Adding series resistance to bass/mid driver to increase Qts

midrange

Member
Paid Member
2012-05-14 3:49 pm
London
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?
 

AllenB

Moderator
Paid Member
2008-10-18 11:31 am
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.
 
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

513783d1447205925-iec-baffles-vs-cabinet-measurements-arpeggio-speaker-jpg


figure1.png
 

TMM

Member
2007-09-01 8:37 am
Australia
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.
 
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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midrange

Member
Paid Member
2012-05-14 3:49 pm
London
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.
 
Any competent simulator will take care of the changed frequency response. The resistor just becomes a circuit component.

Software | Visaton

Boxsim has an item in extras menu for box volume calculations. You can load the T/S parameters and add a 2 ohm presristor to the 6 ohm coil resistance.

I just had a go with a Qts = 0.33 Visaton W200S Reflex Woofer.

The critically damped (0.7) closed box went up from 19L to 44L. And the Fs went down. Which is what you want, I think.