Trying to improve an active LPF

Thread Starter


Joined Jan 6, 2004
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.



Joined Mar 14, 2008
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.

Thread Starter


Joined Jan 6, 2004
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. :(