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I want to drive an ADC with an OpAmp (LT1819). The op-amp is in non-inverting configuration driving the ADC pseudo-differentially.

The filter between opamp and ADC should be ("the specs"):

  • passive
  • -3dB at 100 MHz, 3rd order Butterworth or Bessel (or steeper)
  • not load the Opamp too much (RL >1k)
  • keep component values reasonable for 0402 (pF, nH)
  • not introduce excessive noise (e.g., due to high series source resistance)

The classical 3rd order ladder filter (as found in literature and created by filter tables or programs like Elsie (http://www.tonnesoftware.com/elsie.html)) looks like:

schematic

simulate this circuit – Schematic created using CircuitLab

This is a classical ladder filter described in literature - but it contains the termination resistor.

The first sad thing is that this filter attenuates my signal by 6dB (so I have to increase by apamp gain by a factor of 2).

Second, the signal gets distorted (clipped). It seems the LT1819 can't drive such small resistive loads.

Now I am thinking about the following options:

  1. I could increase the resistor values. But in order to get the linearity to an acceptable level, the load resistance must be >>1k, maybe 10k. That means that the source resistance also needs to be that high and then I'm killed by noise. Besides, L and C values become unreasonable.

  2. If I just take an ordinary RC I only get a simple 1st order rolloff.

  3. The datasheet of the ADA4937 (Fig 67) suggests a solution:

enter image description here

However, using the indicated values (R=33, C1=20p, L=56n, C2=60p), the bandwidth I get is 85.77 MHz (not 125 MHz!). When I normalize the coefficients, they are [2.07 , 2.07 , 0.0326 ] but for a 3rd order Butterworth they should be [ 2 2 1]. Besides I do not know how this filter relates to the ladder filter described above (with resistive termination).

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  • \$\begingroup\$ What's the signal voltage into this (or what do you need out from it?) (Your opamp should be limited to driving no more than \$30\:\text{mA}\$ or so, looks like.) I get an impedance of about 66 Ohms and an angle of about 23 degrees and an attenuation of about Vout/Vin=1/3rd? But I may have made a mistake (I didn't bother with Spice.) \$\endgroup\$ – jonk Mar 1 '18 at 23:21
  • \$\begingroup\$ If you dont have specs, it cant be designed \$\endgroup\$ – Sunnyskyguy EE75 Mar 1 '18 at 23:25
  • \$\begingroup\$ This filter does not drop nearly as fast as mine \$\endgroup\$ – Sunnyskyguy EE75 Mar 2 '18 at 2:06
  • \$\begingroup\$ @TonyStewart.EEsince'75: You mean the filter from the ADA4937 datasheet? That's true but it also doesn't have to drive the 50 Ohm which is what my question is all about. Your proposed filter is the textbook LC ladder filter - the 10 order version of the 3 order version I had originally in my question. The question is really about how to get rid of the load resistor and if yes which implications it has. The ADA4937 DS shows that there is a way but I do not understand what the potential issues are compared to the "classical" ladder filter. \$\endgroup\$ – divB Mar 2 '18 at 6:20
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Have you tried out something like this already? I haven't given the components values much thought, frequency response could definitely be improved, but seems to me it meets your requirements.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Does this circit have a name I can look up and check if there are any disadvantages? Completely without resistor ... doesn't this easily become an oscillator? In this example, setting L1C1=1 and L2C2=1 I obtain 1/(1+(2+L1/L2) s^2 + s^4)$. This is an oscillator. \$\endgroup\$ – divB Mar 1 '18 at 23:16
  • \$\begingroup\$ No special name, just low-pass with reactive elements only. It should not oscillate when considering the existance of damping elements such as the ADC input current, the non-ideal components and the absence of active feedback. If in doubt, a terminating resistor like the Tony's answer should take care of it, but the title of your question implies you do not wish to use one. \$\endgroup\$ – Vicente Cunha Mar 1 '18 at 23:45
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enter image description here enter image description here

(Images from https://www.electronics-tutorials.ws/filter/filter_5.html)

If you didn't specify that the filters have to be passive then you could use something like that and add a few passive filter stages on the output to sharpen the cutoff.

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    \$\begingroup\$ Um, because the OP said it had to be passive. \$\endgroup\$ – Marla Mar 2 '18 at 0:22
  • \$\begingroup\$ 1st order ADC filter is not too great. Do you know what a Nyquist Filter does? \$\endgroup\$ – Sunnyskyguy EE75 Mar 2 '18 at 0:37
  • \$\begingroup\$ @Marla Oh, indeed they did. I wonder why it has to be passive? I still believe that using at least one active filter is a good idea because he's using an op-amp anyways. \$\endgroup\$ – ILoveGit Mar 2 '18 at 0:39
  • \$\begingroup\$ He has a differential input ADC \$\endgroup\$ – Sunnyskyguy EE75 Mar 2 '18 at 0:42
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    \$\begingroup\$ You can see the effect of the load R here tinyurl.com/y9vzgufx \$\endgroup\$ – Sunnyskyguy EE75 Mar 2 '18 at 2:02
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When you calculate a differential Pi filter breakpoint, you convert to a single ended topology and put 2 C's in series with same net equiv value. thus

How to define Nyquist Filter for an ADC to prevent aliasing noise you need to define signal BandPass, BP and noise BandStop, BS with min atten.

Hypothetical Specs

BP ripple & loss = 0 dB
BP f-3dB= 100 MHz
BS atten. @ 250 MHz = -70dB min assuming sampling rate of 2.5x
BS atten @ 300MHz ?

Quick and dirty ladder filter

enter image description here

I might consider LT1886 700MHz 100mA with a gain of 4 with simple compensation. Another variation enter image description here

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    \$\begingroup\$ with rounded values tinyurl.com/y972yjc5 \$\endgroup\$ – Sunnyskyguy EE75 Mar 1 '18 at 23:43
  • \$\begingroup\$ The main reason why I'd like to use the LT1819 is that it has SR 2500V/s (steps appear at input). Also, it is used in the datasheet and the eval board of the ADC (LTC3232) so I'd prefer to use it. I'm still confused about the purpose of the termination resistor. This filter is still loaded with 50 Ohm and results in clipping. If I omit it, I get the peaking. Is it somehow possible to not load the filter but use a series resistor? \$\endgroup\$ – divB Mar 1 '18 at 23:56
  • \$\begingroup\$ Do you know what X/R ratio and Q is? Try my link above and drag R off the schematic \$\endgroup\$ – Sunnyskyguy EE75 Mar 2 '18 at 0:14
  • \$\begingroup\$ Yes I do (and hence I am concerned if there is no R at all). However if you may have a look at the ADA4937 datasheet, Fig 67. This is a 3rd order ladder without the resistive termination. I am looking for something like that. However, when I use the values the BW is 80MHz (not 125MHz) and the normalized coefficients are [ 2.0735, 2.0735, 32.556e-3 ] instead of [ 2 , 2 , 1 ] - and I do not understand how it is related to a classical ladder filter (like yours). \$\endgroup\$ – divB Mar 2 '18 at 0:46
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I think my question got misunderstood since no answer attempted to discuss the actual question (even given in the title) about the load resistance (except Vicente's but a discussion on the practicality was missing)

After some time I got a handle on it and have to answer my own question:

The circuit from the ADA4937 seems to be the way to go. Expanded, the denominator polynomal is:

D(s) = 1 + s R(C1+C2) + s^2 L C2 + s^3 R L C1 C2

This polynomal can be made into Bessel, Butterworth form etc. For Butterworth, the required format is:

D(s) = 1 + 2(s/w0) + 2(s/w0)^2 + (s/w0)^3.

After coefficient comparison, the components can be found given any w0 and R:

C1 = 1/(2 R w0)
C2 = 3 C1
L = 8 R^2 C1^2 / C2 

For 100 MHz and R=20, this results in C1=39.79p, C2=119.37p, L=42.4413n.

R can be used to simplify input matching. But this is only a concern for very high frequencies and/or large traces/circuit. Assuming <100 MHz and short traces, this value can be arbitrary. The lower the R, the better the noise performance. However, very low values may result in unrealistic values (particularly for L). For that reason I pick R=20.

I could not find any disadvantage of this design.

I also found this link which which is the single calculator I found that allows to specify source and load resistance independently: https://rf-tools.com/lc-filter/. This calculator obtains the same results as above when making RL very high (say, 1MOhm).

The only disadvantage is that RF power transfer is not optimal any more: This can be seen looking at S21 in the result of the calculator. For Rs=20 and RL=1Mohm, S21 = 10*log10(1MOhm/20Ohm/2) = -44dB. This makes sense because the whole purpose is to not load the circuit and hence little current will flow, resulting in low RL I^2.

However, this is only relevant for RF applications when maximum power transfer is the objective. For lower frequency, analog circuits where voltage is the quantity of interest, this does not matter.

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