Measuring output impedance

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johnheath said:
.....Moamps gave me an interesting idea if I could make a device on its own… like…just plug and measure?....

Sooner or later you will find that you need some software for complex audio measurement and representing data. In the user manual of Limp module ( Arta software) http://www.artalabs.hr/download/LIMP-user-manual.pdf
you can find all information for the impedance measurement you need right now, IMO.
 
johnheath said:
Thank you all

It has been very interesting to follow your comments back and forth… unfortunately a lot of it is way above my knowledge.

But what I have picked up so far is that the "direct" method is "better" for measuring the output impedance of line level tube devices.

I think it would be more accurate to say it is better for measuring devices that are designed to work into a single specified impedance - this does include some vintage tube line amps which need to be loaded with 600 ohms and also includes most vintage tube power amplifiers. But any line level amp that can drive a 1K load and a 10k load is fair game for the indirect method. All my tube designs, for example, can do this.

Cheers

Ian
 
moamps said:
Do you mean that difference in voltage sags is 0.13dB?
That's exactly what I meant.

For Zg=50 and Zl=545, 600 and 660, voltage sags are 0.76dB, 0.69dB and 0.63dB.
If the output voltage from the DUT is for example 1V, the measurement deviation from Zl=545 to 660 is  13mV what's 38dB below 1V and it is within measurement range of any decent soundcard.
I agree with you, but what precision can you expect in calculating the impedance? As I wrote, you need to measure the sag with a precision of 0.01 dB to have 10% precision for the calculated impedance.
 
abbey road d enfer said:
That's exactly what I meant.
I agree with you, but what precision can you expect in calculating the impedance? As I wrote, you need to measure the sag with a precision of 0.01 dB to have 10% precision for the calculated impedance.

How did you get that?
For Zg=55 (10% of Zg nominal), and Zl=545, 600, 660, sags are 0.84, 0.76 and 0.69 dB so the precision needed  for 10% in Zg variation is 0.06dB min or 7mV if Vin=1V what's still only 43dB lower.
 
moamps said:
How did you get that?
For Zg=55 (10% of Zg nominal), and Zl=545, 600, 660, sags are 0.84, 0.76 and 0.69 dB so the precision needed  for 10% in Zg variation is 0.06dB min or 7mV if Vin=1V what's still only 43dB lower.
0.06dB measurement resolution gives you the nominal value of impedance and the resulting precision would be +50%/-100%. In order to have decent precision on the calculated value, my wet finger says precision has to be about 1/10th of this value; I mentioned 0.01dB precision as a raw approximation. Making measurements with 0.01dB precision is not such an easy task. Temperature variations and connector contacts can easily produce that level of variation. I'm talking about doing this in domestic environment, not in a lab.
 
Monte McGuire said:
Impedance will be a function of frequency, so one number is not a 'right answer'.

I still have to rig up such a device, but the basic idea is that you drive the output of the device under test through a measured resistor, and then measure both the voltage across the resistor and the voltage across the device under test. The ratio of the DUT voltage to the resistor voltage multiplied by the impedance of the series resistor is the impedance of the DUT.

What I wanted to add to the jig were some very large, high quality coupling caps in order to be able to measure DC regulators accurately without worrying about AC coupling the generator. So, I have a pile of 10µF polypropylene caps, an aluminum box, and some connectors waiting to be wired up, but I've been too busy. ;)

However, this jig and a nice test set (like an Audio Precision) that can do sweeps will let you generate a pretty accurate impedance sweep with relatively little hassle. I still have to finish my version, but that's how it would work.

I am very interested in this "Jig"… a bit lazy question would be if you perhaps would like to share how it is built? Some sort of drawing? :)

Best regards

/John
 
abbey road d enfer said:
0.06dB measurement resolution gives you the nominal value of impedance and the resulting precision would be +50%/-100%. In order to have decent precision on the calculated value, my wet finger says precision has to be about 1/10th of this value; I mentioned 0.01dB precision as a raw approximation. Making measurements with 0.01dB precision is not such an easy task. Temperature variations and connector contacts can easily produce that level of variation. I'm talking about doing this in domestic environment, not in a lab.
Back in the -80s I made an audio test with a dB readout. I only offered 0.1dB resolution and that was hard enough (to be accurate).

JR
 
To be honest I have never paid much attention to the direct method as a real way of measuring output impedance. The reason is that when you are taught electronics at University, you soon discover that it can be very difficult to manipulate the equations describing a circuit into a form that describes the output impedance. To get around this there is a trick you are taught which is to imagine a  voltage source driving the output terminal and then derive an equation for the current going into the output. The output impedance is then simply the voltage source divided by this current. This is very similar indeed to the direct method.

It occurred to me there might be a simplification of this that might  be useful as a measurement method. The idea is you send a 0dBu signal via a 1K resistor into the output. You then measure the signal level at the output. Provided you adjust the generator so that the level going into the 1K resistor is exactly 0dBu then there is a direct relationship between the level measured at the output and the output impedance. You can measure the output level and just look up the impedance in a table (there is also a formula you can use but a table is much simpler). What do you think?

Cheers

IAn
 
ruffrecords said:
To be honest I have never paid much attention to the direct method as a real way of measuring output impedance. The reason is that when you are taught electronics at University, you soon discover that it can be very difficult to manipulate the equations describing a circuit into a form that describes the output impedance. To get around this there is a trick you are taught which is to imagine a  voltage source driving the output terminal and then derive an equation for the current going into the output. The output impedance is then simply the voltage source divided by this current. This is very similar indeed to the direct method.

It occurred to me there might be a simplification of this that might  be useful as a measurement method. The idea is you send a 0dBu signal via a 1K resistor into the output. You then measure the signal level at the output. Provided you adjust the generator so that the level going into the 1K resistor is exactly 0dBu then there is a direct relationship between the level measured at the output and the output impedance. You can measure the output level and just look up the impedance in a table (there is also a formula you can use but a table is much simpler). What do you think?

Cheers

IAn


Thank you Ian… I tried your described method since it seemed to be fairly easy to perform and you are a reliable source of information :)

But I tried it on a device that I knew (or understood) had a higher output impedance than the useful 100 ohm for driving a 1k load. I believe it is around 500 ohm…and using your method the voltage drop was so high that the equation gave a negative value… I tried it twice with the same result.

Could this happen or is it just me? :)

Best regards

/John
 
> the equation gave a negative value… I tried it twice with the same result.

Show your data and your math.
 
ruffrecords said:
To be honest I have never paid much attention to the direct method as a real way of measuring output impedance. The reason is that when you are taught electronics at University, you soon discover that it can be very difficult to manipulate the equations describing a circuit into a form that describes the output impedance.
I don't think the equations are that difficult; Z=(voltage measured at the load) divided by (injected current) seems to me a very simple exercise. What is difficult is the implementation; a real AC current source is not without limitations and high impedances present a problem of instrumentation (in order to measure accurately 1 Megohm, one needs to use a very high impedance meter, or apply corrective formulae).


To get around this there is a trick you are taught which is to imagine a  voltage source driving the output terminal and then derive an equation for the current going into the output. The output impedance is then simply the voltage source divided by this current. This is very similar indeed to the direct method.
Agreed. It allow using standard instrumentation. But there are serious limitations to this method, since the source may see an impedance that is too low, resulting in a voltage sag due to current limit rather than simple resistive drop. That's why the same method with a suitable resistor in series is preferred.

It occurred to me there might be a simplification of this that might  be useful as a measurement method. The idea is you send a 0dBu signal via a 1K resistor into the output. You then measure the signal level at the output.
That is indeed the most commonly used method, such as recommendes by Audiomatica (CLIO), Audio Precision, ARTA, RMAA, Klippel and many others, however, the resistor value must be set accordingly to the value of the mesaured impedance; as I mentioned earlier, optimum accuracy is when Zseries=Zload. So 1k would be perfect for measured output Z of most mic/line inputs/outputs, but not adequate for power amplifiers or instrument inputs. The discussion with moamps shows that a low output Z power amp may be necessary in some cases.


Provided you adjust the generator so that the level going into the 1K resistor is exactly 0dBu then there is a direct relationship between the level measured at the output and the output impedance. You can measure the output level and just look up the impedance in a table (there is also a formula you can use but a table is much simpler). What do you think?
I would think an Excel table where one would enter the generator level, series resistance (including that of the generator) and the measured voltage would be a 21st century correct method, that would encompass more real-life applications.
 
abbey road d enfer said:
I would think an Excel table where one would enter the generator level, series resistance (including that of the generator) and the measured voltage would be a 21st century correct method, that would encompass more real-life applications.

That was my first thought, but then I thought, sometimes when you are working on the bench and don't have a PC to hand, a piece of paper you can use to look up the answer is quite handy.

I will create the speadsheet.

Cheers

Ian
 
Simplified direct method output impedance spreadsheet is attached. It contains two version: the one on the left expects you to input measurements in dBu and the one on the right accepts voltages.

You can input the the drive level and the value of the series resistor or use the default values. Then you just input your measured value and the spreadsheet works out the output resistance. V1 and V2 should be left alone. They are just the calculated values of the drive level and the measured level in volts. The formula used to calculate the output resistance is:

Rout = V2 *1000/(V1-V2)

I had to change the extension of the file to .txt in order to attach it. You need to change it to .xlsx

Enjoy

Ian
 
johnheath said:
But I tried it on a device that I knew (or understood) had a higher output impedance than the useful 100 ohm for driving a 1k load. I believe it is around 500 ohm…and using your method the voltage drop was so high that the equation gave a negative value… I tried it twice with the same result.
Maybe the generator you used has a too large output impedance?
 
ruffrecords said:
Most pro gear will drive a 10K load and will also drive a 1K load. With the indirect method you send a tone(typically 1KHz at 0dBu) to the output and measure the actual value of the output, once with a 10K load and then with a 1K load. I have chosen these values because the considerably simplify the maths. Divide the 10K voltage by the 1K voltage level and call this ratio X.  If my math is correct, the output impedance (Rout) in kilohms is:

Rout = 10(X-1)/(10-X)

Cheers

Ian


Some data based on how I understood how to do it… It is probably just me  ::)

A = 0,046V p-p
B = 0,36V p-p

X factor would be 0.1278

Rout=10(X-1)/(10-X) = 10(0.1278-1)/(10-0.1278) = (10 x -0.8722)/9.8722 = 0.883

Which would mean that the output impedance is roughly 883R




According to this the output impedance would be roughly 883R

But if I use the method of measuring according to "radardoug" - using resistor across the output - sending a signal and increase the value of the resistor until the signal drops 6dB the measured impedance would be 270R.


This is confusing to me… please correct me here :)


Best regards

/John
 

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