Chapter 12 Power Amplifier

Definitions In small-signal amplifiers the main factors are: Amplification Linearity Gain Since large-signal, or power, amplifiers handle relatively large voltage signals and current levels, the main factors are: Efficiency Maximum power capability Impedance matching to the output device 2

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 lowresistance loads. 3

Introduction 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. 4

Amplifier Types Class A The amplifier conducts through the full 360 of the input. The Q-point is set near the middle of the load line. Class B The amplifier conducts through 180 of the input. The Q-point is set at the cutoff point. Class AB This is a compromise between the class A and B amplifiers. The amplifier conducts somewhere between 180 and 360. The Q- point is located between the mid-point and cutoff. 5

Amplifier Types Class C The amplifier conducts less than 180 of the input. The Q-point is located below the cutoff level. Class D This is an amplifier that is biased especially for digital signals. 6

Class A Amplifier The output of a class A amplifier conducts for the full 360 of the cycle. The Q-point is set at the middle of the load line so that the AC signal can swing a full cycle. Remember that the DC load line indicates the maximum and minimum limits set by the DC power supply. 7

Class B Amplifier A class B amplifier output only conducts for 180 or one-half of the AC input signal. The Q-point is at 0V on the load line, so that the AC signal can only swing for one-half cycle. 8

Class AB Amplifier This amplifier is a compromise between the class A and class B amplifier. The Q-point is above that of the Class B but below the class A. The output conducts between 180 and 360 of the AC input signal. 9

Amplifier Efficiency Efficiency refers to the ratio of output to input power. The lower the amount of conduction of the amplifier the higher the efficiency. 10

Series-Fed Class A Amplifier This is similar to the small-signal amplifier except that it will handle higher voltages. The transistor used is a high power transistor. 11

Series-Fed Class A Amplifier A small input signal causes the output voltage to swing to a maximum of V cc and a minimum of 0V. The current can also swing from 0 ma to I CSAT (V CC /R C ) 12

Amplifier Efficiency η 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 : h ac dc output input power power P ( ac) o 100% 100% 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. 13

Series-Fed Class A Amplifier Input Power The power into the amplifier is from the DC supply. With no input signal, the DC current drawn is the collector bias current, I CQ. Output Power Efficiency 14

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 I 2 15

Limitation 16

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 CQ 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 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 peak 3 250 10 A) RC (20 ) 0.625W 2 (20V )(0.48A) 9.6W 2 V i 25 17

Transformer-Coupled Class A Amplifier This circuit uses a transformer to couple to the load. This improves the efficiency of the Class A to 50%. 18

Transformer-Coupled Class A Amplifier A transformer improves the efficiency because it is able to transform the voltage, current, and impedance Voltage Ratio Current Ratio Impedance Ratio 19

Transformer-Coupled Class A Amplifier DC Load Line As in all class A amplifiers the Q- point is established close to the midpoint of the DC load line. AC Load Line The saturation point (I Cmax ) is at V cc /R L and the cutoff point is at V 2 (the secondary voltage of the transformer). This increases the maximum output swing because the minimum and maximum values of I C and V CE are spread further apart 20

Transformer-Coupled Class A Amplifier Signal Swing and Output AC Power The voltage swing: The current swing: The AC power: 21

Transformer-Coupled Class A Amplifier Power input from the DC source: Power dissipated as heat across the transistor Maximum efficiency: Note: The larger the input and output signal, the lower the heat dissipation. Note: The larger V CEmax and smaller V CEmin, the closer the efficiency approaches the theoretical maximum of 50% 22

Example Calculate the ac power delivered to the 8- speaker for the circuit of the Figure. The circuit component values result in a dc base current of 6 ma, and the input signal (V i ) results in a peak base current swing of 4 ma.: 24

Solution: the dc load line is drawn vertically from the voltage point: V CEQ =V CC = 10 V For I B = 6 ma, the operating point on the Fig. is V CEQ =10 V and I CQ = 140 ma The effective ac resistance seen at the primary is The ac load line can then be drawn of slope 1/72 going through the indicated operating point. 25

Solution: mark a point (A): Connect point A through the Q-point to obtain the ac load line. For the given base current swing of 4 ma peak, the maximum and minimum collector current and collector emitter voltage obtained The ac power delivered to the load can then be calculated 26

Q For the previous circuit, calculate the dc input power, power dissipated by the transistor, and efficiency of the circuit for the input signal in the previous example The efficiency of the amplifier is then 28