mute switch..simple Q..

[jeth]

Just a quickie...for a mute switch for a channel into a summing amp, which end of the mix R should be shorted to ground, the source side (pre mix R) or the mix input side (post R)? And while I'm here, whats a good value for the R to ground, just the same as the mix R or not?

 

[NewYorkDave]

Well, you left out some specifics in your question. But generally, you'd implement a mute by disconnecting the mix resistor from the source and connecting that end to ground. That way, the summing amp sees the same impedance (and, in the case of active summing, runs at the same gain) regardless of the number of inputs assigned. All of this assumes that the source impedance is small compared to the value of the mix resistor.

In active summing, you can simply disconnect the mix resistor at one end and let it float, and the gain of the summing amp will "self adjust" to compensate. But then you run the risk of the summing stage sounding different depending on the number of channels assigned (see above).

 

[jeth]

Thanks NYD...one thought, if using the first method of shorting to ground, and the gain of the active summing stage is unchanged, then if the voltage at the output is the algebraic sum of the inputs then won't the output level change if the gain is the same but with one input fewer?

 

[NewYorkDave]

Unless you're building a compressor, you want the output level to go down as you mute channels. And this will happen regardless of which of the two muting schemes you use.

 

[bcarso]

If it is a passive summer the resistor's source end should be grounded and as NYD says, disconnected from the source.

If it is an active summer with a virtual ground summing node, the gain from any other connected channel to the output will not change when the resistor is open-circuited. The noise gain will change however---the summing amp's voltage noise will be multiplied less. The current noise from the disconnected resistor will also not have an effect disconnected, reducing the output noise further. This may or may not be desirable depending on the application.

If all you can afford is one shunt switch to ground, you can split the input resistors into two halves (or other proportion) and ground the center tap. Now the loading on the source doubles in mute mode, and the noise due to both noise gain and the resistor noise goes up.

For really high isolation in mute, a T arrangement of switches can be used---two switches in series between source and summing R, and a third from the junction of the first two to ground. The drive to control the grounding switch is inverted relative to the other two.

 

[CJ]

I don't understand.

 

[jeth]

NYD: Thanks again, makes sense now. Obviously the output needs to be the sum of the inputs minus the muted channel, to keep same output the other channels would all be getting louder as the gain increased. Was thinking in sums not sounds me thinks...

Bcarso: I think that I'll try first with the simplest approach and try your suggestions if the noise gain changes make a noticeable difference. Thanks for the insight.

CJ: (For o nce)This time I think I DID understand. :roll:

 

[bcarso]

Maybe I can make it a little clearer.

The gain of an inverting feedback amplifier with an input resistor Rin and a feedback resistor Rf is just -Rf/Rin.

The gain of such an amp with two input resistors Rin1 and Rin2 is -Rf/Rin1 from the source end of Rin to the output, and -Rf/Rin2 from the source end of that resistor to the output. Add more resistors and associated sources and the gain from each source to the output doesn't change, until you start to run out of opamp open loop gain.

However, the opamp voltage noise at the output is the equivalent input voltage noise, call it e.i.n., times (1 + Rf/Rin) for the single (grounded or terminated in a zero impedance source) input resistor*. For the two resistors, taking the simple case of Rin1 = Rin2, the opamp output voltage noise is e.i.n. times (1 + 2Rf/Rin). The more resistors I add to mix more sources the more I amplify my opamp voltage noise. These are examples of noise gain, which in general is different than the signal gain.

If I instead lift the source end and float a given resistor, it's as if it isn't there, and that makes the noise gain go down, so my output noise goes down. Also, the thermal noise of the resistor can't contribute either. So there are advantages in floating the unused source resistors. But in some cases the designer has adjusted the opamp open loop characteristics to work optimally with all the input resistors terminated in a low impedance, so the sound may change if you float them, or even in some cases the whole shooting match may oscillate.



* Neglecting resistor thermal noise and opamp input current noise

 

[jeth]

Another thought on this topic... :?

If I have pots on my mix inputs, feeding the mix resistors, they must are part of the resistance seen by the virtual earth input. Would it therefore make sense to short the input of the pot to ground, rather than putting the mute between pot and mix R? This would seem to me to follow the idea of keeping the resistance seen at the summing amp from the muted input channel the same when muted to ground as with the channel connected.

Also, not directly connected to mute switches...but...

I figure that the pot resistances are also affecting the overall bus source impedance, how crucial is it that this stays constant? Does it just need to stay in the appropriate range for the opamps input?

 

[PRR]

> pot resistances are also affecting the overall bus source impedance

If the pot is driven from a low-Z source, the maximum impedance at the wiper is R/4. If the mix resistor is arbitrarily picked equal to the pot resistance, then the bus "sees" from R to 1.25R.

Consider a 2-input passive mixer. Input A is held full-up, resistance is R. The other input B is silent, but you slide the fader from 0 to 10 (maybe in anticipation of someone using that mike).

With B at zero, source A gain is 0.5. With B at max, source A gain is still 0.5. As B passes through half-way, source A gain is 0.555. So moving knob B through the range causes source A to increase 0.9dB and then back to nominal.

Is this tolerable? Guitar amps sometimes mix the two channels in a way that causes couple-dB change in one channel as the other is adjusted. Obsessive mixer-fixers don't want to see any incidental changes as they move faders.

For more than 2 input, interaction is less.

For active mixing, gain interaction is reduced by feedback. If the feedback is even slightly effective, interaction will be well below 0.25dB. With lots of feedback, 0.05dB-0.01dB.

If you are asking about noise resistance: worst-case, moving both pots, the bus resistance changes from 0.5R to 0.625R. Resistance noise curves are much broader than that.

 

[jeth]

Thanks again PRR...tend to learn more than I intended whenever I read your replies!

I'm still interested to know if it makes sense to mute before the switch, as if the pot does indeed contribute to the resistance "seen" (why do I get the impression you don't like that term?) by the summing input, then this would seem to be in keeping with NYD's explanation of the mute arrangement...

generally, you'd implement a mute by disconnecting the mix resistor from the source and connecting that end to ground. That way, the summing amp sees the same impedance (and, in the case of active summing, runs at the same gain) regardless of the number of inputs assigned.

 

[jeth]

Ok..sorry again...I read through your post a couple more times and I now get why it doesn't make much difference to switch the pot in to ground along with the mix resistor, and that the small changes in bus source impedance are not too big a deal.

So I rephrase my question..

Can I mute in front of the pot? I see no reason why not, and it would actually be convenient for me due to the probable layout i will use.

Sorry.

 

[jeth]

If the pot is driven from a low-Z source, the maximum impedance at the wiper is R/4. If the mix resistor is arbitrarily picked equal to the pot resistance, then the bus "sees" from R to 1.25R.

But if the pot is set either minimum or max the bus looks longingly back at 2R , no?

 

[jeth]

My humble apologies, I neglected to account for the fact that the R to ground at max setting is in parallel with the source output, and therefore insignificant. Still, at minimum setting we have 2R..

 

[jeth]

Or, with the pot set to minimum, is the 10k from the wiper to the top of the pot in parallel with the tiny resistance to ground, therefore making it insignificant?