Chapter 6. POWER AMPLIFIERS

Transcription

1 hapter 6. OWER AMFERS An aplifying syste usually has several cascaded stages. The input and interediate stages are sall signal aplifiers. Their function is only to aplify the input signal to a suitable value. The last stage usually drives a transducer such as a loud speaker, RT, Servootor etc. Hence this last stage aplifier ust be capable of handling and deliver appreciable power to the load. These large signal aplifiers are called as power aplifiers. ower aplifiers are classified according to the class operation, which is decided by the location of the quiescent point on the device characteristics. The different classes of operation are: (i) lass A (ii) lass (iii) lass A ((iv) lass ASS A OERATON: A siple transistor aplifier that supplies power to a pure resistive load R is shown above. et i represent the total instantaneous collector current, i c designate the instantaneous variation fro the quiescent value of. Siilarly, i,i b and represent corresponding base currents. The total

2 instantaneous collector to eitter voltage is given by v c and instantaneous variation fro the quiescent value is represented by v c. et us assue that the static output characteristics are equidistant for equal increents of input base current i b as shown in fig. below. f the input signal i b is a sinusoid, the output current and voltage are also sinusoidal.under these conditions, the non-linear distortion is neglible and the power output ay be found graphically as follows. = c c = c R () Where c & c are the rs values of the output voltage and current respectively.the nuerical values of c and c can be detered graphically in ters of the iu and iu voltage and current swings.t is seen that and c () c ower, ac = (3) This can also be written as, R (4) R

3 ac = (5) D power dc = Q ac dc 8 Q MAXMUM EFFENY: For a iu swing, refer the figure below. & 0 Q & 0 8 Q Q 5%

4 SEOND HARMON DSTORTON: n the previous section, the active device (JT) is treated as a perfectly linear device. ut in general, the dynaic transfer characteristics are not a straight line. This non-linearity arises because of the static output characteristics are not equidistant straight lines for constant increents of input excitation. f the dynaic curve is non-linear over the operating range, the wavefor of the output differs fro that of the input signal. Distortion of this type is called nonlinear or aplitude distortion. To investigate the agnitude of this distortion, we assue that the dynaic curve with respect to the quiescent point Q can be represented by a parabola rather than a straight line as shown below.

5 Thus instead of relating the alternating output current i c with the input excitation i b by the equation i c = Gi b resulting fro a linear circuit. We assue that the relationship between i c and i b is given ore accurately by the expression where the G s are constants. i c = G i b + G i b () Actually these two ters are the beginning of a power series expansion of i c as a function of i b. f the input wavefor is sinusoidal and of the for i b = b ost () Substituting equation (3), into equation () i c = G b os t + G b os t Since os t ost, the expression for the instantaneous total current reduces the for, i = + i c = os t + os t (3) Where s are constants which ay be evaluated in ters of the G s. The physical eaning of this equation is evident. t shows that the application of a sinusoidal signal on a parabolic dynaic characteristic results in an output current which contains, in addition to a ter of the sae frequency as the input, a second haronic ter and also a constant current. This constant ter 0 adds to the original dc value to yield a total dc coponent of current + 0.Thus the parabolic non-linear distortion introduces into the output a coponent whose frequency is twice that of the sinusoidal input excitation. The aplitudes 0, & for a given load resistor are readily detered fro either the static or the dynaic characteristics. Fro fig. 7. above, we observe that When ωt = 0, i c = ωt = π, ωt= π /, i c = (4) i c = y substituting thase values in equation (4) = = (5) = This set of three equations deteres the three unknowns 0, &.

6 t follows fro the second group that 0 = (6) y subtracting the third equation fro the first, = Then fro the first or last of equation (6), = 0 = (7) 4 c The second haronic distortion D is defined as, D = (8) (9) f the dynaic characteristics is given by the parabolic for & if the input contains two frequencies ω & ω, then the output will consist of a dc ter & sinusoidal coponents of frequencies ω, ω, ω,ω, ω +ω interodulation or cobination frequencies. and ω -ω.the su & difference frequencies are called HGHER ORDER HARMON GENERATON The analysis of the previous section assued a parabolic dynaic characteristic. ut this approxiation is usually valid for aplifier where the swing is sall. For a power aplifier with

7 a large input swing, it is necessary to express the dynaic transfer curve with respect to the Q point by a power series of the for, i 3 c Gib Gi b G3i b () f the input wave is a siple cosine function of tie, then i b Then, ost () b i c 0 ost ost 3os3t (3) Where 0,, 3 are the coefficients in the Fourier series for the current. i.e. the total output current is given by i Q ic = Q 0 ost ost (4) Where ( Q 0 is the dc coponent. Since i is an even function of tie, the Fourier series in equation (4) representing a periodic function possessing the syetry, contains only osine ters. Suppose we assue as an approxiation that haronics higher than the fourth are negligible in the above Fourier series, then we have five unknown ters 0,,, 3, & 4.To evaluate those we need output currents at five different value of. et us assue that i iost (5) Hence, iost (6) Q At t 0, Q i, i (7) Att, Q i, 3 i (8) At t, Q, i Q (9) Att, Q i, 3 i (0) At t, Q i, i () y cobining equations (4) & (7) to (), we get five equations & solving the, we get the following relations, 0 - Q () 6

8 (3) 3 Q (4) (5) Q (6) The haronic distortion is defined as, Where D, 3 D3, D n represents the distortion of the 4 D (7) th n haronic. Since this ethod uses five points on the output wavefor to obtain the aplitudes of haronics, the ethod is known as the five point ethod of detering the higher order haronic distortion. OWER OUTUT DUE TO DSTORTON f the distortion is not negligible, the power delivered to the load at the fundaental frequency is given by R () The ac power output is, ac R () = 3 D D (3) Where D, D3 etc are the second, third haronic distortions. Hence, ac D (4) Where D is the total distortion factor & is given by 3 D4 D D D (5)

9 For e.g. if D 0% of the fundaental, then ac 0. ac (6) When the total distortion is 0%, the power output is only %.higher than the fundaental power. Thus, only a sall error is ade in using only the fundaental ter for calculating the output power. THE TRANSFORMER OUED AUDO OWER AMER The ain reason for the poor efficiency of a direct-coupled classa aplifier is the large aount of dc power that the resistive load in collector dissipates. This proble can be solved by using a transforer for coupling the load. TRANDFORMER MEDANE MATHNG Assue that the transforer is ideal and there are no losses in the transforer. The resistance seen looking into the priary of the transforer is related to the resistance connected across the secondary.the ipedance atching properties follow the basic transforer relation. N N and () N N

10 Where riary voltage, Secondary voltage. riary current, Secondary current. N No. of turns in the priary. N No. of turns in the secondary. Fro Eq. () N N N N N N () n As both & are resistive ters, we can write R N R R n N (3) n an ideal transforer, there is no priary drop.thus the supply voltage collector-eitter voltage of the transistor. i.e (4) E When the values of the resistance R (= R R ) and operating point ay be calculated by the equation. E (5) R R appears as the are known, the base current at the

11 OERATNG ONT: Operating point is obtained graphically at the point of intersection of the dc load line and the transistor base current curve. After the operating point is detered; the next step is to construct the ac load line passing through this point. A OAD NE: n order to draw the ac load line, first calculate the load resistance looking into the priary side of the transforer. The effective load resistance is calculated using Eq.(3) fro the values of the secondary load resistance and transforer ratio. Having obtained the value of R, the ac loads line ust be drawn so that it passes through the operating point Q and has a slope equal to R. The dc and the ac load lines along the operating point Q are shown. n the above figure, two ac load lines are drawn through Q for different values of R.

12 For R very sall, the voltage swing and hence the output power, approaches zero.for R very large, the current swing is sall and again approaches zero.the variation of power & distortion wrto load resistance is shown in the plot below. EFFENY: Assue that the aplifier as supplying power to a pure resistance load. Then the average power input fro the dc supply ic. The power absorbed by the output circuit is, R ce, where & are the rs output current & voltage respectively & R is the static load ce resistance. f conservation.of energy, D is the average power dissipated by the active device, then by the principle of R ce D () Since EQ ce c, D ay be written in the for, D () EQ ce c f the load is not pure resistance, then ce e ust be replaced by ce c ust be replaced by ce c os, where os is the power factor of the load. The above equation expresses the aount of power that ust be dissipated by the active device. f the ac output power is zero i.e. f no applied signal exists, then (3) D E

13 acoutputpower % Efficiency, X (4) dcpowerinput n general, R X00% (5) n the distortion coponents are neglected, then % 0 X (6) MAXMUM EFFENY: An approxiate expression for efficiency can be obtained by assug ideal characteristic curves. Referring to above fig., iu values of the sine wave output voltage is, And (7) (8) Q

14 The rs value of collector voltage, Siilarly, ac rs. rs rs (9) (0) The output power is, rs () 8 The input power is, dc. dc Q ac () 8 Q The efficiency of a transforer coupled class A aplifier can also be expressed as, 50 % (3) The efficiency will be iu when 0, 0, cc & Q, substituting these values in eq.(), we get. Q X00 50% (4) 8. Q n practice, the efficiency of class A power aplifier is less than 50% due to losses in the transforer winding. DRAWAKS: () Total haronic distortion is very high. () The output transforer is subject to saturation proble due to the dc current in the priary.

15 USH-U AMFER: The distortion introduced by the non-linearity of the dynaic transfer characteristic ay be eliated by a circuit known as a known as push-pull configuration. t eploys two active devices and requires input signals 80 degrees out of phase with each other. The above figure shows a transforer coupled push-pull aplifier. The circuit consists of two centre tapped transforers T & T and two identical transistors Q and Q. The input transforer T does the phase splitting. t provides signals of opposite polarity to the transistor inputs. The output transforer T is required to couple the ac output signal fro the collector to the load. On application of a sinusoidal signal, one transistor aplifies the positive half-cycle of the input, whereas the other transistor aplifies the negative half cycle of the sae signal. When a transistor is operated as class- aplifier, the bias point should be fixed at cut-off so that practically no base current flows without an applied signal. onsider an input signal (base current of the for ib bost applied to Q. The output current of this transistor is given as, i 0 os t os t 3os3 t () The corresponding input signal to Q is i i b b b os t

16 The output current of this transistor is obtained by replacing t by t in expression for i. i.e. i 0 os t i t) i ( t ) () ( = 0 os t os t 3os3 t -- (3) As illustrated in the above fig., the current i & i are in opposite directions through the output transforer windings. The total output current is the proportional to the difference between the collector currents in the two transistors. i.e. i k i i k ost os3t (4) This expression shows that a push-pull circuit cancels out all even haronics in the output and will leave the third haronic as the principal source of distortion. This is true only when the two transistors are identical. f their characteristics differ appreciably, the even haronics ay appear. ADANTAGES OF USH-U SYSTEM: ecause no even haronics are present in the output of a push-pull aplifier,such a circuit will give ore output per active device for given aount of distortion. Also, a push-pull arrangeent ay be used to obtain less distortion for given power output per transistor. t can be noticed that the dc coponent of the collector current oppose each other agnetically in the transforer core. This eliates any tendency towards core saturation and consequent non-linear distortion that ight arise fro the curvature of the agnetization curve. Another advantage of this syste is that the effects of ripple voltages that ay be contained in the power supply because of inadequate filtering will be balanced out. This cancellation results because the currents produced by this ripple voltage are in opposite directions in the transforer winding and so will not appear in the load. ASS- AMFER The circuit for the class- push pull syste is the sae as that for the class A syste except that the devices are biased approxiately at cut-off. The above circuit (class A) operates in class if R =0 because a silicon transistor is essentially at cut off if the base is shorted to the eitter. ADANTAGES OF ASS OERATON () t is possible to obtain greater power output () Efficiency is higher

17 (3) Negligible power loss at no signal. DRAWAKS OF ASS AMFER () Haronic distortion is higher () Self bias can t be used (3) Supply voltage ust have good regulation OWER ONSDERATON. To investigate the power conversion efficiency of the syste, it is assued that the output characteristics are equally spaces for equal intervals of excitation, so that the dynaic transfer curve is a straight line. t also assues that the iu current is zero. The graphical construction fro which to detere the output current & voltage wave3shapes for a single transistor operating as a class stage is indicated in the above figure. Note that for sinusoidal excitation, the output is sinusoidal during one half of each period and is zero during the second half cycle. The effective load resistance is priary turn to the center tap. R N R N where N represents the nuber of The wavefor illustrated in the above figure represents one transistor Q only. The output of Q is, of course, a series of sine loop pulses that are 80 degrees out of phase with those of Q. The load current, which is proportional to the difference between the two collector currents, is therefore a perfect sine wave for the ideal conditions assued. The power output is

18 () The corresponding direct collector current ion each transistor under load is the average value of the half sine loop since dc for this wavefor, the dc input power fro the supply is, i () The factor in this expression arises because two transistors are used in the push-pull syste. Fro equations () and (), Since 0 X00 4 i 4 X00% X00% 78.5% (3) 4 The large value of results fro the fact that there is no current in a class syste if there is no excitation, where as there is a drain fro the power supply in class A syste even at zero signal. OWER DSSATON The power dissipation dc power input. in both transistors is the difference between the ac power output and dc ac i 0 = = R R (5) This equation shows that the collector dissipation is zero at no signal 0, Rises as is increases and passes through a iu at.

19 MAXMUM OWER DSSATON The condition for iu power dissipation can be found by differentiating eq.(5) wrt and equating it to zero. d d R 0 R R R (6) Substituting the value of in eq.(5), we get, R 4 = R R X = R R (7) Output power, 0 R When 0, R (8) Equation (7) can be written as, 4 R 4, 0, , 0, (9) Equation (9) gives the iu power dissipated by both the transistors and therefore the iu power dissipation per transistor is, 4 0,, per transistor = 0. 0,, (0)

20 f, for e.g. 0W iu power is to be delivered fro a class push-pull aplifier to the load, then power dissipation ratio of each transistor should be 0. X 0W=W. HARMON DSTORTON The output of a push-pull syste always possesses irror syetry, so that 0, & We know that When , &, the above equations reduce to () () (3) Note that there is no even haronic distortion. The ajor contribution to distortion is the third haronic and is given by, 3 D 00% (4) 3 The output power taking distortion into account is given by

21 R 0 D (5) SEA RUTS Fig 3 A circuit that avoids using the output transforer is shown above. This configuration requires a power supply whose centre tap is grounded. Here, high powered transistors are used. They have a collector to eitter output ipedance in the order of 4 to 8.This allows single ended push-pull operation. The voltage developed across the load is again due to the difference in collector currents i i, so this is a true push-pull application. ROEMS. alculate the input power and efficiency of the aplifier shown below for an input oltage resulting in a base current of 0A peak. Also calculate the power dissipated by the transistor.

22 SON: The Q point for the given circuit is detered as follows. R X0 E A 5X9.3A 48. 5A 3 E R X0 X ( peak) ( peak) 5X0A 50A nput power, EAK ac 0. 65W dc 0X 48.5 X W ac 0.65 Efficiency, X00% 6.48% 9.65 dc ower dissipated by the transistor, dc ac W : A class A power aplifier with a direct coupled load has a collector efficiency of 30% and delivers a power input of 0W.Find (a) the dc power input (b) the power dissipation `of full output and () the desirable power dissipation rating of the JT. SON: Given 0W, 30% ac

23 (a) ac dc ac dc (b) Dissipation at full output, W W (c) Dissipation at no output, JT rating = 33.33W dc W 3: A JT supplies 0.85 W to a 4K load. The zero signal dc collector current is 3 A. Detere the percent second haronic distortion. SON: Given 0.85W,, Zero 3A R 4 k,, signal 34A Using the dynaic characteristics of the transistor, i c Gib Gi b We have 0 signal nosignal =34-3=3A R ower, Or or R X X0 4K 0. 6A The second haronic distortions, 3A D X00% X00% 4.6% 0.6A 4. Design a class push pull circuit to deliver 00W to a 4 load. Output transforer efficiency is 70%., E =5, average rating of the transistor to be used is 65Mw at 5 0.Detere, collector to collector resistance R

24 SON: Given ac =00W, R =4, 0. 7, E() =5, trans =65W, at 5 0,R E =0. Assue that the given power delivered to the load is iu. ac 00 ac W 0.7 Maxiu voltage rating per transistor is. et = R acpriary =.5. R X X0 3 acpriary ut n R R R N 8 N R n 4 5 N N 0.5 R on priary of transforer. 5: A single stage, class A aplifier has =0, EQ = 0, Q = 600A, and ac output current is varied by 300A with the ac input signal. Detere the (a) power supplied by the dc source to the aplifier circuit (b) dc power consued by the load resistor (c) ac power developed across the load resistor (d) dc power delivered to the transistor (e) dc power wasted in the transistor collector (f) overall efficiency (g) collector efficiency. SON: Given = 0, EQ =0, Q = 600Ma, R = 6, = 300Ma (a) ower supplied by the dc source to the aplifier circuit is given by dc =. Q =0X0.6=W (b) D power consued by the load resistor is given by dc = ( Q ) R = (0.6) X6 = 5.76 W (c) A power developed across the load resistor is ac.

25 W ac R 7 (d) D power delivered to the transistor dc dc dc W tr 4 (e) D ower wasted in the transistor collected is dc tr 5 dc ac W ac 0.7 (f) Overall efficiency, % dc (g) ollector efficiency, ac tr dc 0.7.5% 6.4