Alexandria High Institute for Engineering and Technology. Power Amplifiers. Sameh M. Selim

Transcription

1 Alexandria High Institute for Engineering and Technology Power Amplifiers Sameh M. Selim

2 Introduction: Power amplifiers are large-signal amplifiers. This generally means that much larger portion of the load line is used during signal operation than in small-signal amplifier. In this chapter, we will cover 4 classes of power amplifiers: class A, class B, class AB, and class C. Class A:The o/p signal varies for as full 360 o of the cycle. Fig. (1-a) shows that this requires the Q-point to be biased at a level so that at least half the signal swing of the o/p may vary up and down without going to a high enough voltage to be limited by the supply voltage or too low to approach the lower supply level, or 0 V in this description.

3 Class B:The o/p signal varies over one half the i/p signal cycle, or for 180 o of the signal. The dc bias point for class B is therefore at 0 V, with the o/p then varying from this bias point for a half cycle. Obviously, the o/p is not a faithful reproduction of the i/p if only one half- cycle is present. Two class B operations: one to provide o/p on the +ve halfcycle and another to provide operation on the ve o/p half-cycle - are necessary. The combined half-cycles then provide an o/p for a full 360 o of operation. This type of operation is referred to as push-pull operation. Class AB:An amplifier may be biased at a dc level above the zero-basecurrent level of class B and above one-half the supply voltage level of class A. Class AB operation still requires a push-pull connection to achieve a full o/p cycle. The o/p signal swing occurs between 180 o and 360 o. The following Figure shows an ac load line with the Q-point moved away from center toward cutoff. The collector current can only swing down to near zero and an equal amount above I CQ. The V CE can only swing up to its cutoff value and an equal amount below V CEQ. If the amplifier is driven any further than this, it will "clip" at cutoff, as shown in the following Figure.

4 The following Figure shows an ac load line with the Q-point moved away from center towards saturation. In this case, the O/P variation is limited by saturation. The collector current can only swing up to near saturation and an equal amount below I CQ. The collector-to-emitter voltage can only swing down to its saturation value and an equal amount above V CEQ. If the amplifier is driven any further, it will "clip" at saturation.

5 Class C:The o/p of a class C amplifier is biased for operation at less than 180 o of the cycle and will operate only with a tuned circuit, which provides a full cycle of operation for the tuned or resonant frequency. (Used for radio or communications) Amplifier Efficiency: defined as the ratio of o/p power to i/p power. The maximum efficiency of a class A is 25% with a direct or series fed load connection and 50 % with a transformer connection to the load. Class B operation, with no dc bias power for noi/p signal, can be shown to provide a maximum efficiency up to 78.5%. The following table summarizes the operation of the amplifier classes. Amplifier Classs A AB B C Operating Cycle 360 o 180 o -360 o 180 o < 180 o Power Efficiency 25%-50% 25%(50%) to 78.5% 78.5% Class A, series fed Amplifier: The transistor used in this circuit is a power transistor that is capable of operatingin the range of a few to tens of watts. DC Bias Operation: V CC and R B fix the dc base-bias current at I B = V CC 0.7 V R B (1) I C = I B (2) V CE = V CC - I C R C (3)

6 AC Operation: When an i/p ac signal is applied to the amplifier, the o/p will vary from its dc bias operating voltage and current. Fig.(a) shows a small i/p signal. As the i/p signal is made larger, the o/p will vary further around the established dc bias untileither the current or the voltage reaches a limiting condition. For the current this limiting condition is either zero current at the low end or (V cc /R c ) at the high end of its swing. For the collector-emitter voltage, the limit is either 0 V or the supply voltage, V CC. Power considerations The power input of the amplifier is provided by the supply. With no i/p signal, the dc current drawn is the collector bias current I CQ. The power drawn from the supply is: (4) Even with an ac signal applied, the average current drawn from the supply remains the same. The power output is provided by the o/p voltage and current varying around the bias point. This ac power is delivered to the load R C, and can be expressed in many ways.

7 Using RMS signals (5) Using Peak Signals (6) Using Peak-to-Peak signals (7) Efficiency %the efficiency of the amplifier is calculated using (8)

8 Maximum Efficiencyfor class A series fed amplifier, the maximum efficiency can be determined using the maximum voltage and current swings. For the voltage swing it is Maximum V CE (p-p) = V CC For the current swing it is Maximum I C (p-p) = V CC /R C Using thepeak-to-peakform of the power output from (7) yields The maximum power i/p can be calculated using the dc bias current set to one-half the maximum value: Using equation (8) to calculate the maximum efficiency: The maximum efficiency of a class A series-fed amplifier is thus seen to be 25%. Since this maximum efficiency will occur only for ideal conditions of both voltage swing and current swing, most series-fed circuits will provide efficiencies of much less than 25%.

9 Example1 Calculate the i/p power, o/p power, and efficiency of the amplifier circuit shown in Figure for an i/p voltage that results in a base current of 10 ma peak. Solution:using equations (1) to (3), determine the Q-point to be: I BQ = 19.3 ma I CQ = A V CEQ = V This bias point is marked on the transistor collector characteristic shown in Figure. The ac variation of the o/p signal can be obtained graphically using the dc load line (by connecting V CE =V CC = 20 V with I C = V CC /R C = 1000 ma = 1 A). When the i/p ac base current increases from its dc bias level, the collector current rises by I C (p) = I B (p) = 25 (10 ma peak) = 250 ma peak Using equation (6) yields P o (ac) = I C 2 (p) 2 R C = ( A) 2 (20 Ω) =0.625 W 2

10 Using equation (4) results in P i (dc) = V CC I CQ = (20 V)(0.48A) = 9.6 W Using equation (8), the amplifier power efficiency can be calculated % η = P o(ac) P i (dc) 100%, check = 6.5% Transformer-Coupled Class A Amplifier Transformer Action: Voltage Transformation V 2 V 1 = N 2 N 1 (9) Current Transformation I 2 I 1 = N 1 N 2 (10) Impedance Transformation R L R L = R 2 R 1 = V 2/I 2 V 1 /I 1 = V 2 V 1 I 1 I 2 If we define a = N1/N2 R L R L = R 1 R 2 = ( N 1 N 2 ) 2 = a 2 (11) The load resistance reflected to the primaryside is: R 1 = a 2 R 2 orr L = a 2 R L (12)

11 Example2 Calculate the effective resistance seen looking into the primary of a 15:1 transformer connected to an 8-Ω load. Solution:Using eq.(12), get R L = (15) 2 (8Ω) = 1800 Ω Example3 What transformer turns ratio is required to match a 16- speaker load so that the effective load resistance seen at the primary is 10 k? Solution: Using eq. (11), check( N 1 ) 2 = R, L = 625, N 2 R L So, N 1 N 2 = 25: 1 Operation of Amplifier Stage DC Load Line:The transformer (dc) winding resistance determines the dc load line for the above circuit. Typically, this dc resistance is very small (ideally 0 ), so a 0 dc load line is a straight vertical line, drawn from the point V CEQ = V CC. Quiescent operating Point: The operating point can be obtained graphically at the point of intersection of the dc load line and the base current set by the circuit. The collector quiescent current can then be obtained from the operating point.

12 Keep in mind that, in class A operation, the dc bias point sets the conditions for the maximum undistorted signal swing for both collector current and collector-emitter voltage. If the i/p signal produces a voltage swing less than the maximum possible, the efficiency of the circuit at that time will be less than 25%. AC Load Line: It is necessary to calculatethe ac load resistance "seen" looking into the primary side of the transformer, then draw the ac load line on the collector characteristic. The reflected load resistance (R' L ) is calculated using Eq. (12). Draw the ac load line so that it passes thro' the operating point and has a slope equal to (-1/R' L ). Notice that the ac load line shows that the o/p signal swing can exceed the value of V CC. It is therefore necessary after obtaining the ac load line to check that the possible voltage swing does not exceed transistor maximum ratings. Signal Swing and Output AC Power:The Figure shows the voltage and current signal swings from the above circuit. The values of the peakto-peak signal swings are The ac power developed across the transformer primary can be calculated as (13)

13 The ac power calculated is that developed across the primary of the transformer. Assuming an ideal transformer, we find that the power delivered by the secondary to the load is approximately that calculated using Eq. (13). The o/p ac power can also be determined using the voltage delivered to the load. For the ideal transformer, the voltage delivered to the load can be calculated using Eq.(9): The power across the load can then be expressed as Eq. (5) Using Eq. (10) to calculate the load current yields With the o/p power then calculated using Example 4 Calculate the ac power delivered to the 8- speaker for the following circuit. The component values result in a dc base current of 6 ma, and the i/p signal (V i )results in a peak base current swing of 4 ma.

14 Solution: The dc load line is drawn vertically fro the voltage point: V CEQ = V CC = 10 V For I B = 6 ma, the operating point on the following graph is V CEQ = 10 V and I CQ = 140 ma The effective ac resistance seen at the primary is R' L = (3) 2 (8) = 72 To help draw the load line, consider the following procedure: For a current swing ofi C = V CE R L = 10 V 72 Ω = 139 ma Mark point A: I CEQ + I C = 140 ma ma = 279 ma along the y-axis Connect point A thro' 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 fro the graph are, respectively, V CEmin = 1.7 V V CEmax = 18.3 V I Cmin = 25 ma I Cmax = 255 ma Using Eq. (13), the ac power delivered to the load can be calculated P o (ac) = (18.3V 1.7V)(255mA 25mA) = W 8

15 Efficiency: The average i/p (dc) power drawn from the supply: P i (dc) = V CC I CQ (14) The dissipated power, P Q by the power transistor, ignoring the dissipated power by the transformer, P Q = P i (dc) P o (ac) (15) Example 5 For the above circuit and results of example 4, calculate the dc i/p power, power dissipated by the transistor, and efficiency of the circuit for i/p signal of example 4. Solution:From Eq. (14), P i (dc) = (10 V)(140 ma) = 1.4 W From Eq. (15), P Q = 1.4 W W = 0.92 W % η = P o (ac) P i (dc) 100%, check = 34.1% Maximum Theoretical Efficiency: For a class A transformer-coupled amplifier, themaximum theoretical efficiency goes up to 50%. Based on the signals obtained using the amplifier, the efficiency can be expressed as (16) The larger V CEmax and the smaller of V CEmin, the closer the efficiency approaches the theoretical limit of 50%.

16 Example 6 Calculate the efficiency of a transformer-coupled class A amplifier for a supply of 12 V and outputs of: (a) V(p) = 12 V. (b) V(p) = 6 V. (c) V(p) = 2 V. Solution: (a)since V CEQ = V CC = 12 V, V CEmax = V CEQ + V(p) = = 24 V V CEmin = V CEQ V(p) = 12 V 12 V = 0 V using Eq. (16), % η = 50% (b) check V CEmax = 18 V, % η = 12.5% V CEmin = 6 V (c) V CEmax = 14 V, V CEmin = 10 V % η = 1.39%

17 Class B Amplifier Operation: A class B operation is provided when the dc bias leaves the transistor biased just off, the transistor turning on when the ac signal is applied. This is essentially no bias, and the transistor conducts current for only one-half of the signal cycle. To obtain o/p for the full cycle of signal, it is necessary to use 2 transistors, each conduct on opposite half cycles, and the combined operation providing a full cycle of o/p signal. Since one part of the circuit pushes the signal high during one half-cycle and the other part pulls the signal low during the other half-cycle, the circuit is referred as a push-pull circuit. Input (DC) Power: The i/p dc power provided by the power supply can calculated using: (17) Where I dc is the average, or dc current drawn from the power supplies. (18) be Using Eq.(18) in Eq. (17) results in (19) Output (AC) Power: Using rms meter to measure the voltage across the load, P o (ac) is (20) If one uses an oscilloscope, the peak or peak-to-peak o/p voltage can be used:

18 (21) The larger the rms or peak o/p voltage, the larger is the power delivered to the load. Efficiency: The efficiency of the class B amplifier can be calculated using the basic equation % η = P o(ac) P i (dc) 100% Using Eqs. (19) and (21) in the efficiency equation results in %η = V L 2 (p)/2r L 100% = π V L (p) 100% V CC [( 2 (22) π )I(p)] 4 V CC [ Using I(p)=V L (p)/r L ]. Equation (22) shows that the larger the peak voltage, the higher is the circuit efficiency, up to maximum value when V L (p) = V CC, this maximum efficiency then being Maximum efficiency = π = 78. 5% Power Dissipation by output Transistors: This is the difference between the i/p power delivered by the supplies and the o/p power delivered to the load, P 2Q = P i (dc) P o (ac) (23) P 2Q is the power dissipated by the 2 o/p transistors. The dissipated power handled by each transistor is P Q = P 2Q 2 (24) Maximum Power Considerations For class B operation, the maximum o/p power is delivered to the load when V L (p) = V CC : (25) The corresponding peak ac current I(p) is then

19 I(p) = V CC R L So the maximum value of the average current from the power supply is maximum I dc = 2 π I(p) = 2V CC πr L Using this current to calculate the maximum value of i/p power results (26) Using (25) and (26), the maximum circuit efficiency for class B operation is then maximum % η = π 4 100% = 78.54% (27) For class B operation, the maximum power dissipated by the o/p transistors does not occur at the maximum power i/p or o/p condition. The maximum power dissipated by the two o/p transistors occurs when the o/p voltage across the load is V L (p) = 0.636V CC (= 2 π V CC) For a maximum transistor power dissipation of (28)

20 Example 7 For a class B amplifier providing a 20-V peak signal to a 16- load (speaker) and a power supply of V CC = 30 V, determine the i/p power, o/p power, and circuit efficiency. Solution: I L (p) = V L(p) R L = 20V 16Ω = 1.25 A The dc value of the current drawn from the power supply is then I dc = 2 π I L(p), check = A The i/p power delivered by the supply voltage is P i (dc) = V CC I dc, check = 23.9 W The i/p power delivered to the load is P o (ac) = V L 2 (p) 2R L, check = 12.5 W The circuit efficiency is calculated, check % = 52.3% Example 8 For a class B amplifier using a supply of V CC = 30 V and driving a load of 16, determine the maximum i/p power, o/p power, and transistor dissipation. Solution: Using Eq. (25), maximum P o (ac) = W (check) Using Eq. (26), maximum P i (dc) = W (check) The maximum % = 78.54% (as expected) Using Eq. (28), maximum P Q = 0.5(maximum P 2Q ) = 5.7 W

21 The maximum efficiency of a class B amplifier can also be expressed as follows: P o (ac) = V L 2 (p) 2R L P i (dc) = V CC I dc = V CC [ 2V L(p) πr L ] From which, we can get: % η max = V L (p) V CC % (29) Example 9 Calculate the efficiency of a class B amplifier for a supply voltage of V CC = 24 V with peak o/p voltages of: a) V L (p) = 22 V. b) V L (p) = 6 V. Solution: Using Eq. (29) gives

22 Class B Amplifier circuits: The i/p signal s to the amplifier could be single signal, the circuit then providing 2 different o/p stages, each operating for one-half the cycle. If the i/p is in the form of 2 oppositepolarity signals, 2 similar stages could be used, each operating on the alternate cycle because of the i/p signal. The Figure shows different ways to obtain phase-inverted signals from a single i/p signal. Fig. (a) shows a center-tapped transformer to provide opposite-phase signals. Fig. (b) uses a BJT stage with in-phase o/p from the emitter and opposite-phase o/p from the collector. If the gain is made nearly 1 for each o/p, the same magnitude results.

23 A- Transformer-Coupled Push-Pull Circuits During the first half-cycle of operation, transistor Q 1 is driven ON, whereas transistor Q 2 is driven Off. The current I 1 thro' the transformer results in the first half-cycle of signal to the load. During the second half-cycle of the i/p signal, Q 2 conducts, whereas Q 1 stays Off, the current I 2 thro' the transformer resulting in the second half-cycle to the load. The overall signal developed across the load then varies over the full cycle of signal operation.

24 B- Complementary-Symmetry Circuits Using complementary transistors (npn and pnp) it is possible to obtain a full cycle o/p across a load using half-cycles of operation from each transistor. One disadvantage of the circuit is the need for 2 separate voltage supplies. Another disadvantage is Crossover distortion which refers to the fact that during the signal crossover from +ve to ve. there is some nonlinearity in he o/p signal. This results from the fact that the circuit does not provide exact switching of one transistor off and the other ON at zero voltage condition.

25 A more practical version of a push-pull circuit using complementary transistors is shown in the following Figure. The load is driven as the o/p of an emitter-follower so that the load resistance of the load is matched by the low o/p resistance of the driving source. The circuit uses complementary Darlington-connected transistors to provide higher o/p current and lower o/p resistance.

26 C- Quasi-Complementary Push-Pull Amplifier In practical power amplifier circuits, it is preferable to use npn transistors for both high current-o/p devices. Since the push=pull connection requires complementary devices, a pnp high power transistor must be used. A practical means of obtaining complementary operation while using the same matched npn transistors for the o/p is provided by a quasi-complementary circuit shown in the following Figure. The push-pull operation is achieved by using complementary transistors (Q 1 and Q 2 ) before the matched npn o/p transistors (Q 3 and Q 4 ). Notice that Q 1 and Q 3 form a Darlington connection that provides o/p from a low-impedance emitter follower. The connection Q 2 and Q 4 forms a feedback pair, which similarly provides a low-impedance drive to the load. Resistor R2 can be adjusted to minimize crossover distortion by adjusting the dc bias condition.

27 Example 10 For the circuit shown, calculate the i/p power, o/p power, and power handled by each o/p transistor and the circuit efficiency for an i/p of 12 V rms. Solution Given: V CC = 25 V, R L = 4, V i (rms)=12 V V i (p)= 2V i (rms)= 16.97V, check P o (ac) = V L 2 (p) 2R L I L (p) = V L(p) R L = V L (p) (the amplifier has ideally a voltage gain = 1) = 36 W, check = 4.24 A I dc = 2 π I L(p) = 2.7 A P i (dc) = V CC I dc = W, check P Q = P 2Q = P i P o 2 2 = W, check % η = P o P i = %, check

28 Example 11 For the circuit of example 10, calculate the maximum i/p power, maximum o/p power, i/p voltage for maximum power operation, and power dissipated by the o/p transistors at this voltage. Determine the max. power dissipated by the o/p transistors, and the i/p voltage at which it occurs. Solution V CC = 25 V, R L = 4, max. P i (dc) = 2V CC 2 max. P o (ac) = V CC % η max. = P o P i 100 = % πr L = W, check 2 2R L = W, check To achieve max. power operation, the o/p voltage must be: V L (p) = V CC = 25 V P 2Q = P i P o = 21.3 W max. P 2Q = 2V CC = W π 2 R L This max. dissipation occurs at V L = V L (p) = 15.9 V 2 Note that at V L = 15.9 V, the o/p transistors dissipate W, Whereas at V L = 25 V they only dissipate 21.3 W.

29 Amplifier Distortion: A pure sinusoidal signal has a single frequency at which the voltage varies positive and negative by equal amounts. Any signal varying over less than the full 360o cycle is considered to have distortion. Nonlinear distortion occurs when the device characteristic is not linear. This can occur with all classes of amplifier operation. Frequency distortion occurs because the circuit elements and devices respond to the i/p signal differently at various frequencies. One technique for describing distortion is by using Fourier analysis, a method that describes any periodic waveform in terms of its fundamental frequency component and its integer multiples (harmonics). For example a signal that is originally 1 khz could result, after distortion, in a frequency component at 1kHz and harmonic components at 2 khz, 3 khz, 4 khz, and so on. The original signal of 1 khz is called fundamental frequency; those at integer multiples are the harmonics. The 2 khz is called a second harmonic, that at 3 khz is the third harmonic, and so on. Harmonic Distortion % D n If the fundamental frequency has amplitude A 1, and the nth frequency component has amplitude A n, the harmonic distortion can be defined as: (30) Total Harmonic Distortion % THD (31)

30 Example 12 Calculate the harmonic distortion components for an o/p signal having fundamental amplitude of 2.5 V, second harmonic amplitude 0.25 V, third harmonic of 0.1 V, and fourth harmonic of 0.05 V. Calculate the total harmonic distortion. Solution Using Eq. (30) yields Using Eq. (31), we obtain

31 Second Harmonic Distortion the Figure shows a collector current waveform I C with the quiescent I CQ, minimum I Cmin, and maximum I Cmax signal levels. The signal shown indicates that some distortion is present. An equation that approximately describes the distorted waveform is i C I CQ + I o + I 1 cosωt + I 2 cos2ωt (32) I CQ the quiescent current which occurs with zero i/p signal; I o is the nonzero average (dc) current of the distorted signal; I 1 is the fundamental component of the distorted ac signal; and I 2 is the second harmonic component at twice the fundamental frequency. a- At point 1 ( t = 0): i C = I Cmax = I CQ + I o + I 1 cos0 + I 2 cos0 i Cmax = I CQ + I o + I 1 + I 2 (a) b- At point 2 ( t = /2): i C = I CQ = I CQ + I o + I 1 cos π 2 + I 2cos 2π 2 I CQ = I CQ + I o I 2 (b) c- At point 3 ( t = ): i C = I Cmin = I CQ + I o + I 1 cos π + I 2 cos 2π I Cmin = I CQ + I o I 1 + I 2 (c) Adding (a), (c), substituting I o = I 2 from (b), we get

32 I o = I 2 = I Cmax + I Cmin 2I CQ 4 Subtracting (c) from (a), we get I 1 = I Cmax I Cmin 2 Using the definition of the second harmonic distortion as D 2 = I 2 I 1 100% 1 2 D 2 = (I Cmax+I Cmin ) I CQ 100% (33) I Cmax I Cmin In a similar manner, D 2 can be expressed in terms of measured collector-emitter voltages: 1 2 D 2 = (V CEmax+V CEmin ) V CEQ 100% (34) V CEmax V CEmin Power of a signal Having Distortion the o/p power delivered to the load resistor R C due to the fundamental component of the distorted signal is P 1 = I 1 2 R C 2 (35) The total power due to all the harmonic components of the distorted signal can be calculated using P = (I I I ) R C 2 (36) Taking I 1 2 as a common factor, the total power can be expressed in terms of the total harmonic distortion "THD", P = (1 + D D )I 1 2 R C 2 = (1 + THD 2 )P 1 (37)

33 Example 13 Calculate the second harmonic distortion if an o/p waveform displayed on an oscilloscope provides the following measurements: a) V CEmin = 1 V, V CEmax = 22 V, V CEQ = 12 V. b) V CEmin = 4 V, V CEmax = 20 V, V CEQ = 12 V. Solution Using Eq. (34), check that a) D 2 = 2.38 % b) D 2 = 0 % Example 14 For a harmonic distortion reading of D 2 = 0.1, D 3 = 0.02, and D 4 = 0.01, with I 1 = 4 ma and R C = 8, calculate the total harmonic distortion "THD", fundamental power component, and total power. Solution Using Eq. (31), check that THD 0.1 Using Eq. (35), check that P 1 = I 1 2 R C 2 = 64 W Using Eq. (37), check that P = W