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matej8888 22nd June 2012 01:50 AM

line-out on guitar amp
 
1 Attachment(s)
So, i would like to add a line out on my guitar amp. I've followed a few forums and found different values of resistors for different amp power as well as speaker impedances. And i've come across the attached picture which is a simplified resistor "signal reducer". Example: on a Marshall 2020 power amp schematic i found 2 resistors (one being 5.1k and the other to the ground or 0V being 680Ohm) on 8Ohm speaker output and on some other brand schematics i found quite different values of resistors on 4Ohm speaker output.
Now, i wonder how is speaker impedance related to the resistors value?
Thinking about having multiple speaker outputs (4, 8, 16Ohm).
Do i need to have speaker connected anyway?
Does the attached scheme work in a way that i can regulate the line level NOT having connected the speaker as well, or should i mount a dummy on a speaker output in any case?

Enzo 22nd June 2012 02:44 AM

The amp produces an output signal voltage, the load draws the current from that. SOlid state amps generally just run outputs to any speaker you like, while tube amps have separate taps for impedance.

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.

That is of course crude, but hopefully it describes the concept for you.

Your Marshall 2020 example serves. You have a 5.1k (5100 ohms) and a 680 ohm in series across the output. 5780 ohms. The line signal is across the 680 ohms. SO 680/5780 = 0.117. SO those two resistances leave about 12% of the speaker voltage on the line jack - a roughly 8 to 1 reduction. COUld you have used 10k and 1200 ohms instead, more or less doubling both resistances? Sure, and the results would be roughly the same. The ratio is what matters. Well as long as you don't go too low. You wouldn't do so well if you tried 51 ohms and 0.68 ohms.

Now look at your drawing. Your control is directly across the output. That would want to work, but it allows you to turn your "line level" output all the way up to the full 20v. Most places you'd want to send that would not be happy with 20v bashing them in the nose. SO add a resistor between the top of the control and the speaker line. Now the control only has a portion of the signal voltage, and your variable resistance samples from THAt rather than the entire signal voltage. You could duplicate the Marshall circuit in your box. Add a 5.1k resistor where I suggested, between the speaker line and the top of the control. A 680 ohm pot would be unusual, but a 1k pot is a common value, so use that. Close enough.

A solid state amp won;t care if you use a dummy load or not. A tube amp MUST have some sort of load on it.

matej8888 22nd June 2012 09:54 AM

Got It Enzo. Thanx for the detailed explanation. I forgot to put in the text that was below the schematic, so i put it now:
"With this one you can control the line out level.
Use a 5K or 10K pot.
If the signal is too hot or need a more refined control, insert a series resistor before the pot, something like 47K or so."

So i guess putting a resiator in series with the pot should do the trick...but i think i'm going to go with the marshall ratio. I must only find a dummy resistor for the load.

Thank you again!

theAnonymous1 22nd June 2012 10:26 AM

Just keep in mind the source impedance with this type of voltage divider will be relatively high, so make sure that's compatible with whatever you're feeding the signal to. Shouldn't be much of a problem in most cases though.

matej8888 22nd June 2012 12:47 PM

I know that. The signal will probably go to a power amp and not to a mixer.
But thank you for the reminder..


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