Help with Passive Summing Theory

[rossco_50]

Hi,

Forssell Tech provide a useful article on passive summing http://www.forsselltech.com/media/attachments/summing_buss.pdf

The article advises that impedance seen by a channel is given by the series resistor + the parallel resistance of all channels that are connected together (including the channel being analysed).

Can anyone explain how this relationship is derived? I am confused by the idea that a single resistor would be both in parallel and in series with other values. It seems it might be explained through superposition, but can't quite get my head round it.

All other articles on passive summing I can find e.g. Rane, DaveNYC either provide universal resistor values or some rule of thumb.

Thanks,

Ross

 

[DF96]

The article advises that impedance seen by a channel is given by the series resistor + the parallel resistance of all channels that are connected together (including the channel being analysed).

The article is wrong in that respect. It is all other channels which are in parallel, not included the one being considered.

Assuming that each signal source acts as a perfect voltage source it will have zero output impedance. This means that the far end of each other resistor is, in effect, connected together. The near ends are all connected at the bus, hence they are ih parallel.

The idea that some components can be viewed in series from one angle while simultaneously in parallel from another view often crops up in electronics.

 

[rossco_50]

Hi,

Thanks for the response.

So with only 2 channels each channel would see the series sum of both resistors?

 

[DF96]

Yes.

 

[rossco_50]

If we look at the rane summing network Why Not Wye? assuming ideal voltage sources, would the resistor to ground therefore create a load of 470 | 470 | 20 K for both channels?

 

[sreten]

The article advises that impedance seen by a channel is given by the series resistor + the parallel resistance of all channels that are connected together (including the channel being analysed).

Can anyone explain how this relationship is derived?

Ross

Hi,

Its true except for including the given channel, i.e. for 5 channels
joined together with 10k each channel will see 10K + 2.5K, not
+ 2.0K, consequently Vout per channel will be 1/5, not 1/6.

Its true the bus impedance is 2K, but only for another input.

Two channels is even more wrong, i.e. each channel will
see 10K + 10K, not 10K + 5K, Vout is 1/2, not 3/2,
even though the bus impedance is 5K. Add another
channel and it is that will see 10K + 5K.

rgds, sreten.

 

[DF96]

If we look at the rane summing network Why Not Wye? assuming ideal voltage sources, would the resistor to ground therefore create a load of 470 | 470 | 20 K for both channels?

No. Each channel sees 470 + ( 470 || 20k ).

 

[sreten]

No. Each channel sees 470 + ( 470 || 20k ).

Hi,

True and for most domestic hifi summing for a sub the R's
should be more like 4.7K and 20K, 4.7K + ( 4.7K || 20K*).

Being unable to drive cable capacitance at h.f.
does not matter for a subwoofer signal.

rgds, sreten.

* Noting 20K should be replaced by 20K || amplifier
input impedance, often 10K = 4.7K + ( 4.7K || 6.6K)

 

[rossco_50]

Thanks both for your replies. Very helpful.

At what point does the input impedance become too high? I was putting a mono amp together which takes the headphone output from an iPhone which is apparently around 5 ohms. Is 4.7 K a common value because that is a good trade off between noise and flexibility or if I know that the iPhone is the only source that will be used is there a (theorectical) benefit in going down as low as 10x output?

 

[sreten]

Hi,

470 is fine for a headphone output, x10 for a line output.

rgds, sreten.