Radio Frequency Integrated Circuit Design - Webs

350 Radio Frequency Integrated Circuit Design
10.2 Power Capability
One of the main goals of PA design is to deliver a given power to a load. This
is determined to a large degree by the load resistor and the power supply. Given
a particular power supply voltage VCC , such as 3V, and a load resistance R L ,
such as 50�, it is possible to determine the maximum power to be
P =
2
V CC
2R =
3 2
= 90 mW ⇒ 19.5 dBm (10.1)
2 × 50
This assumes we have a tuned amplifier and an operating point of 3V, a
peak negative swing down to 0V, and a peak positive swing up to 6V.
10.3 Efficiency Calculations
Efficiency �, sometimes also called dc-to-RF efficiency, is the measure of how
effectively power from the supply is converted into output power and is given
by
� = P out
P dc
(10.2)
where Pout is the ac output power and, assuming voltage and current are in
phase, is given by
Pout = i1v1
2 = i 2 1 R L
2
(10.3)
where i1 and v 1 are the peak fundamental components of the current and
voltage, respectively. These are determined from the actual current and voltage
by Fourier analysis. P dc is the power from the supply and is given by
P dc = 1
T � T
0
VCC iC dt = VCC
T � T
0
iC dt = VCC I dc
(10.4)
where I dc is the dc component of the current waveform.
Power-added efficiency (PAE) is the same as efficiency; however, it takes
the gain of the amplifier into account as follows:
PAE = Pout − P in
P dc
= P out − Pout /G
P dc
1
= ��1 − G�
(10.5)