I have a question about the normalization calculation of multi-order low-pass filters. Thank you!

sxbo88

I have a question about the normalization calculation of multi-order low-pass filters. Thank you! [Copy link]

 

In the above figure, the filter design is calculated according to the parameters of the normalized table on the right to determine the values of resistance and capacitance.
However, the frequency in the normalized table is 1kHZ. If my frequency is 500HZ or other non-1kHZ, how can I
use the data in the table to calculate? Thank you!

maychang

Multiply the capacitor value by the ratio of the normalized 1kHz to the frequency you need. In your case, it is 500Hz, so the capacitor value should be multiplied by 2.

sxbo88

Should I first calculate the capacitance using the formula of 500 and then multiply it by the proportionality factor? Multiply both C1 and C2. Thank you!

maychang

sxbo88 posted on 2022-4-22 21:56 First calculate the capacitance result according to 500 and the formula and then multiply it by the proportional coefficient, multiply both C1 and C2, thank you!

"You should calculate the capacitance using the formula with 500 and then multiply it by the proportionality factor. Multiply both C1 and C2."

The 1kHz capacitor value is given, multiply it by 2 to get the 500Hz capacitor value. Multiply both capacitors by 2.

maychang

sxbo88 posted on 2022-4-22 21:56 First calculate the capacitance result according to 500 and the formula and then multiply it by the proportional coefficient, multiply both C1 and C2, thank you!

From the second equation in the figure, we can clearly see the relationship between frequency, capacitance and resistance. If the resistance remains unchanged and the capacitance is multiplied by 2, the Fc on the left side of the second equation will be half.

sxbo88

Generally, the Fc and Q values are given first, and then the C1 value is determined, and then the R1 and R2 values are determined. If you calculate this way, you must first determine the R1 and R2 values, then calculate the Cf value, and then calculate the C1 and C2 values.

maychang

sxbo88 posted on 2022-4-22 22:18 Generally, the Fc and Q values are given first, and then a C1 value is determined, and then the values of R1 and R2 are determined. If you calculate this way, you must first determine the values of R1 and R2, then calculate the Cf value, and then calculate the C ...

Generally, the Fc and Q values are given first, and then the C1 value is determined, and then the R1 and R2 values are determined. If you calculate this way, you must first determine the R1 and R2 values, then calculate the Cf value, and then calculate the C1 and C2 values.

It's the same thing whether you decide the resistor value first or the capacitor value first. You can do it in the way that's more convenient for you.

Typically, the capacitor value is small and the resistor value is large.

sxbo88

However, the calculated value is not accurate when applied to the actual object. It is possible that the capacitance of the PCB has a greater impact.

maychang

sxbo88 posted on 2022-4-23 08:26 However, the calculated value is not accurate when applied to the actual object. The capacitance of the PCB may have a relatively large impact

"But the calculated values are not accurate when applied to the actual object. Perhaps the capacitance of the PCB has a relatively large impact."

At a frequency of 1kHz, you cannot even measure the impact of the circuit board.

It is common for capacitors sold on the market to have a 20% error. If you want to find a capacitor with a 1% error, the price may be too high for you to afford.

gmchen

The book the OP referred to should be Design of Measurement Electronic Circuits by Akitoshi Toosaka.

First of all, that table is a normalized table. It is not normalized to 1kHz, but normalized to 1rad/s (radian/second), which is 2pi Hz. Don't be confused by the curves and examples in the book.

In the circuit cited by the original poster (Sallen-Key circuit), the premise of the calculation is R1=R2=R. Since the frequency in the normalized table = 1 (in fact, all Butterworth filters are 1), the calculation formula fc=1/(2pi*RC), that is, the ratio of 1/RC to 2pi*fc is 1. When designing, let 1/RC equal to the required cutoff angular frequency (2pi*cutoff frequency), then manually specify a resistance value, and then calculate the two capacitor values according to the following two formulas.

However, the capacitance value calculated in that way is probably not available for purchase, so this is not a reasonable choice.

gmchen

In fact, a more reasonable choice is to first select C1=C2=C in the design, and then calculate R. At this time, the two Rs are not equal.

Still taking the Sallen-Key circuit as an example, the design process is as follows:

1. Artificial selection of C

2. Calculate R1*R2 according to the formula fc=1/[2pi*sqrt(R1*R2)*C]

3. According to the formula Q=sqrt(R1*R2)/(R1+R2), combine the previous result of R1*R2 to calculate R1 and R2

gmchen

Regarding active filters, I have written a series of posts. You can take a look. Talking about active filters - low-pass filters https://en.eeworld.com/bbs/thread-565950-1-1.html

Fred_1977

I haven't paid attention to the basic principles of low-pass filters for a long time. I will review them when I have time.

gmchen

sxbo88 posted on 2022-4-23 08:26 However, the calculated value is not accurate when applied to the actual object. The capacitance of the PCB may have a relatively large impact

Teacher Zhang is right, the distributed capacitance of the printed circuit board can be completely ignored in low-frequency circuits.

If the measured deviation is not too large, then it is a capacitor error problem.

If the actual measurement deviates greatly, then the calculation is wrong.

sxbo88

Thank you everyone, thank you!