Increase total Speaker Impedance Question

The 200 watt power rating of those resistors should be sufficient, but you will then have 3/4.16 or ~72% of your available power being dissipated by the resistors. Only the balance, ~28% of the total power will be available to drive the speakers.

Bill, thanks very much for the nice words. I always find your posts to be particularly knowledgeable, informative, and constructive.

Angelgz2, yes the resistors will certainly affect the sound quality, and most likely for the worse, but the magnitude and character of the effects will depend on the impedance characteristics, damping requirements, and other characteristics of the speakers. If the design of the speakers is such that their impedance and other characteristics are relatively "tube friendly" (for example, if their impedance magnitude doesn't vary greatly as a function of frequency, and if their impedance phase angles are not highly capacitive), their sonics will probably suffer less than they would otherwise.

Also, Bill is right that only a fraction of the amplifier's power capability will be available to the speakers, with the rest being dissipated in the resistors. And his calculation resulted in roughly the right answer, although I think the methodology of the calculation was not quite right.

You didn't say whether the 125 watt rating is for 4 ohms or 8 ohms, but let's assume it is for 4 ohms. And I assume that the receiver is solid state rather than tube-based, at least in its output stage, so that its output impedance can be assumed to be negligible. The receiver would probably be capable of delivering slightly less power than the 125 watts into (3 + 1.6) = 4.6 ohms, but let's ignore that slight difference. 125 watts into 4 ohms corresponds to an output voltage from the amplifier of (square root (125 x 4)) = 22.36 volts. The voltage that would appear across the paralleled speakers under that max power condition would be ((1.6/4.6) x 22.36) = 7.78 volts. That voltage across 1.6 ohms corresponds to (7.78 squared)/1.6 = 37.8 watts. 37.8/125 = 30.2%, close to the bottom line Bill stated.

Good luck. Regards,
-- Al

P.S. to my previous post. Another calculation method, which gives the same 37.8 watt answer but is somewhat simpler:

For many solid state amps, especially those for which the 4 ohm power rating is equal to or close to twice the 8 ohm power rating, the power capability into (1.6 + 3) = 4.6 ohms can be approximated as (4/4.6) x the 4 ohm rating.

125 watts x (4/4.6) = 108.7 watts.

Max power into the 1.6 ohm speaker combination, when in series with a 3 ohm resistor, would be:

108.7 x (1.6/4.6) = 37.8 watts.

Regards,
-- Al

Instead of killing useful power and also useful bandwidth per given load, I'd use single pair of speakers replacing quantity by quality. Another option is to use zone preamp and drive each pair of speakers with separate identical stereo amp.

Hi Bill and Almarg,

Thanks again for your detailed explanation. I understand the power loss in this setup. However, I am still having difficulty grasping why the sound quality would also take a turn for the worse. Will the frequencies be improperly passed or cut off by the resistor? If not, then, all frequencies will still be passed to the speakers, but at lower power, and that shouldn't affect the sound at all theoretically isn't it?

Thanks!!