

Electrical impedance made easy - Part 1 - xd
http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-impedance-made-easy.html

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kwantam
Put simply, impedance is just complex resistance. We get it by taking the
Laplace transform of the differential equation that defines the relationship
between current and voltage in a capacitor or inductor.

    
    
        I = C dV/dt
        Laplace(dV/dt) = V*s
        I = C * V * s
        V / I = 1 / (C * s)
    
        V = L dI/dt
        Laplace(dI/dt) = I*s
        V / I = L * s
    

Another way to look at this: since this is a linear system, we know its
eigenvalues are complex exponentials, i.e., their real and imaginary parts are
sinusoids of any frequency (by the Euler identity). So we can say

    
    
        V = Vm * exp(j*omega*t) -- using j = sqrt(-1) to avoid confusion with current
        dV/dt = Vm * j * omega * exp(j*omega*t)
              = j * omega * V
        V / I = 1 / (C * j * omega)
        | V / I | = 1 / (C * omega)
        arg(V/I) = -pi/2 -- voltage lags current by 90 degrees
    

By a similar argument,

    
    
        | V / I | = L * omega
        arg(V/I) = pi/2
    

We can see that as expected, s=j*omega.

~~~
haberman
All I can say is that if this is the "simple" explanation to you, you and I
think in very different ways. :)

To me, it's _much_ easier to understand the visual where impedance is the
hypotenuse of a triangle whose legs are resistance and reactance.

~~~
jules
That doesn't really _explain_ impedance. First of all, you need to know what
resistance and reactance are to get that explanation, and second it doesn't
explain why it's useful or how to do calculations with impedance.

On the other hand, kwantam's explanation is easy and clear, provided you know
what a Laplace transform is (which you're going to need to understand
reactance anyway). If you set up the system of differential equations for a
linear circuit, how would you go about solving these equations? You take the
Laplace transform, and you get the equivalent algebraic equations of the form
V = ZI for each element. The numbers Z that appear are the impedances.

From a mathematical viewpoint, impedance is a simpler and more natural concept
than resistance and reactance, since it appears naturally when you solve
linear circuits.

~~~
haberman
> That doesn't really explain impedance. First of all, you need to know what
> resistance and reactance are to get that explanation, and second it doesn't
> explain why it's useful or how to do calculations with impedance.

It's amazing to me that we can see this in such opposite ways. To me kwantam's
post doesn't explain impedance at all, it just formally defines it. It doesn't
even begin to explain to me what it measures or how it applies to an actual
circuit. What components of a circuit have an impedance? If I increased the
impedance, what would happen? How would I even do that? kwantam's post doesn't
begin to answer these questions.

I don't get any intuition at all for what this concept _actually means_ from a
formal definition. That's why teaching methods that start with formalisms have
always frustrated me. I bet you could dress up a really simple concept in a
formalism that would be totally indecipherable. I actually think this would be
fun to do sometime: describe some mundane concept in a really formal way and
then ask a bunch of mathematical people "what am I describing?" And it turns
out to be something boring like the rules for driving through an intersection.

I guess my point is, formalisms seem obvious once you understand the concept
intuitively, but are a terrible way to gain that intuition. The basic idea of
resistance is simple (a resistor restricts the flow of electricity), reactance
was new to me but I got a rough idea that it similarly resists the flow of
electricity in an AC circuit and is a function of the frequency.

So if I know that the impedance is a right-triangle combination of those two
things, I've got a great start for getting some intuition: I know that
anything that has resistance _or_ reactance also has impedance, and something
that has _both_ has even more impedance, but in a DC circuit where the
frequency is zero you'd get some edge condition where a non-zero reactance
would, let's see, act as infinite resistance, I think? The formal definition
doesn't plant any of these paths of understanding or inquiry in my brain.

~~~
mturmon
I understand the Laplace transform ideas above (used to TA the class where
it's taught) and I agree with your that they are more definitional than
explanatory.

To provide actual explanatory power, you'd have to explain what the reactive
portion ("imaginary part" in the phasor notation) of the impedance is actually
doing in real terms.

What it's doing is to cause the current to lead or lag the voltage in the AC
signal. In a purely resistive circuit, current and voltage vary exactly the
same way at every instant.

You can actually go a long way with this just by understanding that, e.g., for
an inductor,

V = L dI/dt

so that small current changes will imply very large voltage changes. If you're
in to intuition, this means "voltage leads current" (i.e., the sinusoid of
voltage is 90 degrees ahead of the sinusoid of current).

The above equation is also why lights in your house dim when you turn on a
motor (= inductor). There is a significant current change, and the voltage
changes even more, so the lights dim.

Or, you can be even more intuitive, and note that it's obvious that when you
put a coil of wire (= motor = inductor) across the two sides of your house's
electrical wiring, the coil of wire will pull the + and - conductors together
in voltage at first. Until the magnetic field sets up to provide reactance,
the coil of wire is basically a short circuit.

And finally, a capacitor is just the dual of the inductor.

~~~
flatulent1
"The above equation is also why lights in your house dim when you turn on a
motor (= inductor). There is a significant current change, and the voltage
changes even more, so the lights dim."

Uh, no. Most of the voltage drop in the source occurs in the resistance of the
wiring, and you didn't even figure resistance into the equation. Intuitively,
the surge current more closely matches that of a capacitor charging through
resistances. It isn't even a matter of the motor inductance or the frequency
of the power. In fact the early surge would be seen even if we used D.C. power
and a D.C. motor. The time constant matching the light dimming involves the
mass of the rotor being brought up to speed. The load is actually more
resistive at startup than when an A.C. motor is up to speed and coasting.

A pure inductor starts at zero current the instant voltage is applied and it
ramps up at a rate determined by the inductance, at D.C. or very low
frequencies rising to where it becomes limited by the series resistance. It's
pure capacitors that draw a maximum current at the instant voltage is applied
to the circuit.

I believe that many people get lost in the math and lose a feel for what is
happening. What you said is completely backwards from reality. Mechanical
analogies are far easier for people to get a feel for than Laplace Transforms.
Compare the behavior of an RLC circuit to a machanical system with a shock
absorber, spring, and a mass.

People may get confused over a critically damped circuit, but the idea of a
car oscillating up and down after a bump with no shock absorber is easy to
grasp.

~~~
mturmon
There is no way a motor is well-modeled by a capacitor, either analytically or
intuitively. It's a coil of wire, practically a poster child for an inductor.

Here's a motor's equivalent circuit from Wiki:

[http://en.wikipedia.org/wiki/Induction_motor#Equivalent_circ...](http://en.wikipedia.org/wiki/Induction_motor#Equivalent_circuit)

This does have lumped resistance elements, but by the time you put the two
inductors and two resistors in to the differential equation with forcing

    
    
      sin (2*pi*60*t) u(t), 
    

you're not going to be in a good place as far as intuition goes.

In fact, overall, I'm not seeing much intuitive sense in your summary. Once
you really delve in to it, the subject of starting electric motors (AC and DC)
is very complex.

------
haberman
The only place where I've encountered the term "impedance" is in relation to
the hi-z and lo-z instrument cables that you use to connect eg. electric
guitars to amplifiers. You'd sometimes see these hi-z to lo-z converters. Is
this the same thing?

The video's explanation would lead me to believe that impedance only applies
to AC circuits. But instrument cables don't run AC, do they?

~~~
kwantam
In the case you're talking about, impedance means "characteristic impedance",
i.e., one parameter which (along with speed of light in the given medium and
length) defines the behavior of a transmission line. [1]

It is the same thing in the sense that mathematically both are a complex
(i.e., Real plus Imaginary) relationship between voltage and current. However,
a transmission line is a two-port network while a capacitor is a one-port
network, so these two senses of impedance correspond to different physical
phenomena. In a capacitor, the impedance defines the relationship between
voltage and current at the port (i.e., between the two terminals of the cap;
this is sometimes called the driving point impedance). For a transmission
line, the characteristic impedance defines, among other things, the equivalent
impedance seen at one end of the line when the line is either infinitely long
or an integral number of wavelengths long and terminated at the other end by
an impedance of the same value as the characteristic impedance.

Instrument cables do indeed carry AC signals, inasmuch as the signal on the
line corresponds to the frequency of the note(s) you're playing. AC is just
"non-zero frequency"---any signal which is not a constant, unchanging (DC)
value is considered AC. So, 50/60Hz line power is an AC signal, the 2.4GHz
signals used by your 802.11 network card are AC, and middle C (261.6 Hz) is
AC.

[1] <http://en.wikipedia.org/wiki/Characteristic_impedance>

~~~
lcargill99
At audio frequencies, characteristic impedance is almost always irrelevant.

hi-Z is a crystal mic, guitar pickup or crystal phono pickup source.lo-Z is a
dynamic or condenser mic source. Obviously, a preamp may be used as an
impedance converter to lower hi-Z to loZ. There's such a preamp in the
condenser mic.

The cables are hiZ or loZ only in that they're generally used with one source
or the other. But that's by convention. loZ is usually a balanced XLR
connection; hiZ is generally an unbalanced 1/4" connection.

hiZ is generally at around 1 megohm. loZ is archetypally 600 ohm. Line is
neither; it's 10,000 to 50 or 100k ohm.

------
mrothe
I don't know about the rules in America, but this circuit design wouldn't be
approved for use here in Europe. Circuits are not allowed to draw what is
called 'reactive power', because it stresses the grid. Therefore you have to
compensate it (in the simplest case using an inductor or a capacitor).

------
xd
Part 2:

[http://benkrasnow.blogspot.com/2011/06/tutorial-
electrical-i...](http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-
impedance-made-easy_19.html)

