Input voltage noise (RTI) calculation for preamplifier

Question

I would like to obtain the RMS Input Voltage Noise (RTI) of the AD8428 preamplifier for the following setup

• range of 0.1 Hz to 150 Hz with 1^st order lowpass filter, koeff = 1.57 ---> BW(bandwidth) = 150*1.57 = 235 Hz

From the Fig. 29 (RTI Voltage Noise Spectral Density vs Frequency) in the datasheet I determine the needed parameters for calculation:

• F_L = 0.1 Hz, F_C = 8 Hz, F_H = 235.5 Hz
• with 1.5 nV/√Hz in the flat region and 15 nV/√Hz at 0.1Hz

Calculating the RMS voltage noise

The main idea of the calculation is to integrate the curve in the interesting bandwidth. But if I go into the details I found two ways with different results. All the calculations splits the curve into two regions: a 1/f region (with curve slope = 1/f) and a broadband region (with curve slope = 0) and integrating them separately.

1. Analog Devices MT-048 Tutorial:
   for the 1/f region: $$v_{n,rms}(F_L, F_C) = v_{nw}\sqrt{ F_Cln\left[\frac{F_C}{F_L}\right]} = 1.5 \sqrt{ 8*ln\left[\frac{8}{0.1}\right]} = 8.881nV_{rms} $$ for the broadband region: $$v_{n,rms}(F_C,F_H) = v_{nw}\sqrt{F_H-F_C} = 1.5 \sqrt{235-8} = 22.599 nV_{rms}$$ Together: $$v_{n,rms}(F_L, F_H) = \sqrt{ 8.881^2+22.599^2} = 24.281 nV_{rms}$$ v_nw is voltage noise density in the broadband region
   v_n,rms(F_L,F_C) is the RMS voltage noise in the 1/f region
   v_n,rms(F_C,F_H) is the RMS voltage noise in the broadband region
   v_n,rms(F_L,F_H) is the total rms voltage noise.
   
   This method surprisingly doesn't use the voltage noise density at 0.1Hz which is 15 nV/√Hz

2. The second method I found in this youtube video at 17:00:
   For the 1/f region: $$ E_{nf} = e_{@f}\sqrt{F_L}*\sqrt{ln\frac{F_H}{F_L}} = 15*\sqrt{0.1}\sqrt{ln\left[\frac{235}{0.1} \right]} = 13.215 nV_{rms} $$ for the broadband region: $$ E_{nBB}=e_{nBB}\sqrt{BW} = 1.5 * \sqrt{235-0.1} = 22.989 nV_{rms} $$ together: $$E_{nv}=\sqrt{13.215^2+22.989^2} = 26.516 nV_{rms}$$
   e_@f is voltage noise density at 0.1Hz
   e_nBB is voltage noise density in the broadband region
   E_nf is the RMS voltage noise in the 1/f region
   E_nBB is the RMS voltage noise in the broadband region
   E_nv is the total rms voltage noise.

Conclusion
The main difference in two calculation is that the Analog Devices tutorial dont use (or they neglect) the voltage noise density at 0.1Hz in the 1/f region (15nV/√Hz). They using the noise voltage density for broadband region evrywhere. What do you think which calculation is more correct? Is there a mistake in Analog Devices tutorial?

EDIT

After looking at the answers and a bit of dive to the calculus, I can deliver a final precise analytic solution. The voltage noise density curve you can describe with an expression:

$$ e_n(f) = \frac{1}{f}+e_{n,bb} $$ where f is the frequency e_bb is the noise level in the broadband region.

In order to obtain the RMS voltage noise for desired bandwidth, we need square the voltage noise spectral density to get a power noise density, then integrate it through desired bandwidth finally make a root square to obtain voltage RMS (this is the definition of RMS):

$$ \sqrt{\int_{F_L}^{F_H} {(e_{n}(f))^2} \,df}\ = \sqrt{\int_{F_L}^{F_H} {(\frac{1}{f} + e_{n,BB})^2} \,df}\ = \sqrt{\frac {-1}{2(F_H-F_L)} + 2e_{n,BB}[ln(F_H)-ln(F_L)] + e_{n,BB}^2(F_H-F_L)}$$

For the calculation, we used only the lowest and highest frequency of interest and broadband noise.

Substituting to the expression I get a noise of 23.51 nVrms which is very close to the numbers I get from the approximations.

Answer 1

The first approach doesn't neglect the low f noise. It assumes a 1/f power slope below the corner frequency (through the used formula), which implies 15 nV/rtHz at 0.1 Hz.

The second approach assumes the same slope, but takes a different base point to estimate its contribution, namely at an arbitrary value of 0.1 Hz.

Both methods don't correctly integrate the actual noise curve but make approximations of it. That is why they differ slightly.

The second approach is more correct approximation because it doesn't ignore the low f white noise and also doesn't ignore the high f flicker noise.

Answer 2

Total input noise in a bandwidth is \$ e_{n} \sqrt{BW} \$
where
\$e_{n}\$ = input noise spectral density [\$ V\sqrt{Hz}\$]
\$BW\$ = bandwidth [Hz}

From a practical point, the bandwidth from 0.1 to 10 Hz is a small fraction of the total bandwidth of 235.5 Hz. The simulation plot below is shown with a linear frequency scale to show how little energy is in the 1/f region.
Thus, a first approximation of the total noise is \$ 1.5 \sqrt{236} \approx 23\; nV \$ if we ignore the 1/f noise.

If you want to check your maths against a simulation, you can load up the SPICE model for the device in question and find the total noise using the simulator. The following simulation is done in LTSpice using an AD8429 (LTSpice didn't have the AD8428 in their library) with a gain of about 100. The input noise spectral density is fairly close to noise curve you supplied. In LTSpice, Ctrl+LMB (LMB = left mouse button) on the V(noise) label in the graph will perform the integration giving you total noise of 24.4 nV, pretty close to the first approximation of 23 nV.