Horn loading

[Stig]

Seems to be fact that front loading a woofer will raise the impedence of a given driver.

Does this hold true where only the rear of the cone is loaded?

Example: Jensen Imperial

Thanks,

Stig

[WMcD]

Yeah, it certainly should. I must admit I've never seen the specifics of back loading versus front loading discussed.

I'll talk about technical stuff and math. Some people like it. My own form of rant.

- - -

The closest I've seen about the strucure is in some old midrange drivers. They are front loaded with the horn on the opposite side of the diphragm from the the magnet structure. However, there is a vent at the rear which seems to be open to the air except for some wool. The latter probably acts like an accousic resistence.

Even with this, no one is questioning that the basic engineering applies.

- - - -

There are several electrical analogies to the combination of the rear box, the driver components, the chamber in front of the driver, and the connected to the horn.

As you can imagine, the back loaded horn is very similar to the front loaded horn (the horn is at the back side rather than the front). The difference is that there is no back box, in front. Rather the diaphragm radiates. This way there is mid range frequency sound radiated directly.

The Klipsch Rebel (original ones) and Short Horn, were back loaded, but had a box between the driver and the horn. Otherwise similar to the Imperial in concept. There was radiation off the front of the driver.

- - - -

This is the math and electrical engineering stuff.

It took me quite a while to realize how an increase in overall resistance could lead to a system which is more efficient. I pass it along. The following is more complicated than some readers need, and more simplified than others need.

There is a simplified electrical analogy to the direct radiating diaphragm and the horn loaded diaphragm. Both are same in structure of the analogy. Only the numbers change. The implications are grand.

It is important to realize that in a speaker, some of the electrical energy is used to just heat up the voice coil of the motor (VC), some of it makes sound because it is being delivered into an abstraction called, by me, radiation resistance (RR).

I'll plug in some values which are just for example. The RR value is something I pick out of the air to make math easy.

For the direct radiating diaphragm:

(+)----------(-)

I'm showing the plus (+) and minus (-) electrical input to the speaker as a whole. That is what the amplifier is driving or "sees".

The "VC" is the voice coil resistance which I'm assuming is 8 ohms. The "RR" is the radiation resistance of the diaphragm. The air load is not resisting movement of the diaphragm. So it is given a low resistance.

Essentially, the power into the voice coil resistance results just in heat. The power into the radiation resisance results in sound.

In the above circuit, the total resistance is 8.8 ohms. There is not much power absorbed by the RR because it is only 0.8 ohms. The voice coil resistace is absorbing most of the power.

Let me be honest and do the math. If there are 1.1 volts applied, the voltage across the VC is 1.0 volt. The voltage across the RR is 0.1 volts. (Classic voltage divider.)

We know that Power (watts) = V x V / R.

So for the power to the Radiation Resistance we calculate 0.1 x 0.1 / 0.8 = 0.00125 watts. That is the analogy to the acoustic sound output from the direct radiator.

Putting the diaphragm of the the same driver up against a horn alters things. Now the diaphragm has to pump air into the throat of the horn. It resists mechanical movement more. The circuit values for an optimized horn looks like this:

(+)----------(-)

Please note that the RR value is now equal to the VC value. Otherwise, the circuit is the same.

The total resistance is now 16 ohms. And you might wonder . . . . The overall resistance load has gone up!

In comparison to the first example, we're drawing about half the current with the same voltage. (P = volts x current.) So the electrical power delivered overall must be about half. That is absolutely correct.

However, lets look at the power being delivered into the altered radiation resistance.

Again we apply the same 1.1 volt at the the (+) and (-). Now there is an equal voltage drop across each resister of the analogy.

Therefore we have a voltage across RR of 0.55 volts, and its resistance is 8.0 ohms.

Let's do the power calculation for RR.

We know that Power = V x V / R.

Plugging in the numbers:

0.55 x 0.55 / 8.0 = 0.037 watts into RR.

Wow! Even though the overall resistance seen by the amplifier has doubled, the power into the radiation resistance has increased.

The ratio of the increase of power to the RR (horn loading versus direct radiation) is:

0.037 watts / 0.00125 watts = 29.6

We can calcuate that in dB.

10 x Log(10) of 29.6 = +14.7 dB.

Also, the overall power drawn from the amp is about half. Call that another 3 dB of power gain.

- - - -

I can't crawl into PWK's head, but you can imagine the implications of horn loading to any engineer. There is a grand increase of power transfer, even though the amplifier is being asked to deliver less power!

Regards,

Gil

[Stig]

Gil,

Thanks once again for the in-depth response. I've got to find the book by Olson. So much to learn, so little time.

Stig

[WMcD]

Regarding the book. Check audioxpress.com It is listed there as being on sale.

Your local library can probably get it on interlibrary loan for your, if they can find a copy. It was supposed to be at DePaul University Chicago but was missing from the shelf, and evidently has been for some time.

I finally found a copy at Illinois Institute of Technology Galvin Library, Chicago. Wonder of wonders, it had markings of DePaul Library. Guess it got sold somehow.

Gil