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OPA1679: THD+N & Noise performance at 5V single supply

Hideki Hayashi1
Expert 7940 points
Part Number: OPA1679

Hi team,

The datasheet describes THD+N & Noise performance at +/- 15V supply. How much do you expect the performance degrading in 5V single supply?

Regards,

Hayashi

  • 0
    TI__Guru 64840 points

    OPA1679 specifies THD+Noise at +/-15V  Vout=3Vrms (Vout_peak=+/-4.23V or Vpp=8.46V), which will NOT be possible to do on 5V single supply in G=1 with linear voltage range limited to minimum 0.8V by the output range and maximum limited to 3V (5V-2V) by the input voltage range - see below. 

    Therefore, in order to perform equivalent measurement around mid-supply of 2.5V (Vs=5V), the maximum peak voltage would have to be 0.5V (2V<Vout<3V), thus 8.5 times lower than conditions used in the datasheet.  Since OPA1679 THD+Noise at low frequency in G=1 is dominated by its input voltage noise, THD=~[4.5nV/rt-Hz*sq-rt(GBW*1.57) / Vout]*100%, having 8.5 times lower Vout would result in at least 8.5 times higher THD+Noise.

  • 0 in reply to Marek Lis
    TI__Expert 7940 points

    Hi Marek,

    So do you mean input voltage noise (4.5nV/rt-Hz) not depend on supply voltage, correct?

    Regards,

    Hayashi

  • 0 in reply to Hideki Hayashi1
    TI__Guru 64840 points

    This is correct - to the first-order the input voltage noise is not a function of the supply voltage. However, since the broadband noise is an inverse-function of the current biasing the input differential pair, if it various over process variation by up to +/-25% (see typ IQ of 2mA vs max 2.5mA), its input voltage noise may change as a square-root factor:  +/-(1.2)^.5*100% =~ +/-12%.  So it could be as low as 4nV/rt-Hz for IQ of 2.5mA or as high as 5nV/rt-Hz for IQ of 1.6mA.