Network Design Advice Sought - What Load Impedence Value to Use For Cornwall Bass Bin ?

[radiogram]

This question is for the likes of ALK / Lee and the likewise network experts.

I am planning on building a custom crossover for the woofer section of my Cornsey Speaker (i.e. Cornwall Bass with Crites CW1526C Woofers, Hersey Mid & High - K53/K701 and K77). The mid and high section will be pretty much like stock i.e first order where the K53 is crossed over at about 710 Hz. The only difference between the stock Hersey network and mine will be that the input cap in my case will be 4uF instead of 2uF since for Cornsey the squawker is moved one tap above at the Autoformer to accommodate for the higher sensitivity of the Cornwall bass bin.

The plan for woofer section is a 3 pole PI Filter with two series inductors and a parallel cap in the middle as follows (I want a steeper network for the woofer than 1^st order):

-----L--------L-------

|

C

|

The crossover freq will be around 620Hz or so. In order to compute L & C, I only need the value of Crossover Freq and load Impedence.

The Re of the Crites Woofer is 3.7 Ohms. I measured the actual in cabinet impedence response at various frequencies using a dividing network with a known Resistor value and using voltage division theorem, and I have attached measured impedence plot here.

So my question is to do with what Load Impedence value to use to calculate L & C.

As you can see, if we ignore the two impedence peaks around 20Hz and 65Hz (typical of a Ported speaker), the average impedence between 30Hz – 600Hz is around 5.7 Ohms. At the chosen crossover freq of say 620 Hz is it is about 7.7 Ohms.

So my question is to design the 3 Pole PI network for a Fc of 620Hz, should I use the Load impedence value as Re (i.e 3.7 Ohms) or the Average (i.e 5.6 Ohms) or the impedence at the crossover freq (i.e 7.7 Ohms) ?

Thanks

[radiogram]

Correction - The planed Network for the woofer is a T section and Not PI Section.

[WMcD]

I would use the reading at the crossover point. Your certainly don't have to worry about the peaks.

Builders sometimes use a Zobel to compensate for the rising impedance.

Generally, the lowest impedance above the peaks is actually the voice coil resistance. This is because the mild bottoming out of impedance magnitude signals where the inductive and capacitive reactances balance out -- leaving the resistance. So I'm a little puzzled why it is not a little lower. But I wouldn't worry.

The gentle rise up to 1000 Hz and above is because of the inductance of the voice coil. A Zobel can pull that down flat.

The Zobel is beneficial (at least in theory) mostly when the crossover point is very much higher, as in a two-way system, i.e. when the freq is so high (maybe 2000 Hz) that the XL is significant.

However in this case, at 700 or so Hz and the next half octave, the reactance is fairly low. So you probably don't need a Zobel.

Wm McD

[Al Klappenberger]

Here's how to do what you want to do. It includes a program to do it. BTW: design for 6 to 7 ohms, NOT 3.5! The K33 and Bob's clone is 6-7 Ohms in sereis with 1 MHy voice coil inductance.

http://www.alkeng.com/le/le.html

The Zobel that Gil mentions is a good alternitive to building the voice coil inductance into the filter. It will give you an extra active element. Use 28 Ufd in sereis with 6.8 Ohms. Design the filter for 6 to 7 Ohms.

Al K.

[WMcD]

Like Al was saying. You can compensate for the voice coil inductance in the cross over design. You can assume it is 1 mH and then deduct that from the value of the right-most inductor in the T network -- the one feeding the voice coil. You already have that value built into the voice coil.

There is a bit of an assumption with the above. It is that we can model the voice coil inductance as a lumped (single value) inductor. That is not quite accurate. If you can do the math you can: Determine the voice coil resistance (d.c. or zero Hertz). Determine how much higher the impedance is at 700 Hz or your crossover. Use the added impedance as XL (impedance of the voice coil inductance). Then back figure the value of an inductor with that impedance at 700 Hz. This should be around 1 mH and probably a bit lower.

Then lower the value in mH from the right-most inductor.

Wm McD

[Al Klappenberger]

Gil,

The actual value of the Le is actually rather stable over the woofer frequency range. It does vary, but not by enough to throw things out of wack. I do the measurement my a swept bridge using white or pink noise, It gives me a plot if polar impedance over the entire range ( Z0 plus phase). The big assumption is that the voice coil becomes part of the filtering function. It does NOT! It does become part of the impedance looking into the filter though. If you design for N=3 the slope becomes roughly N=2 but the input Zo looks like N=3. The advantage is that you get rid of one inductor. Using the Zobel, you need all the physical elements and the all function to generate the slope. The Zobel tunes out the woofer Le to form a very nice 6 or 7 Ohms resistive impedance.

Al K.

[Al Klappenberger]

Gil,

I just dug out the Zo plot I did on one of my Belle Klipsch. Here's the actuall values:

Frequency, R component Ohms, Le (Mhy)

300 Hz, 7.16, 1.31 mHy

500 Hz, 7.0, 1.22 mHy

700 Hz, 7.55, 1.11 mHy

800 Hx, 8.75, 1.19 mHy

1000 Hz, 9.52, .88 mHy

So it seems the resistive part of the impedance rises a lot but the Lx only drops a bit as you go up in frequency. I just round it all off to 7 Ohms in sereis with 1 Mhy for most cases. This was my center Belle which was made about 10 years later them my flank speakers. they were closer to 6 Ohms with roughly the same Le.

Al k.

[radiogram]

Alk, Gil:

I appreciate your feedback and sharing info. Thanks.

Are you guys suggesting strongly that I should build a Zobel? If, I do not, what will be the side effect and how bad it will be? BTW, the AL-4 network has a T Section ( I am not sure if it is butterworth/Chebyshev) but does not use a zobel. I am not defending that but this is just FYI.

Thanks

[radiogram]

Determine the voice coil resistance (d.c. or zero Hertz).

Ok. The DC Resisitance as I measured with multimeter is Re=3.7 Ohms

Determine how much higher the impedance is at 700 Hz or your crossover.

At 700Hz based on my measurements and little extrapolation the Impedence @ say 700 Hz will be about 8 Ohms. So, 8 minus 3.7 = 4.3 Ohms by which the Imp is higher at 700Hz.

Use the added impedance as XL (impedance of the voice coil inductance).

Ok, so the extra 4.3 Ohms is attributable to the inductance of the voice Coil.

I am with you so far, but after this step I am stumped. Please help me more.

Thanks

[Al Klappenberger]

Gram,

The use of the Zobel is really your choice. It is required of you use an even order filter (N=2,4,6...). In this case there is no output inductor for the woofer Le to emulate. The output of the filter will always be a shunt cap. You must compensate for it, If you use odd order (N=3,5 7 ..) you can make the Le equal the output inductor. In that case it will function as 1 less element for the transfer function. The impedance will look like the order you design for. This is important to maintaining a flat impedance when you design the highpass section. It should be of the same order as the design of the lowpass. The advantage is that you can build it withOUT the inductor and withOUT the Zobel. It's really your choice. With a higher order filter or one that has a "finete zero" (a notch) to sharpen the skirt, the extra inductor is a moot point. That's why I don't usually uses the Zobel. The extra inductor adds 6 dB/octave to the skirt slope. With a higher order filter, that's small potatoes.

If you will notice, None of the Klipsch networks use a Zobel and most of them are 2 elements. The Le is not compensated for. This is just one of a long list of reasons I conclude the stock network are poor designs. The only exception is the AK-3 network in the Khorn. It just happens to be correct when you include the woofer Le.

BTW: To clarify one thing: There are TWO factors to the resistive component of the impedance. There is the DCR that you can measure with a simple ohmmeter ( 3 to 4 ohms resistance of the wire itself) and there is an additional component to the resistive portion, that is the AC portion. It raises the resistive component to 6 or 7 Ohms. This is in series with the woofer voice coil inductance of about 1 mHy (K33).

Al k.

[Arkytype]

What he said. :>)

Al's step-by-step tutorial is very useful. However, if you aren't blessed with test equipment as Al and I are, you might want to consider a program from Parts Express called WT3. It will measure a boatload of your driver's Thiele-Small parameters as well as accurately measure the inductors, resistors and capacitors in your network.

They have a special that includes the WT3 software/hardware and a precision scale for 100 bucks.

http://www.parts-express.com/pe/showdetl.cfm?Partnumber=390-803

Lee

[Al Klappenberger]

Lee,

I could be wrong, but I betcha that won't show you the reactive (phase angle) of the woofer impedance. It's a factor that loudspeaker people just don't even consider. Below is the sort of plot you really need. It is for the Klipsch K24 driver (Heresy II). The marker is at 137.5 Hz. This is the ONLY frequency where that driver is a resistive impedance and it's 7.98 Ohms. Every frequency higher looks like an inductor in series with a resistance. You need the reactive component in order to properly design the Zobel or to compensate for the Le by absorbing it into the filter output inductor.

--- UPDATE ---

Lee informed me by email that the software he suggests DOES indeed have the Zo and phase capability. I downloaded the demo for it and, a first glance seems to say that he is right! It needs some sort of USB tests leads to work that I don't have, so I can't test it out to be sure though.

[WMcD]

Determine the voice coil resistance (d.c. or zero Hertz).

Ok. The DC Resisitance as I measured with multimeter is Re=3.7 Ohms

Determine how much higher the impedance is at 700 Hz or your crossover.

At 700Hz based on my measurements and little extrapolation the Impedence @ say 700 Hz will be about 8 Ohms. So, 8 minus 3.7 = 4.3 Ohms by which the Imp is higher at 700Hz.

Use the added impedance as XL (impedance of the voice coil inductance).

Ok, so the extra 4.3 Ohms is attributable to the inductance of the voice Coil.

I am with you so far, but after this step I am stumped. Please help me more.

Thanks

Okay, let's go to just beyond were you left off.

We know the standard equation for impedance or X or Z of an inductor L. It is Z of inductor = 2*pi*F*L

EDIT: I gave incomplete calculations in my earlier reply. My error

Attached is a spreadsheet which I believe is accurate. The user has to insert the measured values. The ones I used are dummy values just used to check that results are ballpark.

I apologize for the error.

- - - - - -

I agree with Al when he says that this is part of the load seen by the filer. However must respectfully disagree with Al to the extent he says that it is not part of the filter function. (Smile, for my buddy Al.)

My reason is: If we drive the woofer without a crossover filter, there is a roll off in acoustic output at high freqs and this is in part because of the filtering effect of the voice coil inductance.

Further to this. Let's consider a first order low pass. It is just an inductor in series with the voice coil and adds to the voice coil impedance.

Now I'm gonna say something which agrees with Al in a way as far as load issues. First consider that we're used to thinking that the first order filter inductor acts sort of like a resistor which increases in resistance as freq goes up. Therefore high freqs don't get through.

But by the same token, this increasing reactance (the correct term) prevents the amp from delivering current into the input of the filter and woofer combination. It is the flip side of the same idea. The amp can deliver voltage but not current. So there is no power delivered.

There is something else. The inductive load moves voltage and current going into the filter and woofer (or draw by that load) out of phase with each other. That is a power factor thing and maybe too complicated to go into here.

- - -

You can use a Zobel if it makes you happy. None the less, I think you just deduct 1.0 mH from that right most inductor and things will be good. I don't know what value you're using but it could well be much much higher than 1.0 mH and not make a lot difference.

- - - - - -

Al, thanks for the impedance curve on the Belle. A long time ago I looked at something like that too. I expect that the horn loading of the Belle and its effect on the electrical impedance swamps out the voice coil impedance a bit. The CW's impedance curve is different than a Belle (I think) even though the woofer may be the same. I'm not challenging your good work, but it is something to consider.

Wm McD

Calculate L given Z and R in Series.zip

Calculate L given Z and R in Series.zip

[radiogram]

Al,

Design the filter for 6 to 7 Ohms

I am bit confused here. We have two values 1) Average or Nominal Impedence and 2) Impedence at the Crossover Freq of interest.

Based on my measured impedence (see attached measurement spreadsheet) in 0-800Hz (the bandwidth of interest), the average impedence excluding the two peaks comes to 6 Ohms. At freq of say 630Hz (the chosen crossover freq) the impedence is 7.7 Ohms.

Now, if I design the network for the Nominal Impedence I should use 6 Ohms, on the other hand if I design based on Gil's suggestion to use the value at the Crossover freq, then I have to design for 7.7 Ohms. Can you please clarify why you suggest 6-7 Ohms.

Thanks

CS1526C_Imped_Measurement.pdf

CS1526C_Imped_Measurement.pdf

[radiogram]

None the less, I think you just deduct 1.0 mH from that right most inductor and things will be good. I don't know what value you're using but it could well be much much higher than 1.0 mH and not make a lot difference.

Ok, based on Xover freq=630 and imp=7.7 (imp at the Xover freq), I get the following values:

L1 = 2.92mH; C=43.6uF; L2=.97mH.

If I deduct the 1mH voice coil from L2, L2 goes out the door leaving only L1 and C for the final network.

If I move up the crossover freq then L2 starts getting less and less and for e.g. if I want over at 700Hz L2 comes to .90 and L2-1 = -0.1mH. Upto how much mH difference we can ignore?

[Al Klappenberger]

Gram,

You will never be able to terminate the filter in exactly the correct impedance over the entire frequency range the woofer will operate. You need to design it for the impedance that will best terminate the filter in the most critical area. That is where the filtering function "comes together". That is right in the area of the crossover. The general rule of thumb is an error of 1.5 :1 or less. This is considered acceptable. For a 7 Ohm design you can get away anything from 4.7 to 10.5 Ohms. In the microwave industry this is referred to as "return loss" and represents about 14 dB return loss. 1.5:1 is also "VSWR", a term you are probably familiar with.

Al K.

[Al Klappenberger]

Gil,

The question about the effect of the series inductance of the voice coil on the skirt is one that fooled me too. I needed to do a series of experiments to see if N=3 when the output element was the voice coil inductance performed as N=2 or N=3. By measuring the acoustic output of a woofer with no network, storing it in the analyzer's memory and later subtracting that plot from a similar plot with the network in the circuit, it was obvious that the filter was performing as N=2, NOT N=3. The reactance slope seen at the input of the filter was that of N=3 however. I still have these plots if you would care to see them. I posted the entire research on the Audio Karma forums about 2 years ago. You might want to dig them up if you are a member over there.

Al K.

[radiogram]

Al, Gil:

I am confused yet again.

(For clarity in my terminology below: L2' is some Inductance <> L2; L2" is some inductance <> L2 or L2).

The circuit model for the Load is:

-----------

|

Re

Le

|

-----------

The effective load impedence Rz (freq) = Re + The Inductive Resistance(freq)

In order to compensate for the varying impedence we have two approaches:

1. Implement a Zobel and make the effective impedence to some constant R, forget about the Le and then design the T section for that R.

OR

2. Incorporate Le as part of the T filter by compensating Le from L2 and then the network model becomes as follows:

----L1--------L2(=L2'+Le)--

| |

C Re

| |

---------------------------------

Where, L2 = Some L2' Plus Le.

Since we have incorporated Le in the filter we have taken care of the varying AC Resistance component introduced by the Le. Thus what we have left for the Load is only Re. Should'nt that dictate then if I use the L2'=L2-Le appraoch I have to then design the T Network for Re and not the impedence around the crossover frequency since AC component introduced by Le has been taken into account? Let us say if we design a T Section for Re=3.7 Ohms and get L2 for the right most inductor. So for our case the actual value of the right most inductor then becomes L2' = L2-Le?'.

By designing a T section for say 7 Ohms and then again subtracting Le from L2 seems like a double compensation to me. If we design for 7 Ohms instead of Re=3.7, then by definition we have already taken into account the inductive resistance, so further making L2' = L2-Le seems double error. It seems to me that if we use nominal impedence of 7 Ohms, that being equal to (Re+Some Average Inductive Resistance), then we should forget about Le or use Re for designing the network and then incorporate Le by doing L2'=L2-Le.

Where is my analysis wrong if so?

Thanks

[radiogram]

Reagrding my previous Post that relates to my question on compensating for Voice Coil Inductance (Le) in the filter design, I have made some important corrections for better clarity. So, if you care to respond to that, please re-look at the post and do not go by what you saw before or what you saw in the email you got.

Thanks

[Al Klappenberger]

Gram,

I think you are getting the idea except that there is no need to subtract Le and L1 (actually Ln in filter jargon). As it happens, when you design a Chebyshev filter, the parts values change dramatically with the level of passband ripple you select. For a woofer filter, ripple as high as 2 dB is fine. Your listening room is far worse that that! This means you can select a passband ripple such that the output inductor becomes EXACTLY EQUAL to the measured Le at the crossover frequency. If you do this, the output inductor is totally GONE!The only time you need to subtract the two is in the case where the inductor required for the filter is larger than the Le. That turned out to be the case for my ES400 network. In that case you just add an inductor to the output to make up the difference. The program I provided for download has the ability to keep inputting different ripple values to manually zero in on the correct value to make the inductor equal to Le. That is, a manual iteration. If you choose to use a Zobel to absorb the Le, then you can do virtually anything you like with the ripple value and even go even order.

Al k.