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Multi-Way Conventional loudspeakers with crossovers

Adding series resistance to bass/mid driver to increase Qts
Adding series resistance to bass/mid driver to increase Qts
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Old 14th January 2020, 10:36 PM   #1
midrange is online now midrange  United Kingdom
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Default Adding series resistance to bass/mid driver to increase Qts

I am looking into adding a 2 ohm resistance to a driver (Monacor sph-130). According to my maths this would change the Qt from .32 to .40

The effect this has on a sealed box with a Q of 0.8 is as follows:

Without series resistance Volume 4.1 litres, Fbox 95Hz, F3 85 Hz

With 2ohm series resistance Volume 7.3 litres, Fbox 76 Hz, F3 68Hz

My question is, is the pay off as simple as that (apart from the sensitivity going down) or are there complications and other considerations?
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Old 14th January 2020, 11:57 PM   #2
AllenB is offline AllenB  Australia
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Adding series resistance to bass/mid driver to increase Qts
When you think that you could go as far as passively bringing down the top end until f3 is 20Hz, losing a lot of sensitivity in the process, it may put into perspective what this simple tradeoff will and won't do. I think it's worth remembering that the region of benefit in this case is wholly within the room controlled region and can or will be manipulated in other ways.
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Old 15th January 2020, 02:14 AM   #3
system7 is online now system7  United Kingdom
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You are quite right that this is a valid way of doing things, albeit a bit of power is wasted in the woofer series resistor.

An equivalent process is considering those exotic SET amplifiers that have significant output impedance:

Arpeggio Loudspeaker - diyAudio

Click the image to open in full size.

Click the image to open in full size.
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Old 15th January 2020, 04:52 AM   #4
TMM is offline TMM  Australia
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Since the attenuation mostly occurs in the upper bass and above, with the right sized baffle this may be utilised as a form of baffle step compensation, therefore there is minimal loss of sensitivity compared to no series resistance and an equivalent amount of baffle step compensation.
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Old 15th January 2020, 10:18 AM   #5
forr is offline forr  France
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Quote:
Originally Posted by system7 View Post
You are quite right that this is a valid way of doing things, albeit a bit of power is wasted in the woofer series resistor.
In usual home usage, average power applied to loudspeakers is less much than 1 W, so the wasted power is not a problem.
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Old 15th January 2020, 11:56 AM   #6
Lojzek is offline Lojzek  Croatia
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midrange,

consider postponing your experiment after you make your own measurements. You may not need it at all, very likely.

edit: I suppose it would be ok to congratulate AllenB on his new job!

Last edited by Lojzek; 15th January 2020 at 11:59 AM.
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Old 15th January 2020, 11:57 AM   #7
YSDR is offline YSDR  Hungary
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Because that Qts=0.32 is a manufacturer data, it's even possible that the real Qts is different, can be 0.4 either.
What about the larger box without resistance? F3 would be higher but F6 and F10 are not.
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Old 15th January 2020, 12:27 PM   #8
Azrael is offline Azrael  Europe
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This has also been considered elsewhere:

@Gegentakt & all: Widerstand vor low-Q-Chassis - Visaton Diskussionsforum

Note especially the statements of the user "gegentakt".

As a translation tool I can recommend the page DeepL Translator.

With kind regards
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Old 15th January 2020, 02:17 PM   #9
CharlieLaub is offline CharlieLaub  United States
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Adding a series resistance is creating a voltage divider between the resistor and the impedance of the driver. For simplicity, think of the driver's impedance as pure resistance that is frequency dependent - the familiar "impedance curve" that you see for the driver.

Because of the voltage divider effect, you will also get changes to the frequency response at high(er) frequencies, where the driver has a rising impedance. In essence, by adding a resistor you are implementing some degree of equalization of the input power that roughly follows the contour of the driver impedance magnitude (the impedance curve). The degree of EQ is related to how much series resistance you add.

So make sure to take the effect at higher frequencies into account when you are considering this change. Just like with high impedance amplifiers, many uninformed tweakers think that using it does "wonders" to the sound of e.g. their single-driver loudspeakers, as if the amp itself is the culprit. Instead it is only the "EQ" effect that is responsible for the tonal changes that they hear.
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Last edited by CharlieLaub; 15th January 2020 at 02:20 PM.
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Old 16th January 2020, 12:39 PM   #10
midrange is online now midrange  United Kingdom
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Quote:
Originally Posted by CharlieLaub View Post

Because of the voltage divider effect, you will also get changes to the frequency response at high(er) frequencies, where the driver has a rising impedance. In essence, by adding a resistor you are implementing some degree of equalization of the input power that roughly follows the contour of the driver impedance magnitude (the impedance curve). The degree of EQ is related to how much series resistance you add.
Ah. I was thinking of running the main driver quite high, and I see on the manufacturer's impedance curve that the impedance is about 15ohm at 5k. What you point out will be significant. Thank you.
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