Wattage

Sure. Why is an emitter resistor needed?

Maybe I don't - but then I'm no Electronics "expert".. OTOH, a quick Google search on "output impedance" reveals that a host of highly-respected technology companies including Tektronix and Belden don't either.

eg.

Cameron:-)

I guessed not.

Since you clip my responses (a lie by omission), perhaps you'd rather to argue with a wall. However, you're simply wrong.

Though just to humor Audiophools.

Output Impedance - 10 , 600 or user-defined.

Guess why there is that 10 *ohms* in there? Perhaps because that's the output impedance of most audio gear (and amplifiers are *far* below this)? Naw, you're too "smart" to believe such.

Perhaps you *think* you know something.

Are you talking about an emitter follower with emitter connection only to the speaker? If so, the speaker load is an essential part of the output circuit. In that case, the output impedance is essentially the impedance of the speaker - which is NOT nil.

So, what are you thinking of? What is the emitter circuit you envision?

If you insist, though I would add it's complement to get four- quadrant operation.

Nope. The output impedance of such an amplifier is nil and has "nothing" to do with the load. It's a voltage source.

Exactly as you describe, with the addition of the complement so we can do more than one quadrant supply.

You apparently have no understanding of what "output impedance" is. Hint: It's not the impedance seen at the output. It describes the characteristics of the output (not the load), and in this case is more-or-less the output transistor's base drive (also a low impedance) divided by the beta of the output transistor. ...a rather low number, I.e. nil.

That helps. An ideal voltage source would have 0 internal impedance.

Very likely - certainly less understanding than I would like to have. That's why I ask questions.

So if I'm understanding you right, output impedance is a model representation of how the output behaves, and is not related to circuit impedance which is a real, measurable thing. Do I have that right?

To cut through all the crap consider that there is a source driving a load. The output impedance of the source is the thevenin equivalent impedance of the source. Period! At least until you go nonlinear.

Bill

Sure.

Asking questions from ignorance is laudable. Making statements from ignorance isn't.

Only half right. The output impedance of a circuit is easily measurable. Take the audio amp you propose, draw another amp out of it, if the output voltage drops one volt, the output impedance is one ohm (scale as appropriate). Input and output impedances are "small signal" quantities, so one must be careful to apply them appropriately. They *are* real though.

It's even true when it's non-linear, though perhaps more complicated. ;-)

Assuming ignorance is a mistake. Both questions and statements can come from an effort to understand what the other guy is saying, and have nothing to do with ignorance.

So if I understand you, output impedance Zo = (Ve-Ve2)/(I2-I) with Ve measured with respect to ground. And, to get back to what started this off, Zo does not change with frequency between ~20 to ~20,000 Hz. Do I properly understand what you are saying?

On Thu, 13 Nov 2003 06:30:34 GMT, snipped-for-privacy@bellatlantic.net Gave us:

Keith likes to point inappropriate fingers of falsehood. He does it at every turn. Expect more.

You were clearly ignorant of what (output or any other) impedance means, yet jumped on my case when I called you on it. You didn't' ask, you attemptedd to tell. Sorry, but you were

It doesn't matter what you reference it to. Ground is simple a convenient place of reference. ;-)

No, I never said that (though it is likely true). I stated that the output impedance of most audio gear was nil, which you took exception to, with no understanding what "output impedance" means. You've admitted your (audio)phoolishness, so get along...

Likely not. A 600ohm audio circuit may not have any thing in there that is 600ohms (most certainly not the output). A modern (read as semiconductor) power amp designed to drive 8-ohm speakers certainly does not have an 8-ohm output impedance (likely as close to zero as you'll find). A 50ohm DV port (or whatever) doesn't have a 50 ohm output impedance. Things aren't that simple.

Words mean things. Unfortunately, autiophools think they know what they mean. Yes, that bothers me when these phools tell me that they have some special knowledge that I don't (as an engineer). Ok, they do, they know more about lying that any honest engineer.

You haven't a clue DimBulb! Good grief anyone not believing

You are mistaken. It was Cameron who jumped on your case, not I. I NEVER jumped on your case. If anything I have posted in this thread or any other is mistaken for me jumping on you, I apologize.

I'm sorry to have bothered you.

No, I'm wrong. I didn't notice the change. I do apologize for confusing you two.

No, it is I who is sorry for the confusion.