Reflected Power

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I suspect you are conflating "maximum transfer of power," a linear (small s ignal) concern with that of "efficiency," or "power added efficiency." Eff iciency is simply the available RF power over the DC power input, and power added efficiency also includes the input RF power in the denominator.

For power amplifiers (PAs), maximum transfer of power is irrelevant. The a mp is "large signal" and designed for efficiency. At a first order, the am plifier is "load line matched" to obtain maximum "unclipped" power. Note i t is "matched," but not *impedance* matched in the sense of maximum transfe r of power. The load seen by the PA will have an effect on the symmetry (a symmetry) of the clipping. In PAs, we care about drive levels that approach the supply limits. We don't care in small signal amps, as the supply is n ot part of the "equation."

Load line matching has to do with the attempt to symmetrically hit the plus and minus clipping points, as that is the way to get maximum +/- signal sw ing. It is in the large signal domain, not the linear (small signal) conce rn addressed by maximum transfer of power. While simple load line techniqu e has its analytical limits (pun), small signal analysis does not address r eal limits like clipping *at all*.

Of course, load pull technique is a more sophisticated way of accomplishing the efficiency goal. But to understand the problem, load line analysis se rves us well.

All this said, balanced amplifiers provide good return loss and can be desi gned for efficiency at the same time. But that is a trick, of sorts.

In a slightly related issue regarding LNAs low noise amplifiers..

I still don't understand how an LNA can do these 3 things at the same time:

1) present a good input match (meaning it looks like a 50 Ohm resistive load) 2) have a noise figure better than 3 dB 3) physcially be at room temperature

If the LNA presents a 50 Ohm resistive load and it is at room temperature, it must create a noise power equal to the 50 Ohm source at room temperature and the noise figure would therefore be no lower then 3 dB?

You set the input impedance by negative feedback.

Say, you have an amplifier with high input impedance and low noise, with largish negative gain A, then a resistor of value Z=Z0(1-A) will be seen at the input as a resistance of Z0.

The noise voltage of the feedback resistor sees a constant gain. Of course it also rises with the square root of its value, so the output noise due to this resistor goes up with the square root of the gain. Referred to the input however, its contribution therefore *drops* with the square root of the gain and rapidly becomes negligible.

I've made amplifiers with 50 Ohm input impedance and with measured 260pV/rtHz input-referred noise levels, so this really works.

Jeroen Belleman

Makes sense, thanks Mark