Padding resistor(s), explain this?

[Froste]

A question was posed last night in SpeakerChat:

"Does adding a padding resistor to the source end of a XO effect the crossover point(s)?"

After some debate, Wolf and I tested using the following:

Add 4ohm resistor to the front end of a 2-way circuit, Using PCD(Wolf) and SpeakerWorkshop (Me).

In Wolf's test he found that the overall SPL dropped by 1db and the crossover point raised by 100hzs.

In My test I found the overall SPL dropped by 1db and the crossover point remained the same!

After some more debate, we found that our tests were not the same, PCD will not allow a single padding resistor to be added to the front of the circuit, so Wolf had added a 4ohm resistor to the front (R1,R10) of HP and LP circuits.

I repeated this setup (LP & HP padded) in SW and I got the same results as Wolf's PCD test.

So, the question is WHY?


Here are screen shots of my tests:

Circuits: top standard XO, bottom left single resistor, bottom right dual resistors
http://lh5.ggpht.com/_VxEMw7YIqQM/S5f78LYlQlI/AAAAAAAAAzI/S7kdGSjWE-Y/s912/resistorcompairxo.JPG

Charts from above circuits: 1st - 1459hz cross, 2nd - 1459hz cross, 3rd - 1605hz cross
http://lh5.ggpht.com/_VxEMw7YIqQM/S5f77mubDhI/AAAAAAAAAzE/d2dPAzX2_Ew/s1024/rresistorcompairfrd.JPG

[JustinG]

A question was posed last night in SpeakerChat:

"Does adding a padding resistor to the source end of a XO effect the crossover point(s)?"

After some debate, Wolf and I tested using the following:

Add 4ohm resistor to the front end of a 2-way circuit, Using PCD(Wolf) and SpeakerWorkshop (Me).

In Wolf's test he found that the overall SPL dropped by 1db and the crossover point raised by 100hzs.

In My test I found the overall SPL dropped by 1db and the crossover point remained the same!

After some more debate, we found that our tests were not the same, PCD will not allow a single padding resistor to be added to the front of the circuit, so Wolf had added a 4ohm resistor to the front (R1,R10) of HP and LP circuits.

I repeated this setup (LP & HP padded) in SW and I got the same results as Wolf's PCD test.

So, the question is WHY?


Here are screen shots of my tests:

Circuits: top standard XO, bottom left single resistor, bottom right dual resistors


It is because the XO circuit is seeing a 8 ohm load instead of a 4 ohm load if the resistor is after the main network. In a way you are simulating putting a 8 ohm speaker into a 4 ohm network.

A way you can simulate this in PCD is by putting in an L pad and setting the parallel resistor to 1000 ohms or more and the series resistor to 4 ohms.

EDIT/Additional info: If you use an L-Pad, it keeps the XO network load at 4 ohms but the resistors eat up some of the power causing a lower sensitivity of the system

[Dirk]

It seems sensible that the crossover point will drift slightly because the highpass and lowpass will meet each other at different absolute SPLs. But if the highpass was -12dB down one octave lower it will STILL be -12dB down one octave lower.

[JustinG]

"Does adding a padding resistor to the source end of a XO effect the crossover point(s)?"

No, it will just be a voltage drop across the resistor at all frequencies (assuming it is non-inductive) much akin to using a potentiometer to control the volume of line level.

Noticed I didn't mention that in my first post

[Dirk]

No, it will just be a voltage drop across the resistor at all frequencies (assuming it is non-inductive) much akin to using a potentiometer to control the volume of line level.

Noticed I didn't mention that in my first post

Which is to say that it DOES effect the crossover point, but not the individual slopes of the highpass or lowpass.

[JustinG]

Which is to say that it DOES effect the crossover point, but not the individual slopes of the highpass or lowpass.

Correct, it doesn't effect the F3/F6/slope of the filter but will change the actual cross point of the two driver responses

[curt_c]

All you are really doing is adding resistance to the source impedance, similar to for example, the output impedance of a tube amplifier transformer. Since this ‘source resistor’ is in series with the rest of the network and driver, the voltage dropped by the ‘source resistor’ will vary with frequency if the impedance of the network and driver varies with frequency. (And of course in most cases it will.)

It’s because the voltage at the input of the driver network (after the ‘source resistor’) varies that the resultant crossover summation varies.

I’d be remiss if I didn’t point out that networks 2 and 3 are not equivalent. While network 2 has a single 4 ohm resistor in series R0, network 3 has R0 and R2. This somewhat isolates the impedance of each network from each other so the voltage dropped by each resistor will be different than in network 2. Also the lookback impedance seen by the generator will be 4 ohms plus the complex impedance of the two networks/drivers in network 2, while in network 3 the two 4 ohm resistors will be essentially in parallel and closer emulate a 2 ohm resistor plus the networks/drivers as seen by the generator.

C

And of course we already know the effects of series resistance on woofer Qt' and its changes to enclosure size and/or woofer response in the compliance controlled region...

[Jeff B.]

All you are really doing is adding resistance to the source impedance, similar to for example, the output impedance of a tube amplifier transformer. Since this ‘source resistor’ is in series with the rest of the network and driver, the voltage dropped by the ‘source resistor’ will vary with frequency if the impedance of the network and driver varies with frequency. (And of course in most cases it will.)

It’s because the voltage at the input of the driver network (after the ‘source resistor’) varies that the resultant crossover summation varies.

I’d be remiss if I didn’t point out that networks 2 and 3 are not equivalent. While network 2 has a single 4 ohm resistor in series R0, network 3 has R0 and R2. This somewhat isolates the impedance of each network from each other so the voltage dropped by each resistor will be different than in network 2. Also the lookback impedance seen by the generator will be 4 ohms plus the complex impedance of the two networks/drivers in network 2, while in network 3 the two 4 ohm resistors will be essentially in parallel and closer emulate a 2 ohm resistor plus the networks/drivers as seen by the generator.

C

And of course we already know the effects of series resistance on woofer Qt' and its changes to enclosure size and/or woofer response in the compliance controlled region...

Right. Network 2 and 3 are not equivalent. It is also easy to see why the crossover point shifts up in number 3. It is because the woofer's crossover point is shifting independently of the tweeter's and causing the point where they cross to raise. This is due to the fact that the woofer's is closer to a first order circuit so the added resistance has the affect of raising the crossover point or allowing the woofer's response to extend further before roll-off. The tweeter's on the other hand is a third order circuit so the series resistance does not have the same affect, it mostly lowers the output and creates a bit of ripple near Fc.

[Wolf]

Right. Network 2 and 3 are not equivalent. It is also easy to see why the crossover point shifts up in number 3. It is because the woofer's crossover point is shifting independently of the tweeter's and causing the point where they cross to raise. This is due to the fact that the woofer's is closer to a first order circuit so the added resistance has the affect of raising the crossover point or allowing the woofer's response to extend further before roll-off. The tweeter's on the other hand is a third order circuit so the series resistance does not have the same affect, it mostly lowers the output and creates a bit of ripple near Fc.

Now that makes sense. I couldn't put my finger on it, as the networks seemed to have the same type circuit. I was thinking it was due to the Qes changes on the woofer, but I didn't know for sure.

Thanks for the explanation, Curt and Jeff!
Wolf

[Æ]

Right. Network 2 and 3 are not equivalent. It is also easy to see why the crossover point shifts up in number 3. It is because the woofer's crossover point is shifting independently of the tweeter's and causing the point where they cross to raise. This is due to the fact that the woofer's is closer to a first order circuit so the added resistance has the affect of raising the crossover point or allowing the woofer's response to extend further before roll-off. The tweeter's on the other hand is a third order circuit so the series resistance does not have the same affect, it mostly lowers the output and creates a bit of ripple near Fc.

So, if someone wants to add a 4 ohm resistor to his 4 ohm two way loudspeaker, to make the load the amplifier sees 8 ohms, he is messing up more than just his alignment, he is messing with the crossover/frequency response too. Correct?

[Jeff B.]

So, if someone wants to add a 4 ohm resistor to his 4 ohm two way loudspeaker, to make the load the amplifier sees 8 ohms, he is messing up more than just his alignment, he is messing with the crossover/frequency response too. Correct?

I depends on the crossover, but more than likely - Yes. I thought we covered that already in that thread below.

[Pete Schumacher ®]

So, if someone wants to add a 4 ohm resistor to his 4 ohm two way loudspeaker, to make the load the amplifier sees 8 ohms, he is messing up more than just his alignment, he is messing with the crossover/frequency response too. Correct?

Unless the crossover employs impedance flattening compensation, yes, exactly. If the system impedance ain't flat, the resistor will do more than simply reduce the output across the spectrum.

[Æ]

I depends on the crossover, but more than likely - Yes. I thought we covered that already in that thread below.

It was covered, kind of, but I asked the guys if they would model it for me. Intuitively I know it's wrong to add resistance, but in this instance I wanted a bit more theoretical understanding.

[Pete Schumacher ®]

It was covered, kind of, but I asked the guys if they would model it for me. Intuitively I know it's wrong to add resistance, but in this instance I wanted a bit more theoretical understanding.

If the system impedance varies a lot over frequency, a series resistor will alter the transfer function seen at the input of the system. For instance, a 4 Ohm crossover would see an impedance peak over 8 Ohms at the crossover point for a typical LR4 two-way. If you add a 4 Ohm resistor at the input, the voltage at the 8 Ohm peak would be quite a bit higher than for the rest of the 4 Ohm system impedance. That means a pronounced peaking at the XO point, where the impedance peaks up.

[Æ]

If the system impedance varies a lot over frequency, a series resistor will alter the transfer function seen at the input of the system. For instance, a 4 Ohm crossover would see an impedance peak over 8 Ohms at the crossover point for a typical LR4 two-way. If you add a 4 Ohm resistor at the input, the voltage at the 8 Ohm peak would be quite a bit higher than for the rest of the 4 Ohm system impedance. That means a pronounced peaking at the XO point, where the impedance peaks up.

Thank you.

[Pete Schumacher ®]

Thank you.

No problem mon . . .

http://www.sugarbushsquirrel.com/image/22149502_scaled_360x427.jpg

[Æ]

No problem mon . . .

http://www.sugarbushsquirrel.com/image/22149502_scaled_360x427.jpg

I think I recognize that guy. Wasn't he on the The Jamaican Bobsled Team?

[Jeff B.]

If the system impedance varies a lot over frequency, a series resistor will alter the transfer function seen at the input of the system. For instance, a 4 Ohm crossover would see an impedance peak over 8 Ohms at the crossover point for a typical LR4 two-way. If you add a 4 Ohm resistor at the input, the voltage at the 8 Ohm peak would be quite a bit higher than for the rest of the 4 Ohm system impedance. That means a pronounced peaking at the XO point, where the impedance peaks up.

Right. Think of it in terms of the ratio of the series R to the impedance at each frequency point. As you can see, this ratio varies all over the place, so then would the attenuation.

[Ludo]

A simple example off this is a 500Hz, first-order high pass using an 80uF cap for a 4 ohm driver. Adding 4 ohms to the amp output will increase the attenuation the driver sees by 6dB, and the corner frequency will also shift down to 250Hz.

Louis

[Froste]

Much thanks to one and all. This padding resistor question was throwing me because my tests were proving/disproving what I thought would be.
Looks like the best bet is to model any serial padding resistor setup.

Once again thanks to all who answered.