|
| |
| |
| | Power delivered to antennae - RF Cafe Forums |
Because of the high maintenance needed to monitor and filter spammers from the RF Cafe Forums, I decided that it would
be best to just archive the pages to make all the good information posted in the past available for review. It is unfortunate
that the scumbags of the world ruin an otherwise useful venue for people wanting to exchanged useful ideas and views.
It seems that the more formal social media like Facebook pretty much dominate this kind of venue anymore anyway, so if
you would like to post something on RF Cafe's
Facebook page, please do.
Below are all of the forum threads, including all
the responses to the original posts.
darcyrandall2004 Post subject: Power delivered to antennae Posted: Fri Mar 09, 2007 11:32 pm
Colonel
Joined: Tue Feb 27, 2007 6:16 am Posts: 46 Hello. I am designing a UHF transmitter. I wish to be able to determine the requirements of components on the output stage, after the amplifier and the power delivered to the antennae.
To begin with I am analyzing the application note designed for 380MHz. My final design will be for 450MHz
The RF2175 datasheet claims that the amp delivers 31.8dBm Linear Output Power.
The figure below shows the output stage and where I have attempted to calculate the current in the branches.
BELOW IS THE MATLAB CODE I USED TO CALCULATE THE CURRENTS IN THE BRANCHES, COMPONENT RATINGS AND THE MAXIMUM POWER DELIVERED TO THE ANTENNAE.
format long clear
f=380000000; wo=2*pi*f*i;
%DETERMINE THE RF2175 AMP'S OUTPUT POWER IN WATTS pout=(10^(31.8/10))*10^-3;
%THE CALCULATED RF2175 OUTPUT IMPEDANCE (THE CONJUGATE OF THE LOAD) rf2175_z=1.83712875712576 + 7.16437870763534i
%CALCULATE THE THEVENIN VOLTAGE, P=v^2/r, pout=v^2/(2*real(rf2175_z)) v=sqrt(pout*2*real(rf2175_z))
%CALCULATE THE IMPEDANCES OF COMPONENTS AT 380MHZ XC9=1/(wo*56*10^-12); XL1=wo*3.6*10^-9; XC10=1/(wo*12*10^-12); XC11=1/(10^-9*wo); XL6=wo*12.55*10^-9;
%CALCULATE THE BRANCH CURRENTS I(1,:)=[rf2175_z+XL6,-XL6,0,0,-v]; I(2,:)=[-XL6,XL6+XC9,-XC9,0,0]; I(3,:)=[0,-XC9,XC9+XL1+XC10,-XC10,0]; I(4,:)=[0,0,-XC10,XC10+XC11+50,0];
I=rref(I);
I1=I(1,5); I2=I(2,5); I3=I(3,5); I4=I(4,5);
sprintf('CURRENT RATING FOR L6') [mag,theta]=cart2pol(real(I1-I2),imag(I1-I2)); mag
sprintf('VOLTAGE RATING FOR C9') [mag,theta]=cart2pol(real((I2-I3)*XC9),imag((I2-I3)*XC9)); mag
sprintf('CURRENT RATING FOR L1') [mag,theta]=cart2pol(real(I3),imag(I3)); mag
sprintf('VOLTAGE RATING FOR C10') [mag,theta]=cart2pol(real((I3-I4)*XC10),imag((I3-I4)*XC10)); mag
sprintf('VOLTAGE RATING FOR C11') [mag,theta]=cart2pol(real(I4*XC11),imag(I4*XC11)); mag
sprintf('POWER DELIVERED TO 50 OHM ANTENNAE') real(I4*I4*50)
BELOW IS THE RESULTS PRODUCED
rf2175_z =
1.83712875712576 + 7.16437870763534i
v =
2.35822259134855
ans =
CURRENT RATING FOR L6
mag =
0.25101690661180 Amps
ans =
VOLTAGE RATING FOR C9
mag =
1.82181323340670 Volts
ans =
CURRENT RATING FOR L1
mag =
2.56132106649834 Amps
ans =
VOLTAGE RATING FOR C10
mag =
1.59717311590861 Volts
ans =
VOLTAGE RATING FOR C11
mag =
0.03475316915436 Volts
ans =
POWER DELIVERED TO 50 OHM ANTENNAE
ans =
-0.75495330687175 Watts
»
I understand that the current rating for L6 would be determined by the dc bias current it supplies.
Do these other calculations look correct?
When the data sheet states that the amp has 31.8dBm output power, does this mean that if the load is a conjugate of the output impedance, 31.8dbm will be delieverd to the load? Or is half of this value disspiated in the output impedance of the amp?
Thankyou
Cheers
Top
darcyrandall2004 Post subject: Posted: Mon Mar 12, 2007 11:19 pm
Colonel
Joined: Tue Feb 27, 2007 6:16 am Posts: 46
Not sure why your image isn't appearing. Let's try this: (Kirt Blattenberger 3/13/2007)
Posted 11/12/2012
| |
|
|
|