Quote:

Originally Posted by

**robrob**

After playing around with my spreadsheet I know how Fulvio calculated his chart shown above. He used 0 ohms for amplifier impedance and an open circuit (infinite ohms) for headphone impedance.

I just want to verify the correct amp impedance to plug into the formula. If a tube amp has 4 and 8 ohm speaker terminals I would use 4 or 8 ohms for the speaker impedance in the formula correct?

Not necessarily. Depends on the actual output impedance of the amplifier. If they are matching impedance for the two different loads, then yes, you are correct. If they are using some other scheme, you'll either have to contact the amp manufacturer or measure it with an ohm meter. Of course, that will only give you the answer at DC and may or may not be close enough across the audio spectrum.

Quote:

Originally Posted by

**robrob**

What would be the correct equation to calculate attenuation incorporating the headphone impedance? (R2 + R3) || (R4 + Headphone Impedance) ?

I had to think about this in terms of the voltage divider, so I'll take you through it that way.

DefineV1 as the voltage across V1, V3 as the voltage across R3 and VH as the voltage across the headphones (RH).

so, V3 = V1 * R3 || (R4 + RH) / (R2 + R3 || (R4 + RH))

Then

VH = V3 * RH / (R4 + RH)

The voltage divider ratio V1/VH would then be the voltage attenuation factor.

You can expand all the equations, simplify and remove the voltage references, if you prefer to stay in the impedance realm.