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How does a tube amp "see" a REL sub?

If you connected a REL sub to a tube amp using the high-level inputs (placing the sub in parallel with the speakers), and you graphed the impedance of the speaker "system" (speakers + REL) as seen by the amp, would the impedance of the system be 100,000 Ohms up to the crossover point defined on the sub? Could you assume that the REL would be a good sub to mate with tubes because of the highly-resistive (100,000 Ohms) load the sub presents?
128x128nrenter
seandtaylor99
1,262 posts
 
The impedance of the REL sub is in parallel to the impedance of the speakers, since it is connected to the same speaker terminals as the main speakers. When you add parallel impedances A and B you use the equation :

Imp = 1/(1/A +1/B)

Therefore if we have 10,000Ohms in parallel with 8 Ohms the resultant impedance is 7.9936 Ohms.

In a nutshell the REL's high impedance means that the impedance load that your amp sees will not change from before.
nrenter's avatar
nrenter OP
766 posts
 
That equasion is accurate for simple resistive loads. But a speaker's impedance varies with frequency.
seandtaylor99
1,262 posts
 
Since the REL does not apply a high-pass filter to the signal going to the main speakers, as do some other subs, I would think that the impedance curve of the main speakers would be almost completely unaffected, and that the simple equation would hold.

It doesn't matter what you set the crossover point to : the REL doesn't prevent frequencies below this from going to your main speakers, so if your speakers have an impedance dip at 20Hz, and you set the crossover to 70Hz I would bet you'll still see the impedance dip at 20Hz.

It should be entirely eqivalent to wiring a 10k resistor in parallel with your speaker terminals.
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