Lecture 8: Peak Rectifiers.

Transcription

1 Whies, EE 320 Lecure 8 Page 1 of 10 Lecure 8: Peak Recifiers. he ouu of he recifier circuis discussed in he las lecure is ulsaing significanly wih ime. Hence, i s no useful as he ouu from a DC ower suly. One way o reduce his rile is o use a filering caacior. Consider he half-cycle recifier again, bu now add a caacior in arallel wih he load: (Fig. 4.27a) We exec ha as soon as we urn on he source, he caacior will charge u on + cycles of v I and discharge on he - cycles. o smooh ou he volage, we need his discharge o occur slowly in ime. his means we need o choose C large enough o make his haen, resuming ha R is a given quaniy (he hévenin resisance of he res of he circui) Keih W. Whies

2 Whies, EE 320 Lecure 8 Page 2 of 10 he ouu volage v O will hen be a smoohed-ou signal ha ulsaes wih ime: (Fig. 4.27) Noice he diode curren and he caacior volage. hey dislay behavior much differen han wha one would find in an AC circui. Analysis of Peak Recifier Circuis We ll require ha RC, which means ha he ime consan of he RC circui mus be much greaer ha he eriod of he inu sinusoidal signal:

3 Whies, EE 320 Lecure 8 Page 3 of 10 Now, our ques is o aroximaely deermine he rile volage V r, assuming : v Essenially a sraigh line for V r V v O d Diode on v I No skeched o scale. When D is off, and assuming i is an ideal diode v V e (1) O [If D is no ideal hen O 0.7 v V e.] A he end of he discharge ime, d, he ouu volage equals v V V (2) O d r Subsiuing for v O from (1) a his ime d leads o d d Ve V or 1 e (3) V

4 Whies, EE 320 Lecure 8 Page 4 of 10 his equaion has he wo unknowns V r and d, assuming is known. If we can deermine d, hen we can find V r. Finding d can be done numerically by equaing (1) o he exression for he inu volage vi Vcos (4) and solving for he ime d when he wo are equal as d cos d V d Ve or cos d e (5) his needs o be done numerically since (5) is a ranscendenal equaion. Alernaively, if is small comared o (rue when, as assumed), hen from (3) 1e 1 e (6) V Again, because hen we can runcae he series exansion of he exonenial funcion o wo erms (see Lecure 4) giving ( ) (7) V his simle equaion gives he raio of he rile volage o he eak volage of he inu sinusoidal signal for he half-cycle recifier. I s worh memorizing, or knowing how o derive. Ofen R and are fixed quaniies. So from (7) V ( ) (4.28),(8) RC o obain a small rile volage we need a large C in his case.

5 Whies, EE 320 Lecure 8 Page 5 of 10 Conducion Inerval Lasly, he conducion inerval is defined as he ime inerval in which he diode is acually conducing curren. his ime eriod is skeched in he receding wo figures. he diode conducs curren beginning a ime d and ending a, wihin each eriod. Using equaion (4) a ime d V cos V V V cos V V (9) or d r r We exec he conducion inerval o be small. So runcaing he series exansion of cosine o wo erms, (9) gives 2V r (4.30),(10) V he facor is someimes called he conducion angle,. For V his conducion angle (and conducion inerval) will be small, as execed. Discussion o reierae, he objecive of he eak recifier is o charge he shun C when D is on, and slowly discharge i during hose imes when D is off.

6 Whies, EE 320 Lecure 8 Page 6 of 10 When does D conduc? During he eriods in he revious figure. Also see Fig. 4.27(c). Noe ha his eak recifier is no a linear circui. i D is a very comlicaed waveform and no a sinusoid, as seen earlier in Fig. 4.27(c). here are no simle exac formulas for he soluion o his roblem. he ex only shows aroximae soluions for eak i D : V 2V id 12 [A] ( V R V r V ) (4.32),(11) r Examle N8.1 (similar o ex Examle 4.8). A half-cycle eak recifier wih R = 10 k is fed by a 60-Hz sinusoidal volage wih a eak amliude of 100 V. (a) Deermine C for a rile volage of 2 V. From (8): V C RV r 6010,000 2 or C 83.3 F. For a facor of safey of wo, make C wice as large. Remember, a bigger C ranslaes o smaller rile. (b) Deermine he eak diode curren. Using (11):

7 Whies, EE 320 Lecure 8 Page 7 of 10 V 2V id R V r 10,000 2 or id 638 ma. When secifying a diode for your circui design, you would need o find one ha could safely handle his amoun of curren. Examle N8.2. A half-cycle eak recifier wih R = 10 k is fed by a 60-Hz riangular volage wih a eak amliude of 100 V. (a) Deermine C for a rile volage of 2 V. If you go back and look a he derivaion of (8) you ll find ha here were no aroximaions made ha required a sinusoidal waveform. Consequenly, (8) alies o his riangular waveform as well, rovided. Hence, as before C 83.3 F. (b) Deermine he diode conducion ime,. Referring o his skech of he region near he osiive eak volage for v I :

8 Whies, EE 320 Lecure 8 Page 8 of 10 V r v O v I V Diode on Because he rising orion of he waveform is a sraigh line: rise V vi run / 4 o find, equae 4V 4V vi or V r herefore, for a riangular waveform (12) 4 V In his aricular case, s Comare his ime o a sinusoidal waveform: s 2 V his ime is much longer han for he riangular waveform. Consequenly, we would exec i D for D o be much larger for he riangular waveform han for he sinusoid!

9 Whies, EE 320 Lecure 8 Page 9 of 10 Full-Cycle Peak Recifiers In a similar fashion, we can also add a shun C o full cycle and bridge recifiers o conver hem o eak recifiers. For examle, for a full-cycle eak recifier: he ouu volage has less rile han from a half-cycle eak recifier (acually one half less rile). v O v I he rile frequency is wice ha of a half-cycle eak recifier. Using he same derivaion rocedure as before wih he half cycle, bu wih 2 gives from (7) ( ) (4.33),(13) V 2

10 Whies, EE 320 Lecure 8 Page 10 of 10 Lasly, i can be shown ha he i D for he full-cycle eak recifier: V V id 12 [A] (4.35),(14) R 2V r is aroximaely one-half ha of he half-cycle eak recifier when V.