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December 21st 04, 05:24 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.

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.

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

Thanks,

de KJ4UO

Dear KJ4UO,

Some additional help with your L-pad -

The minimum loss L-Pad for matching 75 ohms to 50 ohms would indeed
have a loss of 5.7 dB. This is from the ITT handbook, pages 10-4
through 10-8. The design values for such an L-Pad would give the 5.7
dB loss based upon the following: Let P1 be the power delivered to a
matched load (50 ohms) from the 50 ohm source. Let P2 be the power
delivered to a 75 ohm load when fed through the 75 to 50 resistive
matching network. Then P1 / P2 will be approximately 3.715.
10 x Log(3.7) = 5.7 dB. You can work that out from the design values
for the L-Pad, which are 43 ohms in the series arm and 86.7 ohms in
the shunt arm.

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