Hi Hooman,

This is in continuation of my previous post on difficulty in observing the main o/p of LMH6518 on oscilloscope, while aux. o/p was coming correctly. I have replaced the part as you advised and now getting the main o/p. But now I am getting low amplitude at the aux. o/p. The aux. o/ps are connected like fig.-68 of data-sheet.

The single ended aux. o/p is 140 mV(pp) riding over a 1.2V CM, the differential aux. o/p is 264 mV(pp) while the diff. main o/p is 480 mV(pp).

All the differential o/ps were measured using a differential probe and all single ended o/ps were measured using a 10X probe terminated at 1MOhm.

Please help.

Regards,

Supratim

  • 0
    TI__Mastermind 22825 points

    Hi Supratim,

    Are you taking into account the LMH6518 internal 50ohm output impedance (see Figure 61) on both Main and AUX outputs? If you are terminating the AUX but not terminating the Main outputs, that could account for the lower amplitude on AUX output.

    Regards,

    Hooman

  • 0 in reply to Hooman Hashemi
    Intellectual 270 points

    Hi Hooman,

     You are right. I missed it. Now I have terminated the main o/p differentially by 100 Ohm. Now the aux. o/p is just half of the main o/p. I am observing single ended aux. o/p (fig. 68) and differential main o/p. I think that explains this factor of 2. Is that correct?

    Regards,

    Supratim

  • 0 in reply to Supratim Majumdar
    TI__Mastermind 22825 points

    Hi Supratim,

    If you are looking single ended at each of AUX output(s) (pin 1 or pin 2), then you would get 1/2 the voltage swing compared to the voltage across the 100ohm differential load on the Main output (pin 15 and pin 14). This is assuming that Aux output is terminated in 50ohm single ended.

    However, Figure 68 loads the AUX output(s) with slightly higher impedance than 50ohm (57ohm is the parallel combination of R1, and R11). Thus, the single ended swing at each AUX output should be close to 0.53 (instead of 0.5).

    Regards,

    Hooman