Hi Nick,
Basically, if the speaker's impedance phase angle becomes significantly negative (i.e., capacitive) across a significant range of frequencies, the amount of current that the amplifier has to supply, and the amount of current that will flow through the speaker cables, will increase at those frequencies.
That will occur for two reasons. First, everything else being equal the efficiency of the speaker will be less than if the phase angle were zero degrees (i.e., purely resistive). The amount of power delivered at a given instant is proportional to the product (multiplication) of voltage (V) and current (I) at that instant. For a purely resistive load, voltage and current are in phase, meaning that they reach their maximum, minimum, and other corresponding values within each cycle at the same time. For a load that is partially capacitive and partially resistive (i.e., with a phase angle somewhere between zero degrees and -90 degrees), voltage and current will be somewhat out of phase (to a degree that increases as the negative phase angle increases), resulting in a reduction in V x I at any instant of time, and therefore less power delivery than if they were in phase. So to deliver a given amount of power and produce a given volume level, the amplifier will have to deliver more current and voltage if the load is significantly capacitive than if it were purely resistive, everything else being equal.
Secondly, the amount of current flowing through a capacitor is proportional to the rate of change of the applied voltage. So if the impedance of the speaker is significantly capacitive at high frequencies, a greater amount of current will have to be provided during high speed (rapidly changing) transients than would otherwise be necessary.
If the impedance magnitude (the number of ohms) reaches low values at frequencies where the phase angle also reaches low (more negative) values, the increased current requirements resulting from those effects will be further compounded.
Best regards,
-- Al
Basically, if the speaker's impedance phase angle becomes significantly negative (i.e., capacitive) across a significant range of frequencies, the amount of current that the amplifier has to supply, and the amount of current that will flow through the speaker cables, will increase at those frequencies.
That will occur for two reasons. First, everything else being equal the efficiency of the speaker will be less than if the phase angle were zero degrees (i.e., purely resistive). The amount of power delivered at a given instant is proportional to the product (multiplication) of voltage (V) and current (I) at that instant. For a purely resistive load, voltage and current are in phase, meaning that they reach their maximum, minimum, and other corresponding values within each cycle at the same time. For a load that is partially capacitive and partially resistive (i.e., with a phase angle somewhere between zero degrees and -90 degrees), voltage and current will be somewhat out of phase (to a degree that increases as the negative phase angle increases), resulting in a reduction in V x I at any instant of time, and therefore less power delivery than if they were in phase. So to deliver a given amount of power and produce a given volume level, the amplifier will have to deliver more current and voltage if the load is significantly capacitive than if it were purely resistive, everything else being equal.
Secondly, the amount of current flowing through a capacitor is proportional to the rate of change of the applied voltage. So if the impedance of the speaker is significantly capacitive at high frequencies, a greater amount of current will have to be provided during high speed (rapidly changing) transients than would otherwise be necessary.
If the impedance magnitude (the number of ohms) reaches low values at frequencies where the phase angle also reaches low (more negative) values, the increased current requirements resulting from those effects will be further compounded.
Best regards,
-- Al