Hello,
I am designing a bandpass filter with a 20dB gain between 10kHz and 1MHz.The Input signal swing is +/- 10mVpp and the desired output swing +/-100mVpp.
Below is my TINA simulation of the BP filter I designed. I am not sure where my component selection went wrong.
Why is the gain response 15.43dB at 1MHz? This gain should be 17dB at 1MHz. Do I need to recalculate my C2 value?
Best regards,
Lucas Pucheta
Hello Ron,
I took your advice and choose a faster op amp (OPA810). However, I still do not get 17dB at 1MHz. Is 16.33 close enough or is it possible to compensate C2 to get 17dB at 1MHz.
I used the following equations to find the component sizes. How would I approach adjusting C2 to compensate for this loss of gain.
Choosing R2 = 1KΩ
-R2/R1 = 10v/v ∴ R1 = 100Ω
1/(2∗pi∗(100)∗C1) = 10kHz ∴ C1 = 159nF
1/(2∗pi∗(1k)∗C2) = 1MHz ∴ C2 = 1.59E-10 = 159pF
Best regards,
Lucas Pucheta
Morning Lucas,
To continue on this a bit, often for these broadband BP designs we do a 2nd order MFB LP with an AC coupled input.
1MHz butterworth 2nd order with gain of 10 is actually quite high, the OPA810 is marginal on GBP margin but can be designed with RC adjustment flows for the relatively low minimum loop gain inside the filter (essentiallyw what Ron is saying about low GBP). Doing that design, then adding a 1st order HP as the input R is just the 1st resistor in the MFB, gives this design hitting things very accurately - in active filter design, I use 2% C0G SMD C's and 0.5% R's to get better spread, these have come way down in price,
the min loop gain thing inside the MFB is where the low GBP errors come in, that is shown here for this design where that <20dB min values is lower than I would normally do, (the OPA810 TINA model has a GBP = 63MHz).
And this file
