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Jan. 28, 2023•0 likes•2 views

Jan. 28, 2023•0 likes•2 views

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I am trying to understand this application note from Microchip about using PWM and a low pass filter to generate an analog output. I have a few questions regarding to the article: In the example they use a 20kHz PWM with K = 5 so we have a 4kHz cut-off frequency. Why choose K = 5? Is it a practical rule? Or is there any reason to choose K = 5? Then, they calculate R and C, I understand that, but after that they calculate how many dB the 20kHz signal from the PWM is cut-off (I don\'t really understand what they mean) and they get -14dB but according to my math I get -10.43dB with: F = 20kHz R = 4K C = 0.01uF Solution As the note says, K should be greater than 1 and an arbitrary value is selected for K. Choosing a different value for K will change fPWM, the frequency of PWM. ie., if K = 6 then fPWM will be 24KHz. That is a printing mistake. The original equation is, G(dB)20KHz=?10 .

- 1. I am trying to understand this application note from Microchip about using PWM and a low pass filter to generate an analog output. I have a few questions regarding to the article: In the example they use a 20kHz PWM with K = 5 so we have a 4kHz cut-off frequency. Why choose K = 5? Is it a practical rule? Or is there any reason to choose K = 5? Then, they calculate R and C, I understand that, but after that they calculate how many dB the 20kHz signal from the PWM is cut-off (I don't really understand what they mean) and they get - 14dB but according to my math I get -10.43dB with: F = 20kHz R = 4K C = 0.01uF Solution As the note says, K should be greater than 1 and an arbitrary value is selected for K. Choosing a different value for K will change fPWM, the frequency of PWM. ie., if K = 6 then fPWM will be 24KHz. That is a printing mistake. The original equation is, G(dB)20KHz=?10