Trying to improve an active LPF

Discussion in 'General Electronics Chat' started by atferrari, Oct 22, 2011.

  1. atferrari

    Thread Starter AAC Fanatic!

    Jan 6, 2004
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    Shift register driven by a 12x clock. The 6 outputs loaded with weighted resistors to produce a sine-like waveform.

    Network output fed, via a unity gain buffer to a 6-pole LP Butterworth active filter designed (FilterLab) for Fc 1150 Hz.

    My intention is to use for a sinewave between 10 Hz and 1 KHz.

    After testing I found that below 200 Hz the output is bad, coming to almost copy the step-like input!!

    Besides revising the design with FilterLab and the hardware, I simulated it, confirming de bad performance at the lower end. LT's .asc file attached.

    The board is already populated and the number of opamps (1 x TL074) thus fixed.

    Any chance to improve it? Otherwise I will have to throw the board (veroboard) which took me ages to assemble. Vessels are too demanding ladies...

    If anyone recall a similar question of mine from some years ago, please note that I never managed to built that circuit so better to forget that thread and start afresh here again.

    Thanks for any help and the comprehension.

    Yes, a follow-up would be asking how to actually center the sinewave at 0V but oviously better to solve this first.

    Gracias again.
     
  2. crutschow

    Expert

    Mar 14, 2008
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    Don't offhand know how to improve the circuit for that wide a frequency range without using a tracking or adjustable filter. A possible choice for that would be a switched-cap filter such as the LMF100 or MF10. The advantage of that is, if you use the same clock (or sub-multiple of the clock) to drive both your circuit and the switched-cap filter, the filter will track the frequency.

    As you see, simulating the circuit before you built saves a lot of time and wasted effort.
     
  3. atferrari

    Thread Starter AAC Fanatic!

    Jan 6, 2004
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    Hola Carl,

    Thanks for replying.

    After thinking of your reply, it appears to be "components" lower than, say, 1 KHz making for that performance.

    Where do they come from? For the extreme case, 10 Hz (with a clock =12*10= 120 Hz), am I getting the harmonic components, like 240, 360, 480, etc in the passband of the filter meaning that they are not attenuated at all?

    In other words, as I increase frequency the harmonics fall more and more out of the passband, isn't it?

    My bad, I am affraid I overlooked something vital for the design. :(
     
  4. crutschow

    Expert

    Mar 14, 2008
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    Your analysis is essentially correct. It's the low frequency harmonics of the steps that are not being filtered.
     
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