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LME49600 Audio Buffer

John Wygonski
Intellectual 870 points
Other Parts Discussed in Thread: LME49600

Can you provide an expression for the gain of the headphone buffer circuit of the LME49600 evaluation board and data sheet?   The non-inverting stage alone should provide a gain of 2 (Gain = 1+ R2/R1) but I measure a gain of 3 on the evaluation board that I purchased.   It seems the resistor in the servo loop has an effect on gain.

Thanks,

John

  • 0
    TI__Expert 3840 points

    Hi John,

    Yes, you are correct. R5 & R6 on the outputs of the servo amps are in parallel with R1 & R2, respectively. Therefore, the impedance off of the inverting terminal is essentially 500 ohms, resulting in the gain of 3 as you have measured.

    By the way, we have noticed an issue where the 1M ohm resistors in conjunction wtih the bipolar input op amp (input offset current) for the servo can cause the output offset to be larger than desired. The recommendation would be to change the 1M ohms resistors to around 100k ohms to ensure that the DC offset is satisfactory. The caps may need to be increased to keep the time constant really slow. This is just a heads up.

    Best Regards,

    JD

  • 0 in reply to John De Celles
    Intellectual 870 points

    Thanks for the reply.   So I can decrease the gain to 2 by simply changing both R1 and R5 to 2k, with no ill effect on the DC servo?

    Also, re the DC offset issue, could I just replace the bipolar op amp with a JFET low input bias current type?

  • 0 in reply to John Wygonski
    Intellectual 870 points

    Sorry, I meant to say can I change R2 and R5 to 2k, leaving R3 at 1k.

  • 0 in reply to John Wygonski
    TI__Expert 3840 points

    Hi John,

    Yes, for the offset issue, you could also use a JFET low input bias for the servos; that would eliminate the issue as well.

    To obtain the gain of two you can get there a few ways.

    1. You could change R1 & R5 both to 2k and leave R3=1k. For the other channel it would be R2=R6=2k and R4=1k.

    OR

    2. You could just increase R5 and R6 to a much larger value (say 100k), so that there isn't any loading on the inverting input nodes, leaving R1=R2=1k.

    OR

    3. Some combination between those two extremes. ie R1=1.1k & R5=10k, so that the inverting leg gain resistance is the parallel combination of the two.

    Vo/Vi = (1 + R3/(R1//R5)) = (1 + 1k/991) = 2

    I think that the last option is probably the most reasonable.

    Best Regards,

    JD