Impedance balanced 1/8" stereo to 2x XLR male (output)

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Potato Cakes

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Hello, everyone,

I'm making cables for various things around the studio and one of them is a nice stereo 1/8" to two (2) male XLR connectors for my Sound Devices Mix-Pre 3 II. I wanted to make the XLRs impedance balanced but I don't know the value of the resistors that is inline with the tip and ring connections on the 1/8" jack on the unit. I emailed the company to find out and they said they didn't have that information and when I pressed further they again said they didn't know, which I thought was weird. On the spec sheet is says that the stereo out impedance is 500 ohms, but I don't know what that means as it applies to what I'm trying to accomplish. To me, that sounds a little high for impedance balancing considering most of what I've seen for accomplishing this task ranges from 47-120R. Can someone break down what a manufacturer is/might be using for the output connection impedance spec?

I know I can just do an unbalanced XLR and call it good, but it's just as easy to add a resistor and have it play nicer with balanced input connections on a console.

Thanks!

Paul
 

dbelousov

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There is 500 Ohm in its specs. Why do you think it's impossible to be true?

You could try to use resistors of this value "implanted" in the output plug and measure the outcome with and without them.
 

Newmarket

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You can effectively measure the resistive impedance by measuring the open circuit signal level - using a constant amplitude test tone. Or more accurately it's loaded by your DMM or oscilloscope etc. Then load with a resistor and calculate the impedance from the resulting signal level.
Easiest illustration is to say that when the value is halved then the impedance is equal to the load resistor value.
But you can calculate it using other resistor values so you don't need a particular resistor value to hand.
 

abbey road d enfer

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Newmarket said:
You can effectively measure the resistive impedance by measuring the open circuit signal level - using a constant amplitude test tone. Or more accurately it's loaded by your DMM or oscilloscope etc. Then load with a resistor and calculate the impedance from the resulting signal level.
Easiest illustration is to say that when the value is halved then the impedance is equal to the load resistor value.
But you can calculate it using other resistor values so you don't need a particular resistor value to hand.
+1.
In addition, one should make sure the output stage is not overloaded. It can happen with some headphone outputs, where the actual output resistance is very low, and halving the voltage implies putting the output stage into protection.
 

Potato Cakes

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Newmarket said:
You can effectively measure the resistive impedance by measuring the open circuit signal level - using a constant amplitude test tone. Or more accurately it's loaded by your DMM or oscilloscope etc. Then load with a resistor and calculate the impedance from the resulting signal level.
Easiest illustration is to say that when the value is halved then the impedance is equal to the load resistor value.
But you can calculate it using other resistor values so you don't need a particular resistor value to hand.

I am little confused on the procedure and I do apologize for my lack of understanding for something that is probably very simple. The Mix Pre 3 would be generating the tone and then I would measure the resistance between + and shield on the output connection while signal is present, is this correct? My thought was to use a 1k multi turn trimmer between pin 3 and pin 1 and adjust while watching pink noise on an analyzer and adjust it till the level and response was optimal. Is this kinda of what you are saying?

Thanks!

Paul
 

JohnRoberts

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Potato Cakes said:
I am little confused on the procedure and I do apologize for my lack of understanding for something that is probably very simple. The Mix Pre 3 would be generating the tone and then I would measure the resistance between + and shield on the output connection while signal is present, is this correct? My thought was to use a 1k multi turn trimmer between pin 3 and pin 1 and adjust while watching pink noise on an analyzer and adjust it till the level and response was optimal. Is this kinda of what you are saying?

Thanks!


Paul
You first measure the no load output voltage, preferably playing a sine wave at steady amplitude.

Then you apply a modest load to the output and measure the voltage sag (drop).. From this data you can generally calculate the effective output (source) impedance...

The output impedance forms a simple voltage divider with the load impedance.

JR 
 

Bo Deadly

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Potato Cakes said:
I am little confused on the procedure and I do apologize for my lack of understanding for something that is probably very simple. The Mix Pre 3 would be generating the tone and then I would measure the resistance between + and shield on the output connection while signal is present, is this correct? My thought was to use a 1k multi turn trimmer between pin 3 and pin 1 and adjust while watching pink noise on an analyzer and adjust it till the level and response was optimal. Is this kinda of what you are saying?
Measuring the output impedance of a single-ended output is a fairly simple voltage divider calculation where R1 is the output impedance of the device and R2 is the load impedance. So if you start with R2 being relatively large like the 10K input of your typical audio interface, you should have little to no drop in voltage and now you know the input voltage. So you know the input voltage and you control R2 and you can measure the voltage drop as you make R2 smaller. So just add a resistor across the output to ground until the voltage drops by ~20-50%. Then take the input voltage, R2 and the output voltage and put it into an online voltage divider calculator like this one:

  https://www.rapidtables.com/calc/electric/voltage-divider-calculator.html

However, there are a few pitfalls here.

One is that you want to use a fairly high level steady signal like a sine wave at say 100Hz. If the level is too low, measuring levels will not be as accurate. If you use something like a noise source as you suggested, the level will not be steady. If you use a DMM, the AC voltage of your typical meter is only good at reading low frequency AC.

Two is if you don't have a DMM or oscilloscope, you might only be able to view dB levels in your DAW or similar. The solution for this is to convert the dB values to a ratio by using a different online calculator like this one:

  http://www.sengpielaudio.com/calculator-amplification.htm

For example, let's say you setup your device to put out a high level sine at 100Hz and adjust the input level in your DAW to make it 0dB. Then you add an R2 across the device output of say 680 ohms and you see that the level in the DAW drops to -4 dB. Now you go to the above calculator and put in 1 for the "Input", -4 for "Level change" and click on "calculate value 2". That gives you a ratio of 1 / 0.63. Now you can go to the voltage divider calculator and use the ratio as "voltages". Specifically, put in Vin = 1, R2 = 680, Vout = 0.63 and then click on "Calculate" to populate R1 which results in 399 ohms. So 399 ohms would be the output impedance of this imaginary a device.

However, I somewhat doubt that the output impedance of the Sound Devices stereo out is "500 Ohms". That's probably just what it can safely drive. So I would make R2 470 ohms and see if you get a drop of 20% or so. It's very likely that you will not and you will need to go down to 200 ohms or lower and that you will find the output impedance is actually more like 100 or lower still.

There is a third pitfall here which is that all of the above does not work if the output is already impedance balanced. Clearly the stereo out on that Sound Devices thing is not already impedance balanced because the sleeve of the jack is grounded to the chassis and it's shared by both outputs so impedance balanced could not work. You need separate resistors to ground (which is precisely what you're trying to do with your special cable in this case). But in all other cases, most outputs of modern devices are already impedance balanced and so the above procedure is insufficient.
 

Potato Cakes

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Thanks for the detailed procedures, squarewave. I will have to read it several times slowly to understand fully what you're saying. I do know about using Ohm's law for voltage drops but for this application I don't know what values to use for two of the variables. I'll mess around with this later this week.

The stereo output is very much unbalanced, being it is a 1/8" TRS.

While I do like learning one more skill to add to the electronics tool belt, I'm equally annoyed that the manufacturer's support doesn't know who to ask for a single resistor value for a product they own. I know I'm not entitled to any information regarding a company's products, but I would have been more satisfied if they would have just told me they can't/won't give out that information.

I'll report back.

Thanks!

Paul
 

computerlen

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Nov 30, 2019
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Hello, everyone,

I'm making cables for various things around the studio and one of them is a nice stereo 1/8" to two (2) male XLR connectors for my Sound Devices Mix-Pre 3 II. I wanted to make the XLRs impedance balanced but I don't know the value of the resistors that is inline with the tip and ring connections on the 1/8" jack on the unit. I emailed the company to find out and they said they didn't have that information and when I pressed further they again said they didn't know, which I thought was weird. On the spec sheet is says that the stereo out impedance is 500 ohms, but I don't know what that means as it applies to what I'm trying to accomplish. To me, that sounds a little high for impedance balancing considering most of what I've seen for accomplishing this task ranges from 47-120R. Can someone break down what a manufacturer is/might be using for the output connection impedance spec?

I know I can just do an unbalanced XLR and call it good, but it's just as easy to add a resistor and have it play nicer with balanced input connections on a console.

Thanks!

Paul
Hi. what about an isolation transformer? Two in this case.
 

Potato Cakes

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The idea is be able to sort it out with resistors and not add the weight or bulk of transformers. Usually, I would do just that. My inquiry was to see how or if I could do this.

I still haven't been able to try the procedures as listed above. I've been covered up with other projects as of late.

Thanks!

Paul
 

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