I would like to use an 8-order LPF switched capacitor: MAX7405 datasheet

The cut-off frequency can be set by simply connecting an external oscillator capacitor, Cosc.

I would like to design a 50Hz cut-off frequency but I have doubts on my calculation for the given formula in the datasheet.

max7405 datasheet

Since I want to have a 50Hz fC, I needs the fCLK to be 5000Hz

From the given equation:

fOSC(kHz) = (K * 10^3) / Cosc; Cos in pF , K = 34 for max7405
5000(k) = (34 * 10^3) / ( C * 1pF)
5000 = 34 / ( C * 1pF)
C = 34 / 5nF
C = 6800M

By using the above equation, I calculated an extreme large capacitor with Mega value which is not reasonable.

Am I calculating it in the right way?

  • \$\begingroup\$ should be internal clock \$\endgroup\$ – jhyap Jul 23 '14 at 7:22
  • \$\begingroup\$ From the description, it says the fc:fclk = 1:100 \$\endgroup\$ – jhyap Jul 23 '14 at 7:24

For 50 Hz cut-off frequency, you would need fOSC = 5000 Hz = 5 kHz, because it says on the data sheet that, there is a 1:100 ratio from corner frequency (or cut-off frequency) to internal oscillator frequency fOSC. Then the equation requires the numerical value of fOSC in kHz units, which we will use fOSC(kHz) = 5 for this. Then K = 34. Our equation will read as:

fOSC( kHz) = (K * 10^3)/ Cosc; Cosc in pF

5 = (34* 10^3)/ Cosc;

Cosc = 6800 solving this equation.

But remember that this value is given in pF. So this means that Cosc = 6800 pF = 6.8 nF. I think you only had a confusion about the (C*1 pF) part, where you could better write (C/1 pF) for the correct unit conversion.


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