On May 16, 3:42*pm, John KD5YI wrote:
On 5/16/2011 6:54 AM, Cecil Moore wrote:









On May 15, 7:54 pm, John *wrote:
100W--50ohm--+--1/4WL 100ohm--200 ohm load

It is a 100 ohm feedline. The impedance at the + point looking toward
the load is 50 ohms. The 100 W source will see the first 50 ohms and
then the impedance at the + point. The total impedance seen by the
source is therefore 100 ohms. A 100 W source will deliver 100V to this
100 ohm combination and 50V of it will appear at the + point. Where does
the 70.707V come from?

I said: "All that matters is that the source is delivering 70.707
volts to the 50 ohm feedline." That would be at the source SO-239
terminal to PL-259 50 ohm coax junction. I did *NOT* say there is
70.7v on the Z0=100 ohm feedline. There is indeed 100 volts on the
Z0=100 ohm line.

The source is Z0 matched to 50 ohms and is
*delivering 70.707 volts and 1.414 amps to the 50 ohm coax*.
70.7/1.414=50 ohms. 70.7*1.414=100 watts.

Here is a better diagram:

http://www.w5dxp.com/enfig1.gif

Pfor1=100w, Pref1=0, Z01=50 ohms, Z02=100 ohms and 1/4WL, Load=200
ohms. Everything except the load is lossless.

I apologize if my one line diagram was confusing. It is extremely
difficult to draw ASCII schematics using a proportional font.
--
73, Cecil, w5dxp.com
"Halitosis is better than no breath at all.", Don, KE6AJH/SK

Okay, so your source really looks like a a 100 W generator with no
internal impedance but which feeds a 50 ohm line internally. Since the
200 ohm load appears as 50 ohms when fed with a 100 ohm 1/4WL line,
there is a match and all the power appears at the load. In that case I
agree with your 70.7 volts since 100 W into 50 ohms will result in that
voltage.

But I have no need of reflections. The fact that a 200 ohm load appears
as a 50 ohm load at the input of a 1/4WL 100 ohm line takes care of all
that for me.

73,
John


Sorry John, you can't cut it without the reflections. They may be
invisible to you, but they're there doing the work. The 1/4wl line
transformer works only with reflections. If you'll give me your email
address I'll send you a graph showing how the reflections perform the
matching. The actions of the reflections from either end of the
transformer is what performs the matching function. The action of the
reflections is as necessary to the function as breathing is to life.
Believe it!!!

Walt