Question on taps of T2A autoformer

[Erik Mandaville]

And:

Disagreements over the cheater plug issue nothwithstanding, I think I have two very good quality 16 ohm L-pads you can have if you would like to try something with them. Worse case is that you don't like them, in which case you can take advantage of one of Bob's autoformers with the greater number of steps in attenuation, and then simply calcuate the input capacitor value based on the multiplied impedance of the driver determined by your chosen crossover frequency. You might like his tweeters, too, by the way. I have the big system playing right now, as we often do for background music while doing other things. The new tweeters were an incredible upgrade, IMO. Some haven't cared for them, but I like them very much. Anyway, you can swamp the input to the autoformer, just as Al does, if you want to.

Erik

[greg928gts]

Gil, can you attach the new corrected spreadsheet?

Thanks,

Greg

[Guest " "]

Gil, can you attach the new corrected spreadsheet?

Thanks,

Greg

Gil updated the original....

[WMcD]

I did, and I edited to take into account the very good remarks by Dean. Mea Culpa.

Gil

[Cut-Throat]

Thanks Gil!

[Erik Mandaville]

You are familiar with your Moondog and DRD schematics aren't you? This low order network is comparatively simple. The whole point is to learn this yourself. Giving someone the answers and doing the work for them doesn't foster learning, so study the schematic you already have for this network (count the number of connections compared to the amps you built), and figure out how to replace the autoformer with a variable L-pad. It's similar to a conventional potentiometer. You just have to remember that the capacitance you need will be different because of the lack of an autoformer. If your driver is 16 ohms, calculate the amount of capacitance needed for the crossover point using that impedance (16 ohms). For the tweeter, it's 8 ohms, but in doing this you will make a discovery.

If you are interested enough, you'll find out what you need to know.

BTW (A): When making a request of someone, the word "Please" or phrase "Would you mind drawing me...." is likely far more effective than a command in this particular situation. I would certainly do the same for you if the situation were reversed.

BTW (: When you start your research, I would suggest looking into a band-pass, which in a low-order network like this consists of two parts: a capacitor and an inductor in series with the load (the voice coil). There isn't a shunt element. It's no more difficult than the connections you made between the voltage divider on the 45 cathode which in turn feeds the plate choke for the 6N1P plate, and is also in common with the grid of the 45. Check the schematic and you'll see what I mean. I don't have one in front of me, but that connection is there, and this crossover with much bigger parts is easier. You've built your own crossovers in the past and already know this. Just learn to draw the map for it using a variable L-pad.

[Erik Mandaville]

Interesting.

Suddenly the request for a schematic vanishes. You asked for a network design that replaces the autoformer (which is a design compromise I would rather do without -- we all choose our own compromises in this...), and I suggested you draw one up yourself.

No scaffold or trap door here. You have a specific driver (I'm guessing its specs are available?), and you want to lower its output a given amount relative to the other drivers you're using. It's not a difficult thing to do, Kevin, and in building your own kit amps already have the manual ability in terms of required soldering. For this type of network, there is literally a small handfull of connections to be made.

So, I'm not going anywhere. Draw something up as I suggested, and we can work it out. I'll let you know if you're on track.

BTW ©: The type A crossover is as simple as simple can get for a frequency divider. There is nothing but two capacitors and an inductor. The autoformer, IMO, is basically an extension of the driver, and is there to create, by way of impedance imbalancing, a poor power transfer between the amplifier and the voice coil of the midrange driver.

BTW (D): Are you even sure this is an appropriate network to use with the driver you're using? I was under the assumption that you were using a K55, and you're not -- at least that's what your former post indicated. Where is that? I can't find it. If you can give me that information, we might be able to see if it's really the best network to use in this situation, or if something else might be better -- or if it has a rise in its impedance that needs to be corrected (do you happen to know the inductance and resistance values of its voice coil?)

Standing by,

Erik

[Erik Mandaville]

Lastly: I never said I wasn't willing to help -- others claimed that to be the case. If all you want are the answers to a problem, which IMO isn't really learning, than I'm not the one to assist you. I will make you work a little harder than that, but what you get in the end is more than just a value of capacitance and resistance. I am happy to put you on the track, but you will have to do the rest, as well as sketch out on paper some possible schematics, based on your choice of autoformer or L-pad. Again, this is for the type A. It is oddly the most simple of designs for the big Heritage speakers, yet the one many prefer to others. All of that without any sort of compensation, notch filters, etc.

So, if you aren't interested in what you proposed yesterday, all you have to do is say so. I'll respect that and move on to some other things I have planned for my own system.

BTW (E): I think it better to not use BOTH a resistive L-pad AND the autoformer, since you start losing some of the benefits of either type. It's like making a roof for your house to keep the rain off, and then adding another one on top just to be sure. If you want to use a swamping resistor, that can be helpful for the impedance matching, but remember, PWK didn't use one on the type A, and lots of people, including me, didn't use one one this particular network. I do use impedance matching on the current network I designed, and just preferred it in this case over not. I tried it both ways.

Regards,

Erik

[Cut-Throat]

Thanks for your offer. At this time, I am not interested in designing crossovers. I don't design amplifiers either. I have buit a few amplifers from kits with a schematic included.

If there were some 'small' adjustments I could make with my existing type 'A' , I was game for that. But apparently it is more complicated than that, so I'll leave it there. If I could buy a Type 'A' that would take into consideration my Altec 288-16K drivers and 1005B horns, I would have ordered it already.

[Deang]

Here you go Kevin. Shoot me an email at home if you have any questions.

[Cut-Throat]

Thanks Dean!

[Deang]

btw, you'll need a pair of Bob's UT3636 autoformers! Shoot him an email at bobcrites@centurytel.net Don't tell him I sent you -- he'll charge double.

[Erik Mandaville]

That's fine, thanks for letting me know. I would say it's still worth studying, and that's something you can do on your own. It's beneficial to know the process rather than just the answer -- but only if you're interested in knowing the 'whys' and 'hows' rather than simply the 'whats'.

Erik

[Erik Mandaville]

A couple of questions, though:

If an 8 ohm tweeter at a crossover of 6000Hz would generally need a capacitor of about 3.3uf, why was the choice in the original type 'A' just 2uf?

Erik

[Erik Mandaville]

And:

We see that the input capacitor is connected to both the crossover input AND the top of the autoformer. So, this puts 13uf in series with the 2uf that's connected to the tweeter, correct? If that IS correct, what then happens if the tap is changed on the autoformer to drop the output, and the input capacitor is changed to something like 3.3uf to compensate for the subsequent change in impedance. 3.3uf is now in series with the 2uf capacitor. Is it the assumption that any change to the input capacitor does not influence the value needed by the tweeter for a 6kHz crossover? The primary purpose of the autoformer is to attenuate the midhorn, but does this mean that the tweeter is completely free of anything associated with it, including the 13uf capacitor that's in front of and connected to the 2uf cap?

13uf and 2uf in series is actually very close to 2uf, so for all intents and purposes would be close enough (it was probably much easier for Klipsch and Associates to find bulk 2uf caps than 1.7uf types -- I would do the same!) But, if we upset that balance by lowering the value of the input cap in the interest of squawker attenuation, what happens to the net capacitance seen by the tweeter for its crossover point? Would those of you who know the answer to this share your thoughts with me?

Thanks in advance!

Erik

[WMcD]

I think you're on the right track.

As far as the tweeter is concerned, it is fed by two high pass filters. The one at the input to the autotransformer is passing everything above the mid crossover point with the 6 dB per octave roll up. Call it 400 Hz. Once we get above, say 800 Hz, this filter cap is sort of acting like a short circuit.

The second filter is the one from the output of the autotransformer to the tweeter. It feeds everything above 6000 Hz with the 6 dB per octave roll up.

The bottom line is that the tweeter filter (cap after the autotransformer) is not affected very much by the mid filter (cap before the autotransformer).

Some of this is analagous to caps in series. If one cap is, say, 10 times larger than the other, you get an effective value close to the smaller one and you can foreget about the bigger one.

- - - -

The situation with the caps is complicated by autotransformer. But it works to our favor. Essentially, a transformer, "transforms" impendance "seen" looking into it on the primary side, and looking into it secondary side. The convention is that the primary is the side where power is applied.

Suppose we have at transformer which steps down voltage and we have Vp = 1 at the input. We pick a winding tap so that it steps down to Vs=0.5 volts.

Because of the transformer action, the current at the secondary is 2 x the primary current. This is necessary if the transformer is lossless. Power is not destroyed. So P = V*I is honored. Therefore Is = 2*Ip.

However. The definition of impedance is Z = V/I. We've agreed on the values and can reckon, Zp = (0.5*Vp) / (2*Ip) = 4 Z.

Here, this means that if we put a 10 ohm load on the secondary (the driver), looking into the primary, we "see" a 40 ohm load.

We see this in looking at the input impedance of the K-Horn (in the Heyser article) and the Heresy (in a Dope from Hope). The impedance seen "looking" into the primary of the transformer (actually at the input terminals of the speaker unit) is up at 40 or even 80 ohms or so (depending on the step down tap used in either). These are rough numbers.

- - - - -

The transformer action has reciprosity. Suppose we put a 10 ohm load on the primary and looked into the secondary. It's effective impendance is 2.5 ohms or 10/4. This is what PWK liked and mentions in "The Trouble with Attenuators" and "That Damping Factor." The autotransformer is making it look like the amp has a lower internal impedance. An L-Pad can't do that.

But this impedance transformation also works on the "effective" impedance of the cap which is at the input to the autotransformer. I know this is getting complicated.

The Zc = 1/(2*pi*F*C). But the one we're considering is the one the primary side. When the tweeter circuit "looks" into the secondary, it sees 1/4 Zc because of the transformation. (Technically called "the impedance transfer function" of the transformer). And note that we call horns, acoustic impedance transformers, which is a somewhat related.

We have to look at the the Zc equation just above. We know Zc (at the input side) is 1/4th of the actual value, If we want that to work, we see that C is 4 times the actual value. Therefore, if there is an actual 13 uF at the input, the tweeter circuit "sees" a 52 uF cap.

But back to the original issue. You are correctly observing that a 2 uF in series with a 13 uF results in mostly a 2uF. But I'm pointing out that from the tweeter's point of view, it is really something like a 2uF (in the secondary circuit) in series with a 52 uF (in the primary). So the 52 uF has even less effect than a 13 uF in a series calculation.

Gil

[Erik Mandaville]

Thanks for responding!

Sure, I understand the how and why of PWK's preference for this chosen approach to attenuation. I see no problem whatever with the 13uf and 2uf in series with one another, which WILL interact, which in the case of these particular values of capacitance isn't much of a problem. I think it's actually a very intelligent solution, despite the fact that I have found through too many hours experimenting with it, that I prefer the tweeter on its own, and entirely independent of the autoformer.

You see, though, that the approximate 2uf (or 1.7ish uf) with the series connection with the 13uf capacitor must combine to form the correct amount of reactance at 6000Hz. From book calcuations and tables, I find that 2uf is really more in line with a 16 ohm rather than 8 ohm voice coil. The calculated capacitance for 8 ohms at this frequency is a little over 3 uf. The autoformer has a very high AC resistance with respect to ground (a good thing), yet nevertheless seems to interact with the behavior of the tweeter. There are two high pass filters, but they are not isolated from one another.

Depending on the amount of attenuation needed, it seems the crossover point to the tweeter will be influenced somewhat by the combined effects of the alterations in the values of the components associated with it. It would thus not seem out of place to compensate for not only values of capacitance associated with the mid driver, but also those of the tweeter.

There may be a correlation with this in regard to what I found with the narrow bandpass/resistive L-pad network I built back whenever that was. Even with an 18dB octave slope, the tweeter seemed to have greater presence and output than the stock type 'A' to which I was comparing it. But there were other significant changes to the network too. I very much respect P. Klipsch's work, and love our Klipschorns and Heresies, but I just don't have his ears with regard to choice of attenuation -- even with the flea watt amps I have. That's just a personal choice based on my own experimentation.

Anyway. I do thank you for taking the time to reply. Were it me (which it's not, so it doesn't really apply), I would also be inclined to keep the value of the tweeter capacitor in mind when making impedance compensation changes associated with stronger steps in attenuation of the squawker.

Thanks again,

Erik

[Deang]

"...it seems the crossover point to the tweeter will be influenced somewhat by the combined effects of the alterations in the values of the components associated with it. It would thus not seem out of place to compensate for not only values of capacitance associated with the mid driver, but also those of the tweeter...I would also be inclined to keep the value of the tweeter capacitor in mind when making impedance compensation changes associated with stronger steps in attenuation of the squawker."

All of the early Heritage speakers, regardless of the crossover point or level of attenuation -- use the single 2uF for the tweeter. Big Heritage with crossover points of 400/6000 and using taps 5 and 4 use the 2uF. The Cornwall with 600/6000 and using taps 4 and 3 use the 2uF, and the Heresy, with 700/6000 and using taps 3 and 2 also use the 2uF for the tweeter.

"...by way of impedance imbalancing, a poor power transfer between the amplifier and the voice coil of the midrange driver."

I've posted the articles several times for you in relation to this, are you ever going to read them -- or did you read them and just decide that PWK didn't know what in the hell he was talking about? In spite of your continued insistance that "the Type A crossover is as simple as simple can get for a frequency divider" -- your questions and assertions show that you fail to grasp even the most basic principles of its design.

[Erik Mandaville]

Hi, Dean!

I've read the articles many times, and am very clear on how it works.

Take care,

Erik