Understanding output impedance

Loquah · 4412

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Offline Loquah

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on: April 25, 2021, 08:09:16 PM
I have an amp that normally uses resistors to tame the output in much the same way as (I think) the early SEX did. The result is a Zout of 230 ohm or so. I've created a mod with an L-Pad off the speaker taps so I now have two outputs with varying Zout.

I'm doing some testing for an upcoming review and have discovered that the high Zout of the stock output is not altering the frequency response of most of my low impedance cans, but is affecting all of the moderate impedance (80-120 ohm) ones.

Does the 1/8 rule work in both directions (i.e. is it about the distance between Zout and transducer impedance?) or is it more likely just to do with the nature of the drivers and their own impedance variations by frequency?

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Offline Paul Birkeland

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Reply #1 on: April 26, 2021, 05:31:58 AM
Are you looking closely at the low frequency response of the headphones?  That is where you are most likely to see lumpiness. 

Paul "PB" Birkeland

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Offline Loquah

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Reply #2 on: April 26, 2021, 03:50:31 PM
Are you looking closely at the low frequency response of the headphones?  That is where you are most likely to see lumpiness.

Hi Paul, yes I was expecting significant deviations in bass which I saw in the 80 and 120 ohm headphones, but not in 2-3 32 ohm headphones!?

For example, check out the 80 ohm Utopia measurement with the expected bass changes and then the 62 ohm AKG K712 with basically no change

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Offline Paul Birkeland

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Reply #3 on: April 26, 2021, 04:13:22 PM
If the impedance curve of the lower impedance headphone is super flat, you wouldn't necessarily see a significant change.

Paul "PB" Birkeland

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Offline Loquah

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Reply #4 on: April 26, 2021, 06:04:09 PM
If the impedance curve of the lower impedance headphone is super flat, you wouldn't necessarily see a significant change.

I wondered if that might be the issue. So, to confirm, the 1/8 rule does not work in reverse (i.e. output higher than transducer impedance) and damping would still be affected even if frequency response is not - right? The sound is a bit smoother from the high output socket, but it also has different (and more) components in the signal path.

By the way, I know this isn't why you guys do what you do, but I'll be thanking you publicly in my upcoming review of this amp for the help you've provided me in understanding the technical aspects of some amp design choices.

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Offline Paul Birkeland

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Reply #5 on: April 26, 2021, 06:18:22 PM
You can do the calculations if you like.  Suppose you have 1V on the amp side of the series resistor going to the headphones.  Find the highest impedance in the impedance curve and the lowest impedance in the curve, then you can fill those in as the bottom leg in the voltage divider and calculate the voltage appearing at the headphones, then take 20*log(Vmax/Vmin) to find how much deviation you should have.

The amp you were looking at I think had a voltage divider at the output also, so it may be helpful to start there and work all the way to the headphone driver.


Paul "PB" Birkeland

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Offline Loquah

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Reply #6 on: April 26, 2021, 06:43:54 PM
You can do the calculations if you like.  Suppose you have 1V on the amp side of the series resistor going to the headphones.  Find the highest impedance in the impedance curve and the lowest impedance in the curve, then you can fill those in as the bottom leg in the voltage divider and calculate the voltage appearing at the headphones, then take 20*log(Vmax/Vmin) to find how much deviation you should have.

The amp you were looking at I think had a voltage divider at the output also, so it may be helpful to start there and work all the way to the headphone driver.

Thanks Paul. You're correct about the amp having both a permanent resistor and also a variable voltage divider. I've applied a really nifty mod (thanks to the input from you and PJ) that bypasses the whole convoluted output circuit and it sounds much more transparent to my ears so I'm just filling in some knowledge gaps now. I was taking measurements to show why the high output impedance was a problem and was surprised at what I saw on a couple of the low impedance cans like the Meze 99 Neo and Ollo S4X & S4R. At first I thought Ollo had some clever impedance matching tech built-in to account for different mixing desks given their focus as a mastering headphone, but then the 99 Neos and K712 kind of ruined that theory.

At this point I think I'm comfortable just discussing the inconsistent results people will get from high output impedances with a couple of measurements and leave it at that. Thank you very much for your input though!!

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Offline Loquah

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Reply #7 on: May 11, 2021, 08:22:48 PM
I ended up buying a portable oscilloscope like the one Dan suggested and I'm now using it to measure a completely different product (Matrix Mini-i Pro 3) that has published output impedance of <11 ohm (single-ended). I must be doing something wrong though because I keep calculating a Zout of about 1 ohm.

Can anyone help me work out my error? Here are my steps:

  • Connect oscilloscope to one channel of the headphone output
  • Play 1kHz tone through device
  • Record RMS voltage from oscilloscope
  • Apply known load (6.8 ohm is what I had at hand)
  • Record new RMS voltage
  • Calculate Zout as (V-VL)/(VL/R)

What am I doing wrong? Is the resistor too low? Should it not be the RMS figures for the calculation?

Here's a photo of the oscilloscope readout in case I'm doing something totally wrong. (By the way, it's set to AC mode, but is showing DC on the screen when connected - not sure what that's about as it shows AC mode correctly when disconnected)

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Offline Doc B.

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Reply #8 on: May 12, 2021, 05:01:55 AM
Could you post the voltage figures you used in the equation?

Dan "Doc B." Schmalle
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Offline Paul Joppa

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Reply #9 on: May 12, 2021, 06:29:21 AM
The scope image is not a sine wave, so the amp is not working correctly with the load and the rms value is irrelevant. That's the reason for using a scope rather than a meter - to confirm that the signal is in fact a sine wave and the amp is working within its limitations.

Paul Joppa


Offline Doc B.

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Reply #10 on: May 12, 2021, 07:31:09 AM
Duh, I didn't blow up the image and see the data. Yup, either the wave form your inputting is not really a sine wave or you need to move the trace up so you can see the entire sine wave.

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Offline Loquah

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Reply #11 on: May 12, 2021, 12:37:27 PM
The scope image is not a sine wave, so the amp is not working correctly with the load and the rms value is irrelevant. That's the reason for using a scope rather than a meter - to confirm that the signal is in fact a sine wave and the amp is working within its limitations.

Thanks for confirming that. I wasn't sure if these little scopes showed you only half of the wave to save on screen space. It must be something to do with it showing me "DC" as the mode when connected, even though it's definitely switched to AC mode and shows "AC" when disconnected.

I'll do some trouble shooting with the scope. Thanks for the help!

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Offline Doc B.

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Reply #12 on: May 12, 2021, 01:15:51 PM
If the settings are available you want to shrink the vertical scale (i.e. use bigger volts/division) so the full wave fits on the screen. Also move the zero crossing point of the waveform (half the wave above, half the wave below) up to the center of the screen. On a "real" scope there is a position control for this.

Dan "Doc B." Schmalle
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Bottlehead Corp.


Offline Loquah

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Reply #13 on: May 12, 2021, 03:57:51 PM
If the settings are available you want to shrink the vertical scale (i.e. use bigger volts/division) so the full wave fits on the screen. Also move the zero crossing point of the waveform (half the wave above, half the wave below) up to the center of the screen. On a "real" scope there is a position control for this.

Thanks Dan - I'm trying to work out how to do this with the very limited instructions. It's concerning to me that it seems to be not holding "AC" mode so I don't know if that's part of the issue too. I'll keep trying...

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Offline Doc B.

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Reply #14 on: May 12, 2021, 05:03:04 PM
If the AC/DC mode switching is automatic it may be that the waveform is not being recognized as AC since it doesn't look like a sine wave. It looks more like a series of pulses. Try to shrink the vertical swing by lowering the signal or by changing the vertical scale to a bigger volts per division setting and see if the mode stabilizes when you have the full sine wave showing on the screen.

Dan "Doc B." Schmalle
President For Life
Bottlehead Corp.