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I want to reduce my amplifier output to line-level.

I found this tutorial, Connecting speaker signals to line level inputs, which is compatible with 50 W output power. There is also mentioned what to do if amplifier is more powerful than that.

So if 10 kΩ resistor handles 50 W and 33 kΩ handles ~300 W, what is compatible with my amplifier which output power is 120 W per line? Or can I just use the 33 kΩ resistor?

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  • \$\begingroup\$ 120W @ 4Ω ⇒ 22k; 120W @ 8Ω ⇒ 33k \$\endgroup\$
    – jippie
    Commented Feb 19, 2014 at 18:48

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Ultimately you want a voltage, so it will be easier to think in terms of the voltage out of your amp. You say your amp has "120 W" output. Presumably that is maximum into a 8 Ω load. Sqrt(120W x 8Ω) = 31 V, which is the maximum RMS voltage from your amplifier. You want to get that down to 1 V or so, so something like this would work:

However, that would require cranking your power amp to maximum gain to get the full line voltage signal. I'd probably attenuate by only half as much, which means making R1 1.5 kΩ or thereabouts. Whatever the line level signal is fed into isn't going to get hurt by 2 V instead of 1 V, just that it will be very loud and may get clipped. That would be a clue to turn down the volume going into the power amp.

Keep the power dissipation of R1 in mind. The voltages accross R1 and R2 will have a ratio of R1/R2. When 1 V is accross R2 in the example above, then 30 V will be accross R1. (30V)2/3kΩ = 300 mW, so a "1/4 W" resistor isn't enough.

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  • \$\begingroup\$ Thanks! This looks so professional for me and is exactly what I needed! =) I will use 2Kohms for R1 to make sure that the sound will not get clipped \$\endgroup\$ Commented Feb 19, 2014 at 19:59
  • \$\begingroup\$ Should I use 8 Ohm resistor or something to make load to the amplifier? \$\endgroup\$ Commented Feb 19, 2014 at 20:04
  • \$\begingroup\$ @nTn: That depends on your amp. Some amps are OK with no load, some not. It should work in all cases with a 8 Ohm load, since that's what a ordinary speaker is. However, it will dissipate significant power and require a physically large resistor, so it's not something I'd do unless I found it was really necessary. \$\endgroup\$ Commented Feb 19, 2014 at 20:15
  • \$\begingroup\$ If I don't use load resistor, can that circuit be used parallel with speaker on both lines (left and right)? \$\endgroup\$ Commented Feb 19, 2014 at 20:16
  • \$\begingroup\$ @nTn: Yes, you need one of these for each separate speaker signal (left and right). A speaker would work as a load for your amp, but of course it will make sound in the process. \$\endgroup\$ Commented Feb 19, 2014 at 20:19
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For a modern A/V receiver 'speaker out' to line level (say you're using Zone 2 out the back of your A/V receiver and wish to feed another amp that handles only outdoor speakers but you wish the outdoor amp volume to be controlled by A/V receiver) then:

  • R1 = 3.9K @ 1watt
  • R2 = 430ohm @ 1/2watt

and

  • R1 connects from Speaker out RIGHT+ to RIGHT/RCA+
  • R2 connects from Speaker out RIGHT- to RIGHT/RCA- AND RIGHT/RCA+(only at RCA)

Repeat for LEFT channel, R3 and R4. -(negative) connections are shared both ends.

Speaker to Line PCB, for home audio - NOT car audio:

Speaker to Line PCB

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According to the rule of thumb on the page you refer to:

  • 120W @ 4Ω ⇒ 22k or
  • 120W @ 8Ω ⇒ 33k.

But essentially you want to check the output voltage of the amplifier and you want to reduce that to 1Veff. Use the voltage divider rule for that.

Not all amplifiers like unloaded outputs. Does your amp have a headphones output, that would be a much safer option to experiment with. You don't want to accidentally sort normal output of your amp.

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  • \$\begingroup\$ I have headphones output, but it switches off speakers output =/ \$\endgroup\$ Commented Feb 19, 2014 at 19:56

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