Calculating summing resistors - some help needed

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Humner

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Sep 22, 2012
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I've been working on a 12 channel summing mixer to use with the CAPI ACA-BO - which as I understand is virtual earth summing?

I'm at the point where I am trying to work out my summing resistors. Most of my knowledge is based on example - I read schematics and test particular circuits I see in the real world and I have a certain level of practical understanding, but there are plenty of gaps in this understanding/knowledge - anyway...

I see schematics for different mixers and a lot of time they are using different summing resistors - how do I pick what summing resistor to use? I thought it might be based on how many channels, but I've seen 33k used for 8  channels, and 22k used for 16. This is where my knowledge based on example falls apart as I think if you have more channels, then you will want higher value summing resistors. So I think that the 16 channel in my example must have a lower signal level for each channel before it gets the summing resistor?

My main question is, should I aim for a particular signal level after the summing resistor? I have left and right bus channels - what is the best way to extrapolate this for 12 channels?

edit side note: is the signal level not that important with virtual earth summing? Am I on the right track?
 
Virtual earth mixing works by using an op amp connected in an inverting configuration. Under these conditions, the gain is just the value of the feedback resistor divided by the input resistor.  The input impedance is equal to the value of the input resistor. It is pretty common to set this gain to unity (0dB) in a summing amp so the two resistor are of equal value.  The beauty of virtual earth mixing is that it does not matter how many separate inputs you feed through individual resistors, the gain for an input  is always the feedback resistor divided by the individual input resistor. For a balanced system there is often two op amps, one for the hot and one for the cold inputs. The outputs of these are then used as a balanced out either directly or via a transformer.

In deciding what value resistors to use there are two things to consider - input impedance and noise.

As a rule, you will be aiming for the input to look like a regular 10K bridging input. You can achieve this simply by setting the hot and cold input resistors to half the desired input impedance so 5K1 or 4K7 are common values. Some people like to have an input impedance of 20K so they use 10K resistors.

Noise is a bit more complex but as a rule, if you set the resistor values for the input impedance you require then you do not really need to worry about noise. With a single 5K1  input and feedback resistors for example, the noise due to the resistors is about -115dBu. As you add more inputs, the noise gain of the op amp increases but the parallel value of the resistors decreases so the noise due to the resistors is approximately constant. However, the op am has it own inout self noise and this is constant so as the number of inputs increases this also increases at the output due to the increased noise gain. Bottom line is noise is principally determined by the op amp not the resistors.

For most applications 5K1 or 10K input and feedback resistors will be fine.

Cheers

Ian
 
Thanks for this.

In my summing amp the signals are unbalanced when summed.

Here is an example of the pan circuit I'm using(the resistor values are not the same, they will change once I work out the summing resistors)

http://www.bmltech.com/refbook/audio/pan_cct.gif

I assume that the summing resistors are made up of the resistor before and after the pot? so in the example I have linked above, the summing resistance is actually 9.4kohms?



 
OK, so this is more your regular unbalanced summing inside a mixer in which case different criteria apply the main one of which is to take care how the bus feed resistors load the pots here are connected to and thus alter their attenuation law.

First a small correction. The input 4K7 resistors and the pan pot form a pot divider. The 4K7 resistors to the right in your schematic are the bus feed resistors . It is common practice to make the feed resistors (the 4K7 on the left) the same value as the pot. This means there is about 9.5dB loss with the pan centred and 6dB loss at the extremes giving around 3.5dB drop with the pan centred.

Since you are suing virtual earth mixing, the bus can be assumed to be connected to ground so the bus feed resistor forms a simple load to ground at the output of the pan pot. Clearly this load alters the loss at the centre and extremes. The idea is to ensure that the law you get with this load is the law you want. If the bus feed resistor equals the pot and the input resistors then the centre loss increase to about 12dB. The loss at extremes increase to about 10dB. So the effect of the bus feed resistor load is to lessen the depth of the pan. The less obvious effect is we have also thrown away 2dB of signal/noise ratio because we need 2dB more gain to restore the signal level.

To reduce the influence of the bus feed resistor we can increase it. Let's make it 10 times the pot/input resistor value. The extreme loss becomes 6.4dB and the centre loss 9.8dB which gives a centre dip of 3.4dB which is  hardly different to the unloaded case. The bottom line is that as long as the bus resistor is between 4 and 10 times the pot/input resistor values, the centre dip of the pan pot will be between 3 and 3.5dB which are generally considered to be within the desired range.

With 4K7 pot/feed resistors you could use 22K bus feeds. The noise due to these at the virtual earth amp output will be about  -112dBu. If you used a 10K pot/feed and 47K bus resistors, the noise increases to  about -109dBu.

Cheers

Ian
 
thanks Ian, looking good so far - I've decided on 10k feed resistors and 10k pot and 47k bus resistors.

Some of the input channels will be dedicated stereo inputs. I take it the best way to do this would be to have a 10k resistor to ground between the feed and bus resistors. That would essentially mimic the pot fully turned to either left or right bus and keep as close to the PAN law I'm using.
 

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