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Joey Joe Joe

Resistance, reactance and impedance

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Hi guys,

I've got midterms coming up at my audio engineering school. I've been studying like mad, and I've got a pretty good idea of what's going on, but there are a few terms I'd like simplified... just so I can have a clear sense of the meanings, rather than just a textbook feel.

 

The first term would be IMPEDANCE. I think where I get stuck is the difference between high, low and very low impedance systems. What is it that causes a device to be high or low impedance? Say a 200 Ohm microphone for example, why is it 200 Ohms? Why not 50 Ohms?

 

I'll post more as I have more questions.... thanks a bunch guys!!

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Originally posted by Bogster

Adios, geeks? Goodbye to Spanish nerds? What the hell are you talkin' 'bout, man???!!!
:p:D;)

 

Those will be my dying words. HAHA!! :D

 

Or maybe Siyonara nerds! Which do you think is better?

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Impedance is a little tricky, actually.

 

You understand resistance, right? It impedes the flow of electrons by converting the energy to heat. It is measured in ohms.

 

Impedance is the vector sum of resistance and reactance.

 

There are two kinds of reactance, capacitive and inductive. In capacitive reactance, the energy is stored in the form of electrostatic fields between the capacitor plates. In inductive reactance, the energy is stored in the form of magnetic fields around the inductor. In both cases, the energy is not converted to heat, but is released back into the circuit after the applied energy is removed.

 

Why can't you just add them together? Because they're out of phase. In a resistive circuit, voltage and current are in phase, tracking together. In an inductive circuit, voltage leads. In other words, you apply the voltage, but there's a time lag before the current starts to flow. The opposite is true in an capacitive circuit. Inductive and capacitive reactance are 180 degrees out of phase with each other, but only 90 degrees ouit of phase with resistance. So, you have to add them together using vectors, or complex equations. (Complex means having both real and imaginary components. They're really not all that difficult.)

 

How's that?

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Hmmm, my electronics class doesn't really go that indepth, so I'm pretty lost on that. I'm off to school now... when I get back I'll see if I can add anything to clear up my question. Thanks!! :D

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Don't blame it on yourself Isaac.

Pretty hard to explain if JJJ doesn't have notion of complex numbers or at least trigonometry.

A nice graphic can explain a lot.

impedance.gif

 

Here Isaac. Work on this, you'll do better than I could.

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Originally posted by Jazz Ad

Don't blame it on yourself Isaac.

Pretty hard to explain if JJJ doesn't have notion of complex numbers or at least trigonometry.

A nice graphic can explain a lot.

impedance.gif

Here Isaac. Work on this, you'll do better than I could.

That's good.

 

Note the horizontal axis (usually called the X axis). It is labelled R. That's resistance. The vector is rotated counter-clockwise, toward Xl, or inductive reactance (the positive Y axis). The negative Y axis is labeled Xc, and represents capacitive reactance.

 

In the example, the circuit has 5 ohms of inductive reactance (j5) and 4 ohms of resistance. The total impedance is given as 4 + j5. Another way is to solve for the length of the vector [/(4^2+5^2)=6.4] and the angle [tan(theta)=5/4 => theta=51.34 degrees]. In audio, we don't worry about the angle, and just call it 6.4 ohms, but in electronics (and crossover design!) it can be important.

 

That seems even more involved than before, but, with the diagram, I hope it makes sense. Thanks, Jazz!

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There are many explainations on how to figure out how many ohms your amp sees when

you have multiple cabinets, with various math formulas and other information

GARANTEED to put the novice musician fast to sleep. So here is a simple table to

help quickly figure out the load your amp will see if all cabinets are paralleled

(BY FAR the most common method, to run cabinets in series, you need special cables

built, etc)

 

You have:.............................................What your amp sees

 

Two 8 Ohm cabs...................................4 ohms total

 

Two 4 Ohm cabs...................................2 ohms total (many amps cannot go

this low)

 

One 4 Ohm and one 8 Ohm cab............2.667 ohms total (many amps can't go this low)

 

Three 8 ohm cabs...................................2.667 ohms total

 

Two 8 ohm and one 4 ohm cab..............2 ohms total

 

Three 4 ohm cabs....................................Meltdown of city services

 

The next thing to understand is that if you have a stereo power amp, the above

applies TO EACH CHANNEL! So if you had four 8 ohm cabinets, you could put two on

each channel and each channel would see 4 ohms (most modern power supplies have no

trouble with a 4 ohm load.) READ YOUR MAUNAL TO SEE HOW LOW A LOAD YOUR AMP CAN

HANDLE!

 

Bridging

Bridging is simply taking a two channel stereo amp and ganging the two channels to

one output. If you decide to bridge your amp, the load the amp can handle is

typically twice as high as the load a single channel can have. In other words, if

one of your amps channels can only go down to a 4 ohm load, when you bridge two

together, you will probably only be able to go down to an 8 ohm load. Once again,

READ THE MANUAL FOR THE FINAL ANSWER to load limits.

 

Formula for final load:

Here's the formula to figure out any load:

 

(Resistanceofboxone X Resistanceofbox2) / (Resistanceofboxone + Resistanceofboxtwo)

Example:

Box 1 is 8 ohms and box2 is 4 ohms so:

 

(8 X4) / (8+4)=

 

32 / 12=

 

2.667 ohms.

Note: in this case, the 4 ohm box will recieve twice the power (watts) as the 8 ohm

cab (this does NOT mean it will be twice as loud, you need 10X the power for that)

 

edit - thanks, lug :)

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