I need to ask this question another way, I think....

[Erik Mandaville]

...but first, thanks again for the help I have gotten so far!

My original belief (which may be incorrect) was that the autoformer was essentially behaving as a voltage divider between it and the squawker. As such, the autoformer was removed, and a 16 ohm L-pad installed in its place. AL has indicated that the autoformer is needed (including the common connection on tap 'O') in order to maintain the desired output impedance of 16 ohms (or slightly less) going into the midrange -- which implies that in addition to voltage division which will lower the output level, the transformer combined with the input capacitor to it, are together functioning to establish both the crossover point AND impedance matching for the driver - which is what I understand him to be saying -- please, please correct me if I'm wrong.

However I was thinking that in most mid-branch networks I have seen the more usual way of doing a first order parallel network is to use an inductor and capacitor in series with the driver, where the coil value is obtained by finding the desired crossover region for the woofer and the value for 'C' from the crossover to the tweeter. No, wait a minute -- that's how the values are found for the tweeter cap and woofer coil, only. I have to do the band pass differently.

So, then I thought that the autoformer may also be providing the required inductance needed for this type of first order network -- which is why I was asking for the value of inductance between input tap 5 and output tap 4.

There was a post on the two channel forum, asking why I am using now using both the autoformer AND L-pad, since he indicates their function is the same. If that is true (now that I'm more confused than ever!)would it not be possible to remove the autoformer and rely ONLY on the L-pad for controlling the output of the driver. Dean has indicated, and what he says makes sense to me, that when one changes the output taps on the x-former, a subsequent change must also be made to the input capacitor -- which may be something that can be done away with by use of a swamping resistor across input and ground to the autoformer -- thus making changes to output taps less complex and involved (again, if I am understanding this correctly).

SOS!

Erik

[Erik Mandaville]

And to complicate things more: If the output of the autoformer is roughly 16 ohms (which will naturally change with the dynamic changes in frequency), it would seem that to use BOTH the autoformer AND L-pad would result in -- what a series connection between the two, which would then suggest a combined impedance averaging 32 ohms as opposed to 16 ohms for one or the other used by itself.

[mikebse2a3]

Hey Eric

First maybe Al will step in and give us a better understanding of exactly how the autoformer functions and its main benefits in this situation.

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No1: So, then I thought that the autoformer may also be providing the required inductance needed for this type of first order network -- which is why I was asking for the value of inductance between input tap 5 and output tap 4.

--------

From my understanding(from reading the design papers of the ALK network)is the Klipsch A & AA networks don't actually roll the upper end of the squawker off they just let it naturally roll off.Where as the ALK and the type A network shown in another post I believe by John Albright is a true bandpass design for the squawker section that also controls the impedence caused by the autoformer as implemented in the klipsch networks designs by the use of the 10 ohm swamping resistor. As you probably know since the 10 ohm resistor is in parallel with the autoformer/driver part of the network it limits the resistance of this part of the squawker circuit to less than the 10 ohm value of the resistor where as the autoformer/driver impedence will increase with frequency without the swamping resistor causing the the driving amplifier to see a more complex impedence load versus say the ALK design.

I remember seeing a paper by PWK about the varying impedance as not necessarly being a bad thing.(maybe I can find it and post it) Naturally the varying impedence will cause some amplifier interaction with the speaker/network causing some frequency response changes which a person might or might not like.

The reason the klipsch network uses the 13mfd value is because that value is necessary because of the varying impedance of there network but if you compensate the network for the varying impedance(10 ohm swamping resistor) the capacitor value must change for the proper crossover between woofer/squawker.

I see no reason why you can't select the coil and capacitor for a bandpass design with a L-pad.

So my question to ALK or anyone else that has a good understanding of this is why is the autoformer better than an L-pad?

Does it offer better coupling to the amplifier?

mike

ps: when you lift the ground side of the autoformer then you basically defeat its ability to attenuate the signal to the squawker.It my as well not be there at that point.

[mikebse2a3]

Here is the Dope From Hope paper for those that might not have seen it.

At least in one instance PWK found a benefit for a rising impedance at least around this time in audio design.

mike

[mikebse2a3]

here is the second page

mike

[mikebse2a3]

Here is a copy(not the best) of the first page of a article by PWK someone posted before.

In the experiment he tells some of the benefits/reasons why he prefered the autoformer method over an L-pad method.Maybe both methods have merit(notice the AK-4 doesn't use the autoformer method but by all I've heard sounds very good) which when the total system design is taken into account one might be prefered over the other

Eric:

take notice of the network in fig. 2

notice the L-2 in the squawker circuit. Its listed as a .5mh

so it looks like early on he designed a network with a bandpass section for the squawker.

I can see two ways of thinking about being able to adjust the balance of a speaker systems.

1: (the ideal and prefered situation to me)It would be hard for us to balance a speaker better than a manufacture like klipsch who has the equipment,facilities and experience that the average person doesn't. So if we take this well balanced system and put it in a room and it doesn't sound good we are left with treating the room and setup to bring out the full potential of the system.

2: Most real world situations are that we are limited in being able to balance the room to the speakers needs and one reasonable compromise might be is to within reason be able to alter the balance of the speaker to make the best of the compromises we have to live with.

mike

[mikebse2a3]

page two of the article for anyone wanting to read the whole article.

especially notice the first and second paragraphs of PWKs steps to reach his crossover designs.

mike

[mikebse2a3]

page 3 of article

Eric

You probably have already seen these articles like I had but it was good to go back and think about why PWK chose the methods he did at the time.I especially wanted you to see the crossover in FIG:2 with the .5mh coil(L2)like you had been thinking about.

mike

[Erik Mandaville]

Hi, Mike -- thanks for taking the time to contribute

John Albright was the one who indicated I was doing it correctly the first time -- with only the L-pad, sans autoformer. He said they basically do the same thing, which exactly follows my own thinking.

The reason I tried lifting the ground was also for just the reason you outlined, because I believed it to be the primary part of the attenutation circuit (which was already being handled by the L-pad. However, based on the measurements taken by Dean and ALK, there is nevertheless the inductance that exists between taps 5 and 4, sinceI was contemplating the possibility of using that value of inductance (which is too high) for the series coil in a true band-pass network. What I need is more in the range of .30-.40mH or so. The capacitor value would also need to be changed.

I found that I already have the 2 coils needed for this, but the capacitors I have are a little high at 47uf. They are also high voltage caps, and I'm saving them for an amplifier project, anyway.

I want to read the articles you have pointed out, which I'm sure would be helpful. I think the difference between an L-pad and autoformer in terms of sound is possibly not very significant. The important thing, is that they both provide the correct load impedance.

Thanks once again for taking the time to respond, Mike. You have been supportive with all of this, and I appreciate it!

Erik

[mikebse2a3]

Hey Eric

just to clarify my thinking on the L-pad/autoformer

The only reason to use the autoformer with the L-pad is because with the L-pad wired in parallel with the squawker the impedance is closely maintained allowing the orginal designs 13mfd to be used( you were orginally looking for a way to add adjustment without changing any other componets).This allows you to be able to adjust the squawker level without changing any other parts of the network. To me this just allows you to basically tweek the orginal design and isn't really a draw back since you already own the autoformer where as if you were designing from scratch you would just buy a larger value capacitor but no need to spend the money on the autoformer and just use the L-pad.I think in your situation for now its a good and simple way to acheive what you set out to do.

If you don't mind changing the 13mfd capicitor value then you can eliminate the autoformer and just use the L-pad.

In another thread you mentioned something about a peak in the K55v driver.

You've been around here so long I fell I'm probably mentioning things you alredy know but just in case: You probably know that the one with the push terminals is the one that suffers with the peak.The solder terminal one doesn't exhibit that problem.Any way if you do add a coil to roll the high freq. off in the squawker circuit maybe that will take care of it if you have the push terminal types.

mike

[Erik Mandaville]

Mike: Thanks for your help. The L-pad does in fact maintain an even 16 ohm impedance for the driver, possibly suggesting that it would be possible to simply run the L-pad inline between the capacitor and driver -- and do away with the autoformer as I first had done. If the capacitor needs to see the 16 ohm ipmedance provided by the autoformer between taps 0 and 4 -- the L-pad is thus achieving that requirement by itself. Moreover, it maintains the 16 ohm impedance regardless of whether the squawker is turned up or down. In this sense, it's much more straightforward to use than the autoformer, which may need capacitor value changes if there is not a swamping resistor across the autoformer.

I was concerned about the combined impedance resulting from the simultaneous use of the autoformer and L-pad, but thinking about that some more this afternoon, I don't think there will be a problem, since the L-pad is the same impedance as the driver, itself.

I do appreciate your sticking with me and contributing to this, Mike! It's hard to bounce ideas around in a room without walls.

I need to take a break from L-pads and autoformers and listen to some music!

Regards,

Erik

[WMcD]

As I usually say, there may be a number of quesions.

One is to observe that if we test the windings of the autotransformer (or plain old transformers) with nothing connected, they do look like inductors. And pretty sizable ones too.

But, do they act like inductors when in a circuit of the type we're working with. That is to say with a load. The answer here is that most or all of the inductance effect disappears. This is because the magnetic field is coupled to the "other" windings which have a resitive, or mostly resistive load. This is the coupling from one set of windings to the other through the magnetic core.

The other issue is whether the autotransformer and L-Pad each act alike. The answer here is no. That is why PWK found trouble with resistive attenuators.

The L pad has two variable resistors, one in series with the input and then one across the output. The overall reason is make sure the input side continues to "see" an 8 ohm load. The crossover network before the L-Pad will have different cross over points depending on the load it sees. So this is a good idea.

But, consider that drivers (hooked to the output of the L-Pad) have different performance depending on the load driving it. This is the whole "damping factor" issue of amps and speakers.

The transformer does step up voltage and step down current, or vice versa (our case). So it can cause a reduction of voltage drive at the output. That is good.

But these effects also cause an alteration of the impedance seen at the input to the transformer, and the output.

Very generally speaking with a 3 dB reduction in voltage, the input of the transformer should see twice the driver impedance. P=V*V/R. So for half power, R must be twice the nomimal resistance of the driver. (And we have to design the crossover network to that.)

There is a symetry to it. The driver looking into the ouput of the autotransformer "sees" one-half of the impedance seen by the input of the transformer. Except for the effect of the crossover circuit, that is just the amplifier output impedance. (An L-Pad can't do this.)

But, is this good? Yes if you want to maintain a low damping factor seen by the driver. Whatever the so called damping factor of the amp, it has now doubled for the driver on the other end of the autotransformer.

I checked out the roll off an AA type crossover at the midrange some years ago and it seems to be at or near what you'd expect. So the inductance in the autotransformer is not that much of an influence. I.e. it does not turn it into a second order crossover.

There are some things going on that an autotransformer does not act like a classic two winding transformer. One is that the resistance of the common windings do act like a resistive voltage divider. On the other hand, I understand the windings are said to be better coupled and may act, in some situations, more like a theoretically perfect transformer.

Best,

Gil

[mikebse2a3]

Gil said:

But, do they act like inductors when in a circuit of the type we're working with. That is to say with a load. The answer here is that most or all of the inductance effect disappears. This is because the magnetic field is coupled to the "other" windings which have a resitive, or mostly resistive load. This is the coupling from one set of windings to the other through the magnetic core.

I checked out the roll off an AA type crossover at the midrange some years ago and it seems to be at or near what you'd expect. So the inductance in the autotransformer is not that much of an influence. I.e. it does not turn it into a second order crossover.

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Thanks Gil for taking the time to explain this.

This was something that I never really tried to understand till Erik started working with the L-pad.

So thanks Erik for giving me a reason to learn more!

mike

[Erik Mandaville]

Gil said: "I checked out the roll off an AA type crossover at the midrange some years ago and it seems to be at or near what you'd expect. So the inductance in the autotransformer is not that much of an influence. I.e. it does not turn it into a second order crossover.

There are some things going on that an autotransformer does not act like a classic two winding transformer. One is that the resistance of the common windings do act like a resistive voltage divider. On the other hand, I understand the windings are said to be better coupled and may act, in some situations, more like a theoretically perfect transformer."

Thank you, Gil!

Erik

[Erik Mandaville]

And thanks Mike, too!

I appreciate any and all input/help/and suggestions!

Regards,

Erik