3.5 Rectifier Circuits

Transcription

1 9/24/2004 3_5 Rectfer Crcuts empty.doc 1/2 3.5 Rectfer Crcuts A. Juncton ode 2-Port Networks - ( t ) Juncton ode Crcut ( t ) H: The Transfer Functon of ode Crcuts Q: A: H: teps for fndng a Juncton ode Crcut Transfer Functon Example: ode Crcut Transfer Functon Jm tles The Un. of Kansas ept. of EEC

2 9/24/2004 3_5 Rectfer Crcuts empty.doc 2/2 B. ode Rectfers H: gnal Rectfcaton Q: A: H: The Full-Wae Rectfer H: The Brdge Rectfer H: Peak nerse Voltage Jm tles The Un. of Kansas ept. of EEC

3 9/24/2004 The Transfer Functon of ode Crcuts.doc 1/5 The Transfer Functon of ode Crcuts For many juncton dode crcuts, we fnd that one of the oltage sources s n fact unknown! Ths unknown oltage s typcally some nput sgnal of the form (t), whch results n an output oltage (t). Juncton ( t ) ode ( ) - t Crcut Q: How the heck do you expect us to determne (t) f we hae no dea what (t) s?? A: We of course cannot determne an explct alue or expresson for (t), snce t depends on the nput (t). nstead, we wll attempt to explctly determne ths dependence of (t) on (t)! Jm tles The Un. of Kansas ept. of EEC

4 9/24/2004 The Transfer Functon of ode Crcuts.doc 2/5 n other words, we seek to fnd an expresson for n terms of. Mathematcally speakng, our goal s to determne the functon: = f ( ) We refer to ths as the crcut transfer functon. Note that we can plot a crcut transfer functon on a 2- dmensonal plane, just as f the functon related alues x and y = f x ). For example, say our crcut transfer y (e.g. ( ) functon s: ( ) = f = 3 2 Note ths s smply the equaton of a lne (e.g., y = 3x 2), wth slope m=3 and ntercept b=2. 2 f ( ) 3 Jm tles The Un. of Kansas ept. of EEC

5 9/24/2004 The Transfer Functon of ode Crcuts.doc 3/5 Q: A functon eh? sn t a functon just your annoyngly pretentous way of sayng we need to fnd some mathematc equaton relatng and? A: Actually no! Although a functon s a mathematcal equaton, there are n fact scads of equatons relatng and that are not functons! The set of all possble functons y f ( x) = are a subset of the set of all possble equatons relatng y and x. = s a mathematcal expresson such that for any alue of (.e., < < ), there s one, but only one, alue. A functon f ( ) Note ths defnton of a functon s consstent wth our physcal understandng of crcuts we can place any oltage on the nput that we want (.e., < < ), and the result wll be one specfc oltage alue on the output. Therefore, examples of ald crcut transfer functons nclude: Jm tles The Un. of Kansas ept. of EEC

6 9/24/2004 The Transfer Functon of ode Crcuts.doc 4/5 Conersely, the transfer functons below are nald they cannot represent the behaor of crcuts, snce they are not functons! Jm tles The Un. of Kansas ept. of EEC

7 9/24/2004 The Transfer Functon of ode Crcuts.doc 5/5 Moreoer, we fnd that crcut transfer functons must be contnuous. That s, cannot nstantaneously change from one alue to another as we ncrease (or decrease) the alue. A scontnuous Functon (nald crcut transfer functon) A Contnuous Functon (Vald crcut transfer functon) Remember, the transfer functon of eery juncton dode crcut must be a contnuous functon. f t s not, you e done somethng wrong! Jm tles The Un. of Kansas ept. of EEC

8 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 1/9 teps for Fndng a Juncton ode Crcut Transfer Functon etermnng the transfer functon of a juncton dode crcut s n many ways ery smlar to the analyss steps we followed when analyzng preous juncton dode crcuts (.e., crcuts where all sources were explctly known). Howeer, there are also some mportant dfferences that we must understand completely f we wsh to successfully determne the correct transfer functon! tep1: Replace all juncton dodes wth an approprate juncton dode model. Just lke before! We wll now hae an EAL dode crcut. tep 2: AUME some mode for all deal dodes. Just lke before! An EAL dode can be ether forward or reerse based. Jm tles The Un. of Kansas ept. of EEC

9 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 2/9 tep 3: ENFRCE the bas assumpton. Just lke before! ENFRCE the bas assumpton by replacng the deal dode wth short crcut or open crcut. tep 4: ANALYZE the remanng crcut. ort of, knd of, lke before! 1. f we assumed an EAL dode was forward based, we must determne --just lke before! Howeer, nstead of fndng the numerc alue of, we determne = f ). as a functon of the unknown source (e.g., ( ) 2. r, f we assumed an EAL dode was reersed based, we must determne --just lke before! Howeer, nstead of fndng the numerc alue of, we determne as a functon of the unknown source (e.g., ( ) = f ). 3. Fnally, we must determne all the other oltages and/or currents we are nterested n (e.g., )--just lke before! Howeer, nstead of fndng ts numerc alue, we determne t as a functon of the unknown source (e.g., = f ). ( ) Jm tles The Un. of Kansas ept. of EEC

10 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 3/9 tep 5: etermne WHEN the assumpton s ald. Q: K, we get the pcture. Now we hae to CHECK to see f our EAL dode assumpton was correct, rght? A: Actually, no! Ths step s ery dfferent from what we dd before! We cannot determne F > 0 (forward bas assumpton), or F < 0 (reerse bas assumpton), snce we cannot say for certan what the alue of or s! Recall that and are functons of the unknown oltage source (e.g., = f ( ) and = f ( ) ). Thus, the alues of or are dependent on the unknown source (, say). For some alues of, we wll fnd that > 0 or < 0, and so our assumpton (and thus our soluton for = f ( )) wll be! correct Howeer, for other alues of, we wll fnd that < 0 or > 0, and so our assumpton (and thus our soluton for ( ) = f ) wll be ncorrect! Q: Ykes! What do we do? How can we determne the crcut transfer functon f we can t determne F our deal dode assumpton s correct?? Jm tles The Un. of Kansas ept. of EEC

11 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 4/9 A: nstead of determnng F our assumpton s correct, we must determne WHEN our assumpton s correct! n other words, we must determne for what alues of s > 0 (forward bas), or for what alues of s < 0 (reerse bas). We can do ths snce we earler (n step 4) determned the f = f. functon = ( ) or the functon ( ) Perhaps ths step s best explaned by an example. Let s say we assumed that our deal dode was forward based and, say we determned (n step 4) that s related to as: ( ) = f = 2 3 Lkewse, say that we determned (n step 4) that our deal dode current s related to as: ( ) = f 5 > 4 Thus, n order for our forward bas assumpton to be correct, = f must be greater than zero: the functon ( ) Jm tles The Un. of Kansas ept. of EEC

12 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 5/9 > 0 f ( ) > 0 5 > 0 4 We can now sole ths nequalty for : Q: What does ths mean? oes t mean that s some alue greater than 5.0V?? 5 > > 0 > 5 A: N! Recall that can be any alue. What the nequalty aboe means s that > 0 (.e., the deal dode s forward based) WHEN > 50.. Thus, we know = 2 3s ald WHEN the deal dode s forward based, and the deal dode s forward based WHEN (for ths example) > 50.. As a result, we can mathematcally state that: = 2 3 when > 5 0. V Jm tles The Un. of Kansas ept. of EEC

13 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 6/9 Conersely, ths means that f < 50. V, the deal dode wll be reerse based our forward bas assumpton would not be ald, and thus our expresson = 2 3 s not correct ( 2 3 for < 50V. )! Q: o how do we determne for alues of < 5.0 V? A: Tme to moe to the last step! tep 6: Change assumpton and repeat steps 2 through 5! For our example, we would change our bas assumpton and now AUME reerse bas. We then ENFRCE = 0, and then ANALYZE the crcut to fnd both f ( ) expresson f ( ) = and a new = (t wll no longer be = 2 3!). We then determne WHEN our reerse bas assumpton s = f > for. For the ald, by solng the nequalty ( ) 0 example used here, we would fnd that the EAL dode s reerse based WHEN < 50V.. For juncton dode crcuts wth multple dodes, we may hae to repeat ths entre process multple tmes, untl all possble bas condtons are analyzed. Jm tles The Un. of Kansas ept. of EEC

14 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 7/9 f we hae done our analyss properly, the result wll be a ald contnuous functon! That s, we wll hae an expresson (but only one expresson) relatng to all possble alues of. Ths transfer functon wll typcally be pecewse lnear. An example of a pece-wse lnear transfer functon s: 2 3 for > 50. = 12 for < Just to make sure that we understand what a functon s, note that the followng expresson s not a functon: 2 3 for > 70. = 12 for < 3 0. Jm tles The Un. of Kansas ept. of EEC

15 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 8/ Nor s ths expresson a functon: 2 3 for > 30. = 12 for < Jm tles The Un. of Kansas ept. of EEC

16 9/24/2004 teps for Fndng a Juncton ode Crcut Transfer Functon.doc 9/9 Fnally, note that the followng expresson s a functon, but t s not contnuous: 2 3 for > 50. = 5 for < Make sure that the pecewse transfer functon that you determne s n fact a functon, and s contnuous! Jm tles The Un. of Kansas ept. of EEC

17 9/24/2004 Example ode Crcut Transfer Functon.doc 1/5 Example: ode Crcut Transfer Functon Consder the followng crcut, called a half-wae rectfer: (t) - (t) (t) R (t) Let s use the CV model to determne the output oltage n terms of the nput oltage. n other words, let s determne the = f! dode crcut transfer functon ( ) AUME the deal dode s forward based, ENFRCE = 0. = 0 (t) _ 0.7 V _ R (t) _ From KVL, we fnd that: ( t) = ( t ) 07.

18 9/24/2004 Example ode Crcut Transfer Functon.doc 2/5 Ths result s of course true f our orgnal assumpton s correct t s ald f the deal dode s forward based (.e., > 0 )! From hm s Law, we fnd that: 07. = = R R Q: m so confused! s ths current greater than zero or less than zero? s our assumpton correct? How can we tell? A: The deal dode current s dependent on the alue of source oltage ( t ). As such, we cannot determne f our assumpton s correct, we nstead must fnd out when our assumpton s correct! n other words, we know that the forward bas assumpton s correct when > 0. We can rearrage our dode current expresson to determne for what alues of source oltage (t) ths s true: > 0 ( t ) 07. > 0 R ( t ) 07. > 0 ( t ) >. 07

19 9/24/2004 Example ode Crcut Transfer Functon.doc 3/5 o, we hae found that when the source oltage (t) s greater than 0.7 V, the output oltage (t) s: ( t) = ( t ) 07. Q: K, e got ths result wrtten down. Howeer, stll don t know what the output oltage (t) s when the source oltage (t) s less than 0.7V!?! Now we change our assumpton and AUME the deal dode n the CV model s reerse based, an assumpton ENFRCEd wth the condton that = 0 (.e., an open crcut). = 0 (t) _ 0.7 V From hm s Law, we fnd that the output oltage s: _ R (t) _ Q: Fascnatng! The output oltage s zero when the deal dode s reerse based. But, precsely when s the deal dode reerse based? For what alues of does ths occur? = R = R (0) = 0V!!!

20 9/24/2004 Example ode Crcut Transfer Functon.doc 4/5 A: To answer these questons, we must determne the deal dode = f ): oltage n terms of (.e., ( ) From KVL: 07. = Therefore: = 07. = = 07. Thus, the deal dode s n reerse bas when: olng for, we fnd: < < < 0 < 07V. n other words, we hae determned that the deal dode wll be reerse based when < 07V., and that the output oltage wll be = 0. Q: o, we hae found that: = 07. when > 07V. and, = 00. when < 07V. t appears we hae a ald, contnuous, functon!

21 9/24/2004 Example ode Crcut Transfer Functon.doc 5/5 A: That s rght! The transfer functon for ths crcut s therefore: 07. for > 07. = 0 for < V Although the crcut n ths example may seem tral, t s actually ery mportant! t s called a half-wae rectfer, and prodes sgnal rectfcaton. Rectfers are an essental part of eery AC to C power supply!

22 9/24/2004 gnal Rectfcaton.doc 1/11 gnal Rectfcaton An mportant applcaton of juncton dodes s sgnal rectfcaton. There are two types of sgnal rectfers, half-wae and fullwae. Let s frst consder the deal half-wae rectfer. t s a = f : crcut wth the transfer functon ( ) deal ( t ) ½ Wae ( t ) - Rectfer 0 for < 0 = for > 0 1

23 9/24/2004 gnal Rectfcaton.doc 2/11 Pretty smple! When the nput s negate, the output s zero, whereas when the nput s poste, the output s the same as the nput. Q: Pretty smple and pretty stupd d say! Ths mght be your most pontless crcut yet. How n the world s ths crcut useful?? A: To see why a half-wae rectfer s useful, consder the typcal case where the nput source oltage s a snusodal sgnal wth frequency ω and peak magntude A: s (t) ( t) = Asnωt A 0 t -A Thnk about what the output of the half-wae rectfer would be! Remember the rule: when (t) s negate, the output s zero, when (t) s poste, the output s equal to the nput.

24 9/24/2004 gnal Rectfcaton.doc 3/11 The output of the half-wae rectfer for ths example s therefore: A (t) 0 t -A (t) Q: That s maybe the lamest result e eer seen. What good s half a sne wae? Why do we een bother? A: Although t may appear that our rectfer had lttle useful effect on the nput sgnal (t), n fact the dfference between nput (t) and output (t) s both mportant and profound. To see how, consder frst the C component (.e. the tmeaeraged alue) of the nput sne wae: T 1 V = ( t) dt T 0 T 1 = Asnωtdt = T 0 0

25 9/24/2004 gnal Rectfcaton.doc 4/11 Thus, (as you probably already knew) the C component of a sne wae s zero a sne wae s an AC sgnal! Now, contrast ths wth the output (t) of our half-wae rectfer. The C component of the output s: T 1 V = ( t) dt T 0 T 2 T 1 1 A = Asnωtdt 0dt T = T π 0 T 2 Unlke the nput, the output has a non-zero (poste) C component ( V = A π )! A (t) V = A π t -A Q: see. A non-zero C component eh? o refresh my memory, why s that mportant?

26 9/24/2004 gnal Rectfcaton.doc 5/11 A: Recall that the power dstrbuton system we use s an AC system. The source oltage (t) that we get when we plug our power cord nto the wall socket s a 60 Hz snewae a source wth a zero C component! The problem wth ths s that most electronc deces and systems, such as TVs, stereos, computers, etc., requre a C oltage(s) to operate! Q: But, how can we create a C supply oltage f our power source (t) has no C component?? A: That s why the half-wae rectfer s so mportant! t takes an AC source wth no C component and creates a sgnal wth both a C and AC component. We can then pass the output of a half-wae rectfer through a low-pass flter, whch suppresses the AC component but lets the C alue ( V = A π ) pass through. We then regulate ths output and form a useful C oltage source one sutable for powerng our electronc systems! A Power upply

27 9/24/2004 gnal Rectfcaton.doc 6/11 Q: K, now see why the deal half-wae rectfer mght be useful. But, s there any way to actually buld ths magcal dece? A: An deal half-wae rectfer can be bult f we use an deal dode. - (t) R (t) _ f we follow the transfer functon analyss steps we studed earler, then we wll fnd that ths crcut s ndeed an deal half-wae rectfer! 0 for < 0 = for > 0 1

28 9/24/2004 gnal Rectfcaton.doc 7/11 f course, snce deal dodes do not exst, we must use a juncton dode nstead: (t) - (t) (t) R (t) Q: Ths crcut looks so famlar! Haen t we studed t before? A: Yes! t was an example where we determned the juncton dode crcut transfer functon. Recall that the result was: 07. for > 07. = 0 for < V Note that ths result s slghtly dfferent from that of the deal half-wae rectfer! The 0.7 V drop across the juncton dode causes a horzontal shft of the transfer functon from the deal case. Q: o, ths juncton dode crcut s worthless?

29 9/24/2004 gnal Rectfcaton.doc 8/11 A: Hardly! Although the transfer functon s not qute deal, t works well enough to achee the goal of sgnal rectfcaton t takes an nput wth no C component and creates an output wth a sgnfcant C component! Note what the transfer functon rule s now: 1. When the nput s greater than 0.7 V, the output oltage s equal to the nput oltage mnus 0.7 V. 2. When the nput s less than 0.7 V, the output oltage s zero. o, let s consder agan the case where the source oltage s snusodal (just lke the source from a wall socket!): s (t) ( t) = Asnωt A 0.7 t -A The output of our juncton dode half-wae rectfer would therefore be:

30 9/24/2004 gnal Rectfcaton.doc 9/11 A (t) 0.7 t -A (t) Although the output s shfted downward by 0.7 V (note n the plot aboe ths s exaggerated, typcally A>>0.7V), t should be apparent that the output sgnal (t), unlke the nput sgnal (t), has a non-zero (poste) C component. Because of the 0.7 V shft, ths C component s slghtly smaller than the deal case. n fact, we fnd that f A>>0.7, ths C component s approxmately: A V 035. V π n other words, just 350 mv less than deal! Q: Way back on the frst page you sad that there were two types of rectfers. now understand halfwae rectfcaton, but what about these so-called full-wae rectfers?

31 9/24/2004 gnal Rectfcaton.doc 10/11 A: Almost forgot! Let s examne the transfer functon of an deal full-wae rectfer: deal ( t ) ( ) - Full-Wae t Rectfer for < 0 = for > f the deal half-wae rectfer makes negate nputs zero, the deal full-wae rectfer makes negate nputs poste! For example, f we agan consder our snusodal nput, we fnd that the output wll be: (t) A 0 t -A (t)

32 9/24/2004 gnal Rectfcaton.doc 11/11 The result s that the output sgnal wll hae a C component twce that of the deal half-wae rectfer! T 1 V = ( t) dt T 0 A T 2 T 1 1 = A snωt dt A snωt dt = T T 0 (t) T 2 2A π V = 2A π t -A Q: Wow! Full-wae rectfcaton appears to be twce as good as halfwae. Can we buld an deal full-wae rectfer wth juncton dodes? A: Although we cannot buld an deal full-wae rectfer wth juncton dodes, we can buld full-wae rectfers that are ery close to deal wth juncton dodes!

33 9/24/2004 The Full Wae Rectfer.doc 1/10 The Full-Wae Rectfer Consder the followng juncton dode crcut: 1 s (t) R (t) Power Lne s (t) 2 Note that we are usng a transformer n ths crcut. The job of ths transformer s to step-down the large oltage on our power lne (120 V rms) to some smaller magntude (typcally V rms). Note the secondary wndng has a center tap that s grounded. Thus, the secondary oltage s dstrbuted symmetrcally on ether sde of ths center tap. For example, f = 10 V, the anode of 1 wll be 10V aboe ground potental, whle the anode of 2 wll be 10V below ground potental (.e., -10V): Jm tles The Un. of Kansas ept. of EEC

34 9/24/2004 The Full Wae Rectfer.doc 2/10 1 =10 V R (t) Power Lne =10V 2 Conersely, f =-10 V, the anode of 1 wll be 10V below ground potental (.e., -10V), whle the anode of 2 wll be 10V aboe ground potental: 1 =-10 V R (t) Power Lne =-10V 2 Jm tles The Un. of Kansas ept. of EEC

35 9/24/2004 The Full Wae Rectfer.doc 3/10 The more mportant queston s, what s the alue of output? More specfcally, how s related to the alue of source = f? what s the transfer fucton ( ) To help smplfy our analyss, we are gong redraw ths crucut n another way. Frst, we wll splt the secondary wndng nto two explct peces: 1 R (t) Power Lne 2 We wll now gnore the prmary wndng of the transformer and redraw the remanng crcut as: 1 2 R (t) Jm tles The Un. of Kansas ept. of EEC

36 9/24/2004 The Full Wae Rectfer.doc 4/10 Note that the secondary oltages at ether end of ths crcut are the same, but hae opposte polarty. As a result, f =10, then the anode of dode 1 wll be 10 V aboe ground, and the anode at dode 2 wll be 10V below ground just lke before! 1 2 =10 R (t) _ =10 _ Now, let s attempt to determne the transfer functon = f of ths crcut. ( ) Frst, we wll replace the juncton dodes wth CV models. Then let s AUME 1 s forward based and 2 s reerse based, thus ENFRCE 1 = 0 and 2 = 0. Thus ANALYZE: 1 = = _ R (t) Jm tles The Un. of Kansas ept. of EEC

37 9/24/2004 The Full Wae Rectfer.doc 5/10 Note that we need to determne 3 thngs: the deal dode current 1, the deal dode oltage 2, and the output oltage. Howeer, nstead of fndng numercal alues for these 3 quanttes, we must express them n terms of source oltage! From KCL: = 1 2 = 1 0 = 1 From KVL: R = 0 Thus the deal dode current s: 1 = 0.7 R Lkewse, from KVL: = 0 Thus, the deal dode oltage s: = 2 2 And fnally, from KVL: = Thus, the output oltage s: = 0.7 Jm tles The Un. of Kansas ept. of EEC

38 9/24/2004 The Full Wae Rectfer.doc 6/10 Now, we must determne when both 1 > 0 and 2 < 0. When both these condtons are true, the output oltage wll be = 0.7. When one or both condtons 1 > 0 and 2 < 0 are false, then our assuptons are nald, and 0.7. Usng the results we just determned, we know that 1 > 0 when: 0.7 > 0 R olng for : 0.7 > 0 R 0.7 > 0 > 0.7 V Lkewse, we fnd that 2 < 0 when: 2 < 0 olng for : 2 < 0 2 > 0 > 0 Thus, our assumptons are correct when > 0.0 AN > 0.7. Ths s the same thng as sayng our assumptons are ald when > 0.7! Jm tles The Un. of Kansas ept. of EEC

39 9/24/2004 The Full Wae Rectfer.doc 7/10 Thus, we hae found that the followng statement s true about ths crcut: = 0.7 V when > 0.7 V Note that ths statement does not consttute a functon (what about < 0.7?), so we must contnue wth our analyss! ay we now AUME that 1 s reerse based and 2 s forward based, so we ENFRCE 1 = 0 and 2 = 0. Thus, we ANALYZE ths crcut: 1 = = 2 0 _ R (t) Usng the same proceedure as before, we fnd that = 0.7, and both our assumptons are true when < 0.7 V. n other words: = 0.7 V when < 0.7 V Note we are stll not done! We stll do not hae a complete transfer functon (what happens when 0.7 V < < 0.7 V?). Jm tles The Un. of Kansas ept. of EEC

40 9/24/2004 The Full Wae Rectfer.doc 8/10 Fnally then, we AUME that both deal dodes are reerse based, so we ENFRCE 1 = 0 and 2 = 0. Thus ANALYZE: 1 = = _ R (t) Followng the same proceedures as before, we fnd that = 0, and both assumptons are true when 07. < < 07.. n other words: = 0 when 07. < < 07. Now we hae a functon! The transfer functon of ths crcut s: 07V. for > 07V. = 0V for 07. > > 07V. 07V. for < 07V. Plottng ths functon: Jm tles The Un. of Kansas ept. of EEC

41 9/24/2004 The Full Wae Rectfer.doc 9/ The output of ths full-wae rectfer wth a sne wae nput s therefore: A (t) 0.7 t A (t) Note how ths compares to the transfer functon of the deal full-wae rectfer: for < 0 = for > Very smlar! Jm tles The Un. of Kansas ept. of EEC

42 9/24/2004 The Full Wae Rectfer.doc 10/10 Lkewse, compare the output of ths juncton dode full-wae rectfer to the output of an deal full-wae rectfer: A (t) 0 t -A (t) Agan we see that the juncton dode full-wae rectfer output s ery close to deal. n fact, f A>>0.7 V, the C component of ths juncton dode full wae rectfer s approxmately: 2A V 07V. π Just 700 mv less than the deal full-wae rectfer C component! Jm tles The Un. of Kansas ept. of EEC

43 9/24/2004 The Brdge Rectfer.doc 1/9 The Brdge Rectfer Now consder ths juncton dode rectfer crcut: 1 2 Power Lne (t) R - (t) _ 4 3 We call ths crcut the brdge rectfer. Let s analyze t and see what t does! Frst, we replace the juncton dodes wth the CV model: Power Lne (t) 0.7 V 1 R - (t) V V _ 0.7 V 3 Jm tles The Un. of Kansas ept. of EEC

44 9/24/2004 The Brdge Rectfer.doc 2/9 Q: Four gul-durn deal dodes! That means 16 sets of dad-gum assumptons! A: True! Howeer, there are only three of these sets of assumptons are actually possble! Consder the current flowng through the rectfer. Ths current of course can be poste, negate, or zero. t turns out that there s only one set of dode assumptons that would result n poste current, one set of dode assumptons that would lead to negate current, and one set that would lead to zero current. Q: But what about the remanng 13 sets of dog gone dode assumptons? A: Regardless of the alue of source, the remanng 13 sets of dode assumptons smply cannot occur for ths partcular crcut desgn! Jm tles The Un. of Kansas ept. of EEC

45 9/24/2004 The Brdge Rectfer.doc 3/9 Let s look at the three possble sets of assumptons: >0 The rectfer current can be poste only f these assumptons are true: 1 and 3 are reerse based. 2 and 4 are forward based. Power Lne (t) _ >0 1 > V 0.7 V - (t) R V 0.7 V Analyzng ths crcut, we fnd that the output oltage s: = 14V. and the f.b. deal dode currents are: 14. = = R Jm tles The Un. of Kansas ept. of EEC

46 9/24/2004 The Brdge Rectfer.doc 4/9 and, fnally the r.b. deal dode oltages are: = Thus, > 0 when: and < 0 when: 14. > 0 R 14. > 0 > 14. V < 0 > 0 Therefore, we fnd that for ths crcut: = 14V. when > 14V. <0 The rectfer current can be negate only f these assumptons are true: 1 and 3 are forward based. 2 and 4 are reerse based. Jm tles The Un. of Kansas ept. of EEC

47 9/24/2004 The Brdge Rectfer.doc 5/9 Power Lne (t) _ <0 < V V - (t) R V 0.7 V Analyzng ths crcut, we fnd that the output oltage s: = 14V. whle the f.b. deal dode currents are both : 14. = = R and the r.b. deal dode oltages are both: = Jm tles The Un. of Kansas ept. of EEC

48 9/24/2004 The Brdge Rectfer.doc 6/9 Thus, > 0 when: and, < 0 when: 14. > 0 R 14. > 0 > 14. V < 14. V < 0 Therefore, we lkewse fnd for ths crcut: = 14V. when < 14V. =0 The rectfer current can be zero only f these assumptons are true: All deal dodes are reerse based! Power Lne (t) =0 0.7 V (t) R 0.7 V 0.7 V _ =0 0.7 V 3 3 Jm tles The Un. of Kansas ept. of EEC

49 9/24/2004 The Brdge Rectfer.doc 7/9 Analyzng ths crcut, we fnd that the output oltage s: = R = 0 whle the deal dode oltages of 2 and 4 are each: 14. = = and the deal dode oltages of 1 and 3 are each: 14. = = Thus, 2 < 0 when: and, 1 < 0 when: 14. < < 0 < < < 0 < 14. > 14. Therefore, we also fnd for ths crcut that: = 0 when both < 14V. and > 14V. (-1.4< < 14V). Jm tles The Un. of Kansas ept. of EEC

50 9/24/2004 The Brdge Rectfer.doc 8/9 Q: You know, that dang Mzzou grad sad we only needed to consder these three sets of dode assumptons, yet am stll concerned about the other 13. How can we be sure that we hae analyzed eery possble set of ald dode assumptons? A: We know that we hae consdered eery possble case, because when we combne the three results we fnd that we hae a pece-wse lnear functon!.e.,: 14V. f < 14V. = 0 f -1.4 < < 1 4V. 14V. f > 14V Jm tles The Un. of Kansas ept. of EEC

51 9/24/2004 The Brdge Rectfer.doc 9/9 Note that the brdge rectfer s a full-wae rectfer! f the nput to ths rectfer s a sne wae, we fnd that the output s approxmately that of an deal full-wae rectfer: A (t) 1.4 t A (t) We see that the juncton dode brdge rectfer output s ery close to deal. n fact, f A>>1.4 V, the C component of ths juncton dode brdge rectfer s approxmately: 2A V 14V. π Just 1.4 V less than the deal full-wae rectfer C component! Jm tles The Un. of Kansas ept. of EEC

52 9/24/2004 Peak nerse Voltage.doc 1/6 Peak nerse Voltage Q: m so confused! The brdge rectfer and the fullwae rectfer both prode full-wae rectfcaton. Yet, the brdge rectfer use 4 juncton dodes, whereas the full-wae rectfer only uses 2. Why would we eer want to use the brdge rectfer? A: Frst, a slght confesson the results we dered for the brdge and full-wae rectfers are not precsely correct! Recall that we used the juncton dode CV model to determne the transfer functon of each rectfer crcut. The problem s that the CV model does not predct juncton dode breakdown! f the source oltage becomes too large, the juncton dodes can n fact breakdown but the transfer functons we dered do not reflect ths fact! Q: Ykes! You mean that we need to rework our analyss and fnd new transfer functons! Jm tles The Un. of Kansas ept. of EEC

53 9/24/2004 Peak nerse Voltage.doc 2/6 A: Fortunately no. Breakdown s an undesrable mode for crcut rectfcaton. ur job as engneers s to desgn a rectfer that aods t that why the brdge rectfer s helpful! To see why, consder the oltage across a reersed based juncton dode n each of our rectfer crcut desgns. Recall that the oltage across a reerse based deal dode n the full-wae rectfer desgn was: 2 = 2 so that the oltage across the juncton dode s approxmately: = 0.7 = Now, assumng that the source oltage s a sne wae = Asnωt, we fnd that dode oltage s at t most negate (.e., breakdown danger!) when the source oltage s at ts maxmum alue A..E.,: = 2A 0 7. mn f course, the largest juncton dode oltage occurs when n forward bas: = 07V. max Jm tles The Un. of Kansas ept. of EEC

54 9/24/2004 Peak nerse Voltage.doc 3/6 A (t) 0.7 t 2A 07. (t) Note that ths mnmum dode oltage s ery negate, wth an absolute alue ( mn = 2A 07. ) nearly twce as large as the source magntude A. We call the absolute alue of the mnmum dode oltage the Peak nerse Voltage (PV): mn PV = Note that ths alue s dependent on both the rectfer desgn and the magntude of the source oltage. Q: o, why do we need to determne PV? m not sure see what dfference ths alue makes. Jm tles The Un. of Kansas ept. of EEC

55 9/24/2004 Peak nerse Voltage.doc 4/6 A: The Peak nerse Voltage answers one mportant queston wll the juncton dodes n our rectfer breakdown? f the PV s less than the Zener breakdown oltage of our rectfer dodes (.e., f PV < VZK ), then we know that our juncton dodes wll reman n ether forward or reerse bas for all tme t. The rectfer wll operate properly! Howeer, f the PV s greater than the Zener breakdown oltage of our rectfer dodes (.e., f PV > VZK ), then we know that our juncton dodes wll breakdown for at least some small amount of tme t. The rectfer wll NT operate properly! Q: o what do we do f PV s greater than V ZK? How do we fx ths problem? A: We hae two possble solutons: 1. Use juncton dodes wth larger alues of V ZK (f they exst!). 2. Use the brdge rectfer desgn. Q: The brdge rectfer! How would that sole our breakdown problem? Jm tles The Un. of Kansas ept. of EEC

56 9/24/2004 Peak nerse Voltage.doc 5/6 A: To see how a brdge rectfer can be useful, let s determne ts Peak nerse Voltage PV. Frst, we recall that the oltage across the reerse based deal dodes was: = so that the oltage across the juncton dode s approxmately: = 0.7 = 0.7 Now, assumng that the source oltage s a sne wae = A sn ωt, we fnd that dode oltage s at t most negate (.e., breakdown danger!) when the source oltage s at ts maxmum alue A..E.,: = A 07. mn f course, the largest juncton dode oltage occurs when n forward bas: = 07V. max Jm tles The Un. of Kansas ept. of EEC

57 9/24/2004 Peak nerse Voltage.doc 6/6 A (t) 0.7 t A 07. (t) Note that ths mnmum dode oltage s ery negate, wth an absolute alue ( mn = A 07. ), approxmately equal to the alue of the source magntude A. Thus, the PV for a brdge rectfer wth a snusodal source oltage s: PV = A 07. Note that ths brdge rectfer alue s approxmately half the PV we determned for the full-wae rectfer desgn! Thus, the source oltage (and the output C component) of a brdge rectfer can be twce that of the full-wae rectfer desgn ths s why the brdge rectfer s a ery useful rectfer desgn! Jm tles The Un. of Kansas ept. of EEC