Electrical impedance made easy - Part 1 | Hacker News Reader

Electrical impedance made easy - Part 1

benkrasnow.blogspot.com

58 pointsxd21 comments

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.

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.

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.

> 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.

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.

"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.

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...

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.

I can certainly see where you're coming from, I have the same problem with many of my physics courses. Too much mathematical formalism and not nearly enough talking about actual physical systems and intuition.

> 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. The formal definition tells me none of this.

This is a rather superficial understanding of impedance. Since impedance is a complex quantity, "more impedance" doesn't really make sense unless you define an order on complex numbers. Reactance is much more complicated than resistance, as you say because it is a function of frequency instead of one number. Impedance has the same difficulty: it is a function of complex frequency s.

Suppose we have an electrical element with impedance Z(s), then we have V(s) = Z(s)I(s). The problem with this equation is that it is an equation in the complex frequency domain, just like you get equations in the ordinary frequency domain with Fourier transforms. What we do with Fourier transforms is to look at what happens at a particular frequency omega. For example if you had V(omega) = Z(omega)I(omega), then you'd know that if you applied a sine wave with frequency omega current, you'd get a sine wave scaled by Z(omega) voltage. Or equivalently, if you applied I(t) = e^(i omega t) current, you'd get V(t) = Z(omega) e^(i omega t) voltage. Laplace transforms just generalize this complex exponential to e^(-st) where s is an arbitrary complex number. So this means that if an electrical element has impedance Z(s), then if you apply e^(-st) current, then you get Z(s)e^(-st) voltage.

For example say we chose s = a - bi, then e^(-st) = e^-a e^(bi), an oscillating wave at frequency b that is exponentially damped with parameter a. Say that at that particular complex frequency, Z(s) = Ae^(i theta). Then if we input the current e^(-st) we get voltage Z(s)e^(-st) = Ae^(i theta) e^-a e^(bi) out. In other words, the electrical element amplifies the current-to-voltage by a factor of A, and it rotates the phase by an angle theta. Or in simpler terms, if you are comfortable with complex numbers: an impedance Z multiplies the current by Z to get the voltage, or V=ZI.

Of course in general the current is a superposition of many damped waves, and the voltage will be a superposition of many damped waves too. But each of the single frequency damped waves at complex frequency s in the current will be multiplied by Z(s) to get the voltage.

Now lets apply this knowledge to a practical situation. Suppose we have a simple circuit with a single inductor L. Then we have Z=sL (the impedance of an inductor is sL). So we have V = sLI. Lets see what happens if we apply an oscillating input signal with frequency omega: s = i omega. Then we have V = i omega L I. In other words, the voltage is the current rotated in phase by omega L.

The nice thing about Laplace transforms is that we can not only analyze pure sine waves easily, we can also analyze decaying input, say I(t) = e^(-2t), here s=2 What voltage do we get? V = sLI = 2LI. That is, the voltage is 2L times the current. So we can see that the faster the current decays, the higher the voltage will be (for example if we take s=3).

In summary: instead of thinking of a signal as a sum of sine waves, think of it as a sum of damped sine waves e^(-st) where s is a complex number. Then the impedance Z(s) gives a relation between the voltage and the current for a particular damped sine wave at complex frequency s: V=ZI. So what an electrical element with impedance Z(s) is "doing": it first takes the current apart as a sum of damped sine waves of frequency s. Then it transforms each of these sine waves by multiplying them by Z(s). Then it sums all the transformed waves back together to get the voltage. Of course you can do the inverse process to get the current from a voltage by dividing every wave by Z(s) instead of multiplying. To see what multiplying by Z does you need basic complex arithmetic: if Z=Ae^itheta, then it scales by A and rotates the phase by theta.

If you understand the Laplace transform you understand impedance. If you understand impedance you understand the Laplace transform.

> 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?

That's all in his post:

>>> 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

Granted, he doesn't take the extra step to say that Z = V/I. Once you see that, you can see that impedance is proportional to inductance and frequency or inversely proportional to capacitance and frequency. (I feel that your participation in this discussion implies that you know what V, I, C, and L stand for.)

> I know that anything that has resistance or reactance also has impedance, and something that has both has even more impedance,

Everything has resistance and reactance. 0 is still a number. It also doesn't follow that two non-zero numbers are greater than one non-zero numbers.

> 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?

There are incredibly few "DC" circuits. Most of what you'd think to call DC circuits are actually switched circuits, which have transient or AC properties. You're right that after settling, a capacitor acts as an open, however an inductor acts as a short. Your intuition failed you here.

His method is a bit more complex, but you eventually need to describe how different frequencies change the characteristics of a system. Initially, it's nice to see impedance in terms of a + bi, but that's not easy to manipulate when you eventually want to know a circuit's behavior at various frequencies.

This is because (a + bi) does not inherently describe the sinusoidal behavior of AC. By changing things into the frequency domain (where writing a phasor implies a driving rotational force at some variable frequency), we can now see how the impedance of an entire system varies as frequency changes.

http://betterexplained.com/articles/intuitive-arithmetic-wit...

http://en.wikipedia.org/wiki/Phasor

It's really just a math abstraction. Maybe this analogy is a stretch, but you interface with stacks using push and pop, and there's an inherent size to the stack that you keep in the back of your mind. You don't care about the individual memory addresses and their values, you just want to manipulate the stack such that it doesn't overflow/underflow. Just like AC - you don't care where in the complex plane an impedance is at a given frequency (which would be in terms of a+bi), you'd rather just see how its behavior changes as frequencies vary.

And why do you care about impedances for a whole range of frequencies? One example: http://www.ecoustics.com/electronics/products/articles/13106...

If you take the term "put simply" and then add the term "Laplace transform" you're doing something wrong.

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?

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

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.

Impedance applies to AC and DC circuits, but at DC the imaginary term is zero, so only resistance is present. However, the term "impedance" still has meaning.

I think your question about AC on instrument cables stems from the popular use of the term, "AC" as in AC power. True, amplifier cables aren't carrying AC power but the signals are still considered AC since they are time-varying.

In theory, sure. But in practice, true DC doesn't exist. All signals will have some components at frequencies other than DC and quite often in electronics design these components are the most important to consider, especially if your planning for your product to pass FCC part 15.

Yes, the 'z' in hi-z and lo-z cables is the standard variable name for impedance (like 'v' for voltage or 'i' for current).

'AC' in this context means 'not DC', which is to say 'varies over time'. Instrument cables certainly qualify under that definition, though it is definitely confusing to call it 'alternating' (which we normally think implies 'goes back and forth between 2 states'), where perhaps 'varying' or 'time-varying' would be more appropriate.

Audio signals are alternating. The voltage corresponds to the sound pressure picked up by the microphone, which typically looks more or less like a bunch of sinusoids.

It's not just typically a bunch of sinusoids: Fourier tells us that it is always the sum of a bunch of sinusoids.

[1] http://en.wikipedia.org/wiki/Fourier_transform

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).

Part 2:

http://benkrasnow.blogspot.com/2011/06/tutorial-electrical-i...

Reply on news.ycombinator.com