drivers/impedence/crossover question on speaker building project

[theone]

I have begun the planning stages for building a pair of 2-way speakers for music and movie purposes. I realize that the project may cost more than I might pay for a pair of pre-made speakers, but money isn't a problem here.

I am wanting to include a 12" rear firing driver as a passive radiator, two midbass/midrange of 10" and 8", and a tweeter, preferable horn. If it turns out to be a dome tweeter, I can build a horn for it.

Now on to the questions. I have an idea of what drivers I want to include, however, all of them are 6 ohm. They are Pioneer drivers and are listed on Parts Express as well as Radio Shack. So, first question. Using three of these 6 ohm drivers and a 6 ohm tweeter, what impedence do I end up with? What kind of amp do I need to power this impedence? I will put links on the bottom to see these drivers. Do you think they are any good? (after seeing them of course)

Next part. The 12" frequency response is 23-1500 Hz, 10" 26-2500 Hz, 8" 29-2500 Hz, and tweeter 2000-20000 Hz. Will the speaker response take care of the crossover portion for me? Or do I need to find out a way to make a crossover for them? I know very little about this subject. if I need a crossover, what would you recommend? How would I set it up?

Links:

8"

10"

12"

Thank you for your help.

[tpg]

I like the looks of the drivers... can't offer anything too constructive, though. I am nto a crossover/impedence guy either. lol

They get me confused... if you take two 8 ohm speakers in the same speaker cab wired together then it is 4 ohm? or 16? and then you can go further to double it and it becomes less? I dunno. Confused... I am gonna stop now.

If the drivers are really this silver shiny color, though, something with them should look pretty cool.

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[WMcD]

I may be able to clear up one issue. It is a piece of the puzzle. I think the question is whether the two drivers present a parallel load to the amp.

The answer is no. The reason is the effect of the crossover filter.

I think this would be called the impedance transfer function across the filter. I.e. if there is an 8 ohm load at the output of the filter, what is the impedance at the input of the filter? The input impedance changes with frequency. More on this below.

To keep it simple, let's just consider a simple two way speaker design with a woofer and tweeter, each driver presents an 8 ohm load at their own input terminals, and we're using a classic cross over design. Say we pick a cross over frequency of 1000 Hz.

Input -> 1000 Hz high pass filter -> tweeter

Input -> 1000 Hz low pass filter -> woofer

The diagram doesn't show it well, but the inputs to the filters are in parallel as far as the output of the amp is concerned. This is just the red and black terminals at the input to the speaker, which is connected by feed wire from the amp.

It is important to realize that in this design the filters are acting independent of each other. The high pass is not effected by the low pass or vice versa. It is just that any frequency lower than 1000 Hz goes to the woofer. Anything higher than 1000 Hz goes to the tweeter.

The purpose of the cross over circuit is to allow the power input to the speaker box electrical terminals to be sent to the proper driver. At the "cross over" frequency, the power "crosses over" to the correct driver, tweeter or woofer.

It is tough to do a simple graph with the typewriter symbols, because there is no elevated line. I'll use ~~~ as a high level. The symbol looks like a sound wave, so it works. Roughly, the power transfer function is:

High pass filter power transfer. ___/~~~

Low pass power transfer. ~~~\ ___

The \ shows the roll off of the woofer power, and the / shows the roll on of the tweeter, power. Each occures at the crossover point of 1000 Hz. And you can see that in a perfect world, the filters dictate that only one driver is receiving power in their band, where they are happy to produce sound.

A tough problem is where the the filters and drivers (woofer and tweeter) are both half turned on at the cross over point, but that is different issue.

An important question is: when a filter is, or is not, sending power to a driver, what is the impedance the filters present to the amp? This is at the parallel inputs to the two filters.

Very simply speaking, in the frequency range when the filter is turned ON, the input to the filter is approximatly the driven driver impedance, or 8 ohms.

However, in the frequency range where the filter is turned OFF (not sending power through), the impedance presented to the amp by the filter is very high, say 100 ohms. (Roughly, again, 100 ohms in parallel with 8 ohms is 7.4) ohms or so. Not a problem for the amp.

Okay, now we have to draw another diagram which is the impedances of the inputs to the two filters in the cross over. It is just the opposite of the power curves.

High pass tweeter filter input impedence = ~~~\___

Low pass woofer input input impedence = ___/~~~

The ~~~ are 100 ohms and the __ are 8 ohms.

So: we see that the cross over impedence transfer issue leads to a result that the amp sees 7.4 ohms across the frequency range. This is true even though there are two 8 ohm driver (the tweeter and woofer) connected, through their respective filters, to the amp.

The following is going a bit far afield but. . . how do the filters stop the power from getting to speaker? The impedance tranfer function tells the story. It is not a matter that the filters absorb the power in an unwanted frequency range and disipate it somehow. Rather, they don't absorb it at all.

Remember that the power delivered by the amp to the load is:

P = Volts x Volts / impedance.

The voltage is constant, but we know the impedance varies from 8 ohms to 100 ohms at the cross over frequency.

By the equation, when the impedance presented to the amp by the filter is 100 ohms, the power the amp can deliver is low. When it is 8 ohms, it is high. Actually 100 may have been a bad number: it goes higher.

In re cap. (1) The filters prevent power from getting from the amp to the wrong speaker, and overloading it. (2) The filters do this by having a high input impedance at their input. (3) Because of (2) the filters' input impedence add so that the amp is presented with a constant impedance across the frequency range. At least in our perfect model.

Please excuse the use of the term "roughly". The real world has many, many more complications.

Gil