I mean.. Its definately THERE... Kinda like saying the graviton is imaginary.. What do you think, dank?

ShadowStar

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Still looking for that CHEAP Thunderdome :D

If the sub moves, use it. If the tweet's metal, lose it.
If you car doesn't have Oz, I just can't excuse it.

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rybaudio@hotmail.com

ohms law, R=V/I, Where R is a constant, and resistence is Linear in nature (for the most part), but resistence is not a constant. Resistence can change based on flow of current. Temperature increases resistence, hence a non-linear element of said resistence.

I know, kinda contradictive, and anyone cares to correct me, by all means, it will help.
What's been going through my head, and most likey has a simple answer, is the fact the resistence can change based on flow of current. hmmm....maybe I'll get some lab time and just do some testing for the heck of it...unless someone wants to give me the answer right off, lol

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http://www.woofers.cjb.net/

Jamming to Pat Boone in a Metal Mood, and my always phat Parapa da Rappa CD

good ol' impedance

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Team "Under Pressure"

Alaska, Get use to it. You’ll see it a lot more through school.

Jc2, Your an EE. Feel free to help me explain. knowledge and explaining knowledge are two different things.

For the rest of the people, to understand impedance, The first thing you have to do is forget all about ohms law. The R is about to be replaced with a Z, which is a complex number. This can make the voltage and current complex, which makes power calculations more difficult. If you try to think of ohms law, what I’m about to say will confuse you.
(Complex number are usually called i in math i = (-1)^1/2)

For inductive loads:

Inductors basically make the electrons feel heavy. So the voltage can kick up right away, but the current takes a second to catch up (A second is an exaggeration). Because of the current not being able to go from zero to its max instantaneously, it starts to lag behind the voltage.

Imagine a sinewave of voltage across a resistor. The voltage will go up and down. Ohms law says that the current will go up and down with the voltage. Now if the electrons feel heavy, they cant react as fast as the voltage so the current is said to be lagging behind the voltage. Its waveform which is the same as the voltage is shifted back between 0 and 90 degrees. So you now have the voltage sine wave, and a current sinewave that is slightly out of phase with the voltage.

If you have a graphing utility, graph the following.

V = cos (x)

I = sin (x)
This is what you would see looking at current and voltage on an O-scope.

Now for the math.

The impedance of a resistor is R.
The impedance for an inductor is jwL.
The impedance for a capacitor is 1/(jwC).

R = resistance in ohms.
w = 2*pi*f
L = inductance in henrys
C = capacitance in Farads
j = (-1)^1/2.

The formula for z is,

Z = R +/- jQ (rectangular form) = Z @x degrees. Q is the reactance.

Say you have a speaker with an inductance of 2mH @ 1khz, and a dc resistance of 2 ohms.

The impedance (z) would be,

Z = 2 + [ j * (2*pi*1000) * (2 * 10^-3)]

= 2 + j12.56 (this is in rectangular form)

To find Z in polar form

First get the magnitude,

= [ 2^2 + 12.56^2]^1/2

Z=12.72 @ some angle

To find the angle

Angle = arctan(12.56/2) = 80 degrees

So the total impedance would be,

Z = 12.72 @ 80 degrees

I think this is enough for now, later I’ll write more about calculating power which consists of apparent power, real power and reactive power.

Plus you're doing a pretty good job by yourself.

Basically what your saying is in a purely inductive circuit, you can use the ELI rule. Where (E) voltage lags (I) current in a (L) inductive circuit and the amount of lag is by 90 degrees.

This is exactly the opposite in a purely capacitive circuit. Where the rule was ELI, it now becomes ICE, where voltage now leads current by 90 degrees. Just wanted to state that because it seems easier to remember this way. I know you haven’t gotten this far yet but I thought I would just throw it out anyway.

Couple of questions, what happens to the phase relationship when you have a circuit with both a capacitive and inductive reactance (RCL circuit)? You might be getting to this (I hope) if you are, I can wait.

What’s the difference between polar and rectangular, just the exponent, but the decimal didn’t move? Sorry, I’m self-edumacated and sometimes it’s a real weakness.

Just to clarify is the formula to find the angle = arctan (impedance/2)? Wanted to make sure you didn’t get the 2 from somewhere.

Good show Mr. D, too bad RC didn’t want to get involved. Oh well.

Need Input……

Since the imaginary part is in polar form (or can be converted to) it gives us an angle. And since voltage leads the current in inductors and current leads the voltage for capacitors (ELI the ICE man...E=voltage, L=inductor, I=current), when you draw a uh, phasor diagram this is where the phases come in. I THINK that is what Mr. Dank said, just tryin to see if i'm thinking the same way as him

just noticed that STACKHOUSE1 already said some of that....
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Team "Under Pressure"

wall w/three shocker15's and a cadence a7hc...in a tiny corolla

quote:

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Originally posted by Snot:
I might as well throw my notes out here, might be helpful, but probabbly not lol

ohms law, R=V/I, Where R is a constant, and resistence is Linear in nature (for the most part), but resistence is not a constant. Resistence can change based on flow of current. Temperature increases resistence, hence a non-linear element of said resistence.

I know, kinda contradictive, and anyone cares to correct me, by all means, it will help.
What's been going through my head, and most likey has a simple answer, is the fact the resistence can change based on flow of current. hmmm....maybe I'll get some lab time and just do some testing for the heck of it...unless someone wants to give me the answer right off, lol

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Hey Snot,
Here's some food for thought.

How does a fuse work?

For rectangular to polar, and polar to rectangular.

Remember the a^2 +b^2 = c^2 formula? This is the same formula to go from rectangular to polar.
let a = resistance (R),
let b = reactance (Q),
let c = impedance (z).

just solve for z = [ R^2 + Q^2 ]^1/2

Now to find the angle

angle = arctan(Q/R)

try the arctan of 1, and if it doesnt equal 45 your calculator is in radian mode. change it to degrees mode.
arctan(1)=45.

Now it is important to be able to go from polar back to rectangular to determine the power your amp is putting out.

polar to rectangular:

say you have,

Z = 12 @ 58degrees.

the real part of Z (the resistance) is,
R = 12 * cos (58)= 6.36

the imaginary part of Z (the reactance) is,
Q = 12 * sin (58) = 10.18

To verify this you can go back to rectangular form

[6.36^2 + 10.18^2]^1/2 = 12

angle = arctan(10.18/6.36) = 58

I should really add this,
In the real world you will measure impedance and resistance, not reactance.
Say you measure the impedance on your speaker. The impedance is almost usless by its self. you need to know its angle, or the resistance.

Say you have a speaker and you measure the impedance at 50hz and it is 10ohms. you also need to measure the dc resistance. say that is 4 ohms. Now you can use the a^2 + b^2 formula to find the reactance, and then you can use the reactance to find the angle.

Z^2 = R^2 + Q^2

10^2 = 4^2 + Q^2
10^2 - 4^2 = Q^2
Q = [100-16]^1/2
Q = 9.17

angle = arctan(9.17/4) = 66.4 degrees

Z = 10 @ 66.4degrees

If you understand all of this, I can show you how to calculate the actual "real" power your amp is putting out to your speaker at the impedance yuo measured. I will also explain and show why people measure the output of there amps and get huge numbers, Its called apparent power, and is the magnitude of the realpower and imaginary power.
Aslo, If you understand this and are good with algebra, and know how to figure out the resistance of resistors in parrallel, series or both, I can show you how to design your own crossovers, and how to caclulate or plot the magnitude and phase at every frequency.

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2-DD 9512's
2-2000X's (for now)
8.0ft^3 box in a civic trunk
One big ass port
?????dB's

[This message has been edited by mr.dank (edited 01-24-2001).]

Trying to keep you on your toes.

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http://www.woofers.cjb.net/

Jamming to Pat Boone in a Metal Mood, and my always phat Parapa da Rappa CD

Mr. Dank, i follow you and am curious as to how you would measure the actual output power of your amp. But i'm unclear on one thing and i'm probably just overlooking something, how do you go about measuring the DC resistance and the impedance? When you use a DMM, which reading is it giving you? This stuff will definitely help me for this next season (i'll be in Minnesota with people who have louder systems than here if i can figure all this out. I am going to retune my enclosure to 51Hz (currently 41Hz) to see if that nets me any more db's (closer to resonant freq hopefully). Also, since i know that the impedance of my sub in the enclosure in my car is NOT the nominal impedance i want to rewire my subs to a nominal .33 ohm load for competition ONLY. BUT, i don't want to do this without some careful measurements and calculations to make sure my amp will be ok.....so spit it all out! thanks
Oh, i think i can follow the math (finished with calc III, taking D.E. this semester...yuck) but i make no guarantees. I was mad because we never covered Fourier Series in Calculus....might have to try and learn it on my own. I got a free copy of the Calc I and II book and the Physics book so the info's there, i just gotta decipher it...it makes more sense when i can apply it to something i already understand (sorta).
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Team "Under Pressure"

wall w/three shocker15's and a cadence a7hc...in a tiny corolla

Also, anytime you measure the resistance with a dmm, it’s the same as dc resistance. Impedance measurements require a frequency sweep and a voltage and current measurement.

O.K. Power,

There is more than one way to measure power.
When dealing with reactive loads, there are three types of power.

Real power- this is the amount of power that is actually doing work. It is rated in watts.
Imaginary power- I don’t know how to explain this. This is rated in vars.
Apparent power – This is the magnitude of the real power and imaginary power.

The apparent power can be calculated using the same formula as impedance.

AP^2 = RP^2 + IP^2

AP = apparent power
RP = Real power
IP = Imaginary power

Notice its similarity with the impedance formula?

Ok, say you know the impedance of the test tone your going to play, and its 50Hz, 6ohms.

You measure the dc resistance to be 1.9ohms.
Find the angel of the impedance,

[6^2 – 1.9^2 ]^1/2 = 5.69

So Q = 5.69

angle = arctan(5.69/1.9) = 71.53 degrees

Now you know the whole impedance and not just the magnitude of the impedance.

Z= 6 @71.53deg

Play your tone and measure the voltage across it.
Say its 75v rms.

Now you can use ohms law for the whole impedance.

I = V/Z
I = 75/(6 @71.53)
I = 12.5 @ -71.53deg

Notice how the angle becomes negative. The rule for this is like logs, just subtract the angle of what you’re dividing by from the angle of what you’re dividing. In this case,

0-71.53 = -71.53.

So now you have the current and its phase with respect to the voltage. The angle represents the phase of the current with respect to the voltage. The current is 71.53 degrees behind the voltage.

Back to power.

Apparent power = V*I (this is magnitude only, so drop the angle)

Real Power = V * I * cos (angle) this is real power in watts.

Imaginary power = V*I* sin (angle) this is in Vars.

So for this case,
Real power = V*I*cos (-71.53) = 297W

Imaginary power = v*I *sin(-71.53) = -889Vars (the negative indicates an inductive load)

There is an easy way to measure real power without doing a bunch of math and you get the same answer. You don’t even need to know the impedance.

Just measure I with a clamp on meter.

I*I * R =12.5 *12.5 * 1.9 = 297W

Most people on the forum claim to be getting almost double the power out of their amps. This is because they confuse impedance with resistance.

V^2 / Z = V * I = apparent power.

Apparent power in this case would be 937 volt-amps (VA).

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Team "Under Pressure"

Thanks, and yeah I even confused myself with the rules..haha.

I'm going to have to digest the information, there is a lot of it, but if have any questions I'll be sure to ask.

Again I as soon as I have time to look this over I'll be hitting you up for those phase angles on crossovers.

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Have meter, Will travel!!