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Question about Power Added Efficiency for Class E power amplifier

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sharethewell

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We all know that power added efficiency is defined as:

PAE=(Pout-Pin)/Pdc

Does Pdc here only means Power supply Power? What if the biasing terminal in the circuit also consumes some power? Is this included in the Pdc as well?

The Pdc here should include all the power supplies if there are separated digital power supply and analog power supply, right?

Thanks a lot!
 

pancho_hideboo

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What if the biasing terminal in the circuit also consumes some power?
Is this included in the Pdc as well?
Usually I include this.

The Pdc here should include all the power supplies if there are separated digital power supply and analog power supply, right?
Usually I exclude dc_power for DVdd.
I include dc_power for only AVdd.
 

vfone

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Theoretically you don't have to include the digital power supply, but if you are a System Design Engineer you have to include.
This mostly to see how the PA efficiency affect the entire system.
To check the efficiency of the PA sometimes is enough just finding the DC to RF efficiency. PAeff = PA_RFout / Pdc
 

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Theoretically you don't have to include the digital power supply, but if you are a System Design Engineer you have to include.
This mostly to see how the PA efficiency affect the entire system.
To check the efficiency of the PA sometimes is enough just finding the DC to RF efficiency. PAeff = PA_RFout / Pdc

So the Pdc in the equation includes power consumption of the biasing terminal in the circuit, right? I just found that the biasing power consumption is almost the same as the input power, which is about 2 mW.
 

vfone

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Yes, the bias power could be comparable to input power.
Bias power consumption is part of Pdc, but usually is much lower than collector/drain DC power, which is the main contributor for Pdc.
 

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