Resistor values for 2x1 unbalanced passive mixer

Hello all -

I'm looking to break into the effects send loop of my chopped Hammond organ, to allow me to pass my Wurlitzer electric piano through the Leslie in addition to the Hammond. Both signals are mono and unbalanced.

It seems that a braindead simple way to do this is through a 2 in by 1 out passive mixer circuit on a card I can fit into the organ. Simply wiring 1 resistor to the tip of each input, tying the other ends of the resistor together, and send that to the output.

I understand that there will be a volume drop. Are there any other audio issues to contend with? In a fixed situation like this, what would you recommend for the resistor values? Are there any other suggestions? I'd like to avoid an active circuit if possible, for simplicity's sake, but if that makes more sense, I can certainly build one.

Many thanks in advance,

Nick

http://electronicdave.myhosting.net/miscimages/mixnetwork.gif

Please pardon my impedance ignorance in advance...

So if I understand Dave's website correctly, each signal in a 2:1 mixer will suffer a 6db loss, irrespective of what value resistors are used.

As far as output impedance, using 10K resistors would result in an output impedance of 15151 ohms. Using 1K resistors would result in an output impedance of 1515 ohms, and 100 ohm resistors would give a 151 ohm output impedance.

I don't know of any easy way to measure the Hammond's output impedance at the send/receive as it currently stands, but must assume it is something that would not freak out normal line level signal processors. Looking at other passive mixer schemos around the net, people tend to use 10K resistors, which would thus create quite high output impedances, yes?

Most of the passive mixer schematics on the 'net are bullshit, quite frankly, and I wouldn't pay them much attention. The web is littered with some really crappy little circuits.

Secondly, I don't understand how you're coming up with those numbers. My chart says "output impedance approximately equals R / N." There's a tacit assumption there that the source impedance of the inputs is low compared to R, which is a fair assumption to make since you generally want Z to be several times the source impedance for minimum signal loss and loading. So, for two "ideal zero impedance" inputs with 10K resistors, R / N = 5k.

But let's assume a more realistic source impedance figure, say 1K. We design the mixing network for a Z of 10K to avoid loading of the 1K sources.
R = (N-1 / N+1)Z = (1 / 3)10K = 3.3k

The impedance looking back into the output of the summing network is
(R + source Z) / N, or
(3300 + 1000) / 2 = 2150, or 2.1K
and that's your output impedance.

If we had used the approximation R / N, our answer would have been 1.6K.

The input impedance of the Leslie amp is rather high, if memory serves, so we don't need to account for additional loss from shunting of the mix network output.

The isolation in a summing network with only two inputs is poor--in this case (assuming 1K input impedances) it's only 16dB. This may very well not be a problem in your application, and the isolation figure is better than that if the source impedances happen to be lower than 1K.

Thanks, Dave. Yep, my math was screwy. This explanation is terrific, and greatly appreciated. I'll go with the 3.3K resistors and see how things work. As far as the poor isolation figures, I'm not concerned with it - this is for live gigging application. I'm just trying to avoid dragging about an extra amplifier for my wurly, so this should be fine.

Best,

Nick

I've been investigating some API schematics and I notice that all of the busses from each input channel are balanced and are using a 49.9k resistor. The schematic that I have is a bit incomplete, so it might be difficult to say what's happening down stream. So, how does this value fit into NYDaves equation? This value seems a bit high doesn't it? Help me to understand, someone??

thanks

Mike