The Full-Wave Rectifier

Transcription

1 9/3/2005 The Full Wae ectfer.doc /0 The Full-Wae ectfer Consder the followng juncton dode crcut: s (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 (20 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 = 0 V, the anode of wll be 0V aboe ground potental, whle the anode of 2 wll be 0V below ground potental (.e., -0V): Jm tles The Un. of Kansas ept. of EEC

2 9/3/2005 The Full Wae ectfer.doc 2/0 =0 V Power Lne =0V 2 Conersely, f =-0 V, the anode of wll be 0V below ground potental (.e., -0V), whle the anode of 2 wll be 0V aboe ground potental: =-0 V Power Lne =-0V 2 Jm tles The Un. of Kansas ept. of EEC

3 9/3/2005 The Full Wae ectfer.doc 3/0 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: Power Lne 2 We wll now gnore the prmary wndng of the transformer and redraw the remanng crcut as: 2 Jm tles The Un. of Kansas ept. of EEC

4 9/3/2005 The Full Wae ectfer.doc 4/0 Note that the secondary oltages at ether end of ths crcut are the same, but hae opposte polarty. As a result, f =0, then the anode of dode wll be 0 V aboe ground, and the anode at dode 2 wll be 0V below ground just lke before! 2 =0 =0 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 s forward based and 2 s reerse based, thus ENFCE = 0 and 2 = 0. Thus ANALYZE: = = Jm tles The Un. of Kansas ept. of EEC

5 9/3/2005 The Full Wae ectfer.doc 5/0 Note that we need to determne 3 thngs: the deal dode current, 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: = 2 = 0 = From KVL: 0.7 = 0 Thus the deal dode current s: = 0.7 Lkewse, from KVL: = 0 Thus, the deal dode oltage s: = 2 2 And fnally, from KVL: 0.7 = Thus, the output oltage s: = 0.7 Jm tles The Un. of Kansas ept. of EEC

6 9/3/2005 The Full Wae ectfer.doc 6/0 Now, we must determne when both > 0 and 2 < 0. When both these condtons are true, the output oltage wll be = 0.7. When one or both condtons > 0 and 2 < 0 are false, then our assuptons are nald, and 0.7. Usng the results we just determned, we know that > 0 when: 0.7 > 0 olng for : 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

7 9/3/2005 The Full Wae ectfer.doc 7/0 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 s reerse based and 2 s forward based, so we ENFCE = 0 and 2 = 0. Thus, we ANALYZE ths crcut: = = 2 0 Usng the same proceedure as before, we fnd that = 0.7, and both our assumptons are true when < 0.7 V. In 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

8 9/3/2005 The Full Wae ectfer.doc 8/0 Fnally then, we AUME that both deal dodes are reerse based, so we ENFCE = 0 and 2 = 0. Thus ANALYZE: = = Followng the same proceedures as before, we fnd that = 0, and both assumptons are true when 07. < < 07.. In 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

9 9/3/2005 The Full Wae ectfer.doc 9/ The output of ths full-wae rectfer wth a sne wae nput s therefore: A 0.7 t A (t) Note how ths compares to the transfer functon of the deal full-wae rectfer: for < 0 = for > 0 - Very smlar! Jm tles The Un. of Kansas ept. of EEC

10 9/3/2005 The Full Wae ectfer.doc 0/0 Lkewse, compare the output of ths juncton dode full-wae rectfer to the output of an deal full-wae rectfer: A 0 t -A (t) Agan we see that the juncton dode full-wae rectfer output s ery close to deal. In 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