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December 21st 04, 04:30 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.

Hi OM,

That is the loss for using the network, and for this network it
appears to be Power lost IN the network (rather than Power not
delivered to the Load). However, this is not the best design for a
matching network, they are usually three resistors in either a PI or T
configuration.

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.

No matching network is required for this scenario.

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.

By your description, you are still not using the matching network. If
you think you are, then the math is incorrect.

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

Of course not. You never accounted for your two matching resistors so
you cannot assign the 5.7dB loss to them.

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.

There are any number of reasons - accuracy and stability being
principal among them. On the other hand, the two resistor variety is
not the most suitable for accuracy (except under known loads).
Another virtue is that it will increase the input power specification
by that same 5.7dB. This very handy when you are putting a flame
thrower to its input. Burnt resistors are cheaper and easier to
replace than the input stage. One very useful 100 Ohm resistor is a
10mA glass fuse (Wollaston wire).

73's
Richard Clark, KB7QHC

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