how do you calculate impedance rise?

[Immacomputer 10+ year member]

Where?

....

It was a rhetorical question. The impedance spikes will usually occur around system resonance where the system is the most efficient. It doesn't mean that it's the loudest here but that it is very efficient.

Take an infinite baffle setup into consideration (or the driver playing free-air). The sub will have a large impedance spike around the Fs of the driver. The driver will also have greatest excursion at this point. How can that be if it's getting so little power??? Well it's being driven into resonance based on the electrical and mechanical parameters of the system. Just like a mass on a spring with a driving force.

When you put the sub into a sealed enclosure, you change where the system resonance is because of the increased air-spring the driver is fighting. The impedance spike will also occur around that point of resonance and usually a little lower than the actual resonance.

Let's consider now a Helmholtz resonator with no sub attached. The system here will have an impedance spike around its resonance as well (the tuning of the enclosure). When you add the sub to the mix though, while the sub experiences the high impedance of the enclosure, it does not always share an impedance spike at this point (tuning). In most cases, the point of resonance in the sub/enclosure system is higher than tuning and in many cases, excursion will peak there.

If you want an enclosure with a minimal impedance spike, build a correctly tuned and stuffed transmission line. This will result in the lowest impedance spike. But don't expect it to be louder than your ported enclosure though. Maybe when you do it, you will realize that power isn't the end all to having decent output.

This part is an educated guess of where the spikes come from without really doing any research on it: I believe that you get the spikes due to the poles in the combined transfer function of the sub/enclosure/amplifier system. If you keep the original transfer functions at a low order, you reduce the number of poles. The more poles, the more spikes. There are ways to shift poles in the transfer function but I don't really understand that stuff well enough to apply it to sub enclosure systems.

 

[headbussaboi 10+ year member]

I guess the main reason I ask is for those of us who really think you are getting your so many thousand watts at whatever ohm load might not be getting what you think if your dual 2 ohm sub wires down to 1 ohm but after box rise, it goes back up to 2 ohms and you have your amp set with a MM to get the full 1 ohm output.

Second of all, I don't get how those of you with say, dual 1.4 ohm coils wire it in parallel to get a .7 ohm load can run that load on an amp that is 1 ohm stable. Is your amp really rated to be stable at .7 ohms?

 

[casper97ta 10+ year member]

I guess the main reason I ask is for those of us who really think you are getting your so many thousand watts at whatever ohm load might not be getting what you think if your dual 2 ohm sub wires down to 1 ohm but after box rise, it goes back up to 2 ohms and you have your amp set with a MM to get the full 1 ohm output.
Second of all, I don't get how those of you with say, dual 1.4 ohm coils wire it in parallel to get a .7 ohm load can run that load on an amp that is 1 ohm stable. Is your amp really rated to be stable at .7 ohms?

A lot of amps that are built real solid can handle lower loads. i.e the kicker zx2500. Its rated for 2500 at 2ohms but can run 1 ohm all day. Given you need to keep voltage up also. The IA 20.1 is rated at 1ohm and people have dropped those things to ridiculous loads,like .175. It all depends on the amp and the electrical.

 

[Immacomputer 10+ year member]

I guess the main reason I ask is for those of us who really think you are getting your so many thousand watts at whatever ohm load might not be getting what you think if your dual 2 ohm sub wires down to 1 ohm but after box rise, it goes back up to 2 ohms and you have your amp set with a MM to get the full 1 ohm output.
Second of all, I don't get how those of you with say, dual 1.4 ohm coils wire it in parallel to get a .7 ohm load can run that load on an amp that is 1 ohm stable. Is your amp really rated to be stable at .7 ohms?

1) Impedance rise only occurs around certain frequencies. Outside of that range, the actual impedance stays close to that of the nominal impedance (the impedance that the sub is rated like dual 2 or dual 1 or whatever). So at many frequencies the sub can be getting all the power that the amp is rated for.

2) When people say this, they are speaking about DC resistance and not actual impedance. They think it's cooler to say .7ohm because that's what the MM said. The impedance will rise to the nominal impedance (1 ohm in this case) once there is an AC signal present.

 

[nineballsafety8 10+ year member]

V x A = WV / A = R

r= impedance

w= watts

v= voltage

a = amperage

actually... you are almost right.

v/a=r is a valid representation of Ohm's law... HOWEVER. r does Not equal impedance... it equals resistance... resistance is not the same as impedance...

Z = Impedance (ohm)

R = Resistance (ohm)

XL = Inductance Reactance(ohm)

where Z=the square root of (R^2+XL^2)

The total impedance is not simply the algebraic sum of resistance and inductive reactance. Since the inductive reactance is 90 degrees out of phase with the resistance and, therefore, their maximum values occur at different times, vector addition must be used to calculate impedance.

and to calculate inductive reactance you need to know the inductive force of the driver in Henry's and the frequency at which you are trying to measure

then you use the formula XL=2(pi)fL

where XL= Inductive Reactance (ohm)

pi=exact number, which is about 3.1416 (constant)

f=frequency

L= inductance in Henry's

so in answer to the OP's question... unless you have a program that will calculate it for you... u might want to let the big dogs handle it.//content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

 

[vehementSPL 10+ year member]

yeah Im a computer. Seems like most of the models I played with in winISD had the imp not peaking but actually at it's lowest near tuning for the most part

 

[helotaxi 5,000+ posts]

actually... you are almost right.v/a=r is a valid representation of Ohm's law... HOWEVER. r does Not equal impedance... it equals resistance... resistance is not the same as impedance...

Z = Impedance (ohm)

R = Resistance (ohm)

XL = Inductance Reactance(ohm)

where Z=the square root of (R^2+XL^2)

The total impedance is not simply the algebraic sum of resistance and inductive reactance. Since the inductive reactance is 90 degrees out of phase with the resistance and, therefore, their maximum values occur at different times, vector addition must be used to calculate impedance.

and to calculate inductive reactance you need to know the inductive force of the driver in Henry's and the frequency at which you are trying to measure

then you use the formula XL=2(pi)fL

where XL= Inductive Reactance (ohm)

pi=exact number, which is about 3.1416 (constant)

f=frequency

L= inductance in Henry's

so in answer to the OP's question... unless you have a program that will calculate it for you... u might want to let the big dogs handle it.//content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

While all of that is true, the fact remains that the peak and RMS current are still proportional to the peak and RMS voltage according to Ohm's law. The driver inductance simply delays the actual flow of the current. The current through the coil is what actually does the work of moving the cone. If the phase relation really mattered, you would have 4 points of zero power applied to the driver (2 points of zero voltage and 2 points of zero current) per sine cycle. Since this isn't the case, it really becomes fairly simple to plot impedance using a clamp meter (or shunt resistor) to measure current and a voltmeter to measure voltage. Do the Ohm's law algebra and you do get the impedance of the driver at that freq. Take multiple measurements at different freqs and plot the results for an impedance graph.

Imacomputer-the width and relative amplitude of the impedance spike of a driver in a sealed box app is defined by the system Q of the driver and enclosure. Same as Qts is determined by plotting the impedance curve of the driver in free-air and doing the math.