OOPS!

Richard,

Now I understand the source of your "lossless resistance" remark in the
class C thread. I respectfully disagree. There is nothing of the sort. I
will admit, however that in a Switched case, the concept of resistance *may*
be thought of this way, but it is not a good way to look at it. I believe
you are taking two things which are not similar and calling them the same.
I will point out the error below. It can be verified by experiment rather
simply if you wish.



"Richard Harrison" wrote in message
...
Old Ed wrote:
"Couild you explain this concept, and/or provide some references?"

...d-c supply to full-scale indication on an external meter.

Therefore we have the situation whe
V out = Ifs x Rm
V out set to the IR drop of the meter at fullscale. All ok so far.


Next, install a chopper (lossless on-off circuit) ... with a 50% duty
cycle...
... The meter reads 50% of full scale.

Still ok so far.


Another way to reduce the scale reading to 50% is to insert a resistor

Yep, still ok.



The resistor which has the same effect ...

Start of the error. I disagree with "same effect". One is a steady
state current (resistor), the other a square wave of current (chopper),
right? The time average current IS the same. ...HOWEVER...



and produces the same scale
indication as the chopper on the effective output current exacts its
loss of 0.0005 amp x 0.5 volt or 0.25 milliwatt during 100% of the time.

The word "effective" here is not standard as it is the same word we
use to denote "producing the same power". Again, it certainly is that the
AVERAGE current is obviously the same for both cases.



The chopper eliminates the power-losing resistor by substituting
off-time in the power source. The source only supplies the power used by
the load. With a resistor limiting power to the load, the source
supplies its loss and the load power.

This is indeed correct. The chopper DOES change produce the same
AVERAGE current as the resistor. The added resistor looses some power, the
chopper does not. Now the zinger.


The power in the load, a meter in our example, is the same using either
the resistor or the chopper.

Here is where the error is. The POWER in the load (meter) in the
chopper case is not the same as it is in the resistor case! It is...Hold on
to your hat... TWICE as much as in the resistor case!

In the meter only case load power is 1 miliwatt.
In the resistor case the load power is 0.25 miliwatt (half voltage x half
current = 25%)
Or P=I^2 * R which is also 0.25 miliwatt.
In the chopper case the load power is 0.5 Miliwatt!

In the chopper case, the RMS current is = Imax x square-root(duty-cycle).
This is 1ma * 0.707 or 0.5 miliwatt.

See
http://www.irf.com/technical-info/an949/append.htm
For verification of the square root of the duty cycle formula.

You can easily verify this since a power difference should be easily
observed by finger-and -clock method. Set up the experiment with a resistor
and power supply which gives some power which can be felt as a temperature
rise within, say 10 seconds. I'm guessing somewhere around 0.2 to 1.0 watt.
This is with the two resistors in the circuit. Then subsitute the chopper
(a relay). Twice the power should be evident in a faster rise in
temperature. It should heat up in half the time. I maintain that if it
doesn't, then you did it incorrectly.



The off-time has the same effective opposition to current to a load as a
dissipative resistance. As the time-limited currented opposition to load
current consumes no power, it is called a dissipationless resistance.

This is what I believe would be called a self conflicting term.
Resistance removes power from the circuit in question in all cases, period.
And YES on the proverbial infinite T-line, the power is gone from the
source, never to return. It looks just like a resistor (Steve says with
great confidence, knowing full well that others may disagree, but
nonetheless firm in his beliefe) It *is* LOST.

Perhaps a concept called "Average resistance" might be a subject to discuss.
In matching a class C power amplifier, there has been myriad of words spent
on just what, if anything, is being matched by the output matching network.
Is there an "output" impedance there or not and if so, what is
it....(rhetorical, of course)

Sorry Richard, but I believe you have it wrong this time. Steady state and
pulsed must be understood with slightly different principles.
--
73, Steve N, K,9;d, c. i My email has no u's.