I am designing a system that will use a phase frequency detector with 400-450 MHz signals. There is a filter required on the output of the phase frequency detector in order to obtain an average value (ie: Low Pass Filter).

I first was planning on using an active low pass filter with just a basic op-amp, but I can see the bandwidth of the op-amp being an issue.

In the answer to one of my questions, Mark recommended I use an LC filter but said:

One issue you do need to consider is the input impedance of the LC filter. You'll need to take it into account when deciding how to treat the transmission line between the filter(s) and the IC.

Also, the phase frequency detector has 4 outputs (D, !D, U, and !U). Initially I was just planning on using D and U and referencing to ground, but I think it might be best to use the differential pair.

The IC specifies that 50 ohm impedance is desired. The filter will ideally provide me an average over about 1 second, so whatever is required to get rid of all switching noise and provide the average.

I plan on feeding the averaged signal to a differential ADC (but can go with single ended).

Size and Cost are not huge constraints in the project but high quality is a must.

What needs to be done to make sure that this is done properly?

  • 1
    \$\begingroup\$ What is the output impedance of the signal you want to low pass filter? What frequency range must the filter pass and not pass? \$\endgroup\$ Jul 12, 2011 at 19:36
  • \$\begingroup\$ @Olin Lathrop I edited the question, let me know if I need to change anything else. \$\endgroup\$
    – Kellenjb
    Jul 12, 2011 at 19:40

1 Answer 1


As I understand it, you need to filter a 400-450 MHz signal to find a much lower frequency signal superimposed on it. The signal has 50Ω impedance, and you are looking for the slow signal to change only on the time scale of a second or so. If this is incorrect, please edit your question to be more specific.

This is a very simple problem. Since you have such a high ratio between the signal you want to block (400 MHz minimum) and the signal you want to pass (a few Hz). A simple passive filter will do very well. I'll assume that the A/D of the micro wants input signals to have 5KΩ or less. Different parts have different restrictions. This one would suite many. You can adjust the values accordingly if you need different.

I would probably use 2KΩ in series, 10uF to ground, another 2KΩ in series, and another 10uF to ground. You don't really need two poles of low pass filter due to your high frequency ratio, but I'm thinking there may be other things in that signal or other noise that would be good to stomp on. The signal to the micro would have a impedance of 4.05 KΩ and frequencies of up to a few Hz would be passed without bother. After that they start getting attenuated. 1 KHz will already be down by over 80 dB with stuff above that another 12 dB every octave.


As Kortuk points out in the comments, at these frequencies parasitic inductances and capacitances can matter. That is another reason I want two poles instead of one, even though a single pole would attenuate the 400 MHz plenty well enough in theory. I also wasn't planning on getting into this level of detail (I have a life and need to get things done. Every answer can only have a finite amount of detail. As a volunteer, I should have the right to decide how much detail I'm willing to get into) until Kortuk essentially called me out on it.

I agree that this filter should be implemented with SMD parts and carefully layed out to minimize stray capacitive coupling from input to output. It should also be physically close to the output pin producing the signal to filter. You don't have to worry about the output being at 50Ω and the first resistor being a mismatch, since the trace is intended to be only a few mm long.

The "down by over 80 dB" I quoted for 1 KHz is still valid. None of the stray stuff is going to matter at 1 KHz. The signal will drop from there another 12 dB per octave for quite a while, at least until well into the MHz range. Eventually the parasitic capacitance accross the resistor and the parasitic series inductance of the capacitor and the leads to it will make the filter work less well, and its gain will actually start to go back up with frequency. You'd have to look at specific part datasheets to get a better idea, but with decent parts and decent layout, I'd expect the bottom to be somewhere (factor of two easily possible) around 100 MHz. The gain at that bottom is so low that the rise in gain from there to 450 MHz should still be well tolerable, and the result good enough. These things are difficult to predict with any certainty, so to get real numbers you pretty much have to build it and see what you get.

However, I'd be real surprised if what I described isn't good enough for the job with significant margin.

  • \$\begingroup\$ So no need for LC or impedance matching? I'm assuming that the no need for impedance matching is since I am LPF it anyways? \$\endgroup\$
    – Kellenjb
    Jul 12, 2011 at 20:17
  • \$\begingroup\$ @Kellenjb: Impedance matching to what? You seem to say there are 4 outputs, each with 50 Ohm impedance. That doesn't mean you have to load them with 50 Ohms unless there is something you're not telling us. The only impedance constraints in what you've said so far is the output shouldn't be loaded below 50 Ohms or so, and the A/D on the micro needs to see a certain minimum driving impedance, very likely much higher than 50 Ohms. \$\endgroup\$ Jul 12, 2011 at 20:36
  • \$\begingroup\$ I guess you are saying that Marks comments have no good reasoning behind them? \$\endgroup\$
    – Kellenjb
    Jul 12, 2011 at 20:39
  • \$\begingroup\$ @Kellenjb Huh? Who is Mark? I don't see any question, answer, or comment here by anyone with the name Mark. \$\endgroup\$ Jul 12, 2011 at 21:27
  • \$\begingroup\$ I quoted and linked to Mark in the question. \$\endgroup\$
    – Kellenjb
    Jul 12, 2011 at 23:16

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