I have a pair of 2 way speakers that are labeled 8 ohms. They have a crossover in them. There are 4 wires from the crossover to the drivers; a + and - to the woofer and a + and - to the tweeter. This looks like parallel wiring. I want to update the woofer and tweeter. The originals have no markings that identify either their impedance or manufacturer. If I go 8 ohms on both the woofer and tweeter and they are wired in parallel to the crossover, then is the speaker 8 ohms at the terminals or 4 ohms??? Why?
If the drivers are 8 ohms and the crossover is competently designed then the impedance at the terminals will be 8 ohms.
Note that what you measure with a multimeter is dc resistance and as there will be a cap in series with the tweeter, you are only measuring the dc resistance of the woofer and any coils in series with it.
AC impedance will vary with frequency, but in the case of both the woofer and the tweeter (assuming they have at least a coil and a cap in series with them) then the filters effectively make the impedance much higher either below or above the filters corner point.
A high impedance in parallel with a low impedance is effectively the same as the low impedance. So whilst the drivers are in parallel, their impedances are not flat and will rise significantly below fc for the high pass, and above fc for the low pass.
Tony.
Note that what you measure with a multimeter is dc resistance and as there will be a cap in series with the tweeter, you are only measuring the dc resistance of the woofer and any coils in series with it.
AC impedance will vary with frequency, but in the case of both the woofer and the tweeter (assuming they have at least a coil and a cap in series with them) then the filters effectively make the impedance much higher either below or above the filters corner point.
A high impedance in parallel with a low impedance is effectively the same as the low impedance. So whilst the drivers are in parallel, their impedances are not flat and will rise significantly below fc for the high pass, and above fc for the low pass.
Tony.
What Wintermute said, plus, there is only a small region where both the woofer and tweeter are active ... at bass frequencies only the woofer is going, at treble only the tweeter. So in effect they never actually appear in parallel.
So, Wintermute, by cap I take it you mean capacitor? So then, the crossover itself is dictating the dc resistance (impedance) at the terminals and is measuring only the woofer when the speaker is disconnected from a power source?
I think the point that has to be made to validate this is that we are only interested in the impedance at audio frequencies. If you apply DC it will only pass through the woofer but as frequencies rise it will also pass through the tweeter. As a thought experiment, it would also be valid to filter the woofer through a large capacitor to block infrasonic signals, then the multimeter would read infinite resistance.
Yes cap = capacitor. Since a capacitor by nature can't pass DC, the tweeter is out of circuit when measuring resistance with a multimeter.
Tony.
Tony.
Substituting drivers with a crossover that was not designed specifically for them is unlikely to produce satisfying results. Impedance is the least of your problems.
I was wondering when that would come up 😉 I probably should have mentioned that in my first post, but was sticking strictly to the question asked......
Tony.
Tony.