#### Chris J

##### Headphoneus Supremus

Quote:

Ok. You'd said "Since you have added a damping resistoryou have done two things..." I thought you meant it was the 220 ohm resistor across the headphones. Sorry.across the headphone

Ok. So if he's got a 27 ohm resistor in parallel with the headphones, that means it's also across the amplifier's output. So the effective output impedance of the amplifier, i.e. the source impedance seen by the headphones, will be much lower still.

If we consider the headphone to be the load, then the headphone is effectively seeing an amplifier with an output impedance of about 25 ohms, i.e. 100 + 220 in parallel with 27.

se

No problem.

If he has the 27 ohm resistor in parallel with the headphones, then the headphones appear to be a 24 ohm load.

For this parallel combination, obviously the 27 ohm resistor will pull more output current than the 250 ohm headphone.

The headphone amp appears to have a 320 ohm output impedance because the 220 ohm resistor is in series with the headphone amp's original output impedance of 100 ohms.

So we have increased the output impedance.

STV014's experiment added a 27 ohm damping resistor across the headphone, he used this damping to counter the effects of the high output impedance (assuming the headphone you are using needs damping!).

Another way to look at this is: assume the headphone amp is an ideal voltage source with ALL the other impedances outside the headphone amp.

The IDEAL voltage source (i.e.headphone amp in this discussion) has Zero output impedance.

Then the load on the IDEAL headphone amp appears to be 100 + 220 + 24 = 344 ohms. So we may now calculate how much TOTAL current is drawn from the amp.

With a little bit of calculation (using Voltage divider principle) and power calcs we can also see that the efficiency of this whole network is very poor, i.e very little output voltage is actually applied across the headphone. And most of the output current is flowing thru a 27 ohm resistor!

I can very easily see how this confuses everyone!

C