In article , Don Pearce
wrote:
On Thu, 17 Jul 2003 10:08:47 +0100, Jim Lesurf
wrote:
[big snip]
I can see that the amplifier having a finite output impedance and
current capability can make the output load-dependent. However I am not
sure that this is best described as the load being part of the loop.
Since I can't view the document you have produced, I don't know if this
is the point you are making.

I always seem to visualise things differently to others - and it has
always served me well. Try a GIF:

http://www.donald.pearce1.btinternet.co.uk/paloop.gif

I haven't bothered including any of the parasitic stuff, or even the
output impedance.

OK. I have now seen the diagram. In order to clarify my own view (which I
believe is the standard one) I can make one comment, then pose one
question.

The comment is that in most conventional audio power amplifiers the
feedback loop is designed to control the *voltage* at the load, and ensure
this is scaled version of the input signal voltage presented to the
amplifier. This is because in normal domestic analog audio systems the
signal pattern nominally required uses votage/time as the analog for
pressure/time.

The question is, for the circuits you show: What effect upon the output
voltage will it have if you change the load resistance by, say, a factor of
two?

In principle, the answer appears to be, "Changing the load as indicated
above, in your circuits, should have no effect at all upon the output
voltage." The loop will ensure that the same output voltage is asserted.
That is its purpose.

Thus on the basis of your circuit and assumptions, I would not say the load
is part of the feedback loop. It just connects to one point in the loop,
and is supplied by current as required in order to maintain the voltage the
loop exists to ensure is asserted.

In the absence of assuming a finite output impedance, etc, the loop is
therefore unaffected by the presence of the load as it is a *voltage*
control loop, not a current loop.


That is another matter of course, and in terms of the block diagram can
be very high - particularly with those amps that use the collectors of
the output transistors as the output point. It is the feedback loop
itself that makes the output impedance look small.

I would say that it does not "look small" but that it *becomes* small. As
can be verified by measurements upon the output impedance. We are concerned
with the behaviour of the *system*, not with how its individual components
might behave if used in isolation, or in a different design.


I have designated the second part of the amplifier as a power amplifier,
but what I really meant of course is that it is a current amplifier.

But used as part of a closed loop designed to control the *voltage* gain of
the system.

I'm happy enough with the idea that once you have a finite output impedance
or some other factor like current limiting then the load has an effect.
However I still don't agree that - with the example you give - that the
load is part of the feedback. It can affect the performance, yes, but that
is not quite the same thing.

Slainte,

Jim

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