On 6 jun, 16:22, "JC" wrote:
Basic question (at least for me) for a very poor antenna matching :
-100 w reach the antenna and 50 w are radiated.
- 50 w are "reflected", what is their fate ?
Are they definitely lost for radiation and just heat the line, the final....
JC

Hello,

If the source has a linear 50 Ohms output impedance (assuming 50 Ohm
cable, no loss), all power will go back into the source. Partly in the
form of heat, partly in the form of power saving. However a PA is not
a 50 Ohms source, so what happens may vary.

It is very likely that some reverse power reflects back to the antenna
(as the transmitter is not a source with 50 Ohms output impedance). So
the reading on an instrument that measures true forward power is the
sum of the real incident power plus the reflected power from the
transmitter towards the antenna. Changing the load may result in a
reduction or increase of total forward power. In case of a real 50
Ohms source, the forward power will not change, no matter the load.

The bad match seen from the transmitter may result in a reduction of
DC input power (input current reduces), but may also result in an
increase of DC input power.

So when you have reverse power reading of 50W, you cannot just say
that the PA has to dissipate an extra 50W, it can be more, less and
even negative. The negative case is when the input power reduces
significantly and the active element has to dissipate less (with
respect to the best match condition). In particular high efficiency
amplifiers (where the active devices are used as switches), show
strong variation in supply current versus load change. Under certain
load conditions such amplifiers may show a strong decrease in
efficiency resulting in high relative increase of device dissipation.

Best regards,

Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me