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Kp115 - Subwoofer or space heater?


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#1 twalkonline

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Posted 08 January 2014 - 02:09 PM

Can anyone tell me why the internal crossover has two 25 ohm resistors in parallel with the woofer? It doesn't look like a Zobel circuit, although I guess it could be an attempt at impedance correction. But If I remember Boylestad correctly, about a quarter of the power going into this thing will be dissipated as heat. No wonder they burn up. Am I missing something?

#2 mustang guy

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Posted 12 January 2014 - 09:46 PM

Those are some massive resistors. I looked and found a couple images of the network:

post-34280-13819689274686.jpg

and a burn transformer:

x.jpg

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.


#3 twalkonline

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Posted 14 January 2014 - 07:28 PM

Thanks for posting the images, although I think they're going to give me nightmares!

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.

Thanks.

#4 pzannucci

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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.



#5 twalkonline

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Posted 23 January 2014 - 11:26 PM

I'm sure that the resistors are there for some sort of impedance compensation, but it is not a proper Zobel circuit. A Zobel circuit would be the series combination of capacitor and resistor(s) in parallel with the woofer. The KP115 resistors are directly across the woofer, after the low pass filter.

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.