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Old 26th July 2006, 04:22 AM   #1
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Default Confused about DVC and Q

OK. My searching has only caused me to be even more confused. One article indicates that putting a resistor across one VC would cause a drop in Qts while another says a rise. One says it changes Qes another says it changes Qms. Nobody seems to define a starting point either.

So what is the story. Say I get a DVC sub with a Qts of 0.4 with both VCs driven in parallel. What happens when I

1) drive only one with the other open
2) drive only one with the other shorted

Can I raise it to say .6 with an appropriately chosen resistor? How would I calculate this.

mike
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Old 26th July 2006, 11:19 AM   #2
sreten is offline sreten  United Kingdom
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Hi,

You can raise it to 0.6 by putting it in a box !

I understand Qts series, parallel, one coil driven / one short, is identical.
One voice coil driven / one open, Qts is double the above.
So a resistor varies Q from 2Qts to Qts.

/sreten.
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Old 26th July 2006, 02:04 PM   #3
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Don't want a box.

Good. Then my hypothetical sub could be raised to .8. Two thumbs up.

thanks.

mike
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Old 26th July 2006, 02:22 PM   #4
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(JPK) This may be more detailed than you want, but Eqs 24 and 25 show how to relate Qts to the values of Qes and Qms to the value of the resistor connected across the second voice coil. In Eq 24 Qes and Qms are the values for a single voice coild conection, with the second VC open. Qms is the same regardless of now the VCs are connected. Qes for the single VC connection is twice the values quoted when both VCs are connected in either parallel or series.

http://www.geocities.com/kreskovs/Dual-VC.html
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Old 26th July 2006, 06:02 PM   #5
sreten is offline sreten  United Kingdom
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http://www.mhsoft.nl/spk_calc.asp

See new Qts with series DCR of inductor

Hi,

Is another way of manipulating Q (and wasting amplifier power).

For an active bass system you can make amplifiers with virtual output
resistances that do not waste power and set Q to what you like.

/sreten.
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Old 27th July 2006, 12:59 AM   #6
Ron E is offline Ron E  United States
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Quote:
Originally posted by john k...
(JPK) Qms is the same regardless of now the VCs are connected.]

Now use that to derive the impedance curve. There is no correct way to think of RDO as anything other than a modification of Qms. That it is electrically sourced mechanical damping is neither here nor there.
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Old 27th July 2006, 04:58 AM   #7
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Quote:
Originally posted by Ron E



Now use that to derive the impedance curve. There is no correct way to think of RDO as anything other than a modification of Qms. That it is electrically sourced mechanical damping is neither here nor there.

(JPK) I would choose to disagree. Ultimately the impedance is derived from Vs/I where Vs is the source voltage and I is the current flowing through the amplifier loop. For the dual VC woofer with one VC loaded by and external resistor this is given by my Eq 6. So Z = (Vs Re)/(Vs-BL X'). The question then becomes how X' (cone velocity) is expressed in terms of the damping of the system. The damping can be considered to be composed of any number of sources; electromagnetic damping associated with the VC back emf generated in the amplifier loop, mechanical damping, fluid damping, and in this case additional electromagnetic damping external to the amplifier loop. Thus in the present case there are the 3 Q's: Qes, Qms and Qds. All of them combine to form the system Qts (Eq 19) which is what will determine the behavior of X'. If you want to combine Qds and Qms into a single Q so the system is described by 2 Q's that is fine, but I would prefer to think along the lines that these are Qes associated with the amplifier loop electromagnetic damping and a second parameter, Qext, which represents the effective damping due to all other sources of damping outside the amplifier current loop. The impedance expression as given by Small would then be expressed in terms of Qext as opposed to Qms. The Q in Small's impedance function needs to represent all external damping. I would personally rather look at it this way since Small never considered sources of external damping other than mechanical resistance and the additional damping due to the resistive loading of the second VC just isn't mechanical. Of course, it is also possible to write the impedance expression in terms of Qts and avoid the argument of what gets modified and how. :-)
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Old 27th July 2006, 05:24 PM   #8
Ron E is offline Ron E  United States
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Quote:
Originally posted by john k...
:-)
Just trying to get a rise out of ya John. As we both know, it is a semantic argument, dependent on definitions. My way of thinking is much more convenient, IMO...
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Old 27th July 2006, 11:41 PM   #9
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Quote:
Originally posted by Ron E


Just trying to get a rise out of ya John. As we both know, it is a semantic argument, dependent on definitions. My way of thinking is much more convenient, IMO...
"Convenient: suited to personal comfort or to easy performance. "

Nothing good ever comes easy. :-)
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Old 28th July 2006, 07:11 AM   #10
sreten is offline sreten  United Kingdom
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Hi,

FWIW I find the concept that the resistor modifies Qms risible.

/sreten.
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