General Category > Technical topics

Understanding output impedance

<< < (2/4) > >>

Paul Birkeland:
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.

Loquah:

--- Quote from: Paul Birkeland 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.

--- End quote ---

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!!

Loquah:
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)

Doc B.:
Could you post the voltage figures you used in the equation?

Paul Joppa:
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.

Navigation

[0] Message Index

[#] Next page

[*] Previous page