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I have been seeking an answer for this question for some time, and i am clearly missing some intuition. I have numerous sources stating that the efficiency is higher when the amplifier is driven close to saturation, which obviously gives troubles in RF applications, when waveforms with a high PAPR is used.

However, there are no arguments stating why - can someone help me answer this?

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  • \$\begingroup\$ The voltage drop across the transistor should be minimal to minimize the dissipation - this applies to all semiconductor amplifiers, not only transistors, and also applies to DC outputs :) \$\endgroup\$ – Kuba hasn't forgotten Monica Jan 19 at 16:18
  • \$\begingroup\$ Kaspar, are you done with this question now or is there still something that is confusing you that needs clarification? \$\endgroup\$ – Andy aka Jan 21 at 12:24
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It makes no difference whether it is a power amplifier or a DC supply feeding a resistive load via a variable resistor. The variable resistor is, in effect, your amplifier. And, as engineers, we learn that when the variable resistor has the same resistance as the resistive load, maximum power is wasted in the variable resistor. See the maximum power transfer theorem.

But we also know that when the variable resistor is set to close to zero ohms, maximum power is transferred to the load resistor hence, efficiency of power transfer is also the greatest.

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  • \$\begingroup\$ When the variable resistor and the load have the same resistance, maximum power is wasted in the variable resistor but also maximum power is transfered to the load. When the variable resistor is close to zero, the efficiency becomes close to 1 (power is transfered to the load with little loss in the variable resistor) but the power transfered to the load is less than is the first case. \$\endgroup\$ – Sacha Jan 20 at 15:14
  • \$\begingroup\$ Maximum power is transferred to the load when the variable resistor is zero. Use your brain @Sacha When they have the same resistance both dissipate the same amount but that doesn't mean maximum power transfer because the variable resistor can be set to 0 ohms. \$\endgroup\$ – Andy aka Jan 20 at 15:15
  • \$\begingroup\$ Maximum power is transferred when impedance matches between the source and the load. That's the tradeoff between power transfer and efficiency. The link is your answer states the same. \$\endgroup\$ – Sacha Jan 20 at 15:18
  • \$\begingroup\$ @Sacha the wiki article says this In electrical engineering, the maximum power transfer theorem states that, to obtain maximum external power from a source with a finite internal resistance, the resistance of the load must equal the resistance of the source - so if the internal resistor (the variable one) can be set to a low value, then max power transfer is when that occurs. \$\endgroup\$ – Andy aka Jan 20 at 15:21
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    \$\begingroup\$ My bad, I was stuck in the typical case in which you can only control the load resistance. Thanks for clarification. \$\endgroup\$ – Sacha Jan 20 at 15:23
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That would be because of power dissipation of a transistor is P=U*I. If it is fully off, I=0, and if it is fully on, U=0, so ideally the transistor dissipates no power. But it dissipates most power when it is halfway on, as there is both voltage over it, and current through it.

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Class A will transfer max current and dissipate least power when the voltage drop is lowest.

This is why Class A is just about 50% efficient on average and Class D 98% or so.

For RF they might use Crest Factor Reduction (CFR) methods to improve efficiency by clipping, filter harmonics or other methods for ODFM.

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The other answers are also correct but I missed the relation to how efficiency is defined, so here's my additional answer:

For most amplifiers, the total powerconsumption consists of two parts, a constant part (for biasing etc.) and a variable part which depends on how much power the amplifier is delivering:

(1) \$P_{total} = P_{constant} + P_{out}\$

If we then define the efficiency as the ratio between the delivered power and the total power consumption:

(2) Efficiency = \${P_{out}}/{P_{total}}\$

Filling in (1) into (2):

(3) Efficiency = \${P_{out}}/(P_{constant} + P_{out}) = 1/(\frac{P_{constant}}{P_{out}}+1)\$

From (3) we can now see that the efficiency becomes higher as \$P_{out}\$ increases.

So for maximum efficiency, you need to make \$P_{out}\$ as large as possible.

The largest "usable" power output of an amplifier is just before the point where it starts to saturate. If you go any further (more power) then the amplifier distorts and it doesn't behave as a "proper" amplifier anymore.

For a small output power, efficiency will be low as the \$P_{constant}\$ part is more dominant. Then most of the power is consumed by the amplifier itself instead of appearing as "usable" power at the output.

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Efficiency of class B amplifier.

Let’s consider a class B power amplifier which we will assume saturates at the supply rails (in reality the amplifier would saturate a few volts below the power rails). The equations below can be used to calculate the power dissipated in the load, the power drawn from the supply rails, the power dissipated in the transistors and the resulting efficiency of the amplifier.

Equations

I’ve used the equations above to generate the values in the table below. I have assumed that the load is a 10 Ohm resistive load and that the supply rails are plus and minus 10V.

Table

Looking at the data in the table above it can be seen that the maximum theoretical efficiency of 78.5% occurs when the voltage across the load is peaking at the supply rails. From here the efficiency reduces linearly as the peak load voltage reduces. The amplifier has a 50% efficiency when the peak load voltage has reduced from 10V to 6.37V when the power dissipated in the load is equal to the power dissipated in the transistors and the sum of the two is equal to the power being supplied by the power rails.

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the PA draws DC power from supply and convert Dc power to AC throug the voltage / current swing. the higher voltage and current swing, the higher DC-AC conversion efficiency.at saurturation, the PA reaches its maximum voltage or current swing, and consequentlich its max. efficiency

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    \$\begingroup\$ You haven't explained why it is more efficient close to saturation and that was the question. \$\endgroup\$ – Transistor Jan 21 at 17:42

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