Using PWM to generate analog output

Question

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:

1. 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?

2. 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

Answer 1

1. 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 \$f_{PWM}\$, the frequency of PWM. \$ie.,\$ if K = 6 then \$f_{PWM}\$ will be 24KHz.

2. That is a printing mistake. The original equation is, $$G(dB)_{20KHz} = -10\times log[1+(2\pi fRC)^2] = -14.19\ dB$$