EPDR

[cjlan01]

How exactly is the EPDR calculated as a function of nominal impedance and phase angle as reported in Stereophile Measurements section? Is it amp class type dependent? Do you have any reference to the actual derivations? Thx

 

[Galu]

I read in the link below that the EPDR derivation is based on a class-B output stage, so it's valid for typical class-AB amps but certainly not for class-A.

https://www.audiosciencereview.com/...ereophile-has-started-calculating-epdr.15497/

Post #19 has a PDF attached. Apparently, you can skip everything and jump to the last equation showing the formula for calculating EPDR from impedance magnitude and impedance phase magnitude.

I hope the link covers your requirements.

 

[cjlan01]

Thank you, Galu. I actually found that information several days ago and cross-checked it with Stereophile. Several figures I checked were quite close, while a handful still showed some discrepancies. I wonder if Stereophile has derived or used different formulations for other types of amplifiers.

 

[Galu]

I suspected you would have already found that information.

I tell you this in order to give your thread a second bump!

 

[cjlan01]

The essence of the remaining responses is about counteracting EPDR and deriving EADR (peak vs. average power). I am more concerned with peak power and want to ensure sufficient headroom when selecting amplifiers. Anyway, I might email Stereophile to see if they would consider releasing such information. Thanks again.

 

[cjlan01]

So I emailed Mr. John Atkinson regarding what exactly the fomula they have used in Stereophile for reporting the EPDR. He kindly repsond promptly as follows:

I use a formula created by editor Jim Austin. after importing the Impedance magnitude and phase data into an Excel spreadsheet.

Data = R (magnitude) ohms & R (phase) degrees

Calculate Vdiss = 1+4.2 x ABS(R Phase)/90

Then EPDR = R(Magnitude)/Vdiss

For example:

If R = 4.0 ohms and Phase = +45 degrees

VDiss= 3.1 and EPDR = 1.2903 ohms

Here Vdiss represents voltage dissipation. As a further analysis, this linear equation for Vdiss provides two exact boundary values: one at 0 degrees (which is obvious) and the other at 90 (-90) degrees, as derived by Mr. Jack Oclee-Brown, shown in the attached file. However, the linear equation overestimates Vdiss values in between, which in turn results in even lower reported effective impedance.

One possible remedy is to fit the data range with an exponential term, as illustrated in the attached figure. Instead of using 1 + 4.2 * abs(phase angle)/90, nonlinear fitting yields 1 + 4.196 * (abs(phase angle)/90)¹.¹⁷⁸, which provides a much closer approximation to Jack’s formulation.