JimL11
1000+ Head-Fier
Well, Dr. GIlmore tends to be terse, whereas I tend to be wordy. So here goes.
So, there are a number of questions here. First, an easy way to determine whether a load is resistive, capacitative or inductive. The impedance describes the relationship between voltage and current as a function of frequency - it is a generalization of idea of resistance for AC frequencies, which includes the ratio between voltage and current, but also includes the phase relationship between voltage and current. A pure resistive load has a constant ratio between voltage and current with frequency (constant impedance), with a pure inductive load the ratio between voltage and current increases linearly with frequency (increasing impedance), and the ratio between voltage and current in a pure capacitative load decreases inversely with frequency (decreasing impedance). This means that for the same voltage, the current through a resistor is constant with frequency, the current decreases linearly with frequency for an inductor and increases inversely with frequency for a capacitor. However, with AC frequencies there is also a phase relationship between voltage and current - with a pure resistive load, the voltage and current are completely in phase, with a pure capacitor or inductor the voltage and current are 90 degrees out of phase. Real components are not purely resistive, inductive or capacitative, but mixtures of two or three of them. So, for example, if you look at the impedance of a real capacitor, it will decrease with frequency up to some point, then may start to increase as an inductive component appears.
A simple way to picture the phase relationship between voltage and current is to take a capacitor that is charging and discharging as you run a sine wave across it. As the voltage rises from zero, charge flows into it (current), as the voltage rises to a maximum, the current flow slows and stops, as the voltage decreases, the charge flows out of it, and as the voltage hits zero, the current flow out is maximum, then as the voltage decreases below zero and hits its negative maximum, the charge flow out slows and stops, then as the negative voltage decreases, the charge starts to flow back in. So the current flow is maximum when the voltage is zero and zero when the voltage is maximum (either positive or negative).
What you are probably thinking of when you say that capacitative loads demand more energy in high frequencies while inductive loads demand more energy at low frequencies, is that the impedance number (the ratio between voltage and current) is lower for capacitors at high frequencies and lower for inductors at low frequencies. But the impedance number is not the same as the energy "demand."
The amount of energy that is "burned up" is voltage x current. For DC this is straightforward, for AC, you have to take into account the phase between voltage and current. For a pure resistor, voltage and current are in phase, so energy is indeed burned up in a resistor. For a pure capacitor or inductor, the voltage and current are 90 degrees out of phase, so when you multiply voltage x current over a full cycle, the result is zero. Again, picture the pure capacitor - as the AC cycle varies across it, the charge sloshes in and out of the capacitor, but because this action is "frictionless" no energy is used up. In fact, of the three, capacitor, inductor and resistor, only a resistor burns up energy.
So, we need to be precise in our terminology. Impedance and energy are two different things. Impedance refers to the ratio and the phase between voltage and current. Energy is the product of voltage x current (taking phase into account). The impedance of a resistor is constant, the impedance of an inductor increases with frequency and the impedance of a capacitor decreases with frequency. If we look at an impedance graph of a headphone, if it is flat with frequency we know that it is behaving like a resistor in that range, and also that the voltage and current are in phase. If we see that the impedance is rising with frequency, we know that there is an inductive component (there may be a resistive one as well) and that the voltage and current are at least partially out of phase, and if the impedance is falling with frequency we know that there is a capacitative component ( again, there may be a resistive component as well) and the voltage and current are at least partially out of phase.
With real headphones, there is always a resistive component to the impedance. That resistive part is largely due to the real wires in the voice coil, which have resistance, but also to some small degree, to the fact that it is making sound, which is energy, and which therefore must be supplied by the amplifier. Usually, that sound energy is only a small proportion of the energy that is burned up due to the voice coil resistance, etc. Which is another way of saying that the efficiency of the headphone is low, as most of the energy supplied by the amp is burned up as heat.
Electrostatic headphones don't have voice-coils of course, but they make sound (which takes energy) and also are not very efficient so again, a significant part of the energy that is supplied by the amp is burned up as heat.
Now, let's get back for a moment to the spectrum of energy in music, which tends to be maximum in the 50-500 Hz range, and look at the implications for electrostatic headphones in particular. What this means is that the bulk of energy that is burned up in the headphones as sound and/or heat occurs in this region. Since electrostatic headphone amplifiers are more or less voltage sources, this means that the output voltages are proportional to the square root of the energy, which in turn means that the maximum voltage and in-phase current occurs at these frequencies (remember, energy is voltage x current). For electrostatic headphones, their impedance is highest in this region, which means that the out-of-phase current demands in this region are relatively low, although, because the absolute voltages are highest, the absolute out-of-phase current may be fairly high. At high frequencies (say, 5000-20 kHz), the amount of musical energy is relatively low, so voltage is lower, in-phase current is lower, but the impedance is relatively low so the relative amount of out-of-phase current is higher (the absolute amount of out-of-phase current may or may not be higher depending on the frequencies involved).
Finally, Dr, Gilmore's statement that the SR007 can pull 1 watt per channel at 20 kHz doesn't have any information on the output voltage or current, or the sound pressure output of the headphones, to make a useful comment. I cannot say how much the headphones will pull at 50 Hz or 500 Hz.
