How to calculate Efficiency in Envelope Modulator (relevant to Envelope Tracking PA)

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Dear All

Hi,
In DC-DC buck converters, we can simply multiply Rload by Iout2 to have Pout (and then divide it by consumed DC power to calculate Efficiency).
In Envelope Modulators, the Input signal (and the output signal, too) are high peak to average power ratio signals (e.g. PAPR= 6.6 dB in LTE) and so, Envelope Modulator's Output Current follows such high PAPR waveform.
Here is the question: What is Iout (as a number) to be employed in Efficiency calculation???

ANY idea to enlighten the ambiguity, is highly appreciated!
 

Iout isn't constant while you're actually modulating the envelope, so there is no simple number to use. Rather you use average Pin and Pout where Pin=Iin*Vin and Pout=Iout*Vout, with all of those being functions of time.
 

Dear mtwieg
Thanks a lot for the reply.
So, how do you calculate the average power while you have Vout on Oscilloscope? or any other instrument is required to find avg(Vout)?


mtwieg said:
Iout isn't constant while you're actually modulating the envelope, so there is no simple number to use. Rather you use average Pin and Pout where Pin=Iin*Vin and Pout=Iout*Vout, with all of those being functions of time.
Click to expand...
 

P(avg) = ∫f(ab(t))d(t))/nT ; n -> ∞

and take 20*log(P) for dB unit. (what's in ratios)
 

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