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23rd December 2012, 12:54 AM  #21 
diyAudio Member

Ok then, just clear this up for me  if there's no correlation between relative impedances and output volumes in the same circuit. How does a resistor on the input, attenuate a driver relative to another driver?

23rd December 2012, 02:40 AM  #22  
diyAudio Member
Join Date: Jan 2008

Design is wandering; slapping tweeter up top without due consideration and measurements is shooting blind.
OB forced close to wall begs undesirable early reflections. Low order cross with 8" woofer at wide baffle spacing produces highly lobed response. Speaker doubtless sounds best with 12P solo, and this is doubtlessly thin. Quote:
Within designed full range application and appropriate box type enclosure to augment/support lower end driver has potential to sound very good. OB is tough territory for ultra light weigh, shallow depth diaphragm. Good luck 

23rd December 2012, 02:48 AM  #23 
diyAudio Member

I've had a bit of a ponder on this: essentially your saying at 20Khz the sensitivities of each driver is at maximum. The 10 Ohm tweeter is at 92db and 15 Ohm A12 is at 89db from spec sheet. That the impedance difference between drivers is irrelevant? Correct?This is simple enough I guess. But I'm not convinced it's as simple as that, just yet.
Have a think about how 8 Ohm and 4 Ohm, but otherwise identical woofers, the 4 Ohm version is nominally 23db more efficient. Also note: the resistor value I've settled on for now is 5.6 Ohms. I didn't calculate this. I started with a 3.3, went up to a 4.7, then to a 5.6. Now it just so happens we have both drivers at roughly 15 Ohms don't we? What about looking at a single driver. The driver is generally loudest, most efficient where the impedance is lowest. Your contention suggests that resistors are really only to correct for a 'sensitivity' imbalance between drivers. But what causes the imbalance? It isn't just 'efficiency'. I reckon somewhere there's an equation/formula to calculate for this. But my maths is terrible. 
23rd December 2012, 03:14 AM  #24 
diyAudio Member
Join Date: Jan 2011
Location: Ladysmith, BC

The resistor in series increases the impedance of that driver, thus reducing current to that driver. The resistor in parallel divides the current. So less power goes to the driver.

23rd December 2012, 03:15 AM  #25  
diyAudio Member

Quote:
Quote:
When I get tired of it, I'll just swap it over to something else. Some other 'boxed' project. I don't take it that seriously. 

23rd December 2012, 03:20 AM  #26 
diyAudio Member
Join Date: Jan 2011
Location: Ladysmith, BC

Didn't notice your last post. It is as simple as your first paragraph.
The 4 vs 8 ohm. They are 3db apart at 2.83V, but they are the same at 1 watt. That's the catch. One is not free SPL. Your example of resistor choice is just a fluke, and not even valid without measurements. Your last point about being loudest at the drivers lowest impedance is also not really true. Look at how effecient a woofer is at Fs. Or way up in the breakup region. Have fun. I can't keep going on about this. 
23rd December 2012, 03:59 AM  #27  
diyAudio Member
Join Date: Jan 2008

Quote:
'by fanatics, for fanatics' and all levels of interest. Enjoy! 

23rd December 2012, 11:36 AM  #28  
diyAudio Member
Join Date: Feb 2008
Location: Sheffield

Quote:
This might seem counterintuitive, but the reasoning goes as follows:  around Fs, the impedance of the speaker increases.  a series resistor acts as a potential divider  at Fs, the speaker sees much more voltage than at other frequencies, so the amount of power it gets in the LF region also increases (compared to the rest of the frequency range), so you get a bass boost at Fs, but not below (where excursion would be large due to the very low frequencies) Try resistor values of 1ohm to 4r7. Yes, you lose some power in the resistor, but in a home hifi situation, it won't be much. 

23rd December 2012, 08:43 PM  #29  
diyAudio Member

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24th December 2012, 12:22 AM  #30 
diyAudio Member

Ahhaaaaa! I THINK I can see what tuxedocivic was trying to explain to me. Voltage and wattage are different. Voltage can remain constant while wattage consumed by driver/supplied by amp will vary according to driver impedance. So an imbalance between drivers in watts doesn't mean an imbalance in output volume. I THINK. I was getting confused by a calculator I was using.
Assuming 2.83 = 1 watt at (x) Ohms At 20Khz amp sees 6 Ohms from 2 x drivers in parallel. At 20 Khz the A12 is 15 Ohms and 89 db efficient. Uses 0.4 watt. At 20Khz the tweeter is 10 Ohms and 92db efficient. Uses 0.6 watt. We still can assume 1 volt input/output. Volume is related to input voltage. Watts is related to power consumed to achieve a given output/voltage. Apologies for my noob/boofheadedness 
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