# Combining 2 channels to one

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#### klitgt

Hi
I have an active x-over with 2 channels in both high pass and low pass lines. Now I want to combine the 2 channels (left & right) in the low pass line after the L & R output buffers to feed one single poweramp that drives one single subwoofer.
As far as I know this is possible if I connect the outputs through resistors as shown in the diagram attached.
But the values of the resistors is a ??? to me.
Does anybody know which value to use?
And am I right with the principle shown in the diagram?

Thank's for inputs !

#### Attachments

• 2ch to 1 ch.gif
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Try 2200 ohms.

#### paulb

Bill Fitzpatrick said:
Try 2200 ohms.
I agree with Bill. It's the bottom resistor in your avatar.
But anything in the same range will work too.

#### klitgt

Thank's guys!

Now that I have a dialog with some knowledgeable persons, can you tell me why it is necessary with these resistors, why not just connect the leads from R and L without resistors?

#### Sch3mat1c

Because if R is on a positive peak while L is on a negative peak, the mixed output shows the average, i.e. zero, but between them, they try feeding each other current. If the outputs don't use current-limited op-amps, it's a good smoke recipie. If they do, it probably won't come out right because the two amps are trying to fight each other (due to negative feedback) for control over the output. Using resistors fixes this.

The value of resistance depends on how much crosstalk/loading you want (i.e., current flowing into and out of each terminal) and the attenuation from feeding the load impedance. Ideally you'd use infinite, followed by infinite gain to counter the attenuation. That doesn't work well in practice, but where an input impedance is tolerable (well, always), a "virtual ground" mixer works well.

Tim

#### klitgt

I found an answer by using the inside of my head: The buffers have low output impedance and if I connect the output leads without output resistors, the output of each buffer looks into a low impedance (the other buffer). This impedance is too low, it must be the same high impedance as if the buffers look into a power amp with an impedance of typically 10-20K.

10-20K for the output resistors would be too much, as the resistors form a voltage divider together with the input impedance of the power amp.....and thus attenuatte the signal too much. So the suggested 2200 Ohms for the resistors is what I will go for, thank's again to all

#### AndrewT

Hi,
connecting those two resistors to an inverting opamp is the normal way to SUM 2 inputs.
The inverting pin on an opamp is a virtual earth and the two outputs each see the resistor as a load to earth with no crosstalk.
The opamp reads each input and adds them together to be passed on to the next stage. It also cancels any stereo information. But into a sub-bass there is probably no real stereo information anyway. Modern electronic mixes could easily put different info irrespective of frequency and if the producer did this you would lose some or all of that as well.

#### klitgt

AndrewT said:
Hi,
connecting those two resistors to an inverting opamp is the normal way to SUM 2 inputs.
The inverting pin on an opamp is a virtual earth and the two outputs each see the resistor as a load to earth with no crosstalk.
The opamp reads each input and adds them together to be passed on to the next stage. It also cancels any stereo information. But into a sub-bass there is probably no real stereo information anyway. Modern electronic mixes could easily put different info irrespective of frequency and if the producer did this you would lose some or all of that as well.

Do you mean the way showed in the upper example (see attachment)?
Should there be a feedback loop (blue lead) with a resistor and if yes which resistor value? And what about the non-inverting input (pin3 at the opamp)?
What about the lower example. Are input resistors needed here and what about feedback from output to pin2? If needed, should a resistor be added? What value?

In my CDP the differential signals from the I/V converters go into pin 3 (positive ½) and pin 2 (negative ½) of the opamp that combines the signals to single ended in each channel. This has inspired med to draw the lower circuit example attached. In the CDP there is a feedback from output (pin6) to pin2 with a 2,2K resistor in parallel with a 180p cap.

#### Attachments

• 2channelsto1+opamp.gif
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#### klitgt

AndrewT said:
Hi,
connecting those two resistors to an inverting opamp is the normal way to SUM 2 inputs.
The inverting pin on an opamp is a virtual earth and the two outputs each see the resistor as a load to earth with no crosstalk.
The opamp reads each input and adds them together to be passed on to the next stage. It also cancels any stereo information. But into a sub-bass there is probably no real stereo information anyway. Modern electronic mixes could easily put different info irrespective of frequency and if the producer did this you would lose some or all of that as well.

Andrew, now I have read your reply a couple of times and I think I have understood it better.
I found an appication note on NE5532/34 and here the non-inverting input is connected to ground if the inverting input receives the signal.
But I do not find any information about the reason for
"connecting those two resistors to an inverting opamp is the normal way to SUM 2 inputs."
Why not use the non-inverting input and connect the inverting to ground?
Again I would appreciate som guidance as to feed-back from output to input etc.

My next stage is a poweramp with only non-inverting input (I guess if there is only one input on a poweramp it is a non-inverting).

So in conclusion I have to find an opamp and do as you suggest. But what are the benefits of your solution as compared with my first suggestion (see picture in the beginning of this thread) just connecting the two resistors to the input of the poweramp, without using any opamp?

#### AndrewT

Hi,
your top diagram is the summing amp. You can have as many inputs as you have resistors connected to the inverting input.

If feedback value =input value then gain=1. Two inputs will give a total output =2 times a single input.
By changing the ratio of feeback value to input value you can select the gain/attenuation you need.

Since your power amp is conventional you cannot use it in the summing manner and you need the summing opamp at its input.

You can connect the non-inverting input direct to ground but this gives a DC offset at the output.

Instead it is better to connect a resistor between non-inv & ground. The resistor value = // resistance of all the connections to the inverting input. In your example diagram if feedback value =2k2 then ground resistor =2k2/3 =733R. THEN 2* 1v INPUT GIVES 2V OUTPUT.

#### klitgt

Hi Andrew
Thank's for fine explanations.
I understand it as shown in the attachment.

You mention that if R to ground is = 1/ Rin1+Rin2+Rfeedback the DC offset at the output will be eliminated.
I did not show in my first diagrams that the output from the electronic crossover output buffers have DC ofsset which is actually blocked by polypropylene caps before the 2K2 resistors.

Do I have to keep them or does an inverting input of an opamp tolerate dc input from the sources?

Will your solution with a resistor of 733R eliminate the dc offset from the output of the summing opamp if there is dc on the inverting input, or must I use a coupling cap after the output from the summing opamp?

Pleas bear with my small knowledge, I appreciate your help!

#### Attachments

• 2channelsto1+opamp2.gif
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#### AndrewT

Hi,
this is a summing opamp and it works down to DC.
Any DC on the input will be added to the other inputs and show on the output.
If you do not block input DC you will offset the opamp's working range and use up valuable dynamic range. I recommend you block DC at the inputs.
The resistor to ground does not eliminate the DC on the output it only reduces the opamps offset. The opamp may still have a residual DC that varies with operating conditions but the offset should be manageable.
I would recomend that the power amp also has a DC block, If it has not then fit a DC blocking cap on the summing opamp output. To allow for low input impedance of any future power amp I suggest a plastic film (polypropylene) about 4uF to allow bass down to about 4 Hz(-3db) into 10K. For a sub this turnover frequency may be a little high, some would try for less than 1Hz which would give very low phase errors at 10Hz.

#### klitgt

AndrewT said:
Hi,
this is a summing opamp and it works down to DC.
Any DC on the input will be added to the other inputs and show on the output.
If you do not block input DC you will offset the opamp's working range and use up valuable dynamic range. I recommend you block DC at the inputs.
The resistor to ground does not eliminate the DC on the output it only reduces the opamps offset. The opamp may still have a residual DC that varies with operating conditions but the offset should be manageable.
I would recomend that the power amp also has a DC block, If it has not then fit a DC blocking cap on the summing opamp output. To allow for low input impedance of any future power amp I suggest a plastic film (polypropylene) about 4uF to allow bass down to about 4 Hz(-3db) into 10K. For a sub this turnover frequency may be a little high, some would try for less than 1Hz which would give very low phase errors at 10Hz.

According to your advice I will block dc at the input of the summing opamp. I assume that it can be done by a small cap, polyprop 1uF or so because the input impedance is high. Or do you recommend a higher value to avoid bass roll off at a too high frequency?
For the output I have a 10uF polyprop, it should be OK!!
Thanks again for valuable help

#### AndrewT

Hi,
your input impedance is NOT high.
It is equal to the first inline resistor. Remember you have a virtual earth input at the inverting pin and the first resistor is the only thing between your source and the virtual earth.
2k2 & 1uF gives 72Hz -3db and about 144Hz -1db.

#### klitgt

AndrewT said:
Hi,
your input impedance is NOT high.
It is equal to the first inline resistor. Remember you have a virtual earth input at the inverting pin and the first resistor is the only thing between your source and the virtual earth.
2k2 & 1uF gives 72Hz -3db and about 144Hz -1db.

I appreciate help from people with knowledge, experience and especially from those who give decent replies and do not laugh at my stupidity
But this forum is "at the end of the day" also for those who need help, right? It is also a good training in using the inside of the head, of course the impedance depends of the resistor values as it is a virtual earth input! You told me that earlier!

#### klitgt

I understand that the gain of the attached circuit depends of the ratio R-input+R-input/Rf, in this case 2,2 + 2,2 / 2,2 = 2.
This means that reducing the value of Rf will increase the gain?
And does that mean that R to earth 733Ohm has to be changed to another value?

#### Attachments

• rf.gif
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#### klitgt

This is what I have done and it works perfectly well!
Sorry for the Danish words in the drawing, I believe it is understandable anyway.
Thank's for help

#### Attachments

• sumamp1.gif
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