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Thread: Fixed L-Pads

  1. #1
    Member dmtp's Avatar
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    Fixed L-Pads

    I have a somewhat general question about L-Pads with fixed resistors. I am still working on my LE14A/LE85 project. I measured the relative sensitivity of the two drivers and find the LE85 about 19dB higher than the LE1`4A (no fixed L-Pad, 16 ohm variable L-Pad set @ 12 o'clock). 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:
    1. The pi speakers article on cross-overs which was excellent http://www.pispeakers.com/Speaker_Crossover.doc shows the 'L' backwards from the JBL (which I believe is standard) version with the series resistor coming AFTER the parellel. It also tends to have a series resistor larger than the parellel one. Unfortunately, they do not give formulae to calculate values.
    2. Vanc Dickerson's Cookbook gives the formula for a fixed L-pad. Unless the labels of R1, R2 are swapped (which is why I refer to parellel and series rather than 1-2), the series resistor comes out much larger than the parellel one. (For 16 ohm, 16dB attenuation, the numbers are sereis= 13, parellel = 3)
    I tried the L-pad as calculated in the cookbook and the attenuation seems about right, but I am wondering what advantages there might be to using the pi speaker approach (they claim increased damping of the horn) and how one might calculate values. Also I wonder that JBL uses a small series resistor and larger parellel resistor in most of their designs.
    Finally, pi speakers bypasses their series resistor with a small cap (which varies between 0.47 and 5 uF). Not sure how this value is calculated. 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.
    Thanks for all your help. This forum is really great!
    MarkT

  2. #2
    Administrator Robh3606's Avatar
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    Hello Mark

    If I remember correctly you use a resistor value that 2x the drivers to help minimize impeadance changes that the networks see's. As far as the bypass circuit it was used in the 4430/4435 as well. If you look, it is connected at a point before any attenuation takes place. You have a capacitor, inductor and resistor in series. You get a resonant circuit with the resistor changing the circuit Q. With no resistance you get max output. The higher the resistor value the more you drop and broaden the peak just like the Q/Width adjustment in a parametric EQ. I would go with the L200B network values to start and you could put another potentiometer in series, only connect 2 legs, adjust it for the best balance. Then you can either leave it in so you can adjust the level like the 4430 or measure the value and drop in a fixed resistor.

    Read this it will help explain things better than me. It's a great reference. The Profile for the 4430/35 also explains the HF Bypass as well.

    http://www.audioheritage.org/vbullet...6021#post76021

    http://www.audioheritage.org/html/pr...bl/4430-35.htm

    Rob

  3. #3
    RIP 2011 Zilch's Avatar
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    L-Pad Basics

    Click #122 in the right column here:

    http://www.bcae1.com/

    Are you trying to eleminate the adjustable L-Pad, or just want more attenuation before it?

    Either case will require knowing the actual impedance of the component combinations to calculate accurately and retain the original conditions.

    *****

    See Fig. 7.155 and related text in Dickason for what the bypass cap does in filter contouring.

    The LCR resonant circuit is tuned like a notch filter to establish a more specific range of frequencies to be bypassed. The calculations should be similar. Instead of shunting a notch to common, you're passing it on to the driver.

    I've only recently come to understand that the resonance of that LCR generates some amount of actual boost. What the limits of that are, and how useful it is, I don't know.

    The fundamental compensation comes from contouring the differential SPL available from the HF driver/horn combination, attenuating more it at the lower frequencies.

    ******

    I'll re-read the Pi Speakers documents. Been a while. Maybe I'll "get" more of it this time....

  4. #4
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    Quote Originally Posted by dmtp
    Finally, pi speakers bypasses their series resistor with a small cap (which varies between 0.47 and 5 uF). Not sure how this value is calculated. 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.
    - A capacitor strapped across a series resistor ( the resistor is used to attenuate the mids below the reactive impedance of the cap ) forms a HF shunt around the resistor. The cap, allows frequency specific signals to flow around the resistor ( signal flow increases with an increase in frequency ).

    An example ;

    Stick a 16 ohm resistor in front of a 16 ohm driver and the total resistance the amplifier now "sees" is 32 ohms. This higher resistance will limit/reduce all signal flowing into the driver .

    Add a 1 uF cap paralleled to the 16 ohm resistor and you have formed a "HF signal bypass", routing directly to the driver . This "HF bypass" will become ineffective once the sum of the reactive capacitance ( measured in ohms ) along with the drivers' impedance, becomes greater than the sum ( of resistance of the inline/series resistor added to the drivers' impedance ) . This is because voltage, like water, follows the path of least resistance . Once the total resistance of the capacitor "bypass" becomes more than the resistance of the simple resistor, then signal will "prefer" to flow through that route.
    - ( I'm using "sums" here because only then can one use "vector addition" to properly account for the phase rotations that are inherent within Xc . These phase differences prevent the straight addition of the impedances / though I've conviently ignored this fact for the purpose of these demonstrations )

    Xc = 1 / ( 2 * pi * f * C )

    Xc is stated in ohms'
    Frequency is stated in hertz
    C is stated in Farads ( so divide uF by a million to restate your cap value into Farads )

    Also , from the above formula ;

    f = 1 / [ 2 * pi * ( Xc + R ) * C ]

    R = drivers resistance
    Solving for "f", means a 1 uF cap strapped overtop of a 16 ohm resistor and feeding a "real" 16 ohm driver will start conducting signal around the 16 ohm resistor, somewhere around 4974 hz .

    If you think you know where you want the HF boost to start, one can solve for C by using ;

    C = 1 / ( 2 * pi * Xc * f )

    - In reality, matching the boost point exactly to where the horns' frequency response starts to droop is pretty tough.

    - In my experience ; it's easier to "oversise" the bypass cap by a factor of 2 ( thus lowering the boost point ). Then carve out ( all the excess midrange that sneaks in with this larger cap ) by using a fairly broadband ( series LCR strapped in parallel ) that is targeted solely at the excess midrange content ( while leaving the HF content ) . This approach will usually allow for a couple of extra db in the HF frequencies . Another additional bonus to this approach is that the large midrange frequency area effected by the LCR notch has "frequency specific damping" applied to it ( while leaving the damping of the HF & UH,F untouched . This is a voicing trick that helps one think he has a tweeter in the system.

    Quote Originally Posted by dmtp
    ,,, snip 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.
    - The N200b , HF bypass that you have asked about has an extra wrinkle from the simple RC bypass. That 0.16mH coil is in place to form a resonant boost circuit ( a series LCR , strapped in series with the load ) . My calcs. show the "boost" to be centered around 12,600 hz . The "Q" of the boost is indeterminate to me , because I'm not sure what value to use as "R" in the denominator. Using a "pure" 5 ohms in the denominator gives a higher "Q" boost as compared to using the "loop" impedance formed from the driver back to the 2 conjugate resistors ( 1 conjugate being the ground leg of the variable Lpad ) .
    - The use of these resonant LCR circuits to achieve boost in certain frequency bands seems to have fallen from favour ( with JBLs' network designers ). The 4430 used this same ( resonant LCR ) approach / but that circuit is now 25 years ??? old and was the last example of its' genre that I can find . A study of Greg Timbers network designs will show that he alternatively, uses other types of "compensations" to achieve "extra" HF boost.

    regards

  5. #5
    RIP 2011 Zilch's Avatar
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    Quote Originally Posted by Earl K
    -This is a voicing trick that helps one think he has a tweeter in the system.
    [Heh, heh.... ]

  6. #6
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    Quote Originally Posted by Zilch
    Are you *** Mark*** trying to eliminate the adjustable L-Pad, or just want more attenuation before it?

    Either case will require knowing the actual impedance of the component combinations to calculate accurately and retain the original conditions.
    - Zilch, why don't you do Mark a favour and measure the "working circuit" impedance of a le85 with everything "driver-side" of the fixed Lpad ( ie; none of the 3 pole/passives in place ) ? ( You could have done that for Todd when he was looking for help It would have nicely rounded out the study of that circuit )

    - Depending on the actual rotation of the variable Lpad , one might have a ( MF and lower ) load of ( say ) 15 ohms ( from the comp driver/conjugate & variable Lpad ) with the 5 ohm conjugate ( parallel ) with a 2.5 ohm buildout ( series ) resistor. That gives me a working impedance of around 6.25 ohms (to be used by the 3 passives working in the crossover region ).

    - Once the actual working impedance is known for that part of the circuit / then a new fixed Lpad ( with more attenuation ) can be designed that still mantains the "correct" circuit impedance .



  7. #7
    RIP 2011 Zilch's Avatar
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    O.K., tonight, probably.

    I calculated 6.3095 Ohms, so we're close....

    What we're talking about:
    Attached Images Attached Images  

  8. #8
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    Quote Originally Posted by Zilch
    What we're talking about:
    - Almost! Run a WT2 impedance study of the network / without the 3 passives in place.
    - Ie ; get rid of ( for the impedance test ) ; the 16.5 uF and 24 uF caps, plus lose the .8mH inductor .

    - The "HF compensation" bypass network is also superfilous to the needs of this exercise .

    <>

  9. #9
    RIP 2011 Zilch's Avatar
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    Yup. I get it.

    [Might use CLIO, tho, depending on how complicated it becomes.... ]

  10. #10
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    Quote Originally Posted by MarkT
    Does anyone really understand how this circuit works or how the values are calculated?
    (a) The resonance point for your series LCR is found by ;

    Fo = 1 / 2 * Pi * [ squareroot of ( L*C ) ]

    Fo is expressed in hertz
    L is expressed in Henries ( so divide a mH value by one thousand )
    C is expressed in Farads ( so divide a uF value by one million )

    example;

    Fo = 1 / 2 * Pi * [ squareroot of ( 0.00016 * 0.000001 ) ] is
    Fo = 1 / 2 * Pi * [ 0.000012649 ] is
    Fo = 1 / 0.000079477 is
    Fo = 12, 582.30303 hz

    (b) The "Q" of a series LCR is found by this equation ;

    "Q" = [ squareroot of ( L / C ) ] divided by "R" ( or the circuit resistance ) .

    "Q" is expressed as a number
    L is expressed in Henries ( so divide a mH value by one thousand )
    C is expressed in Farads ( so divide a uF value by one million )
    R is expressed in Ohms

    example;

    [ squareroot of ( 0.00016 / 0.000001 ) ] ÷ R , is
    [ squareroot of ( 160 ) ] ÷ R , is
    [ 12.64911064 ] ÷ R ,
    if "R" = 5 ohms then "Q" = 12.64911064 ÷ 5 giving a resonant boost "Q" of @ 2.53

    12,583 hz divided by this "Q" of 2.53 gives a filter bandwidth ( with 3 db down points ) , 5080 hz wide ( @ centered at 12,583 hz ) .

    (c) Mark , theoretically one could choose a different Fo ( center ) point for this resonant LCR circuit / but that would require choosing a different capacitor value and then a different coil value .
    - The capacitor value would need to be chosen first , since the capacitor has the dual job of determining the lowest frequencies allowed through the LCR bypass & "boost" circuit .
    - The coil value is then chosen with 2 other criteria in mind .
    (i) That coil is still a coil and after the resonance of this circuit has died down , the coil will actually start to attenuate the HF going through this circuit . So one needs to keep that in mind
    (ii) An appropriate value coil must be chosen that will cause a resonance with the previously chosen capacitor at the desired frequency .

    That formula is

    L = 1 / [ 4 * pi^ * frequency^ *C ]

    C is expressed in Farads
    ^ is a short form for "squared"


    Quote Originally Posted by MarkT
    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.
    - See above if you really want to get into it . There's a bit more math to learn before you can get started .

    - Personally, I'd get rid if the 5 ohm resistor and the .16mH coil in that circuit and then take note of what happens.
    - Going to a capacitor that is valued in the range of 1.5 to 2 uF , will move the F3 point down to 8490 or 6366 respectively . You'll then need to adjust your notch filter accordingly / once you start allowing more MF to HF content into the bypass circuit . The good thing here is , that you've already played around with notch filters / so / correcting anamolies ( in the presence range ) with that notch filter won't be such a daunting task .

    - I'd also "graduate" to something more accurate than a RS SPL meter when measuring relative levels. IME, they aren't very reliable beyond 10 or 12 khz .


  11. #11
    Member dmtp's Avatar
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    Click #122 in the right column here:

    http://www.bcae1.com/
    Great little tool! It comes up with the same 13 ohm series and 3 ohm parellel that the formula does (but much easier). Assuming the 13 ohm is "towards" the input, that will give 16 ohm to the preceding circuit, what happens when you put it "after" (or toward the horn) ala pi speakers? Obviously need different numbers!
    See Fig. 7.155 and related text in Dickason for what the bypass cap does in filter contouring.
    Ooops! I have 5th edition and the highest figure is 7.112!
    MarkT

  12. #12
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    Wow! Earl K really DOES understand this circuit! The numbers certainly seem good - boost from 10k-15k. Of course, I put a variable resistor (1/2 L-pad) instead of the 5 ohm so I guess that as I turn that up (lower resistance), the boost gets higher, but the bandwidth gets narrower, right? (I think this is part of that "no free lunch" thing??)
    I'll try the same circuit with just the 1 uF and see what I get.
    MarkT

  13. #13
    Member dmtp's Avatar
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    (a) The resonance point for your series LCR is found by ;

    Fo = 1 / 2 * Pi * [ squareroot of ( L*C ) ]
    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?
    MarkT

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    Quote Originally Posted by Mark
    Of course, I put a variable resistor (1/2 L-pad) instead of the 5 ohm so I guess that as I turn that up (lower resistance), the boost gets higher, but the bandwidth gets narrower, right? (I think this is part of that "no free lunch" thing??)
    You've surmised correct . Reducing "R" gives a higher Q resonance / resulting in less bandwidth of boost at the centre frequency ( boost frequency ) .


  15. #15
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    Quote Originally Posted by me
    (a) The resonance point for your series LCR is found by ;

    Fo = 1 / 2 * Pi * [ squareroot of ( L*C ) ]
    Quote Originally Posted by dmtp
    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?
    Yes, you can, but ;

    - First you need to determine what value of "R" will give the correct amount of attenuation ( for the desired frequencies ) within that part of the circuit .

    - That means learning how to make "AC impedance measurements" for that specific part of the circuit . I call that , measuring the "Loop Impedance" .
    - How to "Take AC impedance Measurements" will be covered in your "Loudspeaker CookBook " ( somewhere ) . You should plot out the impedance curve onto graph paper .

    - To calculate "R" you need to know a couple of things ( apart from the loop impedance ).
    (i) The "Q" of the bump you are trying to flatten.
    - You need to measure and then plot out the FR of the whole horn circuit / so that you can easily "see" the centre of the "bump" along with the bumps' 3 db down points . Once they are identified; determine the bandwidth of the bump by subtracting the F"low" ( low frequency point that approximates a 3 db down point from the highest peak ) from the F"high" point ( same criteria ).
    - These 2 points need to be approximately centered on either side of the bumps peak .
    - Divide this number into the frequency of the peak and you'll get the "Q" of bump.

    example ;

    - There is a 5 db high peak centered at 6500 hz . The upper 3 db down point on this bump occurs around 10000 hz . The lower 3 db down point occurs around 3200 hz . 10000 minus 3200 = 6800 . 6500 ÷ 6800 = .9559
    - The "Q" of the bump is @ .96 . Slightly less than 1 . A "Q" of 1 equates to an octave wide bump.

    (ii) How many db of cut are needed to flatten the bump.

    - The good visual response chart ( made or obtained from an RTA ) tells one that 5 db of cut are needed ( hypothetically, within this example ).
    - Lets say that you measured a loop impedance of about 8 ohms for this frequency area. Then the question becomes what value of "R", in parallel with 8 ohms, gives one 5 db of attenuation in the "bump" area . The answer is @ 10.3 ohms. I would start with 10.5 ohms as the "R" in my series LCR . Zilch has a nice simple formula ( if you need it ) for determining db attenuations ( when only the load resistance is known / as well as the desired attenuation in db ) Hopefully he'll share it and save me the typing .
    - The you need to multiply this derived "R" by the "Q" of the bump . The product of these two will give you a number ( slightly larger than 10 in this example ) that represents the electrical "Q" of the LC filter that you need to build .
    - You can get an idea of the coil size by this formula ;
    L = ( our newly derived ) Product ÷ 2 ÷ pi ÷ frequency ( the Fo )
    L is an answer in Henries ( multiply by 1000 to get mH )

    Example ;

    10 ÷ 2 ÷ pi ÷ 6500 = 0.000244854 Henrys ( Henries ? ) or 0.2448 mH . This is the ballpark size of the coil needed for this "Q" of notchfilter to give 5 db of attenuation within an 8 ohm circuit loop . I'd start the design process of my LCR notch by using a .25 mH coil .

    To find "C" use this formula ;
    ( again, ^ is used as a marker to denote that "something" is squared )

    C = 1 / [ 4 * pi^ * freq.^ * L ]

    C is stated in Farads
    L is stated in Henrys
    ( pi^ = 39.4784176 )
    example ;

    C = 1 / [ 4 * 39.4784176 * 6500^ * .00025 ]
    C = 1 / [ 4 * 39.4784176 * 42250000 * .00025 ]
    C = 1 / [ 416990.7859 ]
    C = .000002398 Farads, ( multiply by one million to get an answer in uF ).
    C = 2.4uF

    So; a 5 db cut, centered at 6500 hz, about 1 octave wide, in an 8 ohm circuit, should be obtainable with an LCR made up of a 10.5 ohm resistor / a .25mH coil / and a 2.4uF capacitor / all wired in series to each other and wired in parallel across the driver / after some inline padding of sorts ( such as an Lpad or a simple inline resistor ) .


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