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Old 13th October 2004, 10:27 PM   #1
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Default Trying to understand PLLXOs

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'Nuther newb question:

Been poring over Dave's PLLXO Primer, and I have a question.

According to the primer, a first-order lowpass is formed by a series R followed by a cap shunt to ground. I can see the need for the R to set a minimum load for the preamp.

However, in a first-order highpass, why is a series cap followed by an R shunt to ground? The R seems extraneous to me. Is it there because the amp doesn't like to "look back" through more than a certain impedance?

Dart
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Old 13th October 2004, 11:21 PM   #2
Pan is offline Pan  Sweden
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The combo of C and R values is what gives a specific x-over point.

Change the R to another value and the x-over point moves with it.

/Peter
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Old 14th October 2004, 12:53 PM   #3
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Thanks, Peter.

I guess I'm still trying to understand why it's different from a speaker-level 1st-order XO where all you need is a series cap.

If I used a series inductor for a lowpass, would I still need the resistor to ground?

Also, what about line-level 2nd-order LC filters? Do they need resistors too?
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Old 14th October 2004, 01:59 PM   #4
Pan is offline Pan  Sweden
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In a passive, the series cap makes the x-over function WITH the series resistance in the VC of the tweeter.

IOW, tweeters that are identical (as far as possible) with the exception of one being a 4ohm and the other a 8ohm driver, will have different x-over functions with the same value of cap.

To end up with same x-ing the 4ohm driver needs to have a cap that is double the size of the 8ohm driver.

All x-overs needs a defined resistance to work in.

/Peter
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Old 14th October 2004, 02:45 PM   #5
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Ok, I think I'm beginning to see.

I was thinking the interaction of the cap with the amp input impedance was what determined the frequency--but this is not so?

So any line-level filter, of whatever order, needs a resistor either in series or to ground?
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Old 14th October 2004, 02:57 PM   #6
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Quote:
Originally posted by Dartagnan
I was thinking the interaction of the cap with the amp input impedance was what determined the frequency--but this is not so?
Basically the input impedance of the amp is so high that the impedance of the filter is dominant. This is why passive filters are a bit tricky -- you need to make sure that the load impedance is going to have negligible effect on the filter.

A good rule of thumb is to load the filter with no less than 10x the filter impedance. However, this can present problems if say an amp input impedance is fixed at 20k, as then you use 2k for the filter resistor, but then this loads your source too much.

Also, the source impedance is important as it will contribute to the filter effects. Again, the rule of thumb is that the source impedance should be 1/10 or less of the filter impedance.

/richie00boy -- likes active filters
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Old 14th October 2004, 03:42 PM   #7
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Oookay, the light is beginning to dawn... Thanks for the informative posts!

I'm beginning to see that my tube preamp (Zout 1kohm) and a sixpack of SS amps (Zin 24.3kohm) leaves no room for PLLXOs. Is there no hope for this combo short of a buffer?

Quote:
Basically the input impedance of the amp is so high that the impedance of the filter is dominant.
So if the amp Zin is low enough, can you lowpass with just a series choke? I have a pro amp with a 10kohm Zin...
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Old 14th October 2004, 05:53 PM   #8
Pan is offline Pan  Sweden
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Dont be to afraid of buffers. They can be made very transparent Id say. If you use batteries and only the best ICs, or even discrete, you should be satisfied.

Tube output with your 1k driving 3 channels with PLLXO is not something Id recommend, especially with that 22k load.

One solution is to use a passive solution for the tweeters HP section, and use opamps/buffers for the mid and lows. Typically its the highs that suffers most from an active solution if not good enough.

/Peter
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Old 14th October 2004, 06:20 PM   #9
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For 1st order filters, you can use an R approximately equal to the input impedance of the amp but you have to include the amp's impedance in the calcs. Art Ludwig gives the formulas.

http://www.silcom.com/~aludwig/Sysde....htm#Bi_amping
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Old 14th October 2004, 08:33 PM   #10
Pan is offline Pan  Sweden
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Just about any resistor value can be used as long as you calculate with the amp resistance, no need to use 1/2 the amp value.

/Peter
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