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-   -   Question on driver impedance compensation (http://www.diyaudio.com/forums/multi-way/228820-question-driver-impedance-compensation.html)

Patnio 28th January 2013 11:01 AM

Question on driver impedance compensation
 
Hello

Just a newbie trying to do the first build, so some of the questions here may be pretty basic:

I "inherited" some old and crappy Sony speakers which I am trying to re-vitalize by replacing the drivers. This is more of a learning project than a real build that will act as a stepping stone for a nicer build in the future.

The speakers were 3 way: a 8 ohm woofer, a 6 ohm "midrange" and a third speaker whose impedance is unknown. Sony had marketed these as "6 ohm" speakers. I plan to convert this into a 2 way arrangement.

Questions:

In a loudspeaker taken as a whole, arent the 3 speakers in parallel with each other typically? So how does a 8 ohm woofer in parallel with 6 ohm midrange result in a "6 ohm speaker"? (It is unlikely that the third speaker has high values)

I plan to pair a 4 ohm tweeter driver with a 8 ohm woofer driver when I replace. My understanding is that this will reduce the impedance of the loudspeaker as seen from the amplifier's point of view. Is this correct? (I assume so..)

I also need to compensate for sensitivity for the tweeter. This will be done using a L-Pad. Can I put a 2 ohm series resistor in series with the L-pad to compensate for the 4 ohm tweeter? So effectively it will look like:

High pass Filter circuit -> 2 ohm resistor -> L pad -> Tweeter

The high pass filter will be designed assuming a 6 ohm value for the tweeter rather than 4 ohms.

Is the above approach daft? What are the disadvantages or other ways to compensate for a 4 ohm tweeter ( I do not want my amplifier to see a 4 ohm tweeter).

I am aware impedance is a function of frequency, I am being simplistic here.

turk 182 28th January 2013 11:30 AM

Patnio
i don't know which model Sony's you've got but even the cheapest will have at least a cap or two which is why the overall impedance isn't a simple "paralleling" of three speakers.
the L pad will effect your tweeter sensitivity so the 2 ohm resistor may not be needed.
designing your high pass filter with a different value than what will actually be used may change your x-over frequency

Kjeldsen 28th January 2013 11:48 AM

X-over is parallel not the speakers. With no x-over the speakers will be parallel, and resulting impedance would be very low. For three 8 ohm speakers the resulting impedance would be 2,667 ohms - not very healthy for the amplifier.

With a parallel x-over the resulting impedance would still be 8 ohm. IF the impedance was a straight line of the frequency. But impedance is a function of the frequency and varies heavily with the frequency. An 8 ohm speaker is just an average.

No easy task to understand, but with a little reading on the topic you will learn.

Patnio 28th January 2013 12:37 PM

Thanks for the replies
Quote:

Originally Posted by Kjeldsen (Post 3345554)

With a parallel x-over the resulting impedance would still be 8 ohm. IF the impedance was a straight line of the frequency. But impedance is a function of the frequency and varies heavily with the frequency. An 8 ohm speaker is just an average.

No easy task to understand, but with a little reading on the topic you will learn.

I am not sure I understand:

The impedance seen by the amplifier will be the impedance of the crossover in parallel with the impedance of the speaker assuming a parallel crossover. Both impedances are a function of frequency. Now if I change a nominal 6 ohm speaker to a 4 ohm speaker, surely the overall impedance seen by the amplifier will reduce.

Do I not need to compensate for the changed speaker value? I understand that there is in reality no such thing as a purely resistive 4 ohm speaker: it is more of an equivalent RLC circuit. But surely when one changes from a 6 ohm nominal to a 4 ohm nominal, there is greater strain placed on the amplifier due to higher current draw? Do I need to compensate for that , by using a 2 ohm resistor mentioned above? If not, why not?

Please note I am not disputing what you are saying, just trying to understand this a bit better.

Pano 28th January 2013 01:41 PM

We really need a FAQ with this in it. It's a basic question that gets asked again and again.
It a logical question, of course, but one that get asked regularly.

turk 182 28th January 2013 01:43 PM

patnio
in the case of your tweeter which would be after a capacitor the "nominal" impedance is no longer "seen" by the amp.
you've removed part of the frequency spectrum(the cap)and as you stated impedance is a function of frequency.
your "compensating" resistor would reduce tweeter level and serve to not deviate the x-over frequency but recalculating the x-over with the final (actual) value may be better

turk 182 28th January 2013 01:58 PM

pano
i agree a set of "primers" for newbies would indeed help steer and educate.
as a matter of fact i probably would benefit reviewing basics meself!

Patnio 28th January 2013 02:24 PM

Quote:

Originally Posted by turk 182 (Post 3345694)
patnio
in the case of your tweeter which would be after a capacitor the "nominal" impedance is no longer "seen" by the amp.
you've removed part of the frequency spectrum(the cap)and as you stated impedance is a function of frequency.
your "compensating" resistor would reduce tweeter level and serve to not deviate the x-over frequency but recalculating the x-over with the final (actual) value may be better

Just so I am sure I get this correctly:

From an electrical circuit perspective, a speaker can be modeled as

http://sound.westhost.com/lrp-f36.gif

(Ignore the series RLC circuit at left, thats the impedance compensation)

When I replace the old tweeter with another one, all the parameters represented at far right by the resistors, inductors, capacitors change. Obviously the impedance of these are all a function of frequency. However, if the new tweeter is nominally "4 ohms", the equivalent impedance of the new tweeter has reduced. Exactly by how much is difficult to state since it is a function of frequency.

Adding a L-pad type of circuit + a 2 ohm resistor (or combining it with the series resistor of the L-pad) will somewhat compensate for the drop in the impedance, though this will not be exact.

The result of adding the 2 ohm resistor will necessitate a re-calculation of the crossover assuming a 6 ohm tweeter (rather than the actual value of a 4 ohm tweeter) to ensure crossover frequency remains correct.

Effect of this will be a reduction in tweeter sensitivity and possibly more power being dissipated in the series resistor of the L-pad.

Am I Correct?

turk 182 28th January 2013 03:19 PM

ummm!
i had a look at the link and i think you may have that a little wrong.
i may be mistaken but i believe that the components on the left are intended to represent the equivalent of a loudspeaker and the components on the right are the compensation.
i'm a little unclear as to the L pad configuration your planning and where your putting the two ohm resistor so i'll leave that alone.
"Am I Correct?" close but no cigar! i'm not sure how to help you at this point not knowing your level of knowledge or background i'm at a loss as to how to help clarify this for you.
sorry.

Inductor 28th January 2013 03:46 PM

Patnio, You are fine.
Quote:

Originally Posted by Patnio (Post 3345760)
The result of adding the 2 ohm resistor will necessitate a re-calculation of the crossover assuming a 6 ohm tweeter (rather than the actual value of a 4 ohm tweeter) to ensure crossover frequency remains correct.

Not if you placed it between the crossover and amp.
I think your problem is the tweeter/amp. I don't have experience with amps but I don't know of a problem with that. Usually is with low impedance in the lower frequencies. (Check that first). In this case you can use the 4R tweeter. Calculate the crossover and attenuate. In case you want 8R go with a 6 or 8 Ohm tweeter. Look at real values of Re in spec sheets.


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