"Ron Johnson" wrote in message
...
Dave Plowman (News) wrote:
In article ,
Arny Krueger wrote:
"Eeyore" wrote in
message

"Dave Plowman (News)" wrote:

Sigh. Find a basic DC coupled design and drive it *very*
hard - we're talking severe overload here - and look at
the output. It will approach the power rails.

That's what saturated output devices do.

Indeed.

It'll be virtually a square wave with almost twice the
sinewave power of the amp but not 'DC' though.

As far as the speaker coil is concerned it might as well just be. And if
you measure the signal with an analogue meter DC is what you'll see.

Right, there will be a sort of sloppy square wave whose spectral
content
will depend on the waveform driving the amplifier.

Agreed.

The peak amplitude will be the rail voltages (which will have sagged if
there is a significant load), minus the saturation voltage of the
output
devices. If you look closely, there may be some funny stuff where the
output stages snap out of saturation.

This square wave will of course have more energy in it than a sine wave
with a similar peak amplitude.

Of course you could argue about what practical DC is. Most will have
some
form of ripple etc.


I`m getting a terrible sense of DejaVu about this thread



Yeah, it's been hashed a dozen times, clipping ain't DC, and people with
high levels of knowledge don't say it is.

Just salesmen and and fader jocks. Not real engineers.

In practice, the severely clipped amp won't do a perfect square wave, it
will sag. so the "virtual DC" argument falls flat on it's face...most supply
rails do not have enough storage to sustain a full square wave...some of the
real heavy old iron may.

But I don't talk about old amps to make a point, nobody uses them any more.