Roger wrote:
Roy Lewallen wrote:
Roger wrote:

The power output of the Thévenin equivalent circuit follows the load.

Sorry, I don't understand this. Can you express it as an equation?


There seems to be some confusion as to the terms "Thévenin equivalent
circuit", "ideal voltage source", and how impedance follows these
sources.

Two sources we all have access to are these links:

Voltage source:
http://en.wikipedia.org/wiki/Voltage_source
Thévenin equivalent circuit:
http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem

As far as I can see, I've used the terms entirely consistently with the
definitions and descriptions in the linked articles. If you can find any
instance in which I haven't, please point it out so I can correct it.

Three important observations about the ideal voltage source:

1. The voltage is maintained, no matter what current flows through the
source. Presumably, this would also mean that the voltage would be
maintained if a negative current flowed through the source.

Actually, the wikipedia article correctly states this explicitly.

Thus, if we
set the voltage to 1 volt, the voltage would remain 1v even if we
supplied an infinite amount of power to the source.

You have to be very careful with using infinite amounts of anything
because the underlying mathematics runs into problems. For example, look
up "impulses" in any circuits text, and you'll find situations where a
pulse has zero width and infinite height yet finite area. So let's just
say the voltage stays at 1 volt if you apply any finite amount of current.

2. The idea voltage source can ADSORB an infinite amount of power while
maintaining voltage.

Yes, that's correct.

In this capacity, it is like a variable resistor,
capable of changing resistance to maintain a design voltage, no matter
what current is supplied to it.

No, that's not correct, unless you're willing to work with negative
resistances. V/I is a negative number when you supply current to a
voltage source. So it doesn't in any way act like a resistance, changing
or not.

3. The idea voltage source has an infinite impedance at the design
voltage, not a zero impedance as many have suggested.

No, that's not correct.

Zero internal
resistance is assumed to reasonably allow the ideal voltage source to
supply or adsorb current without changing voltage internally.

Yes, that's correct.

It is not
zero impedance with the result that voltage drops to zero if external
power FLOWS INTO the ideal source.

Sorry, I don't understand that statement. The voltage never changes,
regardless of the current flowing in or out. Review the wikipedia
article or any circuits text.

Current flows into the ideal voltage source when the applied voltage
exceeds the design voltage.

You can't "apply" a voltage to an ideal voltage source, except through
an impedance. If you did, for example by connecting it to another
voltage source of different value, the conditions defined for the
voltage sources are contradictory and can't exist at the same time. If
you do apply a voltage through an impedance, the current is simply the
voltage across the impedance divided by the impedance.

At the point the current reverses, we have
voltage/zero current, which is infinite impedance.

It sounds like you're trying to calculate some sort of "instantaneous
impedance". This isn't a generally accepted concept, and you'll have to
develop a fairly complete set of rules and mathematics to cover it.

Two things to notice about the Thévenin equivalent circuit:

1. It contains an ideal voltage source "IN SERIES" with a resistor.
This has important implications when externally supplied voltage exceeds
the design voltage.

Not really. The current simply equals the voltage across the resistor
divided by the resistance, in accordance with Ohm's law.

Any returning power would not only reverse the
current flow in the ideal voltage source, it would develop voltage
across the internal series resistor. The output Thevenin voltage would
be the design voltage plus the voltage developed in the resistor.

Your problem here, and I suspect elsewhere, is that you've embraced the
idea that there are waves of flowing and reflecting power. Your
otherwise reasoned questions and arguments are a good illustration into
the traps this concept leads a person into. You're seeing that in order
to support this mistaken concept, you have to reject some very
fundamental principles of electrical circuit operation. So you end up
with only two choices:

1. Re-write a great deal of fundamental circuit theory and reject the
foundation which has consistently given demonstrably correct results for
over a century, or
2. Realize that the flowing-power concept is flawed and abandon it.

The choice is yours.

2. The impedance of the Thevenin equivalent source would be infinite at
the design voltage because a voltage will existed from the ideal source
but current does not flow. This is true no matter what the design
impedance is for the Thevenin source.

You're trying to define the impedance of a source as the ratio of its
voltage to the current being drawn from it. That's not the impedance of
the source, it's the impedance of the load. The two are not the same.
The impedance of the source is its open circuit voltage divided by its
short circuit current. Another way to determine the impedance of the
source is to apply various loads and see how much the voltage drops. Any
test you run will confirm that the ideal voltage source has a zero
resistance, as my postings, the wikipedia article, or any circuits text
tell you.

Please notice in the link about the Thevenin circuit, a reference to a
"Thévenin-equivalent resistance".

This is the resistance component of the Thevenin equivalent circuit.
It's equal to the resistance of the circuitry for which the Thevenin
equivalent is being substituted.

This resistance uses the ideal
voltage source set to zero.

Can you explain this? There is no requirement that the source be set to
zero in determining or using the Thevenin equivalent resistance.

This appears to be the circuit that entered
the discussion at some point, justifying a negative 1 reflection factor.

My analysis did not contain a Thevenin equivalent to any circuit. It
contained only an ideal voltage source.

Therefore, when the load delivers power, the Thévenin equivalent
circuit adsorbs power. Right?

Certainly, any energy leaving the transmission line must enter the
circuitry to which it's connected. Is that what you mean?

Roy Lewallen, W7EL

If the rules found in the links are acceptable, I could agree that
energy be allowed to enter the connected circuitry. I think we will
find that the returning power from a reflected wave is either completely
reflected with no change in energy, or adds to the voltage, thus
increasing the current flow and power contained in the system.

You're building a logical argument from a flawed premise. I'm afraid
you've only just begun to see the problems you'll be having as you
pursue this path.

Roy Lewallen, W7EL