- #1

- 598

- 0

[tex]D=\frac{R_B}{R_A+2R_B}[/tex]

[tex]f=\frac{1.44}{\left(R_A+2R_B\right)C}[/tex]

where C is my capacitance value and the two R's my resistance values. This should be a strait forward calculation right?

then from the above equations

[tex]R_B=\frac{D 1.44}{C f}[/tex]

and[tex]R_A=\frac{D 1.44}{C f}\left[\frac{1}{D}-2\right][/tex]

right?

I get [tex]R_B=36k\Omega[/tex] and [tex]R_A=72k\Omega[/tex]

This is a capacitance value of [tex]0.01\mu F[/tex]

Does this look correct?

Thanks