SPL loss with passive XOs

Rocky · 15th June 2004, 01:33 PM

Is there any mathematical way to predict the SPL loss a passive XO causes on a drivers rated SPL (in dB/W/m)?

As an example, take an 8 ohms driver with a 97dB/W/m sensitivity and add a 2nd order low pass filter.. What would be the resultant sensitivity, and how can it be calculated? Is there any precautions one can take in selection of crossover components to minimize this loss (coil resistance etc.)?

Svante · 15th June 2004, 01:48 PM

The loss for a 2nd LP order filter in the passband would approximately be a simple voltage divider between the filter coil resistance and the driver impedance. For most applications this loss would be well below a dB. Example: 0.5 ohm coil resistance 8 ohm speaker; 8/(8+0.5)=0.94 or 0.5 dB.

As the impedance of the driver varies, so will this attenuation, and at the fs impedance peak, the attenuation will be even smaller. In effect this will lead to a relatively higher sensitivity at fs. Another way of understanding the same thing is that the driver, with the series resistance included, has a higher Qes (and Qts).

Rocky · 15th June 2004, 01:53 PM

That is indeed what I call a quick and precise response, Svante.

Thank you for kindly sharing your infinite wisdom

15th June 2004, 01:33 PM	#1
Rocky diyAudio Member Join Date: Apr 2004 Location: Kongsberg/Oslo	SPL loss with passive XOs Is there any mathematical way to predict the SPL loss a passive XO causes on a drivers rated SPL (in dB/W/m)? As an example, take an 8 ohms driver with a 97dB/W/m sensitivity and add a 2nd order low pass filter.. What would be the resultant sensitivity, and how can it be calculated? Is there any precautions one can take in selection of crossover components to minimize this loss (coil resistance etc.)?

15th June 2004, 01:48 PM	#2
Svante diyAudio Member Join Date: Feb 2004 Location: Stockholm	The loss for a 2nd LP order filter in the passband would approximately be a simple voltage divider between the filter coil resistance and the driver impedance. For most applications this loss would be well below a dB. Example: 0.5 ohm coil resistance 8 ohm speaker; 8/(8+0.5)=0.94 or 0.5 dB. As the impedance of the driver varies, so will this attenuation, and at the fs impedance peak, the attenuation will be even smaller. In effect this will lead to a relatively higher sensitivity at fs. Another way of understanding the same thing is that the driver, with the series resistance included, has a higher Qes (and Qts). __________________ Simulate loudspeakers: Basta! Simulate the baffle step: The Edge

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15th June 2004, 01:53 PM	#3
Rocky diyAudio Member Join Date: Apr 2004 Location: Kongsberg/Oslo	That is indeed what I call a quick and precise response, Svante. Thank you for kindly sharing your infinite wisdom