How can I direct out of a Mesa Dual rec for monitoring? (Also splitting OHMs question

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dialtone

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I want to direct out into a mixer, so I can monitor out of my mixer with headphones. But the Mesa Dual Rec, has no "direct out". On the back of my head I have different OHM options and 2 of them are 8 OHMs, 1 of the 8 OHMs "outs" is going to my cab. Would it be possible to direct out into my mixer from the additional 8 OHM, or any other OHM option? i would perfer not to mic the amp, as all the mics are currently being used.

Thanks
 
I have no experience with that particular amp, but if you don't have a "line out" or "direct out," you cannot do what you are asking. Running a speaker level singal into your mixer is a very very bad idea, for both the amp and the mixer.
 
even if you did have one, it would suckola. I have one on the dc-5 and its nothing to mention about soundwise.

Don't fight it, just mic it!
 
dialtone said:
I want to direct out into a mixer, so I can monitor out of my mixer with headphones. But the Mesa Dual Rec, has no "direct out". On the back of my head I have different OHM options and 2 of them are 8 OHMs, 1 of the 8 OHMs "outs" is going to my cab. Would it be possible to direct out into my mixer from the additional 8 OHM, or any other OHM option? i would perfer not to mic the amp, as all the mics are currently being used.

Thanks
Click to expand...
DO NOT CONNECT A SPEAKER OUTPUT TO A MIXER INPUT. Does your amp have an effects loop? You can use the send to go to a mixer input, but do not operate the head without speakers attached unless you can disable the power section of the amp by plugging a dummy jack into the FX loop receive port.
 
Nope, can't do it without a line out or direct out.

Pete
 
You guys check my work, but this seems feasible:

Assuming it's a 100 watt amp and he's got two speaker outputs, one of which is occupied by an 8 ohm load, solving P=(V^2)R for V, we get 28 volts across the load. OK, double that for a cushion to 50 volts.

If he builds a 56:1 voltage divider from a 560K resistor in series with a 1k resistor and connects it to the other (parallel) output of the amp, the amp would still see an 8 ohm load and the voltage divider would dissipate less than .005 watts, plenty low enough power to use 1/4 watt resistors. If he takes a signal from the terminals of the 1 k resistor to feed the mixer (one terminal of the 1 k resistor would have to be ground common to the amp and mixer), he should get less than 1 volt of signal, and 1 volt is the standard for 0 VU. The 1 k resistor is large enough that there should not be any impedance mismatch probs going into the mixer, and his amp would not notice the minuscule change in load.

Any reason why this would not work?
 
Don't the Dual Rectos have the "Slave Out" on the back?

I had assumed (though possibly/probably incorrectly) that this could run to a power amp/mixer.... :confused:
 
The Dual Rectum has a slave out on the rear of the amp. It runs a pre-amp signal so you could probably hook that up to a mixer.
 
cellardweller said:
Don't the Dual Rectos have the "Slave Out" on the back?

I had assumed (though possibly/probably incorrectly) that this could run to a power amp/mixer.... :confused:
Click to expand...

Yes, a slave output can be run into a mixer, just don't run the amp without speakers.
 
ggunn said:
You guys check my work, but this seems feasible:

Assuming it's a 100 watt amp and he's got two speaker outputs, one of which is occupied by an 8 ohm load, solving P=(V^2)R for V, we get 28 volts across the load. OK, double that for a cushion to 50 volts.

If he builds a 56:1 voltage divider from a 560K resistor in series with a 1k resistor and connects it to the other (parallel) output of the amp, the amp would still see an 8 ohm load and the voltage divider would dissipate less than .005 watts, plenty low enough power to use 1/4 watt resistors. If he takes a signal from the terminals of the 1 k resistor to feed the mixer (one terminal of the 1 k resistor would have to be ground common to the amp and mixer), he should get less than 1 volt of signal, and 1 volt is the standard for 0 VU. The 1 k resistor is large enough that there should not be any impedance mismatch probs going into the mixer, and his amp would not notice the minuscule change in load.

Any reason why this would not work?
Click to expand...

Because anyone who needs to ask this question probably doesn't have the know how to do this properly!
 
Just as a side note, I've heard some pretty nice sounding guitar tones from pulling a line out from a decent amp and then applying a guitar cabinet impulse response using a convolution reverb. Rest assured it will sound like ass without it.

But yeah, heed the advice given here. Don't run a tube amp without speakers.
 
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