Power delivered to antennae - RF Cafe Forums

The original RF Cafe Forums were shut down in late 2012 due to maintenance issues - primarily having to spend time purging garbage posts from the board. At some point I might start the RF Cafe Forums again if the phpBB software gets better at filtering spam.

Below are the old forum threads, including responses to the original posts.

-- Amateur Radio
-- Anecdotes, Gripes & Humor
-- Antennas
-- CAE, CAD, & Software
-- Circuits & Components
-- Employment & Interviews

-- Miscellany
-- Swap Shop
-- Systems
-- Test & Measurement
-- Webmaster

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