The difference between impedance and resistance


My Dad would be ashamed for me for writing this thread (he had a background in electrical engineering) but can someone explain the differnce between impedance and resistance (other than the former is for ac and the latter dc) as it pertains to audio circuitry?
russellrcncom
Hi Bob,

The 2pi factor converts the units of frequency from cycles per second (Hertz) to radians per second, one radian equaling 360 degrees/2pi = approximately 57.3 degrees.

The need for using radians per second is inherent in the calculus (differentiation and integration) that I'll get into below, but those mathematical subjects were not exactly my forte and I can't explain that aspect of it any further :-)

The 90 degree phase shifts basically stem from the relations between voltage and current for inductors and capacitors. An inductor resists abrupt changes in current, and the voltage across an inductor and the current through it are related by the equation:

V = L(di/dt)

Where di/dt is the rate of change of current, or the derivative of current to use the calculus term. V is voltage, of course, and L is inductance.

That's the basis of the theory behind spark coils, btw. If the rate of change of current is near instantaneous, as would happen if some part of the path of the current flowing through the inductor were to suddenly be opened, the resulting voltage will be extremely large.

Capacitors resist abrupt changes in voltage. The relation between voltage and current is:

i = C(dV/dt)

Where dV/dt is the rate of change of voltage, or the derivative of voltage, in calculus terminology. C is capacitance, of course.

Based on standard calculus and differential equations:

The derivative of a sine wave is a cosine wave, which is the same as a sine wave except that the sine wave lags by 90 degrees.

The integral of a sine wave is an inverted cosine wave plus some constant. A sine wave leads an inverted cosine wave by 90 degrees.

For an inductor, rearranging the equation V = L(di/dt) to (di/dt) = V/L, and taking the integral of both sides of the equal sign, we have:

i = integral(V/L)

So if V is a sine wave, V will lead i by 90 degrees

For a capacitor we have:

i = C(dV/dt), so if V is a sine wave, per the above relationship that the derivative of a sine wave is a cosine wave, V will lag i by 90 degrees.

Whew! And all we want to do is to listen to well reproduced music :-)

Best regards,
-- Al
Al covered lots of good details but I would like to summarize it for those that might have gotten lost.

The fundamental difference has to do with frequency, as in frequencies of the recorded music, for example. There are innate qualities of capacitance and/or inductance in certain components (like capacitors and inductors) but they are not called into action until there is a signal composed of frequencies (such as music) interacting with them.

As the frequency of the signal increases, the way the signal sees every electrical part in the system changes dramatically. The higher the frequency, the more inductance and capacitance, quite literally, appear. Eventually you get to a point where even a resistor acts like a capacitor and an inductor! And then after that, even a wire exhibits inductive and capacitive effects! It is a crazy world - and it's no wonder cable and component break-in occurs.

Since music is not a DC signal (frequency of zero), technically speaking there is no such thing as a pure resistance in a stereo system. Everything has some sort of reactance associated with it, be it capacitive or inductive, which requires the use of the term "impedance." This is ever more accurate as frequencies go above human hearing, but there is no mathematical lower limit for this effect. This is why it pertains to our stereos.

As far as a "bad load" is concerned, it is tougher to drive a capacitive load than an inductive or resistive one. Capacitive loads can be seen in impedance plots as negative phase angles, meaning the current is ahead of the voltage. Not only does the efficiency of the circuit decrease if the load is capacitive, but the current demands during transients can be significantly more than looking at resistance alone. That double whammy is hard on an amplifier.

Arthur

Aball

As the frequency of the signal increases, the way the signal sees every electrical part in the system changes dramatically. The higher the frequency, the more inductance and capacitance, quite literally, appear. Eventually you get to a point where even a resistor acts like a capacitor and an inductor! And then after that, even a wire exhibits inductive and capacitive effects! It is a crazy world - and it's no wonder cable and component break-in occurs.

I don't see what any of that has to do with break-in.
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