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14th February 2006, 05:04 AM  #1 
Wizard of Kelts
diyAudio Moderator Emeritus

Resistor Values For L Pad?
We are crossing over a Vifa D27 tweeter at 3200 Hz. We have need of an L pad.
The inductance is negligible, and the voice coil resistance is 4.6 ohms. Could someone calculate the resistor values to cut the SPL A) 4 dB B) 6 dB? Also, can someone calculate the values for those two SPL cuts which would leave the tweeter with an ohmage of 8 ohms instead the present 6 ohms? Thanks in advance for any help on this.
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14th February 2006, 06:31 AM  #2 
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Join Date: Nov 2004
Location: Québec, Québec

Z = 6 Ohms
A = 4 dB R1 = 2.21 Ohms R2 = 10.26 Ohms Z = 6 Ohms A = 6 dB R1 = 2.99 Ohms R2 = 6.03 Ohms http://www.lalena.com/audio/calculator/lpad/
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14th February 2006, 07:07 AM  #3 
diyAudio Member
Join Date: Nov 2004
Location: Québec, Québec

Now to have 8 ohms load...
You add 2 ohms so it will take 25% of the power. You lose 1.2493873660829995313244988619387 dB. Z = 6 Ohms A = 2.75 dB + 1.25 dB R1 = 1.63 Ohms + 2 ohms = 3.63 ohms R2 = 16.11 Ohms  Z = 6 Ohms A = 4.75 dB + 1.25 dB R1 = 2.53 Ohms + 2 ohms = 4.53 ohms R2 = 8.24 Ohms Tell me if my math is incorrect.
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14th February 2006, 12:16 PM  #4 
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Hi Simon,
could you post the spreadsheet and/or formulae you are using? How do you calculate to allllll those sig. fig.?
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14th February 2006, 03:47 PM  #5 
Wizard of Kelts
diyAudio Moderator Emeritus

Thank you both for answering.
I tried the calculator here. the top one works, apparently the bottom one does not. Here. Adding a 2 ohm resistor to the 6 ohm D27 tweeter yielded a final impedance of 8 ohms and 2.5 dB attenuation. So for 4 ohms total attenuation, I needed 1.5 dB additional. Remembering I am already added a series resistor of 2 ohms, I get this for 4 db attenuation: Rseries=1.27 ohms Rparallel=43 ohms. Zout= 8 ohms For the 6 db attenuation, I needed 3.5 db additional attenuation. Remember that I already added a 2 ohm resistor in series. To achieve this, I got the following values: Rseries= 2.65 ohms Rparallel= 16.12 ohms. Zout= 8 ohms Is this approximately correct?
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14th February 2006, 09:05 PM  #6 
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14th February 2006, 10:49 PM  #7 
Wizard of Kelts
diyAudio Moderator Emeritus

Well, that seems to be the key. and I had that program, forgotten, on my hard drive all along. I only remembered I had it when I downloaded from the link and I was asked if I wanted to overwrite an existing file.
The good part about this program is that instead of dealing with fractions of ohms in order to get the attenuation absolutely correct, instead you deal with real whole resistor values in order to get the attenuation and Z out approximately correct, which is what I wanted. Getting 0.27 of an ohm resistor is a little silly, lol. Here are the values I got using this calculator: 4 dB Cut R Speaker = 6 ohm R Series = 3 ohm R Parallel = 50 ohm Final R total = 8.36 ohm Attenuation = 3.86 dB 6 dB Cut R Speaker = 6 ohm R Series = 4 ohm R Parallel = 10 ohm Final R total = 7.75 ohm Attenuation = 6.31 dB And, just for good measure because it is so easy, 9 dB Cut R Speaker = 6 ohm R Series = 5 ohm R Parallel = 5 ohm Final R total = 7.73 ohm Attenuation = 9.05 dB Thanks, Lionel. This is just what I needed.
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"A friend will help you move. A really good friend will help you move a body." Anonymous 
14th February 2006, 11:51 PM  #8  
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Join Date: Nov 2004
Location: Québec, Québec

Quote:
To modify the LPad to end up with a 8 ohms load, I used usual formulas. Power formula dB to linear : x dB = 10 * log y Power formula : W = V * I Voltage formula : V = R * I Now you play a bit ... W = R * I^2 W = 8 * I^2 but 2 ohms are passive so W = 6 * I^2 in the tweeter, so you lost 25% of the power in the resistor. With full power you get : 10 * log 1 = 0 dB With 75% power you get : 10 * log 0.75 = 1,2493873660829995313244988619387 dB Now I safely approximated it to 1.25 dB so I used the calculator like before but for 4.75 dB cuts at 6 ohms and 2.75 dB cuts at 6 ohms since I already cut 1.25 dB and already added 2 ohms in series. Calculator give me a LPad with a 6 ohms total impedance. I add my 2 ohms resistance to the resistance in series. I have now a 8 ohms impedance LPad. BTW, yeah with a spreadsheet it would be easy... If I understood your last question : Why I got so many numbers after the dot? calc.exe (windows xp calculator in scientific mode will give you that many)
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16th February 2006, 02:41 AM  #9 
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Join Date: May 2004
Location: Sydney

I would use SW to model it.
Adding resistor(s) in series and parallel WILL alter the roll off characteristic in a significant way on top of attenuating dBs in a crossover network. In other words, you may add in one resistor in series with the driver which attenuates some dBs. You can also add in one resistor in series and another one in parallel (i.e. LPad) to attenuate some dBs. But the frequence response of each of them is very different from the other. A safer way to do it is to forget about the parallel resistor or LPad (in my network the parallel resistor causes the roll off to happen much earlier even though attenuation at high frequences remains the same while a series resistor keeps the response curve similar to that without the resistor but lower). Buy a number of inexpensive noninductive wirewound resistors at (0.22R, 0.5R, 1R, 1.5R, 2.2R, 2.7R) then start from adding in the 1.5R in series with the driver first. You then listen to your speaker while adding or removing resistors (don't forget to unplug your speaker cables from the amplifier's end first before soldering your XO or you can smoke your amp). Of course you can also use a LPad as a volume pot, i.e. use two legs only for your test. Hope this helps. Regards, Bill 
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