Fixed L-Pads

[Earl K]

,,,, snip,,,,The JBL L200B network (which I am using as a starting point) uses an L-Pad composed of 2.5 ohm series and 5 ohm parellel. This does not provide enough attenuation for my horn. In researching options, I have found two resources with different options:

- Forget the "pi" forums' option for now. Reversing the order of the buildout resistor to the conjugate ( parallel ) will only confuse you at this stage . The 20 ohm resistor that is already across the driver provides quite a lot of broadband damping .

- The existing fixed Lpad provides around 4.5 db of attenuation .You can make up a fixed Lpad ( with slightly more than 10 db of loss ) by using a 2.5 ohm parallel resistor , preceeded by a 4 ohm buildout resistor . ( 4.1 or 4.2 ohm buildouts ) would be better if you can conjure them up .

- In case you haven't realized it yet, the working load impedance of your network that your CLC passives look into, is somewhere in the area of the low 6s' ( 6.25 to 6.5 ohms ). Zilch is working on that answer right now .
- It's not 16 ohms .
- The inline fixed Lpad drops the working circuit impedance

[Zilch]

http://audioheritage.org/vbulletin/s...615#post104615

I'll measure stuff after dinner. Made a run to the City to meet up with a forum member.

[THAT had priority.... ]

[Zilch]

With PE 16-Ohm 50W L-Pad.

Z800:

L-Pad -2 dB 17.36 Ohms
L-Pad @MID 14.78 Ohms
L-Pad +2 dB 11.68 Ohms

If we take "MID," it's calculated to be 6.2461 Ohms.

[That was a LOT of work just to prove Earl off by 0.0039 Ohms .... ]

[Zilch]

Ooops! I have 5th edition and the highest figure is 7.112!

You KNOW what you gotta do, then....

Can I use these same formulae to calculate values for a series LCR to put across the voice coil to attenuate the med freq as Earl K was suggesting?

Zilch will take credit for suggesting that @ #3.

[Earl won't mind one whit.... ]

[Robh3606]

Looking through this what horn are you using???

Rob

[Earl K]

Hi Rob,

Looking through this what horn are you using???

Assuming you're talking to Zilch ;

- The answer to your question is in the header portion of each graphic .
- ie; It's a H91 .

<>

Thanks Zilch !

PS ; Unfortunately ( for the sake of clarity ) these impedance graphs were made without the "fixed" 4.5 db Lpad in place . The average 15 ohm value used , ( which Zilch refers to ) is used to calculate the working circuit impedance.

ie; Adding a 5 ohm parallel resistor and 2.5 ohm buildout resistor onto the 15 ohm load ( amplifier-side of the variable Lpad and driver/horn combo ) gives the lowish 6.xx ohm value that was previously referenced as the working circuit impedance. ( This 6.25 ohm value is the impedance value that must be used to correctly calculate the values of the hipass network .)

[dmtp]

Looking through this what horn are you using???

I am actually using a homemade 500Hz cut-off Tractrix horn.

[dmtp]

- In case you haven't realized it yet, the working load impedance of your network that your CLC passives look into, is somewhere in the area of the low 6s' ( 6.25 to 6.5 ohms ). Zilch is working on that answer right now .
- It's not 16 ohms .
- The inline fixed Lpad drops the working circuit impedance

Yes, I had just (independently!) figured that out - that explains why none of the values made any sense when calculated against 16 ohms! Any idea (or speculation) on WHY JBL would do that? It would have been just as easy to create fixed L-pad that maintained 16 ohms.

[dmtp]

Referring to Zilch's graphs (Thanks for all the work - this thread is really turning into a textbook on XO design!), I see the Le for the LE85 is a NEGATIVE number in each case. How do you get a NEGATIVE Le? Is that, like an imaginary coil?

[Earl K]

,,,, snip ,, Any idea (or speculation) on WHY JBL would do that? It would have been just as easy to create fixed L-pad that maintained 16 ohms.

- My speculation ( for one reason ) is that by changing impedances within separate portions of the same network, JBLs' crossover designers, can economize on the overall price of parts.
- Ie; Smaller Coil sizes can be used when "passives" are working into lower impedances ( though cap sizes do need to go up ). All in all , the price of copper ( by the pound ) likely overshadows the extra cost of the necessary larger size ( Mylar or Electrolytic type ) caps .

- There may also be a "damping" preference . I know that a low value parallel resistor does offer more circuit damping ( at least to my ears ) when used with a compression driver .

[Earl K]

Yes, I had just (independently!) figured that out - that explains why none of the values made any sense when calculated against 16 ohms! Any idea (or speculation) on WHY JBL ,,,,, snip, snip

(A) Okay, now that you realize that the actual load impedance ( that the crossover portion of the N200b ) "sees" is actually in the neighbourhood of 6.3 ohms / calculate ( through your text-book formulas ) what the F3 point is for the following ;

(i) 16.5 uF cap = gives an F3 of ??? hz into a 6.3 ohm load
(ii) .8 mH coil = gives an F3 of ??? hz into a 6.3 ohm load
(iii) 24 uF cap = gives an F3 of ??? hz into a 6.3 ohm load

(B) Then calculate what the textbook values are ( caps & coils ) of a 3-pole Butterworth network ( HP portion only ), crossing at 800 hz / or 900 hz or 1000 hz ( ??? ) .

(C) Once you've finished point (B) , do the same exercise of calcs. for F3 ( as in point "A" ) for the new values of "textbook-derived" passives .

[Earl K]

Hi Mark

The L200B accomplishes this bypass with 1 uF cap and 0.16mH coil and a fixed 5 ohm resistor. Does anyone really understand how this circuit works or how the values are calculated? I would like to add a little more high end boost (above 10k) but at this point it would be blind guessing at alternative values.

- Okay, back to this original post .

- Increase the value of the 1uF cap ( in the LCR bypass circuit ) to as high as 3uF . During this exercise ( in empirical study ), bypass your 0.16mH coil with a piece of wire. Your variable pot. will effect the F3 point of the cap in question when you rotate it . Increasing the cap to as high as 3 uF will likely mean you will need to broaden/deepen your 6K or 7K notch filter to get rid of excess level in this area .

- Once you have done this, and are happy with the amount of extra Hf your getting / then feel free to add back in an appropriate resonating coil in your series LCR. Be warned though ; to maintain a 13K center point for your boost ( with a 2.5uF cap , for instance ) leads to a tiny coil . @ 0.06 mH for a 2.5uF cap . If the R = around 5 ohms, the "Q" of this boost is around 1 octave wide. Such a low "Q" will likely result in little real boost / but go ahead and try it .

[Earl K]

Hi Mark,

The next trick to getting a tiny bit more HF & UHF into that horn is to create a ;
"Frequency-Dependant, Fixed Pad ".

How to Construct this sort of pad ?

- It's simplistic to implement. One just buries a small inline coil ( JBL likes to use .2mH ) underneath the parallel resistor of a high loss , fixed Lpad.
- One end of the coil is connected to the parallel resistor / while the other end of the coil is connected to the common / ground buss .
- The 10 db pad I suggested you build will be a starting point / though 15 db pads are better for this exercise .

How does it work ?

- Any coil builds up Inductive Reactance ( resistance ) as AC frequency is increased.
- The formula is as follows;
XL = 2 * pi * L * Frequency

L is stated in Henries
Frequency is stated in Hz .

- This ( inductive ) resistance is "added" ( though it's not a straight addition ) to the value of the parallel resistor ( in the fixed pad ). This higher value conjugate ( parallel ) resistor, essentially lessens the overall attenuation of the pad, with increasing input frequency. At 10K, ( and if the addition was not vector based ), we would add 12.6 ohms to the value of that 2.5 ohm parallel resistor ( found in the 10 db pad ).

- So roughly , at 10K the pad is attenuating only 3.8 db ( versus the original 10 db ) . 15K would be around - 3.4 db .
- At 5K, attenuation is around 4.77 db ( versus the original 10 db design of the straight pad ).
- At 1250hz , attenuation, calcs out to be - 7.16 db .

As stated, the above numbers are misleading, because I haven't used vector based addition to arrive at the real value for the combined resistance of the coil and resistor combo . The real resistance will be somewhat less .

Anyways, this is a "cheap an cheerful" way to steal a few more db back for the horn driver ( when compared to 1000 hz ) .

The down side ? The per octave db increase is small; Here about a single db per octave. Though a comparison of 15K compared to 1K looks better since it's a gain in the 4db range.
- The other downside is that this slight increase in HF also pulls with it a bit more midrange / which will likely need to be further notched out by your midrange LCR "trap" filter .

[Robh3606]

The other downside is that this slight increase in HF also pulls with it a bit more midrange / which will likely need to be further notched out by your midrinage LCR "trap" filter .

Why would you use an LCR trap in this case with this driver and horn combo??? Why wouldn't you use a simple series cap to do the midrange attenuation to flatten the curve above the mass break point?? These are not like the newer horns and drivers where you need to use LCR traps to do basically parametric EQ on the driver horn combo. Are Tractrixs???(spelling??) CD type horns or a mix between Exponential where the directivity changes with frequency so the EQ is not as straight forward as a true CD horn????

Rob

[Zilch]

Unfortunately ( for the sake of clarity ) these impedance graphs were made without the "fixed" 4.5 db Lpad in place.

Hadta scrounge up the correct R's this morning.

Z800 = 6.2417 Ohms, measured.

The impedance the actual filter faces is quite stable with this topology. Compare to that of just the driver and horn in #18, above, top.

Le is calculated at 1 kHz, and the impedance is dropping at that frequency, so it's negative.

Le at all sampled frequencies is given in the WT2 full sweep data file.

[Seems no matter what, Fs remains the same.... ]