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I had a look at another Exchange question, and couldn't quite understand.

I have a set of cheap speakers, 8ohms each with a combined output of 6 watts.

If sqrt(6w/8Ω) = 0.866 v, and the post says line level is approximately 1 volt, what should my amperage be?

How can I calculate for an appropriate resistor?

I have about 10 150Ω resistors at home... will wiring them half and half between speakers attenuate the signal? How many do I need of each?

EDIT (CLARIFICATION):

I want to know how to calculate the resistance needed to reduce a powered signal (6w) to line level.

yeeeeee
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  • I think your calculation's a bit wonky. 0.866V into 8 ohms won't give you 6W. – Simon B Jun 01 '17 at 13:59
  • It ought to be \$\sqrt(3W * 8R)\approx5V\$, \$I\approx0.6A\$ per speaker. However, no idea what you are asking re the resistors.....question is unclear. Voting to close. – Trevor_G Jun 01 '17 at 14:10
  • You seem to be confusing all sorts of nomenclature. Line level is not used to drive speakers, it is used to send audio signals to devices expecting line-level (nominally 1-2Vpp) inputs, like power amps (used to drive speakers), audio interfaces, or studio effects units. – Chris M. Jun 01 '17 at 14:57
  • @ChrisM. Hence my question says reduce *to* line level, not *from* line level. I have speakers that are powered, but I want to cut the wires to them and output them to a jack *at line level*. – yeeeeee Jun 02 '17 at 06:24
  • @Trevor, I don't come from any electrical background at all. Don't "vote to close" because of my naivety. – yeeeeee Jun 02 '17 at 06:25
  • Even if I am confusing all sorts of nomenclature, I've come on this site to learn. As I said, *no electrical background*, however I don't see how it's hard to gather the basic facts from what I've written… 6 watt speaker output; 8 ohm independence; what resistance do I need to reduce *to* line level... – yeeeeee Jun 02 '17 at 06:29
  • @SimonB, that's the maths I took from another post to get to an approximate voltage… I don't know what it does, but I agree, 0.866/8 ≠ 6. – yeeeeee Jun 02 '17 at 06:32
  • Without an electrical background, you can't see why your question is confusing. I'm not blaming you for misusing terms, I was simply pointing out what I perceived to be an error in your understanding. @replete that's the post OP referenced as confusing. – Chris M. Jun 02 '17 at 07:21

1 Answers1

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schematic

simulate this circuit – Schematic created using CircuitLab

Something like the circuit shown should work. I'm assuming that the amplifier will be happy driving a 18 ohm load, rather than 8 ohm. Doubling the resistance halves the power dissipated. Even so, R1 needs to be something chunky, such as a 2W resistor.

I chose the relative values of R1 and R3 so that you don't have to run the amplifier at full volume. Turning down the volume a bit significantly reduces the distortion.

If the impedance is important, you'd have to halve the values of both resistors, and use an even bigger resistor for R1, such as 5W rated.

Simon B
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  • In a random turn of events, this circuit (with halved impedances) helped me salvage a 1995 Tiger Electronics "Talkgirl Jr." today. You are my Wife's hero, Simon B. LOL! – iwolf Jun 10 '20 at 23:17