Home |
Search |
Today's Posts |
#9
![]() |
|||
|
|||
![]()
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. It appears that you forgot to put the resistive matching network into the circuit. There must be a series resistor in there between the source impedance and the load impedance to obtain that 5.7 dB of isolation. -- 73, Cecil http://www.qsl.net/w5dxp -----------== Posted via Newsfeed.Com - Uncensored Usenet News ==---------- http://www.newsfeed.com The #1 Newsgroup Service in the World! -----= Over 100,000 Newsgroups - Unlimited Fast Downloads - 19 Servers =----- |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
please need help with delta loop antenna better matching system than gamma match | Antenna | |||
Clemens match modelling | Antenna | |||
Problem with Gamma Match? | Antenna | |||
Gamma match question 6-meter yagi | Antenna | |||
Gamma match: Inherently inferior to balanced match systems? | Antenna |