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Old 14th February 2006, 04:04 AM   #1
Wizard of Kelts
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Default 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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Old 14th February 2006, 05:31 AM   #2
simon5 is offline simon5  Canada
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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

Click the image to open in full size.

http://www.lalena.com/audio/calculator/lpad/
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Old 14th February 2006, 06:07 AM   #3
simon5 is offline simon5  Canada
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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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Old 14th February 2006, 11:16 AM   #4
AndrewT is offline AndrewT  Scotland
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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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Old 14th February 2006, 02:47 PM   #5
Wizard of Kelts
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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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Old 14th February 2006, 08:05 PM   #6
Lionel is offline Lionel  France
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Passive CrossoverDesign Calculator

A usefull Excel spreadshit.
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Old 14th February 2006, 09:49 PM   #7
Wizard of Kelts
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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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Old 14th February 2006, 10:51 PM   #8
simon5 is offline simon5  Canada
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Quote:
Originally posted by AndrewT
Hi Simon,
could you post the spreadsheet and/or formulae you are using?

How do you calculate to allllll those sig. fig.?
I used the little calculator to calculate the cuts with 6 ohms impedance.

To modify the L-Pad 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 L-Pad with a 6 ohms total impedance.
I add my 2 ohms resistance to the resistance in series. I have now a 8 ohms impedance L-Pad.

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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Old 16th February 2006, 01:41 AM   #9
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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. L-Pad) 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 L-Pad (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 non-inductive 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 L-Pad as a volume pot, i.e. use two legs only for your test.

Hope this helps.

Regards,
Bill
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