How to calculate effective output impedance of something like this?

[Joe Bloggs]

From here http://www4.head-fi.org/forums/showt...threadid=10548

It's not 20ohm, is it? The 2ohm in parallel affects the calculations, right? But how?

Say I wanted to design an adapter with effective output impedance of 80ohm for my etys ER4P, what values should I use?

 

[kevin gilmore]

It is slightly less than 2 ohms.
(actually 2 ohms in parallel with 20 ohms)

You would be better off leaving this alone and adding
80 ohm series resistors on the way to the headphones.

You could also bump the 2 ohms to 80 ohms, and the 20
ohm to 800 ohms, but some power amps really like to drive
lower impedance loads, so you would still have to add a
20 or so ohm resistor across the power amp.

 

[Joe Bloggs]

How does it come out to be equivalent to the 2 resistors in parallel?

Well, if you say so... (an understandable explanation would be welcome though

)

Quote:

but some power amps really like to drive lower impedance loads,

Oh, how's that? Would they distort with a high impedance load? I'm demanding less current, and I'm not demanding more voltage swing...

 

[braillediver]

To calculate Parallel Resistance

formula RT = 1 / ((1 / R1) + (1 / R2) + (1 / R3))

http://www.csgnetwork.com/parallelresistcalc.html

 

[Joe Bloggs]

Yeah, I know how to calculate parallel resistance--my question is, why does the circuit look like the two resistors are in parallel in front for the headphones? It's really just the 2ohm in parallel with the headphone and then the 22ohm in series in front...

Oh, and today I tried powering the Senn HD580 (300ohms) directly out of the satellite output of the subwoofer. I just used an RCA male to mini female plug. When I both channels all the way in, only the left channel makes sound. The funny thing is, when I pull the left channel plug halfway out, the right channel would start making sound about as much as the left channel; and also the supplied satellites are working just fine plugged all the way in.

Has this got something to do with what Kevin Gilmore said about power amps wanting low impedance loads? (I'm honoured to have Kevin Gilmore reply to my thread!

)

 

[tailspn]

The 2 Ohm and 20 Ohm resisters are effectivly in parallel because the amplifier's source impedence is effectivly 0 Ohms at audio frequencies. Therefore the earphones see the two resistors in parallel with one another as a source impedence.

Hope this helps...

 

[Joe Bloggs]

That doesn't really answer my question... sorry

Or should I say I don't understand your answer

 

[AssafL]

I think tailspn is referring to that some amplifiers are voltage devices (and therefore have low output impedance). Examples are devices used to power electrostatic drivers.
Regular power amplifiers for magnetic voice coil systems are generally trans-addmittance stages (i.e. Voltage to Current devices) and therefore have a relatively high output impedance.

Are you interested in the impedance of the device plugged into the power amplifier (about 22 Ohms) or the impedance of the amplifiers as the headphone plug sees it (slightly less than 2Ohms - i.e. a trans impedance stage)?

 

[Rpell]

Quote:

Regular power amplifiers for magnetic voice coil systems are generally trans-addmittance stages (i.e. Voltage to Current devices) and therefore have a relatively high output impedance.

Actually, most power amplifiers are designed to act as voltage sources (not current sources), and as such usually have a very low output impedance. (While the input stage of a typical power amplifier is a voltage-to-current transconductance (or transadmittance) amplifier, it is then followed by a current-to-voltage transimpedance stage, and usually after that by a voltage-follower stage.)
Quote:

That doesn't really answer my question...sorry

Because the output impedance of a power amplifier can (usually) be assumed to be zero ohms, it effectively appears as a short circuit to ground from the load's perspective. Thus, looking back into the amplifier's output through the above circuit, the headphone "sees" a 2-ohm resistor to ground in parallel with a 20-ohm resistor to ground.

 

[Joe Bloggs]

*tosses hands up in despair*

I don't know how to phrase my problem... look, it just looks like any electric current HAS to travel through the 20ohm resistor before reaching the headphones. A lot of current passes through the 2ohm resistor instead, but that current would never pass through the headphones anyway, so I don't see how that affects the impedance that the headphone 'sees'?

 

[tailspn]

Here is a link to an explanation of Thevenin's Voltage Equivilent theorem:

http://hyperphysics.phy-astr.gsu.edu.../thevenin.html

Basically it says that ANY circuit can be replaced by an equivilent source voltage and internal series resistor. In the case of the amplifier that those banana plugs plug into, the equivilent resistance is, for all practical purposes, zero Ohms. So, if you look at your diagram at the begining of this thread, and draw a wire across the two bananna plug prongs, for each channel, the 2 and 20 Ohm resistors become in parallel. That is the source impedence your headphones see.
Kevin Gilmore's response then answers your origional question.

 

[Joe Bloggs]

I still don't understand at all X_X

Although I've figured out that if you use this method of measuring output impedance:

1. Measure open-circuit voltage across headphone jack, with speaker amp playing a tone

2. Measure voltage across headphone jack with a resistor of x ohms connected across the gap

3. Calculate impedance using some formula

You'll get the result of output impedance < 2ohms.

But does this circuit yield the equivalent of an output with impedance < 2 ohms in terms of damping factor? I'm especially worried that there seems to be an easy path for back-EMF to travel, in a loop around the 2 ohm resistor?

 

[Rpell]

Quote:

But does this circuit yield the equivalent of an output with impedance < 2 ohms in terms of damping factor?

Yes, by definition.
Quote:

I'm especially worried that there seems to be an easy path for back-EMF to travel, in a loop around the 2 ohm resistor?

The "easy path" provided by the 2-ohm resistor is largely responsible for ensuring that the source impedance in the circuit is low relative to the load impedance. The lower the source impedance--i.e., the easier the path--the less likely that "back EMF" will have any significant effect on the audio signal.