Politecnico
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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!
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!