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December 21st 04, 04:39 PM

On 21 Dec 2004 14:23:10 GMT, (PDRUNEN) wrote:

Hi All,

I was reviewing a 75 to 50 ohm resistive matching network using two resistors,
the insertion lost was 5.7 db.

If we have a 100Vrms source with 50 ohm source impedance and it is driving a
matched 50 ohm load then the load takes 1A and the power in the load is 50
watts.

This is correct, if the 100 V rms source is the Thevenin emf - not the
loaded terminal voltage.

If the load is replaced with 75 ohm, then 0.8 amps will flow and the power is
48 watts. (I*I*R) == (0.8)*(0.8)*75.

Under those condx the load voltage is 60 V rms and the load power is
48 watts, as you indicated.

I guess I must be not be taking something in account, but 2 watts does not
equal 5.7 db.

Since the original load voltage was 50 volts rms and now it's 60
volts, one could argue that the load voltage is up by 1.6 dB. However,
the load power at 48 watts is in fact down by 0.17 dB

I know there must be a good reason to put the matching pad in line for the
sprectrum analyizer but I don't under why.

Interestingly enough, none of this has anything to do with your L-pad
matching network. All of your calculations above relate only to the
hypothetical situation in which you are actually "changing" the load
resistance and computing the new load power - nothing whatsoever to do
with the situation you would have when you actually insert an L-pad
between the real source and the original 50 ohm load.

A 75 to 50 ohm resistive matching pad is designed to work with a 75
ohm load and to present a 50 ohm load to the source, which it will do.
When you calculate the amount of power in the 75 ohm load before and
after inserting the L-pad you will get a significant loss, depending
upon the exact design of the two-resistor network. 5.7 dB sounds a bit
high, but it depends upon how the L-Pad losses are defined.

As to the reason for using the L-Pad, it's probably for the purpose of
minimizing the SWR on the cable between the L-Pad/Load and the source.
Since you haven't specified anything about the source, the load or
what the purpose of the circuit is, one can only guess.

Bob, W9DMK, Dahlgren, VA
http://www.qsl.net/w9dmk

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