Bus Source Impedance (w / opamps)

[bjoneson]

Alright... I'm trying to wrap my head around some theory stuff.

Let's take a simple case...

Passive summing bus. 24 channels with 10k bus resistors. Inputs (high side of bus resistors) are fed via opamp to a SPDT assign switch. (Legs are opamp output / ground, common is high side of bus resistor).

I think I understand the bus impedance / loss part. 24 channels of 10k bus resistors yields a bus source impedance of (10k / 24) Apx. 417 Ohms.

Bus gain (loss) is therefore (417 / 10k) Apx. .042 Assuming the bus is feeding a high impedance input like an opamp input, and ignoring the NFB load (if any).

Ok... so here's where my understanding of theory breaks down. All that makes perfect sense when the channels are in the off position. That is, with the high side if the bus resistors switched to ground. The bus source impedance is literally impedance to ground in that case.

But when the channels are switched on, the high side of the bus resistors are now connected to an opamp output. I understand that in "theory" an opamp output has a near zero "source / output impedance". But is this actually an impedance to ground? An ompamp has no concept of ground. Why do we get to treat the output of the opamp as equivalent to a low impedance path to ground (or I suppose 0V)?

Maybe I just kind of answered my own question. Am I understanding correctly, or am I missing something?

 

[JohnRoberts]

You may be overthinking it...

If the resistors are either connected to lowZ opamp or back grounded impedance of interest is 10k/24

JR

 

[bjoneson]

I think now it makes sense once I consider the opamp as a voltage source, which will source or sink whatever current is necessary to maintain 0V at the output pin in steady state. In this way, it behaves just like a 0V node would from an analysis perspective.

I think....

 

[john12ax7]

The opamp does indeed have a concept of ground. And the dynamic (AC)  output impedance is referenced to ground (DC could be different). Look up small signal opamp models,  they are dependent sources with an output Z.

It might help to look at things in terms of delta (change).

Z = V / I, but also Z = (V2 - V1) / (I2 - I1)

So when the opamp is asked to supply two different currents to the load, for example driving 1k Vs driving 10k load,  how much does the output voltage change? You can actually measure this and plug in the numbers.  Ideally the output voltage changes very little,  delta V is small,  therefore output Z is also small.

 

[bjoneson]

Makes perfect sense guys. Thanks for being a sounding board.

 

[Monte McGuire]

The bit you're missing is the idea of 'superposition'. With a linear circuit of amplifiers and resistors, you can solve it by suppressing all of the sources except one, solving for that source, then solving for all of the rest of the sources individually in the same way, and then adding all of the solutions together.

Suppressing a source such as an op amp is basically making its output zero volts, and it's easy here to see that its output will then behave almost exactly like a ground. This behavior is no different when the op amp is producing a signal, except that it's signal gets superimposed over what's happening from the rest of the circuit.

I hope that helps!

 

[bjoneson]

That is exactly how I've come to understand it in my head. Thanks again for all the great perspective.

 

[PRR]

The naked opamp does not know "ground".

The way you are surely using it, the input and feedback networks are normally ground-referenced. If you put 1V DC from ground to a unity-gain stage, the output WILL be 1V DC.

 

[bjoneson]

Yes, that was a horrible misnomer on my part. It's always referenced to something.

 

[benb]

It was so many decades ago that not only did I learn about superposition, I also learned about Thevenin's Theorem and Norton's Theorem, and learned to manipulate resistor networks and sources of DC voltage and current to reduce them to either a voltage source with a series resistor or a current source with a parallel resistor. That really helped with understanding the op amps course later on.

For extra fun, in the next course I did this with AC sources and impedances as represented by complex numbers.

 

[ruffrecords]

Just as an aside,  when working out bus loss, if you have N channels feeding the bus and the gain make up amp has a relatively high input impedance then the bus loss is simply 1/N.

Cheers

Ian

 

[5v333]

2channels = 6db loss
10channels = 20db loss

Right?

 

[ruffrecords]

2channels = 6db loss
10channels = 20db loss

Right?

Exactly.

Cheers

Ian

 

[5v333]

Exactly.

Cheers

Ian

This is actualy very interesting.
Is it even possible to sum a number of sources with out the loss of gain. In the analogue domain.

 

[ruffrecords]

This is actualy very interesting.
Is it even possible to sum a number of sources with out the loss of gain. In the analogue domain.

Not passively no. If you assume the output impedance of the amps is close to zero, or at least very small compared to the bus resistor then;

If you have two channels, each one is connected to the other by 2  bus resistors. Since the output of the (inactive) one is at 0V then the two resistors in series form a potential divider so the active ones output is halved (6dB loss). Some simple maths demonstrates that if you extend this to N channel, the loss is 1/N.

I just tried writing this using regular text but it becomes confusing. If you are really interested I will write it by hand, scan it and post it.

Cheers

Ian

 

[ruffrecords]

My last answer is incorrect. It is possible to sum several sources passively without loss of signal. The method is to have a transformer at each output and wire the secondariness in series.

Cheers

Ian

 

[Monte McGuire]

Yes, the virtual earth summing amp sums channels directly without gain loss. However, the noise gain of the summing amp still increases with a greater number of channels summed, so the requirement of gain to sum analog signals is probably more of a fundamental requirement than signal level loss.

Edit: transformer summing does not require gain... forgot about that!

 

[JohnRoberts]

One more time around the block... Back in the 80s I developed a bus structure using synthesized current sources so I could sum any number of stem with no insertion loss or noise gain build up... 

but yawn... digital summing is arbitrarily perfect.

JR

 

[5v333]

I get that the summing resistors presents voltage dividers among them. Just wanted to know if this method is universal law.

Jr that would be cool to have a look at!

Transformers you say... Have to take a look at that. Any old desks using such methods?

Im studying summing in digital domain and its funny all this came up.

 

[abbey road d enfer]

Just as an aside,  when working out bus loss, if you have N channels feeding the bus and the gain make up amp has a relatively high input impedance then the bus loss is simply 1/N.

Cheers

Ian

Actually 1/(N-1)
Does not make a difference for high channel count though.