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Old 19th January 2004, 03:49 AM   #1
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Default Question on L-pad theory/formula

For the L-pad:

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from this site: http://lalena.com/audio/calculator/lpad/help.shtml

I see that R2 = function (impedance, attenuation) and
R1 = function (impedance, R2).

Does this mean that varying R1 has relatively little impact on attenuation?

The reason I ask is that I am trying to understand how to estimate the amount of padding present in a published crossover design, which of course often have very different R1 values or no R1 resistor at all.
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Old 19th January 2004, 04:44 AM   #2
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R1 is the the constant impedance to the amplifier example "8 ohms"
R2 is the "variable" and provides attenuation to the driver while R1 maintains the impedance necessary for the amplifier
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Old 19th January 2004, 04:54 AM   #3
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R2 does the attenuation by providing a parallel current path. Regardless of the value of R2, the total impedence for R2 and Z will alway be less than Z. Therefore R1 is used to bring the total circuit impedence back equal to Z to theoretically provide a constant impedence to the X-over or amp.

Since the Z of a driver varies with frequency, the L-Pad is at best an approximation of a constant impedence. Therefore, the actual amount of attenuation will be frequency dependent.
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Old 19th January 2004, 05:10 AM   #4
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Thanks for the replies.

So, as you didn't actually say it in these terms, does this mean that for a published crossover, I can estimate the attenuation merely by knowing driver impedance Z and resistor R2, and ignore R1?
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Old 19th January 2004, 05:19 AM   #5
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Quote:
Originally posted by leadbelly
Thanks for the replies.

So, as you didn't actually say it in these terms, does this mean that for a published crossover, I can estimate the attenuation merely by knowing driver impedance Z and resistor R2, and ignore R1?
Theoretically, almost. As long as you realize that the drivers impedance is changing with frequency and that R1 is droping part of the voltage (therefore some of the power).
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Old 19th January 2004, 06:19 AM   #6
sreten is offline sreten  United Kingdom
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Quote:
Originally posted by leadbelly
Thanks for the replies.

So, as you didn't actually say it in these terms, does this mean that for a published crossover, I can estimate the attenuation merely by knowing driver impedance Z and resistor R2, and ignore R1?
No. Not at all. There's some misinformation in this thread.

Without R1 R2 would simply reduce the impedance as its
in parallel with the driver, R2 would not attenuate at all.

R1 forms a voltage divider with R2||driver.

Both R's affect attenuation.
The series R is necessary for attenuation but used on its own
the impedance always rises. R2 is introduced with a lower value
of R1 such that the attenuation is the same and the impedance
is ~ the same.


To estimate attenuation estimate R2||driver, then estimate
the the effect of the voltage divider R1 and R2||driver.

As an example consider a tweeter crudely modelled by a 6R.

R1 = 6R only, attenuation = 6dB, Z =12R.

(R2 = 6R only, attenuation = 0dB, Z =3R.)

R1 = 3R, R2 = 6R, attenuation = 6dB, Z = 6R.


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Old 19th January 2004, 03:13 PM   #7
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Quote:
Originally posted by sreten
To estimate attenuation estimate R2||driver, then estimate
the the effect of the voltage divider R1 and R2||driver.
Thanks.

Since there can often be no R1 resistor at all in some crossovers, does this mean that there is little attenuation and only impedance adjustment, or that for a proper estimate, the DCR of all inductors must be considered in the calculation?
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Old 19th January 2004, 04:08 PM   #8
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Quote:
Originally posted by sreten


No. Not at all. There's some misinformation in this thread.

Without R1 R2 would simply reduce the impedance as its
in parallel with the driver, R2 would not attenuate at all.

R1 forms a voltage divider with R2||driver.

Both R's affect attenuation.
The series R is necessary for attenuation but used on its own
the impedance always rises. R2 is introduced with a lower value
of R1 such that the attenuation is the same and the impedance
is ~ the same.


To estimate attenuation estimate R2||driver, then estimate
the the effect of the voltage divider R1 and R2||driver.

As an example consider a tweeter crudely modelled by a 6R.

R1 = 6R only, attenuation = 6dB, Z =12R.

(R2 = 6R only, attenuation = 0dB, Z =3R.)

R1 = 3R, R2 = 6R, attenuation = 6dB, Z = 6R.


sreten.
Sorry for being misleading. Yes, R1 needs to be considered as it is used to make the total circuit impedence = to the Z of the speaker (Zs). But once you assume that the total impedence is = Zs than the current will be Vg/Zs and will be split between R2 and the speaker. The attenuation will be = the square of the the ratio R2/Zs.
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Old 19th January 2004, 06:39 PM   #9
sreten is offline sreten  United Kingdom
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Quote:
Originally posted by leadbelly

Since there can often be no R1 resistor at all in some crossovers, does this mean that there is little attenuation and only impedance adjustment, or that for a proper estimate, the DCR of all inductors must be considered in the calculation?
In terms of attenuation R2 is much more often omitted than R1.

If R1 is missing R2 is not for attenuation.
Its probably there for impedance correction only.

If R1 is missing the tweeter has the correct nominal sensitivity.
This is common in commercial speakers where a low sensitivity
tweeter allows savings in the tweeters magnet size.
And saves the cost of a resistor in the crossover.

And yes high resistance coils are often used in commercial
speakers to allow the coils to to be cheaper when they are
used in combination with a series resistor or they are the
series resistor as well.

Quote:
But once you assume that the total impedence is = Zs than the current will be Vg/Zs and will be split between R2 and the speaker. The attenuation will be = the square of the the ratio R2/Zs.
Sometimes R2 is selected to suppress the tweeters Fs impedance
peak and R1 used for attenuation, the target being a higher Z than
the nominal speaker rating, as high Z at HF is good for amplifiers.

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