Driving your LED s. LED Driver. The question then is: how do we use this square wave to turn on and turn off the LED?

Transcription

1 0//00 rng your LE.doc / rng your LE s As we hae preously learned, n optcal communcaton crcuts, a dgtal sgnal wth a frequency n the tens or hundreds of khz s used to ampltude modulate (on and off) the emssons of a Lght Emttng ode (LE): co( t ) τ mn T t O A LE LE rer The queston then s: how do we use ths square wae to turn on and turn off the LE? rst, we must use the precse language and nomenclature of electronc deces. An LE as ts name makes clear s a dode. Jm Stles The Un. of Kansas ept. of EES

2 0//00 rng your LE.doc / Howeer, most LE are not slcon juncton dodes but nstead are made of Gallum Arsende (GaAs). Ths makes them somewhat dfferent than the p-n juncton dodes we studed n EES 3, but the bascs are the same.. An LE emts energy (t s on ) when the dode s forward based. or slcon p-n juncton dodes, sgnfcant but plausble (.e., nondestructe) forward bas current results n a forward bas dode oltage drop (from anode to cathode) of between 0.5 and 0.9 olts. Thus, we typcally approxmate ths forward based oltage as 0.7. Howeer, for GaAs LEs ths forward bas oltage s between.0 and.0 olts when the dode current s sgnfcant (.e., > ma) yet plausble (.e., < A).. An LE does not emt energy (t s off ) when the dode s reerse based. n ths state, the dode current s essentally zero. There s of course a transton regon between the forward and reerse bas states (.e., ther defntons are a bt subjecte), but we can safely say that an LE s reerse based ( 0) when ts dode oltage s less than (say) 0.5. Jm Stles The Un. of Kansas ept. of EES

3 0//00 rng your LE.doc 3/ LE Q: So, just how do we take the output of the O and use t to place the dode n one of these two state? A: The answer s the LE drer. As we look at the dgtal sgnal from the O, we see t has one of two oltage states ether hgh oltage or low oltage mn. Our job s to map each of these two oltage states nto a dode bas state of ether reerse or forward. An excellent drer crcut for accomplshng ths s: The desgn problem s ths: gen a specfc LE and JT, as well as oltages mn, and, what should be the two resstor alues and? To begn, let s do a lttle crcut analyss. rom KL: and rom KL: n or n mn LE E E and 0 E 0 n E Jm Stles The Un. of Kansas ept. of EES

4 0//00 rng your LE.doc 4/ And fnally from Ohm s Law: and Now, t s reasonable to decde that when n mn the LE should be reerse based (off) and when n, the LE should be forward based (on). Let s look at the reerse bas case frst. We know that for ths to be true, the dode current s zero ( 0) and the dode oltage s less than about 0.5 ( < 500m ). Note ths means that the current through resstor s zero, so that 0, as well. ut, by KL we know that the s also the collector current. Thus, to make the dode reerse based, we must bas the JT to the proper mode. Q: see! We must make the npn JT ether on or off, rght? n mn LE 0 0 E E A: NO!!!!!! Transstor modes are not on or off. There are three count em three specfc and unambguous transstor modes. or polar Juncton Transstor (JT), these modes are Acte, Saturaton, and utoff. Jm Stles The Un. of Kansas ept. of EES

5 0//00 rng your LE.doc 5/ Hopefully, the correct mode for the reerse based transstor s edent. We requre that 0, and all transstor currents are zero f the JT s n cutoff! f the JT of our crcut s n cutoff mode, the LE wll be reerse based. So, let s ASSUME that our JT s n cutoff. ecall that we ENOE the equaltes 0. E f we now ANALYZE ths crcut we fnd: E and E n We now HEK to see f ths analyss s consstent wth the nequaltes assocated wth our orgnal cutoff assumpton. ecall these nequaltes are: E > 07. and < 0 E Now f the LE s reerse based, we hae establshed that < 05., meanng that for ths crcut assumpton < 05. > 05. E E n mn LE E E Jm Stles The Un. of Kansas ept. of EES

6 0//00 rng your LE.doc 6/ Thus, n order for > 07., the oltage source must be: E 05. > 07. >. Ths means that must be greater than. for the JT to be n cutoff. Ths of course s not much of a restrcton, as s always much greater than. olts! rom the second nequalty, we can conclude that n order for the JT to be n cutoff, the nput oltage mn must be negate (.e., mn < 0 )! Q: Ykes! m not sure that ths wll be the case. Although mn wll lkely be zero (or at least ery small), don t thnk t wll actually be negate!?! A: Well, the nequalty E < 0 s actually a lttle bt too restrcte. emember, the mportant thng here s that the ase-emtter Juncton (EJ) of the JT s reerse based. Agan, ths defnton s a lttle nebulous. learly, the EJ wll be reerse based f E < 0, but t lkewse wll exhbt almost no dffuson current f E s poste but small. A less restrcte, but nearly as accurate nequalty would be < 03.. Meanng that < 03. s E requred for the JT to be n cutoff. Ths restrcton s qute realstc. mn Jm Stles The Un. of Kansas ept. of EES

7 0//00 rng your LE.doc 7/ Thus we can conclude for our drer crcut, the LE wll be reerse based (off) f both these condtons are satsfed: >. and < 03. n mn Q: Hey wat a mnute! Nether of these desgn statements hae anythng to do wth resstors and. Where do they come n?? A: f the JT s n cutoff, they don t! f the two nequaltes aboe are satsfed, then the JT wll be n cutoff and the LE reerse based for any (reasonable) alue of and. To see how the resstors affect the crcut, we must consder the case where the LE s forward based! ecall that f the LE s forward based, the dode current wll be poste, wth a sgnfcant but plausble alue (e.g., ma < < A ). Lkewse, the dode oltage wll be n the range of one to two olts (.e., < < ). LE endors call the oltage across a forward based LE the forward oltage and ge t the arable. n LE E E Jm Stles The Un. of Kansas ept. of EES

8 0//00 rng your LE.doc 8/ Snce now we hae a case where > 0, the JT s clearly not n cutoff. Of course t could be n ether ) saturaton or ) acte mode. Let s see f we can desgn a drer for each mode! rst, we must determne what dode current we desre when the LE s forward based (endors typcally refer to ths as forward current ). Of course, the hgher the current, the brghter our LE lght. rom that standpont, we wsh to make that current as large as possble. Q: an we just make t really large lke, say, 0 Amps? A: Unfortunately no. We attempted to put that much current through an LE, we would surely melt t. Eery dode has a mum power ratng P. The power absorbed by the dode s smply the product of the oltage across and the current through t. Of course, these alues wll be changng wth tme as co ( t ) toggles between and mn. The tme-aeraged power dsspaton, howeer, can be determned and not surprsngly t depends on the duty t : cycle τ T of sgnal ( ) co Jm Stles The Un. of Kansas ept. of EES

9 0//00 rng your LE.doc 9/ τ P T Snce we wsh to aod meltng, we want P > P meanng: τ P T < P < T τ Once we hae selected a sutable we can desgn the LE drer. Let s frst ASSUME that the JT s n saturaton. We ENOE equaltes: E 07. and 0. Now we ANALYZE ths crcut. rom Ohm s Law: E ( 0). earrangng, we can determned the requred alue of resstor : ( 0). Lkewse: 07. n LE E E Jm Stles The Un. of Kansas ept. of EES

10 0//00 rng your LE.doc 0/ Now we must HEK our results to see f/when they are consstent wth the nequaltes assocated wth JT saturaton. Specfcally, the nequalty <. Snce, ths nequalty leads to: < < Thus, we conclude that the JT wll be n saturaton, wth collector current, f: ( 0). and 07. < The problem wth ths desgn could be the resultng base current: 07. t s possble that the O cannot prode that much current. Thus, an alternate desgn can features the JT n acte mode. or ths mode, we ENOE the equaltes: E 07. and And so now ANALYZE ths crcut: Jm Stles The Un. of Kansas ept. of EES

11 0//00 rng your LE.doc / and as before: combnng: 07. Lkewse: Now HEK to see f/when these results are consstent wth acte mode nequaltes, specfcally > 07.. E E > 07. E 0 7. < auton: don t make too small!!! n LE E 07. E Thus, the JT wll be n acte mode, and the LE current wll be, f: and < Jm Stles The Un. of Kansas ept. of EES