RF Linear Power Amplifiers

Power amplifiers(PA) are broadly classified into two categories, linear PAs and switch-mode PAs. In linear power amplifiers the active device works as a transconductor, whereas in switching mode power amplifiers it act as a switch.

Below are the assumptions made to simplify the following analysis:
– active device is ideal VCCS or ideal switch
– input is single tone

Introduction

The schematic of a linear power amplifier is shown in Figure 1. The same amplifier works in class-A, class-AB, class-B or class-C mode depending on conduction angle of the active device.

For $i_D>0$ , MOS transistor MO behave as a linear transconductor. Inductor L1 acts as DC short. Capacitor C1 blocks DC current flowing to load and acts as short to ac component.

$L_t$ and $C_t$ form the tank circuit whose resonant frequency $\omega_c =1 / \sqrt{L_t C_t}$ . The tank circuit is required to suppress harmonics in the output. The nonlinearity of active device generate harmonics, which are suppressed by high Q tank circuit.

Rendered by QuickLaTeX.com

Figure 1. Linear RF Power Amplifier

Class-A Power Amplifier

In class-A amplifier the active device conducts for full cycle. Therefore conduction angle ( $\theta$ ) = 360 degrees

Under queiscent operating conditions,
$V_{DS}= V_{DD}$ ; $V_{C1}= V_{DD}$ ; $I_{D} = I_{DC}$
Due to voltage signal at the gate input, the drain current is
$i_D = I_{DC} + i_{rf} \sin{\omega_o t}$
voltage across load, $v_{out} = - i_{rf}\sin{\omega_o t} R_L$
drain to source voltage of $M_0$ , $v_{DS}= V_{DD} - i_{rf}\sin{\omega_o t} R_L$

For the transistor to operate in saturation region, $v_{DS} \geq 0$ .
Therefore peak current is
$V_{DD} - i_{rf,p}\sin{\omega_o t} R_L \geq 0$ $\quad \Rightarrow i_{rf,p} \leq {V_{DD} \over R_L}$
Maximum output voltage, $V_{o,p}$ = $V_{DD}$

Maximum power delivered to load,

(1) $\begin{equation*} P_{o,avg} = {V_{o,p}^2 \over 2R_L} \end{equation*}$

Power drawn from source or DC biasing power,

(2) $\begin{equation*} P_{dc} = V_{DD} I_{DD} = {V_{DD}^2 \over R_L} \end{equation*}$

Maximum drain efficiency,

(3) $\begin{equation*} \eta_{max} = {P_{o,avg} \over P_{dc}} = 0.5 \end{equation*}$

Class-B Power Amplifier

In Class-B mode the active device ( $M_o$ ) acts as a transconductor with conduction angle of 180 degrees.

Under quiescent conditions,
$v_{DS}=V_{DD}$ , $V_{C1} = V_{DD}$ and $I_{D}=0$ .

Due to small signal at the input of the transistor, the drain current is given by

(4) $\begin{eqnarray*} i_D(t) = \begin{cases} I_D \sin\omega_c t & 0 < t < {T \over 2} \\ 0 & {T \over 2} < t < T \end{cases} \end{eqnarray*}$

By Fourier series,

(5) $\begin{equation*} i_D(t) = {I_D \over \pi} + {I_D \over 2} \sin\omega_c t - {2I_D \over 3\pi} \cos2\omega_c t - {2I_D \over 15\pi} \cos4\omega_c t + \ldots \end{equation*}$

The DC part of drain current flows only through inductor $L_1$ , where as fundamental and higher order harmonic current find the low impedance path through capacitor $C_1$ . The resonant tank offers low impedance path or appears as short to harmonics and high impedance to fundamental component. Therefore only fundamental flow through load resistor $R_L$ .

Swing limits
$v_{DS} = V_{DD} - {I_D \over 2} R_L \sin\omega_ct$
As the drain current increases and device enters into cutoff.
To keep the device in saturation, $v_{DS} > 0$ .
The current is maximum under this condition, and is given by
$\Rightarrow$ ${I_{D,max}\over 2} = {V_{DD}\over R_L}$
Max drain efficiency,
Maximum power delivered to load,

$\begin{equation*} P_{o,avg} = {1 \over 2}V_{DD} {I_{D,max}\over 2} \end{equation}$

Power drawn from source under maximum output power,

$\begin{equation*} P_{dc} = V_{DD}I_{dc}=V_{DD} {I_{D,max}\over \pi} \end{equation}$

Maximum drain efficiency,

(6) $\begin{equation*} \eta_d = {P_{o,avg} \over P_{dc}} = {\pi \over 4} = 78.5\% \end{equation*}$

Class-C Power Amplifier

Conduction angle : $0 < \theta < 180$
From the Figure,
$\theta=\pi-2\phi$
Transistor drain current,

(7) $\begin{eqnarray*} i_D(t) = \begin{cases} I_D \sin\omega_c t & \phi < \theta < \pi-\phi \\ 0 & \mbox{otherwise} \end{cases} \end{eqnarray*}$

By Fourier series, DC and fundamental components are given by

(8) $\begin{equation*} i_D(t) = {I_D \over \pi} \sin({\theta \over 2}) + {I_D \over 2\pi}(\theta+\sin\theta) \sin\omega_c t + \ldots \end{equation*}$

Swing limits
Voltage at the drain
$v_{DS} = V_{DD}-i_L R_L = V_{DD}-{I_D \over 2\pi}(\theta + \sin\theta)R_L \sin\omega_ct$
Maximum fundamental current is limited by voltage swing at drain
$I_{D,max}={V_{DD}2 \pi \over (\theta + \sin\theta)R_L}$

Maximum average output power,

$P_{o,avg} = {1 \over 2} V_{DD} i_L = {1 \over 2} V_{DD} {I_{D,max}\over 2\pi}(\theta + \sin\theta)$

Power drawn from source at max output power,

$P_{dc,max} = V_{DD}I_{dc} =V_{DD}{I_D \over \pi}\sin({\theta\over2})$

Max. Drain efficiency,

$\eta_d = {P_{o,avg}\over P_{dc}}={1\over 2}\left({\theta+\sin\theta \over \sin({\theta/2})}\right)$

The drain efficiency is maximum, when $\theta \rightarrow 0$ .


	Class-A	Class-AB	Class-B	Class-C
Conduction Angle( $\theta$ )	$360^o$	$180^o-360^o$	$180^o$	$0-180^o$
Drain Efficiency	50%	50% – 78.5%	78.5%	78.5% – 100%
Linearity	Best	Fair	Good	Poor

Analog/RF IntgCkts

A RFIC designer's notes