Speaker Impedance: Resistance and Phase Question

Al, I think you touched on my question in your post above about "[t]he phase angle of the impedance of a speaker, at a given frequency, is the angle of the "vector sum" of the magnitudes of the resistive, inductive, and capacitive components of the impedance." But let me try to break this down a little bit.

Let's focus on what a speaker's phase angle might look like at a x-over point. Say for example the x-over point is where the woofer rolls off and the midrange picks up. Would the designer use an inductor in series in the woofer driver to roll off frequencies over a certain point. If so, I would surmise the phase angle might be inductive, or positive.

But at the same time, the designer might want to filter low frequencies from reaching the midrange below a certain point by using a capacitor in series with the midrange driver. Again, I surmise the phase angle might be negative.

So both a inductor and a capacitor might be in the circuit within the same frequency range. How would the interaction of these two passives plot as a phase angle?

P.S. My example is intentionally oversimplified in order to make a point. I suspect that the drivers themselves could be inductive or capacitive at certain frequencies. Plus I also suspect that modern x-overs don't simply insert one passive in series with a driver to achieve the desired roll-offs.

Thanks

Bruce

Bruce, I'm not totally certain, but I believe the answer to your question is that at the crossover frequency the phase angle of the speaker's impedance will be zero degrees, i.e., it will appear to the amplifier to be resistive. Per your well stated question, this is based on the hypothetical assumptions that the driver impedances are purely resistive and are equal in value, and that the rolloffs of the low and mid frequency parts of the speaker are symmetrical.

Best,
-- Al

The phase angle is zero at resonance. That is when the capacitive load and the inductive load components cancel each other out at a specific frequency and you are left with only a resistive load.

enjoy