Chapter 6: Power Amplifiers

Chapter 6: Power Amplifiers

Contents Class A Class B Class C

Power Amplifiers Class A, B and C amplifiers are used in transmitters Tuned with a band width wide enough to pass all information sidebands without distortion As narrow as feasible To reduce harmonics and other spurious signals

Block Diagrams

Power Amplifier (Class A) Introduction of Power Amplifier Power and Efficiency Amplifier Classification Basic Class A Amplifier Transformer Coupled Class A Amplifier EE 414 Lectures Series_1 By Dr. Muhammad H. Rais

Introduction Power amplifiers are used to deliver a relatively high amount of power, usually to a low resistance load. Typical load values range from 300W (for transmission antennas) to 8W (for audio speaker). Although these load values do not cover every possibility, they do illustrate the fact that power amplifiers usually drive low-resistance loads. Typical output power rating of a power amplifier will be 1W or higher. Ideal power amplifier will deliver 100% of the power it draws from the supply to load. In practice, this can never occur. The reason for this is the fact that the components in the amplifier will all dissipate some of the power that is being drawn form the supply.

Amplifier Power Dissipation The total amount of power being dissipated by the amplifier, P tot, is V CC I CC P tot = P 1 + P 2 + P C + P T + P E I 1 I CQ The difference between this total value and the total power being drawn from the supply is the power that actually goes to the load i.e. output power. P 1 = I 1 2 R 1 P 2 = I 2 2 R 2 R 1 R 2 I EQ RC R E P C = I 2 CQ R C P T = I 2 TQ R T P E = I 2 EQ R E Amplifier Efficiency h I 2 7

Amplifier Efficiency h A figure of merit for the power amplifier is its efficiency, h. Efficiency ( h ) of an amplifier is defined as the ratio of ac output power (power delivered to load) to dc input power. By formula : ac output power P ( ac) o h 100% 100% dc input power P( dc) i As we will see, certain amplifier configurations have much higher efficiency ratings than others. This is primary consideration when deciding which type of power amplifier to use for a specific application. Amplifier Classifications 8

Amplifier Classifications Power amplifiers are classified according to the percent of time that collector current is nonzero. The amount the output signal varies over one cycle of operation for a full cycle of input signal. v in A v v out Class-A v in A v v out Class-B v in A v v out Class-C 9

Efficiency Ratings The maximum theoretical efficiency ratings of class-a, B, and C amplifiers are: Amplifier Maximum Theoretical Efficiency, h max Class A 25% Class B 78.5% Class C 99% 10

Class A Amplifier v in A v v out output waveform same shape input waveform + phase shift. The collector current is nonzero 100% of the time. inefficient, since even with zero input signal, ICQ is nonzero (i.e. transistor dissipates power in the rest, or quiescent, condition) 11

Basic Operation Common-emitter (voltage-divider) configuration (RC-coupled amplifier) +V CC I CC I 1 R 1 I CQ R C R L v in R 2 R E 12

Typical Characteristic Curves for Class-A Operation 13

Typical Characteristic Previous figure shows an example of a sinusoidal input and the resulting collector current at the output. The current, ICQ, is usually set to be in the center of the ac load line. Why? (DC and AC analyses discussed in previous sessions) 14

DC Input Power +V CC The total dc power, P i (dc), that an amplifier draws from the power supply : P ( dc) V I i CC CC I 1 R 1 I CQ I CC R C R L I CC I CQ I 1 I I ) CC I CQ ( I CQ 1 v in R 2 R E P i ( dc) V CC I CQ Note that this equation is valid for most amplifier power analyses. We can rewrite for the above equation for the ideal amplifier as P( dc) 2V i CEQ I CQ 15

AC Output Power AC output (or load) power, P o (ac) i c P ( ac) i v o c( rms ) o( rms ) v 2 o( rms ) R L v in v ce r C R C //R L v o Above equations can be used to calculate the maximum possible value of ac load power. HOW?? R 1 //R 2 Disadvantage of using class-a amplifiers is the fact that their efficiency ratings are so low, h max 25%. Why?? A majority of the power that is drawn from the supply by a class-a amplifier is used up by the amplifier itself. Class-B Amplifier 16

I C(sat) = V CC /(R C +R E ) I C(sat) = I CQ + (V CEQ /r C ) DC Load Line ac load line I C I C (ma) V CE(off) = V CC V CE(off) = V CEQ + I CQ r C V CE V CE I C ac load line Q - point P ( ac) o VCEQ 2 I CQ 2 1 V 2 CEQ I CQ 2 VPP 8R L dc load line h P o( ac) P i ( dc ) 100% 1 V 2 2V CEQ CEQ I I CQ CQ 100% 25% V CE 17

Limitation 18

Example +V CC = 20V Calculate the input power [P i (dc)], output power [P o (ac)], and efficiency [h] of the amplifier circuit for an input voltage that results in a base current of 10mA peak. R B 1k I C R C 20 V o I I V I V I BQ VCC VBE 20V 0.7V 19.3mA RB 1k I 25(19.3mA) 482.5mA 0.48A c( sat) CE( cutoff ) C ( peak) P P CQ CEQ o( ac) i( dc) V V R I P h P B CC V I V o( ac) i( dc) ICR CC C b( peak) 2 C( peak ) CC CC 2 I 20V 20V 20V 20 CQ C 25(10mA 100% 6.5% (0.48A)(20) 10.4V 1000mA 1A peak) 250mA 3 25010 A) RC (20) 0.625W 2 (20V )(0.48A) 9.6W 2 peak V i 25 19

Transformer-Coupled Class-A Amplifier A transformer-coupled class-a amplifier uses a transformer to couple the output signal from the amplifier to the load. Z 1 +V CC N 1 :N 2 R L The relationship between the primary and secondary values of voltage, current and impedance are summarized as: R 1 Z 2 = R L N N 1 2 N N 1 2 V V 2 1 2 Z Z I 1 2 I 2 1 Z 1 R L Input R 2 R E N 1, N 2 = the number of turns in the primary and secondary V 1, V 2 = the primary and secondary voltages I 1, I 2 = the primary and secondary currents Z 1, Z 2 = the primary and seconadary impedance ( Z 2 = R L ) 20

Transformer-Coupled Class-A Amplifier An important characteristic of the transformer is the ability to produce a counter emf, or kick emf. When an inductor experiences a rapid change in supply voltage, it will produce a voltage with a polarity that is opposite to the original voltage polarity. The counter emf is caused by the electromagnetic field that surrounds the inductor. 21

Counter emf SW1 + - + - 10V 10V + - 10V 10V - + This counter emf will be present only for an instant. As the field collapses into the inductor the voltage decreases in value until it eventually reaches 0V. 22

DC Operating Characteristics The dc biasing of a transformer-coupled class-a amplifier is very similar to any other class-a amplifier with one important exception : the value of V CEQ is designed to be as close as possible to V CC. The dc load line is very close to being a vertical line indicating that V CEQ will be approximately equal to V CC for all the values of I C. R 1 Z 1 +V CC N 1 :N 2 R L Z 2 = R L The nearly vertical load line of the transformercoupled amplifier is caused by the extremely low dc resistance of the transformer primary. Input R 2 R E V CEQ = V CC I CQ (R C + R E ) The value of R L is ignored in the dc analysis of the transformer-coupled class-a amplifier. The reason for this is the fact that transformer provides dc isolation I C between the primary and secondary. Since the load resistance is in the secondary of the transformer it dose not affect the dc analysis of the primary circuitry. DC load line I B = 0mA 23 V CE

AC Operating Characteristics 1. Determine the maximum possible change in V CE +V CC N 1 :N 2 Since V CE cannot change by an amount greater than (V CEQ 0V), v ce = V CEQ. R 1 Z 1 R L Z 2 = R L 2. Determine the corresponding change in I C Find the value of Z 1 for the transformer: Z 1 = (N 1 /N 2 ) 2 Z 2 and i c = v ce / Z 1 Input R 2 R E 3. Plot a line that passes through the Q-point and the value of I C(max). I C I C(max) =?? DC load line I C(max) = I CQ + i c 4. Locate the two points where the load line passes through the lies representing the minimum and maximum values of I B. These two points are then used to find the maximum and minimum values of I C and V CE Q-point ac load line I B = 0mA ~ V CEQ ~ V CC ~ 2V CC 24 V CE

+V CC N 1 :N 2 Z 1 R L R 1 Z 2 = R L I C I C(max) =?? DC load line Input R 2 R E I CQ Q-point ac load line i c v in v ce I B = 0mA Z 1 v o V CE ~ V CEQ ~ V CC ~ 2V CC R 1 //R 2 25

Maximum load power and efficiency The Power Supply for the amplifier : P S = V CC I CC Maximum peak-to-peak voltage across the primary of the transformer is approximately equal to the difference between the values of V CE(max) and V CE(min) : V PP = V CE(max) V CE(min) Maximum possible peak-to-peak load voltage is found by V (P-P)max = (N 2 / N 1 )V PP The actual efficiency rating of a transformer-coupled class-a amplifier will generally be less than 40%. V PP N 1 : N 2 R L V (P-P) max 26

There are several reasons for the difference between the practical and theoretical efficiency ratings for the amplifier : 1. The derivation of the h = 50% value assumes that V CEQ = V CC. In practice, V CEQ will always be some value that is less the V CC. 2. The transformer is subject to various power losses. Among these losses are couple loss and hysteresis loss. These transformer power losses are not considered in the derivation of the h = 50% value. 27

One of the primary advantages of using the transformer-coupled class-a amplifier is the increased efficiency over the RC-coupled class-a circuit. Another advantage is the fact that the transformer-coupled amplifier is easily converted into a type of amplifier that is used extensively in communications :- the tuned amplifier. A tuned amplifier is a circuit that is designed to have a specific value of power gain over a specific range of frequency. 28

Class C Amplifier Ic Class A Class AB Class B Vce Class C

Class B Amplifier Review The Class B amplifier is biased at cutoff; in other words there is no bias. The amplifier relies on the input signal to bias the transistor. The first 0.7 volts of the signal in is lost Because the transistor does not turn on until the base voltage reaches 0.7V. Amplifier is very efficient but provides crossover distortion. Amplifier is used to amplify power in situations where waveform is not important (power supplies) 12V +V 1k TP1-5/5V NPN 1.0 khz 1k

Class C Biasing The Class C amplifier has a negative voltage at the base, and zero volts at the emitter. The Base-Emitter junction is reverse biased The transistor will turn on at a signal voltage of 4.7V. The transistor will only conduct a small pulse through to the collector output. Amplifier is very efficient but has very little usefulness. Unless. 12V +V 1k TP2-5/5V About 0.3V NPN 1.0 khz 1k -4V DC

1uF Class C Application Collector resonant circuit responds to an impulse by ringing at its resonant frequency. This is like pushing someone on a swing. A sharp short push each trip allows the swing to oscillate back and forth at its resonant frequency Collector impulses occur by design at the resonant frequency of the tank circuit. Resonant circuit continues to ring and restores the sine wave at the output Class C Amplifier is used in radio/rf applications and is also called a tuned amplifier. 12V +V Resonant (tank) circuit C1 1uH Positive alternations filled in by tank circuit ring TP3-5/5V 1.0 khz NPN 1k -4V DC

Triode Vacuum Tube EE 414 Lectures Series By Dr. Muhammad H.

Impedance matching network One of the possible surprises in PA design is the realization that output impedance matching is not based on maximum power transfer criteria A impedance matching system may be a wideband transformer that is used for broad band matching or inductor-capacitor or transmission line structure EE 414 Lectures Series By Dr. Muhammad H.

Impedance matching network The purpose of the impedance matching network is to transform a load impedance to an impedance appropriate for optimum circuit operation The network are two pole LC band pass or low pass resonant circuits to minimize noise and spurious signals harmonics EE 414 Lectures Series By Dr. Muhammad H.

EE 414 Lectures Series By Dr. Muhammad H. AM Block Diagram

Basic AM Modulation Amplitude Spectrum Carrier Sideband v( t) (1 m cos t) C C m m E C cos t C Modulation index 0 < m < 1 How to measure? Modulated carrier vs baseband signal What happens when m>1?

Spectrum Double Side Band Full Carrier Spectrum LSB, USB, Carrier

Waveforms Base Band and Modulation m = 100%

Efficiency Efficiency in Using Bandwidth? Two side bands Efficiency in Using Power? Carrier P c V R 2 E 2 c (sinct) R 2 2 Ec 2R Side band P lsb P usb 2 m Pc 4 Information in side band Carrier for synchronization

Crucial Components RF Amplifier High dynamic range Autodyne Oscillator and Mixer Tuning Loop Antenna LC resonator Intermediate Frequency Amplifiers 455 khz Tank circuit (double tuner) with flat center band Diode RF Rectifier Envelope detection Automatic Gain Control Strong vs weak stations Audio Amplifier Speaker

Incoming RF RF amplifier Bandpass filter Local Oscillator Mixer Heterodyne v RF ( t) (1 m cos t) E Nonlinear device or balanced mixer Bandpass filter v t) (1 msinmt) cos t Signal proportional to local oscillator strength Intermediate Frequency (AM at 455 khz) No further tuning EMI from Local Oscillator m vlo ( t) ELO cos( C IF ) t C IF ( ECELO cos( IF ) t C

Baseband Signal Coherent Detection AM modulation Local oscillator Mixer LP filter AC coupling High sensitivity for low S/N Detection v M v RF v v RF LO ( t) (1 m A( t)) E ( t) E LO cos t C C cos t ( t) (1 m A( t)) E E cos t cos t v v LP O ( t) m A( t) C LO ( t) (1 m A( t)) ECELO ( t) m A( t) ECELO / 2 C C / 2 C

AM Modulator Circuits