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The "low noise" LDO TI TPS7A20 specifies output noise as 7μV(rms) at 100kHz, whereas the Diodes AP2210 specifies output noise as 260nV/√Hz.

This SE answer compares the two mathematically as:

  • (nV/√Hz) ⋅ √(bandwidth) = μVrms

If I plug the numbers in then I get each LDO uV(rms) and nV/√Hz as:

  • TPS7A20: 7μV(rms) at 100kHz == 0.022 nV/√Hz
  • AP2210: 82,215 uV(rms) at 100kHz == 260 nV/√Hz

The nV/√Hz and the uV(rms) values are widely different between the two considering both are boasting "low noise".

  • Did I not understand something, or is the TI component really that much better than the Diodes component?
  • Did the referenced answer waive over the uV to nV conversion in their math, thus being off by a 1000 from uV to nV, such that the AP2210 is really 82.2uV(rms)—which is still dismal compared to the TI?
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    \$\begingroup\$ "This SE answer compares the two mathematically as: (nV/√Hz) ⋅ √(bandwidth) = μVrms" - where does it state that? \$\endgroup\$
    – Bruce Abbott
    Commented May 19, 2021 at 2:21
  • \$\begingroup\$ Thank you. Please correct your formula to match what the answer actually stated. \$\endgroup\$
    – Bruce Abbott
    Commented May 19, 2021 at 2:35
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    \$\begingroup\$ Both of those datasheets give plots of the noise power vs frequency using identical units. Rather than look at the headline number (which depends on various factors and is anyway not likely to be the same bandwidth you care about), look at the noise plots. You'll see that (ignoring PSRR), the tps7a20 is much less noisy. \$\endgroup\$
    – user1850479
    Commented May 19, 2021 at 4:02
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    \$\begingroup\$ You are mixing up uV and nV : the TPS7a20 calculation should give you 22nV/rtHz. \$\endgroup\$
    – user16324
    Commented May 19, 2021 at 11:37

2 Answers 2

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The formula from the referenced SE answer is oversimplification. If total noise in a given bandwidth is desired, one must integrate the noise over a bandwidth, and the spectral noise densities at, say, 10 Hz vs 100 kHz are different figures. You cannot just multiply the spectral noise density at 100 kHz by a square root of the bandwidth and have a total rms noise figure for this frequency range. The product gives only valid result for a flat PSD, i.e., for a white noise frequency distribution. Also, when applying the general formulas to DUTs, you should include all the circuit components and usage into your analysis.

You can make an attempt at such research for the designated components: the datasheets you cited contain the graphs of spectral noise density vs. frequency.

The Texas Instruments Application Report AN-104 Noise Specs Confusing? provides a detailed guide on all sorts of terms like signal-to-noise ratio, noise figure, noise factor, noise voltage, noise current, noise power, noise spectral density, noise per root Hertz, etc.

Some documents use a parameter called something like 'accumulated spectral output noise' which is not a measured spectral noise distribution value but rather a calculated parameter that can be used to estimate the rms noise applying the formula 'noise-per-root-Hertz by sqrt-bandwidth product'. Looking into Diodes' datasheet, page 21, the graph 'Output Noise vs. Frequency', I see the output noise value of 0.1 uV/√Hz at 100 kHz. The table AP2210-ADJ Electrical Characteristics, page 19, indicates the output noise e_no of 260 nV/√Hz. Maybe it is an 'accumulated' value, and you can use it to calculate the rms noise at 100 kHz by the simple multiplication formula, but still this value does not include the contribution of higher frequencies and may be inconsistent with the setup for the rms noise measurement.

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    \$\begingroup\$ Specifically you must integrate noise power which is the square of voltage noise spectral density. If the relevant noise bandwidth has a negligible 1/f part, the basic multiplication is a good enough estimate to compare the parts. \$\endgroup\$
    – tobalt
    Commented May 19, 2021 at 3:59
  • \$\begingroup\$ @tobalt 1/f is not the unique source of non-uniform frequency dependence, see the AN-104 report, for example, for an (incomplete) enumeration. But you are right, there exist scenarios where "the basic multiplication is a good enough estimate to compare the parts", I added one of those to my answer. \$\endgroup\$
    – V.V.T
    Commented May 19, 2021 at 4:46
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The conversion equation needs to balance with the units as well, so if you use nV on the left side, you should on the right as well.

TI component 7000nV at 100kHz leads to 7000/SQRT(10^5) or 22nV/SQRT(Hz) while the Diode's Inc is 260 nV/SQRT(Hz), or 10x worse voltage noise figure.

The self-noise is one component of the overall output noise, the other large component is the PSRR, which relates how noise on the input gets rejected at the output. Filter capacitors (input and output) can have a large effect on the noise as well.

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  • \$\begingroup\$ "so if you use nV on the left side, you should on the right as well." So 1000 nV = 1 μV is not a valid equation? :) \$\endgroup\$
    – endolith
    Commented May 19, 2021 at 2:56
  • \$\begingroup\$ +1: PSRR is one of clues for accurate comparison \$\endgroup\$
    – V.V.T
    Commented May 19, 2021 at 3:55
  • \$\begingroup\$ Mathematics does not have physical units, so if you drop the units, 1000 = 1 is not true :) In physics you need the same physical units on both sides, so using unit conversion 1000 nV = 1 μV * 1000 nV/μV which balances mathematically and physically :). \$\endgroup\$
    – filipmu
    Commented Dec 19, 2021 at 14:29

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