Solving for 3db frequency of RC filter

[uoficowboy]

I'm having a senior moment. I know the 3db frequency of an RC LPF is 1/(2*pi*R*C). My memory is that you can solve for that by setting R = C. So:

R = magnitude(1/JwC)
R = 1/(wC)
R = 1/(2*pi*F*C)
F = 1/(2*pi*R*C)

So that all seems right. But if R = magnitude of C doesn't that mean your gain is 0.5? (voltage divider where both top and bottom elements are the same). 20*log(0.5) = 6db, not 3db!

I'm getting the right answer but it doesn't feel to me like it should be right. What am I missing? Thank you!

 

[scopeprobe]

I'm having a senior moment. I know the 3db frequency of an RC LPF is 1/(2*pi*R*C). My memory is that you can solve for that by setting R = C. So:

R = magnitude(1/JwC)
R = 1/(wC)
R = 1/(2*pi*F*C)
F = 1/(2*pi*R*C)

So that all seems right. But if R = magnitude of C doesn't that mean your gain is 0.5? (voltage divider where both top and bottom elements are the same). 20*log(0.5) = 6db, not 3db!

I'm getting the right answer but it doesn't feel to me like it should be right. What am I missing? Thank you!

Gain = power(10,(dB/20))

-3db = Attenuation of 0.707946
-6dB = Attenuation of 0.501187

Not sure where your getting the R=C from but your math is correct to find the 3db point. 3dB would need an attenuation of 0.707946 as shown above

 

[FvM]

The condition for -3 dB point is R = Xc, Xc being capacitive reactance magnitude. But why is attenuation of RC voltage divider 0.7 rather than 0.5 for equal magnitude of both impedances? That's due to phase shift. Resistor and capacitor voltage are 90 degree phase shifted, relation of voltages can be visualized by phasors, both phasors form a right angled triangle, their magnitudes add as squares. |Vr + Vc| = 1.41 * |Vr| = 1.41 * |Vc|.

 

[KlausST]

My memory is that you can solve for that by setting R = C

This is not correct.

It should be: R = Xc

and Xc = 1/( 2 * Pi * f * C) = 1 / ( w * C)

Klaus