Yes, this is
absolutely correct for a
first-order crossover at 8Ω, at least for standard inductor values. The exact frequency would be 12.7 mH but that value requires two inductors, and it isn't that critical.
Ummm, no, that's incorrect.
JimPA is
completely correct.
Consider the formulas (cut-and-paste from a posting I elsewhere made):
Given:
R is Nominal Speaker Impedance (Ω)
C is Capacitance (Farad)
L is Inductance (Henry)
f is Frequency (Hertz)
NB:
L is inductance in Henry.
To convert from Henry (H) to milliHenry (mH) multiply by 1,000.
To convert from milliHenry (mH) to Henry (H) divide by 1,000.
C is capacitance in Farad.
To convert from Farad (F)to microFarad (µF) multiply by 1,000,000.
To convert from microFarad (µF) to Farad (F) divide by 1,000,000 or multiply by 0.000001.
Inductor Formulas:
L = R / (2π × f) (Henry)
f = R / (2π × L) (Hertz)
R = 2π × L × f (Ω)
Capacitor Formulas:
C = 1 / (2π × f × R) (Farad)
f = 1 / (2π × C × R) (Hertz)
R = 1 / (2π × C × f) (Ω)
Running the math for an
8 Ω speaker clearly yields the correct values as above set forth by both
JimPA (in No. 19) and by myself (in No. 11):
L = R / (2π × f) (Henry)
= 8 Ω / (2π × 100 Hz)
= 12.7 milliHenry
C = 1 / (2π × f × R) (Farad)
= 1 / (2π × 100 Hz × 8 Ω)
= 199 µF
Random internet calculators may yield different values, of course, but one can always verify the math as per the above and see that any other value would be incorrect.
So unless your speaker is 16 Ω, and I looked up the JBL 136A which is listed as 8 Ω, your crossover points are incorrect, leaving a large hole: twice the frequency for the 100 µF capacitor (now 199 Hz), and half for the 20 mH inductor (now 64 Hz). Halving the capacitor doubles the frequency, while doubling the inductor halves the frequency. This is why tweeter capacitors are small and woofer inductors are large. I suspect you misremembered the values, as you would have noticed such a hole in your frequency response.