Passive Sum two signals - Resistor Value Help

[moley]

Hello folks,


I am putting together a pedal board for my two acoustic guitars which each have a pair of pickups in them.

Basically I want to passively sum each pair to mono and have been doing some reading on the values keeping in mind impedances of the chain.

The two signals to sum would be coming from two separate preamp/EQs with output impedances of 1Kohm and 100 ohms respectively.

After summing they would be going into a TC Electronic Hall of Fame Reverb which has an input impedance of 1 meg Ohm, if true bypassed using it's foot switch, I guess the summing network would "see" the DI box placed after the Reverb which also has input impedance of 1Mohm.

Going by the 10:1 rule that would dictate that I need 10Kohm resistors to sum the two channels prior to the TC Reverb pedal given that one of the EQs has an output impedance of 1K.

I understand that 10Kohm resistors would give a 10Kinput impedance at the summing network and roughly a 6db drop in signal.

Is 10K the best value to use or is there a gremlin in these details that I am missing?

Perhaps it would be best to sum after the reverb given that the reverb pedal has a lower output impedance (100 ohms) and accepts a stereo signal. Bypassed of course though I would be in the same scenario where the two preamps would see the summing network impedance of 10Kohms.


Any help or advice greatly appreciated!!

Many thanks in advance.

 

[rayma]

I understand that 10Kohm resistors would give a 10Kinput impedance at the summing network and roughly a 6db drop in signal. Is 10K the best value to use or is there a gremlin in these details that I am missing?

The 10k resistors should be fine, give it a try.

 

[AndrewT]

Ignoring the 1Mohm of the Receiver, the output of one Source sees the output of the other Source as it's input.
say you have the 100ohm output feeding into the 10k and then into another 10k feeding into the 1k output impedance of the other Source.
The 100ohm outputs sees 10k+10k+1k as it's load.
The 1kohm output sees 10k+10k+100 as it's load.
There is a small reduction in load when the 1Mohm is paralleled into the loading at the 10k junction.
Both Sources will be able to drive those loads.

 

[moley]

Thanks vey much for the replies.

Ignoring the 1Mohm of the Receiver, the output of one Source sees the output of the other Source as it's input.
say you have the 100ohm output feeding into the 10k and then into another 10k feeding into the 1k output impedance of the other Source.
The 100ohm outputs sees 10k+10k+1k as it's load.
The 1kohm output sees 10k+10k+100 as it's load.
There is a small reduction in load when the 1Mohm is paralleled into the loading at the 10k junction.
Both Sources will be able to drive those loads.

Thank you, OK so I guess I have learnt something here. The input impedance of my proposed summing circuit in my rig is of course higher than the 10Kohm I was expecting.

While the sources can drive the load calculated above, is there any benefit at all to reducing those two summing resistor values down from 10K?

Many thanks again.

 

[moley]

I am trying to work out the db loss in the proposed summing network.

Am I correct that the minimum signal loss of summing two signals passively will always be at least be 6db but the actual result depends on the summing resistor values used and the impedance of the input following the summing network.

Am I correct in my understanding that the summing circuit can be seen as a simple voltage divider where the input is the summing resistor at 10K and the value of the shunt is the input impedance of the next stage (D.I. Box 1Meg Ohm) in parallel with the other 10K summing resistor (9.9K).

9.9K/(9.9K + 10K) = 0.4975 is this just over a 6 db drop, i.e. 2.01 times reduction?

Thanks!

 

[rayma]

Am I correct that the minimum signal loss of summing two signals passively will always be at least be 6db
but the actual result depends on the summing resistor values used and the impedance of the input
following the summing network. Am I correct in my understanding that the summing circuit
can be seen as a simple voltage divider where the input is the summing resistor at 10K and the value
of the shunt is the input impedance of the next stage (D.I. Box 1Meg Ohm) in parallel with the
other 10K summing resistor (9.9K). 9.9K/(9.9K // 10K) = 0.4975 is this just over a 6 db drop, i.e. 2.01 times reduction?

Yes, that's right. This method is called superposition, using Thevenin equivalent circuits.
This mixing circuit works better using an active device to provide a virtual ground at the junction
of the mixing resistors. Then there is no loss (other than higher noise gain in the op amp).

 

[AndrewT]

if the two signals are the same and in phase, then each is reduced by 0.4975 and then summed to come back to virtually the same as one signal.

If you are summing very low bass from a stereo source there is a high chance that both signals are the same.