By selecting R and C values based on a certain formula you must have a particular circuit in mind.From what I understand I need to select my R and C values such that my f_n = 1/(2*pi*sqrt(R1R2C1C2)) is less than or equal to that 4.5 kHz,
You are correct here, that is one problem I had in my first implementation of this filter. I had a 1mV/bit resolution but a 17mV ripple which caused an output voltage that would oscillate between +-1V around my desired level.Residual ripple with your configuration will be considerably higher than 1 LSB, so we could guess
Thank you for the quick response.
I am using a passive RC Low Pass filter I have attached an schematic.
Its a ladder network configuration of resistors and capacitors.
Here is a calculator and analysis tool for 2nd order RC low pass filters that might help you: **broken link removed**. It supports many other filter types too.
J
That is correct, I know ideally I want 0.707 < d < 1, however with this configuration d cannot go below 1 so I aim to get it as close to 1 as possible.If I understand the definition of your damping ratio correctly, d = 1 refers to two decoupled cascaded first order low passes. That means, you can't achieve it in a purely passive second order filter, only approximate it by setting R2 >> R1, and R1C1 = R2C2.
With no load on the output, it should be -3dB at 510 Hz.What I have found is that if I use:
R1 = 2k
R2 = 1M
C1 = 0.1u
C2 = 200p
Then I get a 3dB frequency of about 800 Hz......
With no load on the output, it should be -3dB at 494 Hz.Alright, so I changed my 1M to 40k and the 200p to 5n in my simulation, and you are right there is almost no change in the output, although this does put my 3dB point at 25 kHz.....
How did you come up with this value? The equations I have and the website listed above each come up with 795.77 HzWith no load on the output, it should be -3dB at 510 Hz.
I don't think the source impedance is significant, but the load impedance I believe to be about 380k (it has a maximum capacitance of 18pF and a resistance of 320 ohms) at a frequency of 23.4375 kHzFor accurate calculations you need to take the load impedance into account (and the source impedance, if it's significant).
I cheated and used a simulator, but it looks reasonable.How did you come up with this value?
That's correct for a first order filter. With a second order filter, the response is about -6dB at that frequency because both filter sections reduce it by 3dB.The equations I have and the website listed above each come up with 795.77 Hz
I have set the system clock at 24 MHz and I have 10 bit resolution on my PWM so it has a frequency of about 23.4375 kHz.
Here's one way to do the calculations:I have concluded however that the ripple is much more important than the settling time
That formula isn't really useful. It just gives the mean of the Fc for each section. It also doesn't take into account the interaction between them.What am I doing wrong?
Ha, I'd say so. Guess I should have previewed that one.P.S. Something went wrong with your LaTeX.
by this you mean that I am reducing my PWM signal by a factor of 1000 correct?reduction by a factor of about 1000.
Yes, but I don't know where you got 2268 from.So for each stage fc should be at 10.577 kHz / 67 = 167.23 ? 158 Hz
Is that correct?
I don't know. It depends what the input and output will be connected to.I then get the following:
R1 = 11k
R2 = 110k
C1 = 0.1u
C2 = 10n
would those values be acceptable or is the output impedance too high?
I'm not quite sure what you mean by this. The settling time of the filter is not too important, I have actually been letting it slip to around 500 ms.the bandwidth respectively settling time required by the application.
I get the 2268 because I found out that 1V on my high voltage side corresponds to 1.46mV out of my PWM.Yes, but I don't know where you got 2268 from.
Both input and output are connected to the microcontroller that I am using for this applicationI don't know. It depends what the input and output will be connected to.
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