BTW, I have been meaning to ask you about something. I want to hook up 2 tweeters to the one L-pad. I could do it 2 ways. 1st--- (in parallel) hookup both tweeter positive wires to L-pads output or 2nd---- (series) hookup a wire from l-pad output to 1st tweeter positive input and then another wire from 1st tweeter to 2nd tweeter positive input. I think this would give different impedances. Comments ??
Either approach is 100% guaranteed to
FAIL. I suggest reading a good book on speaker design which will explain all of the issues with which you are wrestling. Deeper understanding is required to not make a mess out of the speakers.
Ohm's Law remains the law of the land:
If the two drivers are series wired, the total resistance is increased: 8 Ω + 8 Ω = 16 Ω
If the two drivers are parallel wired, the total resistance is decreased: 8 Ω || 8 Ω = 4 Ω
The crossover is another name for an
electrical filter, which was, of necessity, designed for a particular resistance and
only that resistance. I suggest reviewing the formulas below set forth, which will should clearly render the issues.
As you may below see, given that for an low-pass filter (inductor)
fc ≈ R, and for a high-pass filter (capacitor)
fc ≈ 1/R. So for the high-pass tweeter filter, doubling
R with the high-pass filter will reduce the the corner frequency by half. The overlap with the midrange will not only sound terrible — excessive volume, comb artifacts from geographically distant drivers, etc. — depending upon the tweeter type the lower frequencies may result in damage or destruction.
Here is what I have previously posted about the actual filter calculations, since most people are using calculators which mask understanding the effects of changing values for the filter components.
Formulas for crossover filters:
Given:
R is Nominal Speaker Impedance (Ω)
C is Capacitance (Farad)
L is Inductance (Henry)
fc is Corner Frequency (Hertz)
Inductor Formulas:
L = R / (2π × fc) (Henry)
fc = R / (2π × L) (Hertz)
R = 2π × L × fc (Ω)
Capacitor Formulas:
C = 1 / (2π × fc × R) (Farad)
fc = 1 / (2π × C × R) (Hertz)
R = 1 / (2π × C × fc) (Ω)
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 (uF) multiply by 1,000,000.
To convert from microFarad (uF) to Farad (F) divide by 1,000,000 or multiply by 0.000001.
The easiest way to run these calculations with a calculator, be it physical or computer, is to compute the invariant portion, store in memory, and then tinker with values. For example, the inductor calculations for the low-pass filter can be performed as:
fc = R / (2π × L) (Hertz)
R = 8 Ω
L = (unknown value) in milliHenries, not Henries
fc = 8 / (2π × 0.001 × L) (Hertz)
Compute without L and then divide by the mH value (not scaling to Henries) as needed, essentially:
fc ≈ 1,273 / L (Hertz)
Example:
L = 1.59 mH
fc ≈ (memory recall) / 1.59 (Hertz)
L = 1.60 mH
fc ≈ (memory recall) / 1.6 (Hertz)
So, mechanically, compute the entire equation
without L, and store in memory. Then recall that value and divide by L as mH. For example, to compute the frequency for, say, 1.59 mH
versus 1.6 mH, compute everything except L, store in memory, then do a memory-recall and divide by 1.59 and then a memory-recall and divide by 1.6, since the 0.001 factory is already entered. Makes it far less time consuming.
A similar trick can be done with capacitance calculations for the high-pass filter.
f = 1 / (2π × C × R) (Hertz)
R = 8 Ω
C = (unknown value) in µF
f = 1 / (2π × 0.000001 × C × 8) (Hertz)
Compute without C and add as needed, essentially:
f ≈ 19,894 / C
Example:
C = 8.2 µF
f ≈ (memory recall) / 8.2 (Hertz)
C = 8 µF
f ≈ (memory recall) / 8 (Hertz)
So if you wanted to compute the difference between 8 µF and 8.2 µF (standard value) all that would be needed is to divide the pre-computed value by 8 or 8.2.
Greatly simplifies the calculations and saves considerable time.
