Passive LP filter design

[ImperfectSeven]

Hi,

I'll check that out. In the mean time ignore that equation and you can use the first equation or check this next one out which is much simpler.

But take another look at your plot. Remember you have to look at the straight line segment to see the -40dB/decade. It actually is -40dB/decade.

Also, in view of the fact that you dont mind using the second resistor being equal to 10 times the first and the second capacitor 10 times less than the first cap, we can get to an estimate this way which is much faster...

R1C1=sqrt(sqrt(18081*A^2+4000000*Vin^2)-241*A)/sqrt(200*A*w^2)
where
A is the peak to peak voltage,

So once we calculate R1C1 we divide that by C1 and we get the value of the first resistor:
R1C1/C1=R1

and since R2=10*R1 and C2=C1/10 we have all four components. The only value we have to choose is C1.

Here is the w just 2*pi*f or does it have a different meaning?
If that is the case and I use
A = 1.46mV
w = 2*pi*10.577kHz = 66.46k
Vin = 3.3
then I get an RC value of 583.17m
this would require a large resistor and/or a large capacitor

Oh and that was a typo in my last post, it was suppose to say 1st equation, I think there might be a 1/2 missing in your first equation.

 

[MrAl]

Hello again,


Yes the Vin value was normalized, so instead change that Vin^2 to just Vin:

R1C1=sqrt((sqrt(18081*A^4+160000*Vin^2*A^2)-241*A^2)/(200*w^2*A^2))
R1=R1C1/C1

With Vin=5v and A=0.002 we get close to 11k for R1.

 

[ImperfectSeven]

Alright, yes that equations gets me something much better.
Again, thank you for all your help.

 

[MrAl]

Hello again,


I went over it again and have a better estimation here:

R1C1=sqrt((sqrt(18081*A^4+160000*Vin^2*A^2)-241*A^2)/(200*w^2*A^2))
R1=R1C1/C1

I checked it for Vin=1, 5, and 10 volts and it always gives a good estimate for R1.
These are in the class of frequency domain estimates. They are based simply on the response to the fundamental frequency and that the higher harmonics will never amount to an increase by more than 2 times but will most likely be much less than that.

 

[MrAl]

Hello again,

I went over this again and found some interesting information.

First off, using the equations for the circuit and about 160 digits of precision the peak to peak ripple voltage comes out to:
1.291mv
That's using R1=11k and C1=0.1uf and R2=10*R1 and C2=C1/10.
That result comes from running the equations half cycle by half cycle for 1000 cycles, then switching to a smaller time increment of 100 times smaller than a half cycle and hunting for the min and max (the peaks) and then subtracting the min from the max, all while using about 160 digits of precision.
Decreasing the increment time resulted in better and better results:
1.2911mv and 1.29111mv with 1000 times smaller and 10000 times smaller respectively.

Second and most surprising, the result is not that much different using R1=11k and C1=0.1uf and R2=R1 and C2=C1. At least not looking at the peak to peak ripple. However, the time constant to charge to 99 percent of the final value gets longer by about 3 times. This might make a difference if you need fast settling time, but if you dont then it lowers the impedance of the output which is always nice.
The ripple peak to peak result was 1.2906 which is close to 1.291. The settling time is longer though as noted.

The reason the second item above is so is because this is not a harmonic filtering application so the 10x resistor rule does not apply. This is a simpler fundamental filtering application where we want the fundamental down as low as practical.

Also, here is another estimation formula which is probably more accurate than the others although this is for the filter with R2=R1 and C2=C1:
R1=sqrt((sqrt(45*pi^2*A^2+64*Vi^2)-7*pi*A)/(2*pi*w^2*A*C1^2))
where
A is the peak to peak ripple desired (0.002 for example),
Vi is the input voltage pulse amplitude (5v for example).

And for the R2=10*R1 and C2=C1/10 circuit:
R1=sqrt(sqrt(18081*pi^2*A^2+640000*Vi^2)/A-241*pi)/(5*2^(3/2)*sqrt(pi)*w*C1)

These forms allow calculation of R1 directly.