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.

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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

 



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