Kp115 - Subwoofer or space heater?
Posted 08 January 2014 - 02:09 PM
Posted 12 January 2014 - 09:46 PM
Those are some massive resistors. I looked and found a couple images of the network:
and a burn transformer:
They would get warm, and they would increase the standing resistance by 12.5 ohms. Since this is an 8 ohm speaker, I don't know how adding 12.5 ohms would even work, even though we are talking about impedance instead of resistance. I don't believe a resistor cares what the frequency is or even whether it is DC or AC. This would make the resistance increase by 12.5 ohms at every frequency. This is confusing me...
I hope somebody answers you twalk...
Edited by mustang guy, 12 January 2014 - 09:47 PM.
Preamp: Integra DHC-80.3
Main amp: Integra DTA-70.1
Speakers: LCR, Sides and Rear center (6 total) La Scalas, wides Heresys
Subs: 2 Bill Fitzmaurice THT pwrd by Crown XLS-202
BluRay: Oppo BDP-93
Turntable: Rega P5 w/Ortofon 2M Blue
Projector: Epson 5020UB Screen: DaLite 130" HC Cinema Vision
Also in shop Fisher 80-T, 90-T, or KM-60 connected to a pair of Fisher 80-AZ monoblocks powering the front LaScalas
Front Scalas switched with Niles SPK-1
THT's Subwoofers controller Paradigm X30 and powered by same Crown XLS-202 above
Posted 14 January 2014 - 07:28 PM
Here's what I know: any woofer will have a nominal impedance, but the exact impedance that the amplifier sees will vary with frequency. Thus there will be a curve that describes the woofer's impedance versus frequency. This impedance will also be affected by the change in impedance mechanically when the woofer is mounted in an enclosure. The K-48 is a nominal 4 ohm rated woofer, but the KP 115 and other speakers that use it are rated at a nominal 8 ohms. I get that. What I don't get is why the resistors are in parallel (not in series) with the woofer. If I remember my circuit analysis 101 correctly, current divides in inverse proportion to the values of the resistors in parallel. Therefore, if the K-48 were a true 4 ohms, the 12.5 ohms in parallel with it would see approximately 1/4 of the current and thus 1/4 of the amplifier power would be dissipated in the resistors. If the K-48 were a true 8 ohms, it would be worse and approximately 2/5 of the amplifier power would be dissipated in those resistors.
Again, if I'm missing something I would appreciate someone setting me straight. Otherwise, I would appreciate someone explaining how those resistors are a good idea.
Posted 21 January 2014 - 08:50 PM
If those are capacitors under the resistors, it could be a zobel network to allow the filter to work more correctly and have the proper cut off frequency. That is the key to the zobel, allow you to use more of a text book filter arrangement not bothered by the rise in the impedance as frequency goes up.
Posted 23 January 2014 - 11:26 PM
I'd still like to know how burning approximately one quarter of the power in those resistors could be a lesser evil than a network without those resistors.