How to measure output impedance?

[watchnerd]

What's the best way to measure output impedance of a commercially made electronics device?

In this case, I want to measure the output impedance of my amp RCA jacks, but I might want to do it for my old headphone amp, too.

If I have a DMM, is that enough?

Or do I need more equipment?

And what's a good methodology?

 

[mansr]

You need a variable load. For line-level outputs and headphone amps, a resistance decade box will do. For power amps, something more heavy-duty is required. With the output impedance in series with the load, we observe that

Vout = Iout * (Rout + Rload)​

which can also be written as

Vout = Iout * Rout + Vload​

Vout is unknown but presumed constant. Measuring/calculating Iout and Vload for two different load settings, we get

Iout1 * Rout + Vload1 = Iout2 * Rout = Vload2​

Solving for Rout, we get

Rout = (Vload1 - Vload2) / (Iout1 - Iout2)​

For better results, measure at more than two load settings and do a least-squares line fit to obtain Rout.

 

[pma]

What's the best way to measure output impedance of a commercially made electronics device?

In this case, I want to measure the output impedance of my amp RCA jacks, but I might want to do it for my old headphone amp, too.

If I have a DMM, is that enough?

Or do I need more equipment?

And what's a good methodology?

This is the best way

1kHz

5kHz

The output impedance is nonlinear

 

[solderdude]

What's the best way to measure output impedance of a commercially made electronics device?

In this case, I want to measure the output impedance of my amp RCA jacks, but I might want to do it for my old headphone amp, too.

If I have a DMM, is that enough?

Or do I need more equipment?

And what's a good methodology?

An easy way is to use a multimeter that has at least 200mVAC or 2VAC settings.
You will also need a potmeter of say 1kOhm or a log type 10kOhm when trying to measure R out from tube circuits.
This requires no calculations but is not suited for all types of sources, only some high resistance line level sources.

1: Measure the output voltage without a load (400Hz sine for instance)

2: Now connect the potmeter with the wiper and the low fixed part in parallel to the source (it will be a variable load) and rotate the potmeter till you get exactly half of the output voltage of that without a load.

3: Remove the potmeter and measure the DC resistance (without accidentally adjusting it) and voila.. the reading you get is the output resistance.

Only suited for equipment with a high output R... never do this with speaker amps and low output R headphone outs. It won't give the correct results.

 

[pma]

Beware of usual simple and pointless numbers. They do not reveal nonlinear issues. One cannot take 2 same oversimplified numbers of 2 amps and state that the output impedance of both amps is the same. This makes just another useless comparison.

 

[mansr]

If the output impedance is non-linear, it results in harmonic distortion. Frequency-dependent impedance is still linear, though it will contribute to any deviation from a flat frequency response. In practice, other factors will dominate both of these effects.

 

[Jimbob54]

Sorry for reviving an old thread and please excuse my ignorance in advance

Is it possible to calculate or even estimate headphone amp output impedance using the measurements in amir's reviews? (when he doesn't measure R out in the review of course)

Specifically this amp https://www.audiosciencereview.com/forum/index.php?threads/ifi-zen-can-review-headphone-amp.25224/

Another member has had a go and is suggesting the balanced output impedance could be way more than the spec of 2ohms

Thanks in advance.

 

[amirm]

Yes. Look at at the last graph:

Pay attention to where I have circled. If the graphs all start at the same point, then the output impedance is essentially zero. Here it varies a tiny bit so it is a hair above but still negligible.

 

[Jimbob54]

Many th

Yes. Look at at the last graph:

View attachment 197335

Pay attention to where I have circled. If the graphs all start at the same point, then the output impedance is essentially zero. Here it varies a tiny bit so it is a hair above but still negligible.

Many thanks!

 

[solderdude]

Yes. Look at at the last graph:

View attachment 197335

Pay attention to where I have circled. If the graphs all start at the same point, then the output impedance is essentially zero. Here it varies a tiny bit so it is a hair above but still negligible.

I think this is a measurement error. The current limit is shown at the right side (the knee).
On the left side there is almost no current output as we are looking at 10mV (stimulus) and the actual measured voltage (the trace) is in the 50 millivolt range (= 0.05V). Even at 12 ohm the output current is 0.8mA at this level. Also note that the 12 ohm trace, unlike the 16, 20, 300 and 600 ohm trace do not have the weird 'bend' on the left side.

The right side cannot be used either as current limit and output resistance are not the same thing.
A simple opamp circuit (like a C'Moy) for instance can have an output R of 0.1 ohm (due to the feedback loop) yet have a current limit of 50mA.
This will clip the output voltage at a lower level (as seen in the quoted plot) at a lower voltage into a low impedance load as when less current is drawn.

So the only way is as described earlier by looking at the voltage drop between unloaded, or 2 different loads, when the amp is still within normal operation limits (so not reaching current limits).
Not reaching current limits is paramount and requires a scope to make sure this isn't happening. When this is not checked the calculated output resistance may not be correct.

 

[xnor]

I think this is a measurement error. The current limit is shown at the right side (the knee).
On the left side there is almost no current output as we are looking at 10mV (stimulus) and the actual measured voltage (the trace) is in the 30 microvolt range (= 0.00003V). Even at 12 ohm the output current is 0.8mA at this leve. Also note that the 12 ohm trace, unlike the 16, 20, 300 and 600 ohm trace do not have the weird 'bend' on the left side.

Isn't that completely irrelevant? And I don't see how current (on the left side) is relevant either.

The THD+N vs. V plot is not ideal for judging output impedance for these reasons:
1) the stimulus signal may be a sine wave at some fixed frequency, for example 1 kHz, but output impedance of the amp could be low at 1 kHz but rise quite a bit towards 20 kHz (we'd also see this in a low impedance load FR measurement, but that wasn't measured either)

2) the plot above does not give us individual measurement points at fixed voltage levels - that would allow for comparison of voltages between different loads - but plots lines

The second point means there are only two data points left that we can use: the start of the lines and the end of the lines.
The end of the lines are cut off, amps run into limits, etc. so we also cannot use that.

That only leaves the start of each line as the only point where we have a reference voltage and can make comparisons .
(And that is all you need to calculate output impedance. The output current is irrelevant. The bend is irrelevant as we're not comparing THD+N. Also I don't know how you came up with microvolts here. m = milli.)

Yes. Look at at the last graph:

[graph]

Pay attention to where I have circled. If the graphs all start at the same point, then the output impedance is essentially zero. Here it varies a tiny bit so it is a hair above but still negligible.

Does not seem negligible:

Black = 12 ohm
Blue = 600 ohm

Vertical grid lines range from 7 .. 8 mV

So about 8 mV into the 600 ohm load and about 7.5 mV into the 12 ohm load (this is based on assuming subdivisions are the same as in the rest of the graph, if subdivisions were linear instead then we'd be closer to 7.7 mV).

If we solve the system of those two voltage divider equations we get Zout = ~0.8 ohm.

Disclaimer: this is only a rough estimate since the resolution of the graph is poor and the exact voltages are unknown

 

[Ken Tajalli]

Hi, I was the one who did a quick mental arithmetic calculation that Jimbob54 is talking about.
The issue was that a certain amp could pump out 15.1V into 600R load, but it dropped to 11V with a 64R load.
I did a quick mental math and came up with 23R for output impedance. But I did mention that, that figure includes PSU current limitation, so that had cut into it.
He was trying to check my method.
BTW , I too think that curve I'd irrelevant .

 

[freemansteve]

Are you keen to measure output resistance or output impedance?
Impedance is usually loosely used to mean resistance for amp outputs, but as everyone knows, one is scalar and one is vector.

 

[xnor]

@KenTajalli You cannot compare the values like that as I've explained. You'd need to compare the voltages before you hit any limiters or clipping points which is hard to do in a line chart because you don't see individual data points and therefore don't know how far each point is shifted along the x-axis, so to speak.
The only point of reference is that start of each line.
It could only be worse if the measurement device automatically compensated for output impedance, because then the lines would always start at the same voltage.

The clipping point could only be used if it was known that the amp clips at some fixed voltage independent of the load, which typically is not the case.
If you don't know that then the numbers are meaningless. It's effectively modeling the amp's complex load-driving behavior with a single impedance, not a calculation of the actual output impedance.

 

[solderdude]

Hi, I was the one who did a quick mental arithmetic calculation that Jimbob54 is talking about.
The issue was that a certain amp could pump out 15.1V into 600R load, but it dropped to 11V with a 64R load.
I did a quick mental math and came up with 23R for output impedance. But I did mention that, that figure includes PSU current limitation, so that had cut into it.
He was trying to check my method.
BTW , I too think that curve I'd irrelevant .

The right side can not be used to calculate output resistance. It can only be used to see at which output voltage level clipping occurs.
In the case of this amp it is current limited clipping.

At 12 ohm we see distortion starting to rise at 2.1V = 175mA
At 16 ohm we see distortion starting to rise at 2.8V = 175mA
At 32 ohm we see distortion starting to rise at 6V = 180mA (but folds back possibly heat related as the level sweep continues)
At 50 ohm we see distortion starting to rise at 9.5V = 190mA (and folds back a little as less heat is dissipated in the output devices and less current is drawn).
So current limit of the amp is about 175mA and the reason the output voltage clips (and shoots up the distortion).

The 300 and 600 ohm load is too little to reach current limiting so the voltage rail is the reason for clipping here. This is about 16V, but not distorted up to aboue 13V.

Undistorted max. output power can be calculated from this plot as well.
13Vx190mA = 2.47W into 13/190mA = 68 ohm.


This is not an indicator for output resistance as that is determined by the feedback and open-loop gain at any given frequency as well as ohmic losses between the feedback point and HP output socket.

On the left side we see noise + signal (mostly noise) and this is not high-res enough to say something accurate.
The 16 and 20 ohm traces on the far left deviating from the other lines is measurement error. Not related to output resistance.

XNOR has a point about the few mV on the left between 12 and 300 ohm but we are looking at mostly noise, not a reliable indicator as it can vary over time.

These plots can not tell you anything definitive/accurate about output R.
They can tell you about output power, distortion (+ noise) current and voltage limits at 1kHz.

And I don't see how current (on the left side) is relevant either.

It isn't. It is about 0.8mA in 12ohm and much less at higher frequencies (sorry.. impedances/resistances) which was my point.

 

[ahofer]

I have no electrical engineering training (which may already be clear).

This conversation makes me despair, though, of performing blind listening tests that properly control for specs. If there are amps out there with wildly variable output impedance under special conditions, then they will, indeed, sound quite different under those specific conditions (ie right music and level). How would you control for that?

 

[solderdude]

The only amps with 'wildly' varying output R might be tube amps with little to no overall feedback and small amps with under-dimensioned output capacitance. Also some amps with very little open loop gain and overall feedback will rise in output R but not vary wildly like speaker impedances do.

For headphone amps the variance over the entire audible range usually is very little and not consequential in most cases. With speaker amps and loads like electrostats it is another matter.

 

[Ken Tajalli]

The right side can not be used to calculate output resistance. It can only be used to see at which output voltage level clipping occurs.
In the case of this amp it is current limited clipping.

At 12 ohm we see distortion starting to rise at 2.1V = 175mA
At 16 ohm we see distortion starting to rise at 2.8V = 175mA
At 32 ohm we see distortion starting to rise at 6V = 180mA (but folds back possibly heat related as the level sweep continues)
At 50 ohm we see distortion starting to rise at 9.5V = 190mA (and folds back a little as less heat is dissipated in the output devices and less current is drawn).
So current limit of the amp is about 175mA and the reason the output voltage clips (and shoots up the distortion).

The 300 and 600 ohm load is too little to reach current limiting so the voltage rail is the reason for clipping here. This is about 16V, but not distorted up to aboue 13V.

Undistorted max. output power can be calculated from this plot as well.
13Vx190mA = 2.47W into 13/190mA = 68 ohm.


This is not an indicator for output resistance as that is determined by the feedback and open-loop gain at any given frequency as well as ohmic losses between the feedback point and HP output socket.

On the left side we see noise + signal (mostly noise) and this is not high-res enough to say something accurate.
The 16 and 20 ohm traces on the far left deviating from the other lines is measurement error. Not related to output resistance.

XNOR has a point about the few mV on the left between 12 and 300 ohm but we are looking at mostly noise, not a reliable indicator as it can vary over time.

These plots can not tell you anything definitive/accurate about output R.
They can tell you about output power, distortion (+ noise) current and voltage limits at 1kHz.



It isn't. It is about 0.8mA in 12ohm and much less at higher frequencies which was my point.

As I said, it was a quick calculation from top of my head, assuming certain things to get a rough idea of what output impedance appears to be. Jimbob54 asked to see my calculation, so I jot it down on a sticky.
To save myself, I did say to him that the figure is rough, and because of current limiting, this is what it appears to be at.
Remember, Rough, Mental maths, what it appears to be!
BTW , I used advertised spec. figures.
take it with a fistfull of salt.

 

[xnor]

On the left side we see noise + signal (mostly noise) and this is not high-res enough to say something accurate.
The 16 and 20 ohm traces on the far left deviating from the other lines is measurement error. Not related to output resistance.

XNOR has a point about the few mV on the left between 12 and 300 ohm but we are looking at mostly noise, not a reliable indicator as it can vary over time.

Sorry to be so frank but you still don't seem to understand. You're still looking at the y-axis, which is irrelevant. It's as irrelevant as the comments about current.
We're not looking at noise at all.
We're only looking at the x-axis here, since we're only interested in the stimulus signal level across the load.

Let's make this as simple as possible: If the load is 10 ohms and the output impedance is 10 ohms then a 1V stimulus signal will drop to 0.5V across the load. So if the measurement device started its measurements at 1V then the resulting line would start at 0.5V. Ok? This is pretty basic stuff.

The distortion products will be relative to that, but again, they are irrelevant here. Y-axis = irrelevant. So we're not looking at noise and I also don't see anything varying over time here.
The main issue that prevents a more accurate calculation of the output impedance is the poor resolution of the graph.

You're a smart guy but if this is already a challenge (I honestly don't understand why it would be) then it's a safe bet that 99.9% of ASR readers don't have the slightest clue what they're looking at when they look at any of the review graphs.

It isn't. It is about 0.8mA in 12ohm and much less at higher frequencies which was my point.

I don't get it. If it's irrelevant then what was the point? And why say your point was about higher frequencies when you introduced that into your point just now?

 

[solderdude]

'frequencies' should have been 'impedances'

Let's make this as simple as possible: If the load is 10 ohms and the output impedance is 10 ohms then a 1V stimulus signal will drop to 0.5V across the load. So if the measurement device started its measurements at 1V then the resulting line would start at 0.5V. Ok? This is pretty basic stuff.

Yes, see post #4.

You're still looking at the y-axis, which is irrelevant

Why is the y-axis irrelevant when explaining that current limiting and output R have no relation as soon as current limiting occurs

We're only looking at the x-axis here, since we're only interested in the stimulus signal level across the load.

You are only looking at the X-axis when it comes to determining the output R. And that is correct you only need to look at the starting points from the traces (on X axis).

I was explaining that the amp is current limited in order to explain why the simple math from Kenji can't be used for determining the output R.
A completely different thing.

Hoping you understand we are explaining different things here.

Disclaimer: this is only a rough estimate since the resolution of the graph is poor and the exact voltages are unknown

Indeed.
Using that graph to get an accurate value is like measuring sub millimeters using a ruler with rather thick mm markings.
Better to use actual measured values instead of the plot.
The plot does give an indication but the circle Amir drew seemed to include the 2 traces that deviated and that could be confusing as it suggests that those lines were indicators. They aren't they are measurement errors.
It would have been better to draw the circle around the starting points of the black and blue trace that you handily enlarged and put 2 vertical stripes in it.