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TPA6120A2: THS6012 / TPA6120A2 Source Resistance

Jonah Strauss
Prodigy 240 points

Part Number: TPA6120A2
Other Parts Discussed in Thread: THS6012

I was looking at some of the subtleties of the TPA6120A2 by combing the THS6012 datasheet. I came across Figure 42, Source Resistance vs Noise Factor, which is delightfully confusing.

Why is ~150R the optimal Rs for THS6012, vs anything lower?

Is the optimal Rs for TPA6120A2 also ~150R?

That graph refers to the non-inverting input, correct?

Is there any advantage (in noise terms, not offset) to matching the source resistance of the inverting input to that of the non-inverting input? For instance, in a gain of 5, one might choose Rf=402R and Rg=100R for best balance of bandwidth and peaking. But with ~150R in series with the non-inverting input, one could choose Rf=806R and Rg=200R - if it did indeed provide a noise advantage.

Thanks!

 

  • 0
    TI__Mastermind 20680 points

    Hi Jonah,

    I believe the curve looks this way is because as the Rs value continues to increase, the numerator in the NF equation above Figure 42 will only grow at a faster pace. At lower values, en is dominating which is why reducing Rs only does so much. Reworking the formula and splitting the fraction helps get a better idea of the shape of the graph. For the TPA6120A2, the value of the optimal Rs value would be different since en and In+ (non-inverting current noise) is different for this device so I would assume the optimal Rs would shift.

    As for your last question, I cannot think of a reason why this would offer any benefit to your noise performance. If anything, it would technically hurt your noise performance as you increase resistances. However, another consideration is that your total Vpp noise depends on your bandwidth, so increasing Rf would decrease bandwidth so you would see an overall lower Vpp noise due to this change in values. However, from a matching perspective I am not sure this would have any sort of impact. 

    Best Regards,

    Ignacio