On May 11, 7:15*am, Wimpie wrote:
On 10 mayo, 23:37, walt wrote:

[deleted]



Now Wim, your math is very challenging: You state 100w delivered by
the source, but at one point you also state 111w is delivered to the
100-load--at another point you state that 100w is delivered to the 100-
ohm load. Which is it?

You also state that voltage out of the source is 212.1v--sum ting wong
here. 212.1v across 50 ohms yields 899.73w, and 212.1v across 100 ohm
yields 449.86w. These power values are nowhere near the values
appearing in you statements.

In my calculations, with 100w the voltage across a 50-ohm line is
70.71v, and across a 100-ohm line the voltage is 100v.

With a 50-ohm line terminated in 100 ohms, the vrc is 0.3333 as you
stated, thus the power-reflection coefficient is 0.1111. This means,
as I've stated continually, that with 100w delivered by the source,
the power reflected at the 2:1 mismatch is 11.111w, which when added
to the 100w supplied by the source, makes the forward power 111.11w.
Now with 11.111w of power reflected at the mismatch, this leaves 100w
delivered to the 100-ohm load, not 111w. Please tell me where the 111w
came from. You also haven't told us why you calculate 130w forward and
15w reflected. Someone else may have made these calculations, but it
was you I asked for an explanation, because you repeated those
calculations.

When using the correct physics and math in this example, how can the
results *be so different?

To close, let me present the procedure I use to calculate the total
forward power--I guess it is related to Ohm's Law:

With the power-reflection coefficient as prc, the forward power PF = 1/
(1 - prc).

So Wim, can you clarify the confusion you appear to have made?

Walt

Hello Walt,

The source (that is the PA in this case) produces an EMF (as mentioned
before) of 212.1Vrms, Source impedance is 100 Ohms (so forget the
"conjugated match" thing for this case).

When terminated according to the numbers above the socket ("100W into
50 Ohms"), this results in 100W *( Vout = 212.1*50/(50+100) = 70.7V,
just a voltage divider consisting of 100 Ohms and 50 Ohms ).

70.7V into 50 Ohms makes 100W.

Now we remove the 50 Ohms load and create a mismatch (referenced to 50
Ohms), by connecting a 100 Ohms load (VSWR = 100/50 = 2)

Now the output voltage will be:

Vout = 212.1*100/(100 +100) = 106V

106V into 100 Ohms makes 112W (forgive me the one Watt difference).

This all without any transmission line theory.

Now the forward and reflected power balance (this is exactly the same
as in my previous posting, except for using 112W instead of 111W):

100 Ohms equals |rc| = 0.3333 (50 Ohms reference).

Hence "reflected power" = 0.333^2*"incident power" * = 0.111*"incident
power".

"net delivered power" = "incident power" – "reflected power" = 112W..

"net delivered power" = (1-0.11111)*"incident power" = 112W.

"incident power"= 126W, "reflected power" = 14 W

As far as I can see, there is nothing new with respect to the previous
calculation, except for the truncation/rounding.

Similar reasoning can be applied to my class-E PA. It is designed for
500W into 4.5 Ohm, but the output impedance is below 1 Ohms. If you
try to achieve conjugate match, you fry the active devices. *Same is
valid for most audio amplifiers, they may mention: *"80W into 8 Ohm",
but that does not mean that Zout = 8 Ohms.

Awaiting your comment,

Wim
PA3DJSwww.tetech.nl

Of course, with pretty much any decently designed power amplifier, no
matter the frequency, included protection circuits will limit the
dissipation and voltage in the critical areas (e.g., the output active
devices). Loads become disconnected or shorted sometimes; winds blow
stray wires or tree limbs across antennas. Through a piece of
transmission line, such a change can reflect back to become any
possible phase angle, and any of a very wide range of magnitudes, at
the amplifier's output.

Though it's a red herring typical of audio-speak, most modern high
fidelity audio amplifiers have a very low output impedance, a small
fraction of an ohm, so they can claim a high damping factor. Off
topic: the reason it's a red herring is that the impedance of the
speaker connected to the amplifier must be included to figure the
damping, and that impedance (even just the DC resistance of voice
coils) changes by considerably more than the amplifier's output
impedance (resistance) just because of heating on audio peaks.

Especially when reactive components (inductors and capacitors) couple
power between a source and a load, you can get stresses--voltages and/
or currents--well beyond what's safe when you try to operate the
source into a load it's not intended to handle. That's true even when
the net power delivered to the load is considerably LESS than the
rated output power of the source. Wim's example of the class-E
amplifier is true enough, but it's not necessary to ask the source to
deliver more net load power than it's rated to deliver, to establish
conditions that cause trouble. Thus, even sources that have an output
impedance at or very close to the rated load impedance will have
circuits to protect against loads that could destroy things inside the
amplifier.

Cheers,
Tom