How to calculate a differential + common mode LPF circuit?

I'm trying to validate my understanding of a differential + common mode LPF circuit using the AK4499EX datasheet (019001308-E-00 2022/07 version, p. 47). It presents the following circuit:

and its frequency response table:

The problem is that the frequency response values I'm getting using any RC filter frequency response calculators differ from the table in the datasheet.

For simplicity, let's focus on the 20kHz value.

As I understand it, the differential frequency response of this circuit is equal to an RC LPF with the following values: R = 1kOhm, C = 1nF. Then, the frequency response at 20kHz is ~-0.07dB:

The table indicates quite a different value of -0.18dB.

1. Could you verify, whether my calculations are wrong or if it's an error in the datasheet frequency response values?
2. Does the common mode LPF which follows the differential mode LPF in the circuit affect the differential frequency response (I suppose) the table represents?
3. What's the purpose of the 100 Ohm resistors in this circuit?
4. In TI filter design guides, I read that CM LPF capacitors should have 10 - 20 times less values than the differential LPF capacitor value. Could it be the reason why my calculated 20 kHz value of the RC LPF simplification differs from the datasheet value?
• Why do you ignore the 100 Ohm resistors and the 1.2 nF capacitors in your description of the differential mode calculation? The differential mode signal will see them... Commented Jun 30 at 22:09
• @JosBergervoet Well, that's the point of the questions #2 - #4. I'm not quite sure how the values of the elements of the common mode LPF affect the differential mode LPF parameters :) Commented Jun 30 at 22:38
• Seems to be just a RCRC network for differential mode: the R = 1kOhm, C = 1nF that you mention followed by R = 200 Ohm, C = 0.6 nF. (The two RC pairs do influence each other so you cannot compute them as two independent LP filters of course. You just need to compute, or simulate, the whole RCRC network.) Commented Jun 30 at 23:02
• @JosBergervoet I believe you're right. With the values of R2 = 200 Ohm and C2 = 0.6 nF, a 2-nd order LPF Butterworth filter calculator produces results which comply with the AK datasheet. Looks like indeed the circuit's differential frequency response could be calculated using this simplification. Commented Jul 1 at 5:22

The circuit would be more complicated to analyze if it were not symmetric. But it is symmetric. So that helps simplify a straight-forward analysis, a lot.

Here's the schematic with names added. Despite my statement about symmetry, I'll keep all the names distinct for now for generality's sake:

Using Sympy/Python/SageMath:

ev1 = Eq( v1/r1 + v1/r3 + v1*s*c0, vip/r1 + vop/r3 + v2*s*c0 )    # KCL v1
ev2 = Eq( v2/r2 + v2/r4 + v2*s*c0, vin/r2 + von/r4 + v1*s*c0 )    # KCL v2
eop = Eq( vop/r3 + vop*s*c1, v1/r3 )                              # KCL vop
eon = Eq( von/r4 + von*s*c2, v2/r4 )                              # KCL von
ans = solve( [ ev1, ev2, eop, eon ], [ v1, v2, vop, von ] )       # matrix solution
tf = simplify( (ans[vop]-ans[von]) / (vip-vin) )                  # transfer function


The transfer function is just the output difference divided by the input difference, as shown above.

I could show that function. But it's a long one, algebraically. And also, we know there is symmetry to it. So let's just apply the symmetry and see what results:

tfsymmetric = simplify( tf.subs({ c2:c1, r2:r1, r4:r3 }) )
tfsymmetric
1/(2*c0*c1*r1*r3*s**2 + 2*c0*r1*s + c1*r1*s + c1*r3*s + 1)


That's a 2nd order transfer function with $$\\omega_{_0}\approx 2886751.346\$$, $$\\zeta\approx 2.48261\$$, and $$\A=1\$$ and where $$\G_s=A\frac{1}{\left(\frac{s}{\omega_{_0}}\right)^2+2\zeta\frac{s}{\omega_{_0}}+1}\$$.

Let's plug all this into LTspice:

You can see that all three curves are on top of each other, confirming that the analysis is likely correct.

Let's look at the table values now:

The two underlined values match the table. (I checked that the third also matches, but LTspice doesn't give me three cursors at once and posting another large image really isn't necessary here.)

• Thank you for confirming the AK datasheet numbers! As to my other ('filters theory') questions, is it correct to say that this circuit is an equivalent to a 2-nd order LPF Butterworth filter with the values: R1=1kOhm, C1=1nF, R2=200Ohm, C2=0.6nF (like @Jos Bergervoet mentioned)? When I used these values, the calculator produced all the values equal to your LTspice simulation (and AK datasheet as well). Commented Jul 1 at 4:38
• @AlexanderAbakumov I don't think I'd call it a Butterworth. No matter what the order is, a Butterworth always has a $-3.0103\:\text{dB}$ crossover frequency exactly at $\omega_{_0}$ and the specific damping factor of $\zeta=\frac12\sqrt{2}$. Here, $f_{_0}\approx 459.441\:\text{kHz}$ but the $-3.0103\:\text{dB}$ crossover frequency is pulled back all the way to $f_\text{c}\approx 96.431\:\text{kHz}$ by the very high damping factor. As far as the values you mention, are you referring to my own numbering? (Keep in mind I have a $C_0$.) Commented Jul 1 at 6:20
• This wouldn't be a 2nd-order Butterworth filter, because it has two real poles, while a 2nd-order Butterworth has two complex poles. Commented Jul 1 at 6:23
• @AlexanderAbakumov When reading from The Photon, please note that it still wouldn't be a 2nd order Butterworth even with complex poles. A 2nd order Bessel has complex poles. And it isn't a Butterworth. A Butterworth has specific complex poles. It's true that real poles do exclude a Butterworth (and exclude a Bessel, etc.) But real poles mostly mean that it's neither under-damped nor critically damped, but instead over-damped. For a lowpass, this means the -3 dB crossover is at a lower frequency than expected. For a bandpass, this means a wide bandpass and not a narrow one. Commented Jul 1 at 6:42
• @periblepsis As far as the values you mention, are you referring to my own numbering? Not really. The equivalent values I mentioned (R1=1kOhm, C1=1nF, R2=200Ohm, C2=0.6nF) correspond to a conventional 2nd order R1C1R2C2 LPF filter. Feeding them into a 2nd order RC LPF calculator gives me the correct results, i.e. the same as your simulation. That's why I thought it could be the correct approximation for the differential frequency response of the circuit given that it's symmetrical. Commented Jul 1 at 17:19