adding line out to a speaker output

[Randall]

I want to add a line out to a 100 watt SS modeling amp with an 8Ω speaker. The following is from another forum, Enzo in fact.

"All in all, the impedance shouldnlt much matter to a line out. A line out from a speaker line is just a voltage divider. The line out will not draw any current, so higher resistances will be invisible to the amp. The ratio of resistors is what determines the voltage division, not the resistors themselves. IN other words, If I put two 10k resistors in series then across the speaker output, the junction of resistors would be at half the signal level of the speaker. If I did the same thing with 22k resistors, the junction point would STILL be half the speaker voltage. SO if you see something like a 10k resistor from the speaker hot joining a 1k resistor to ground, at their junction there would be about 1/10 the speaker signal at the junction. Feed that junction out as a line signal.

Let's make up an example: 50 watt solid state amp into an 8 ohm speaker, full output. SOlving for voltage, we find 20v of signal across 8 ohms is 50 watts. SO we have 20 volots at the speaker, and you want to drop that down to about a volt - line level. SO you;d need a 20/1 voltage divider. COmes to mind easily a 22k and a 1k resistor. Put them in seriess, then 22k free end to speaker, and 1k free end to ground. 1v signal then is across the 1k resistor."

The old fella who plays the amp will never get near 100 watts, so why can't I use the above values for a 50 watt amp to make sure I get a good signal level at lower volumes? If someday someone does drive it harder, they could just trim down the input on whatever board or whatever they are feeding, right? The owner said he doesn't care about a level control, and would rather not pay extra for it.

My other question is about polarity. Is the junction between the two resistors in phase with the speaker, and why?

[Dave H]

The old fella who plays the amp will never get near 100 watts, so why can't I use the above values for a 50 watt amp to make sure I get a good signal level at lower volumes? If someday someone does drive it harder, they could just trim down the input on whatever board or whatever they are feeding, right? The owner said he doesn't care about a level control, and would rather not pay extra for it.

My other question is about polarity. Is the junction between the two resistors in phase with the speaker, and why?

27k, 1k would give a 1V line out from a 100W amp driving 8 ohms but I don't see why you can't use 22k, 1k for a little more output.

Yes, the junction of the two resistors is in phase with the speaker. It's a resistive divider so there's no capacitance or inductance to cause phase shift.

[Randall]

Thanks Dave, now one more question, my math tells me 1 watt resistors would be plenty, but can someone confirm?

[Chuck H]

Thanks Dave, now one more question, my math tells me 1 watt resistors would be plenty, but can someone confirm?

Consider that the load from the line out voltage divider will be at 23,000 ohms. If you divide that by 8 ohms you get 2875. So 1:2875th of the wattage will be across those resistors. If the guy is cranking the amp and there are peaks to 150W (hypothetical) then 150/2875 = .052 watts. You could use 1/8W resistors for this circuit. I generally just use the 1/2W resistors I have on hand. And, FWIW... The impedance of this circuit DOES matter because the user will be running cables to whatever it's fed into. Since there is truly a lot of leeway here in the impedance margin I would suggest going to a 470 ohm dropping resistor and a 15k series resistor for the divider. The lower impedance will lose less top end when longer cables are used. Not that a 1k would be that bad, but why not give the circuit the extra edge if you can? It's not a big concession and I'll bet you have 470 ohm and 15k resistors on hand.

[Randall]

Chuck, I am down with lowering the impedance, but wouldn't 15K/470 be taking it from 22:1 to 32:1? Wouldn't that drop signal as well? What about 10K/470, that would be closer, no?

[Leo_Gnardo]

Chuck, I am down with lowering the impedance, but wouldn't 15K/470 be taking it from 22:1 to 32:1? Wouldn't that drop signal as well? What about 10K/470, that would be closer, no?

You guys! Splittin' hares again I see.



Jeepers creepers it's just a line out. As Enzo reminds us, in a guitar amp! Pick a combination & go. In the time you take typing you could wire up 50 of 'em.

Whilst we're splitting, with these values you can use half watt resistors. Split that!

[Randall]

I know, but it is to get my head around the theory to make sure I get it. So maybe splitting hairs, but is my question correct?

[Enzo]

10k instead of 15k? Sure, go ahead.

[Justin Thomas]

I know, but it is to get my head around the theory to make sure I get it. So maybe splitting hairs, but is my question correct?

Yes. Some of us like to try a few hypotheticals in order to make sure we know WHY it works and can we work it out outselves in the future with different parts, and not just want to know IF it will work. So to everyone else, forgive us for the "check my math!" posts!

Thanky Thanky!

Jusrin

[Randall]

+100!

[Chuck H]

re: the 15k or 10k series resistor value with a 470 ohm dropper,..

With the 22k/1k values being discussed there was already a concession being made toward the low ratio side. For a 100W amp the correct ratio would be about 28/1. 10k/470, at 21/1 would be another tad lower in ratio where the 15k/470 at 33/1 is actually closer to the ideal ratio. YMMV because as you said, your guy is not likely to ever push the amp to it's full output. I think 10k would be fine and make for a better noise floor in use when the amp isn't making many watts.

EDIT: For that matter, consider that the EQ of the line out is not affected by the frequency limitations of a typical guitar speaker. Since a guitar speaker will roll off considerably at either end of the spectrum you can make things more manageable for whatever is being used as an interface with the line out by running the circuit through a 470n capacitor and paralleling a 33n cap across the dropper. This would approximate the sound from the speaker better for re-amp at a board or processing in a recording unit. The typical peaks and dips in a guitar speaker response can only be achieved with inductors and that's more complicated than anyone wants to get. I'm sure.

[Randall]

Well, I wired it up today, and the cold slap of reality hit me in the face. It doesn't work. It seems in this amp the speaker neg is not grounded, so the sound goes really low if the line out jack is touching the chassis, even with nothing plugged into it. Also, with the jack isolated the sound goes very weak if a cable is connected to a mixer. Clearly this is a problem. I thought this would be pretty straight ahead, but I'm stumped.

http://www.fmicassets.com/Damroot/Or..._schematic.pdf

[g1]

The Mustang III phones output is designed so it can be used as a stereo line out. If you don't want the speaker muted, use the stereo FX send jacks instead.

[Randall]

"If you don't want the speaker muted, use the stereo FX send jacks instead. "

I have come to the same conclusion. Man, that was a long way to get there, a hole added to the back, and no charge. I'm going fishing.

[vintagekiki]

https://drtube.com/schematics/marshall/5213psu.gif

It works convincingly the best.
Insulated jack (JS3) and R27 (100 ohm) to avoid ground loops (ground loop buzzing)
Instead of VR1, it can have a trimmer pot or a fixed resistor.