Would not this imply a maximum efficiency of 50% ?

P-T



On 1 August 2014 00:40, Matt VK2RQ <[email protected]> wrote:

> I think we have hit the posting limit for this topic, so I'll just finish
> with this remark: the maximum power transfer between a power amplifier and
> its load will occur when the source and load impedances are complex
> conjugates (this can be seen using the voltage divider equation and some
> differential calculus). In the case of a purely resistive load, this
> reduces down to the source and load resistances being equal.
>
> 73,
> Matt VK2RQ
>
> > On 1 Aug 2014, at 8:24 am, Per-Tore Aasestrand <[email protected]> wrote:
> >
> > Hi list,
> >
> > Isn't an RF transmitter or PA actually more like a current generator? Its
> > internal resistance is kept low in order to reduce internally dissipated
> > power, with consequent reduction in efficiency. It is usually designed to
> > work efficiently into 50 ohms.
> >
> > It is important to know the difference between these two statements:
> > (A) The transmitter has a 50 ohm output impedance.
> > (B) The transmitter is designed to work into a 50 ohm load
> >
> > A transmitter will be able to put its rated power into a 50 ohm load, but
> > this says nothing about its output impedance. That is why I don't think
> > conjugate matching will be appropriate.
> >
> > (Hope this is not getting too much OT now)
> >
> > Per-Tore / LA7NO
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