understanding inductive loads in circuit analysis

DaveH

Joined Jan 1, 2009
53
If you take the example of a loudspeaker with nominal impedance of 8 or 80 ohms. Is this the ohmic resistance of the coil (measured with a dc voltage), or is this the reactance of the coil at some particular frequency?

I was thinking it's purely the ohmic resistance of the coil and therefore you should add on the reactance of the coil inductance (2∏fL), so it's like a resistor in series with an inductor, or may be in parallel? Perhaps at audio freqencies the reactance of the coil is not significant, but I have no idea what inductance levels speakers have.

In the example of motors can they be analysed in circuits the same way?

KL7AJ

Joined Nov 4, 2008
2,208
If you take the example of a loudspeaker with nominal impedance of 8 or 80 ohms. Is this the ohmic resistance of the coil (measured with a dc voltage), or is this the reactance of the coil at some particular frequency?

I was thinking it's purely the ohmic resistance of the coil and therefore you should add on the reactance of the coil inductance (2∏fL), so it's like a resistor in series with an inductor, or may be in parallel? Perhaps at audio freqencies the reactance of the coil is not significant, but I have no idea what inductance levels speakers have.

In the example of motors can they be analysed in circuits the same way?

In general, it's an AVERAGE impedance of the coil in mid audio frequency range. Actually, the equivalent circuit is quite complicated when you add the mechanical compliance in with the coil inductance and resistance. You have both parallel AND series equivalents.
eric

KL7AJ

Joined Nov 4, 2008
2,208
If you take the example of a loudspeaker with nominal impedance of 8 or 80 ohms. Is this the ohmic resistance of the coil (measured with a dc voltage), or is this the reactance of the coil at some particular frequency?

I was thinking it's purely the ohmic resistance of the coil and therefore you should add on the reactance of the coil inductance (2∏fL), so it's like a resistor in series with an inductor, or may be in parallel? Perhaps at audio freqencies the reactance of the coil is not significant, but I have no idea what inductance levels speakers have.

In the example of motors can they be analysed in circuits the same way?
Here's an excellent overview of the matter:

http://sound.westhost.com/tsp.htm

DaveH

Joined Jan 1, 2009
53
Hi thanks for that link.

I think therefore for my audio circuit analysis purposes I can assume the speaker is a resistor with value equal to it's nominal impedance.

In terms of circuit design where the inductive load is a motor (and it's a small ac motor in my case where the impedance rating is unknown - the only rating is 12 - 20 v. ac rms I assume) I measured the coil resistance with a multimeter and funnily enough it was 80 Ω. I'm going to power it with a 50Hz wave, through experiment I'll see if the voltage dropped across the device changes much from what it should be if it was an 80 Ω resistor and from that I'll learn whether the inductance is significant at that frequency.

thatoneguy

Joined Feb 19, 2009
6,349
I have a small "Universal Motor", which runs on AC or DC, 120VAC 50/60Hz, or 120VDC, 90Watts, 7k RPM.

The measurements are with the motor not turning:

Field windings have an inductance of 31mH, DCR of 46 Ohms, and Z at 120Hz of 21 Ohms. There are two of these, one for each side/pole.

For the rotor, there are a dozen poles, each with an Inductance of 106 mH, DCR of 120 Ohms, impedance of 66 Ohms at 120Hz. The rotor is in series with the field coils.

As a unit, the motor has a DCR of 280 Ohms, Impedance of 124 Ohms at 120Hz, and inductance of 184mH.

The numbers are just an example, and your motor may be wound differently, series or shunt (parallel), and may have permanent magnets instead of field coils. However, the above should give you an idea of impedance vs. DC Resistance, it would be lower at 60Hz (static shaft/not turning), and the reactance will change when the motor is turning.

i.e. Speaker coil measured while not moving doesn't give a lot of information about impedance, which varies widely over the frequencies it is fed, as well as how fast the cone is moving in relation to the magnetic field.

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studiot

Joined Nov 9, 2007
5,003
If you take the example of a loudspeaker with nominal impedance of 8 or 80 ohms.
This is the reactive impedance, which you want. As KL7AJ etc said, this is nominal and varies over the frequency range in question.

Ideally the resistive component is zero, as it simply wastes power as heat. The heat, of course, causes eventual failure of the voice coil assembly. Further it does nothing to produce sound.

You want the reactive power which drives the speaker piston (cone whatever) as a motor and need to know that it is 8 ohms or whatever to have a basis to calculate upon.

steveb

Joined Jul 3, 2008
2,431
This is the reactive impedance, which you want. As KL7AJ etc said, this is nominal and varies over the frequency range in question.

Ideally the resistive component is zero, as it simply wastes power as heat. The heat, of course, causes eventual failure of the voice coil assembly. Further it does nothing to produce sound.

You want the reactive power which drives the speaker piston (cone whatever) as a motor and need to know that it is 8 ohms or whatever to have a basis to calculate upon.
We need to be careful here. Reactive power does not do work, while resistive power does. The goal of a speaker is to generate real outgoing acoustic power which appears as a resistive load on the circuit side. Yes, the resistance may be frequency dependent, but that does not mean it is necessarily reactive. Of course, there is a reactive component to the speaker impedance, but not all of the resistive part is heat loss.

We see a similar thing when we make equivalent circuits for motors. We introduce a frequency dependent equivalent resistor which simulates the mechanical power (torque times angular velocity) the motor is generating.

I've not studied speaker impedance to know the typical ratio of reactive/resistance, but I think the above principle is important to keep in mind.

I have seen that motors appear mostly inductive when they are not outputting mechanical power (locked rotor, or spinning fast with low friction and no torque load), and appear mostly resistive when generating mechanical power (spinning fast at maximum torque output).

studiot

Joined Nov 9, 2007
5,003
We need to be careful here. Reactive power does not do work, while resistive power does.
Yes Steve my terminology was rather lax here. I was trying for an informal chatty response.

However in the case of the loudspeaker it is the reactive power (electrically speaking) which eventually does the work of moving the speaker piston. This is done against the electromechanical resistance, measured in mechanical ohms.

The following ignores the effect of heating upon both the dimensions and resistance of the wire.

Consider a length of voice coil wire, of resistance R, stretched out straight. If I apply an alternating voltage V to the ends and pass a current I, there is no inductance in the circuit and I = V/R.
The power is I$^{2}$R and appears a heat in the wire.

If I now coil the wire neatly many times around a former I find that I need to increase the voltage substantially to drive the same current, I through the same wire.

I can prove by immersing both wires in colorimeters that the only work done is the heating effect and is identical for both the long wire and the coiled one at the same current.

However the product VA is now much greater than the heating effect. This is because the coil also has inductance. The resistance and inductance of the coiled wire combine to give a higher impedance than R alone. Thus

Z$^{2}$ = R$^{2}$ + X $^{2}$ and

I is now equal to V/Z where Z > R

If I rearrange the coils to be more spread out X and therefore Z will change as will the voltage needed to achieve a given I. But the heating effect will remain constant regardless.

As matters stand the coil makes no attempt to move and no other work is done.

If however I introduce a suitable magnetic field the coil tries to move and further work is extracted from the system, as it pushes against the attached piston, although the resistance of the coil does not alter. The piston and the air on the other side of it are an inertial system so "push back" with a force which increases with increasing push from the coil. It is against this force that the work is done.

From here on the maths gets a bit hairy as it involves the solution of two simultaneous equations which I will list, to see if there is any interest in more detail.

Mechanical Equation

mD''+ rD'+ sD = FI

Electrical Equation

LI' + RI + FD' = V

Where The voice coil has resistance R and inductance L.
F is the force per unit current in the voice coil
m is the mass of the vibrating system, r the mechanical resistance, s a system constant.
D is the displacement amplitude.

In both equations differentiation is with respect to time i.e is d/dt.

studiot

Joined Nov 9, 2007
5,003
For reference we are aiming to explain the attached plot of impedance v frequency for a typical loudspeaker.

Note the dashed line showing the constant 'electrical resistance' of the voice coil, which generates the heat.

thatoneguy

Joined Feb 19, 2009
6,349
Going back to motor/heat.

The difference in heat dissipation is the amount of impedance that is in the reactive vector, rather than the resistive. This is illustrated above with the motor having a DCR of 280Ω, but an impedance of only 120Ω at 120Hz. The majority of the resistance is "bypassed" by the reactive "imaginary" impedance, which doesn't generate heat. If the DC Resistance creates continual heat, the motor would generate heat at the same rate when rotating freely compared to loaded/stalled.

In practice, we all generally know that an overloaded or stalled motor heats up very quickly, even small motors where rotation isn't turning an internal fan for self-cooling.

That said, the topic is extremely complicated, motor design is a major art in itself, and there are instances where what is observed doesn't match what one would think should be happening. This goes hand in hand with the "mystic" of inductance and odd effects thereof, including some who have mixed magnetism with coils to the extent they have confused themselves and others into believing overunity has been acheived.

steveb

Joined Jul 3, 2008
2,431
However in the case of the loudspeaker it is the reactive power (electrically speaking) which eventually does the work of moving the speaker piston. This is done against the electromechanical resistance, measured in mechanical ohms.
Perhaps I'm misunderstanding, but I think i would like to respectfully disagree here. If real acoustic power is generated, this must be reflected into the electrical side as a resistive load. This is a basic requirement. Reactive loads have current and voltage 90 degrees out of phase, and can't provide real power (i.e. work done at a certain rate).

Please explain what I'm missing. I'm not trying to get into one of those famous discussions about definitions. But reactance is considered either capacitive or inductive. The speaker is a complicated impedance function, but the resistive part of the electrical load is either heat or sound power.

Now, I would like to stress again that I've not studied speaker impedance in detail. I'm just talking about principles. Perhaps the sound power is too low to be significant, I don't know. However, I do know that this is a critical issue with motors. Real mechanical work looks like a resistive electrical load, and this thread had discussed both speakers and motors. So I think my comment is relevant and quite important here.

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thatoneguy

Joined Feb 19, 2009
6,349
Please explain what I'm missing. I'm not trying to get into one of those famous discussions about definitions. But reactance is considered either capacitive or inductive. The speaker is a complicated impedance function, but the resistive part of the electrical load is either heat or sound power.
I agree with the above. A speaker, when all is stripped away, is a linear motor that compresses/expands air. Compressing air is work, and in the active/real power domain.

• Power that does work or produces heat is active power, measured in watts. By definition, active power is the product Vrms*Irms when they are in phase.

• Power that oscillates back and forth each cycle is reactive power and is measured in VAR (Volt-Ampres-Reactive). Reactive power is the product Vrms * Irms when they are 90° out of phase.

Reactive Power is absorbed by the load, then fully returned to back to the source. Due to this definition, no work can be done with reactive power, otherwise less would be returned to the source than absorbed, which is the real power/work done.

This PDF on Power Factor starts out with an excellent example, using a horse on a rope.
http://www1.eere.energy.gov/industry/bestpractices/pdfs/mc60405.pdf

studiot

Joined Nov 9, 2007
5,003
If real acoustic power is generated, this must be reflected into the electrical side as a resistive load.
Yes that's correct and I did actually say something like that. Resistance is opposition to flow of current. A common approach is to regard the back EMF of any motor (a speaker is a form of motor) as providing this resistance and to regard the increase in driving voltage as that necessary to overcome this back EMF.

The (complex) solution to the differential equations I showed adds a resistance (r in my first mechanical equation) to the real part of the solution. This resistance is the one which associates with acoutic power generated.

This is the magic of electromechanical machines. They can transform electrical energy into something other than heat by virtue of the interaction between the electrical and mechanical working parts. Without both being present this cannot happen.

studiot

Joined Nov 9, 2007
5,003
The difference in heat dissipation is the amount of impedance that is in the reactive vector, rather than the resistive. This is illustrated above with the motor having a DCR of 280Ω, but an impedance of only 120Ω at 120Hz. The majority of the resistance is "bypassed" by the reactive "imaginary" impedance, which doesn't generate heat. If the DC Resistance creates continual heat, the motor would generate heat at the same rate when rotating freely compared to loaded/stalled.

In practice, we all generally know that an overloaded or stalled motor heats up very quickly, even small motors where rotation isn't turning an internal fan for self-cooling.
I would be interested to see your explanation of how the following well known equation is satisfied.

Impedance = √{R$^{2}$ + (2πfL)$^{2}$}

Presumably you are talking about a motor with shunt wound coils as well?

I hope you are not seriously suggesting that the heating effect of passing a current I through a resistance R depends upon whether the motor is spinning or not? Or that this effect is not I$^{2}$R in all cases.
The motor is self cooling when spinning, but not when stalled.

thatoneguy

Joined Feb 19, 2009
6,349
It is a shunt wound universal motor. No permanent magnets.

The readings are from an LCRZ meter at 120Hz. My text was oversimplifying the math for brevity.

studiot

Joined Nov 9, 2007
5,003
It is a shunt wound universal motor
Well there you are then you have two or more reactances in parallel, which obviously have a reduced combined parallel impedance, at AC.

The loudspeaker only has one coil so is not directly comparable.

thatoneguy

Joined Feb 19, 2009
6,349
I hope you are not seriously suggesting that the heating effect of passing a current I through a resistance R depends upon whether the motor is spinning or not?
No, I apologize for that statement which reads misleadingly. I was over-simplifying the concept of reactive power vs real power, at a level I perceived the average reader might comprehend. In doing so, technical aspects were necessarily glossed over, causing confusion between a few people. I believe all that have posted know and agree what is happening, but are expressing the concept in different ways. Similar to getting caught up on if a color is called rouge, red, or crimson. Without filling the page with TeX formulas/theory that most don't want to read/understand, this is a common problem.

My statement was this: When a motor is moving without a load, either electric motor or speaker without damping/enclosure, the reactive impedance is typically at its peak, resistive element is at it's base, outside of special cases such as resonant frequency in a speaker.

When a motor is moving with a heavy load, or a heavily damped speaker, such as in a very small sealed box, the load tends toward more real power consumed, generating more heat as well as power in the form of torque or air pressure.

The reason I put measurements for a motor in this thread was that a motor was mentioned that was an AC motor running at the same frequency of one I had beside me, so I gave the values for that motor. In hindsight, I could have posted the inductance of a 12" sub, but as speakers vary widely, I did not want an exact inductance value to be used as "L of all woofers".

Also, I may be off track in understanding the actual question from the original poster. I understand it to be "What is used to simulate a speaker when designing amplifiers". This is an excellent question. Usually, a resistor of equivalent impedance will suffice for some measurements, but when analyzing for effects from reactive loads, what "standard component" should be used to represent all speakers?

I apologize for adding confusion with data from a motor without magnets, as it wasn't much of a help, in hindsight.

studiot

Joined Nov 9, 2007
5,003
It suggests using Smith charts to plot speaker impedance
From article

The main drawback compared to the normal, two-plot method is that we lose the frequency dependence of the impedance.
As I understand Smith charts you would have to draw a separate one for each frequency under consideration, a daunting task for speakers when we consider the range concerned, so I'm not sure I see the advantage over a plot such as the one in my post #9. Unlike the author of the article I don't have any trouble interpreting this.

However the idea that someone actually uses a Smith chart is interesting. What does he derive from this as the article doesn't say?

steveb

Joined Jul 3, 2008
2,431
As I understand Smith charts you would have to draw a separate one for each frequency under consideration, a daunting task for speakers when we consider the range concerned,
Actually, this is not true. You basically trace the impedance for every frequency onto one Smith chart. This is why there is a trace/curve on the picture he shows on the second page of the article.

The author claims that frequency information is lost, but actually he is showing a little lack of creativity here. All one needs to do is identify the low frequency point, which will be one end of the trace. Then place a tic-mark every octave so that the frequency information is readily apparent.

As the author points out, the nice thing about the Smith chart is that the 8 ohm resistive impedance point will be at the center of the chart, and the further the trace is from the center point, the less desirable it is.

I agree that the standard mag/phase or real/imaginary plots are not that hard to interpret, but this just provides another way to look at it. I'm not necessarily endorsing this method, but personally I think it is noteworthy.