EE3310 Class notes Part 2. Solid State Electronic Devices - EE3310. Class notes. p-n junctions

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1 EE3310 Class otes Part Verso: Fall 00 These class otes were orgally based o the hadwrtte otes of Larry Overzet. It s exected that they wll be modfed (mroved?) as tme goes o. Ths verso was tyed u by Matthew Goecker. Sold State Electroc evces - EE3310 Class otes - juctos U to ths ot, we have looked at sgle materals that are ot touchg aythg else. Such a system does ot make a devce. We ow wat to ut a -tye ad a -tye materal together ad see what haes. Ths s our frst true devce! Eergy E c E E F E F E V -tye -tye What wll hae? Posto 1) Holes the -tye wll dffuse to the -tye (because of the desty gradet). ) Electros the -tye wll dffuse to the -tye (because of the desty gradet). 3) The Ferm levels the two materals have to match as we are equlbrum > There wll be a electrc feld that forms betwee the materals to OPPOSE dffuso ad also so that the Ferm level across the devce s costat. UT EE3301 otes art Page 80 of (49+79) Last udate :3 PM 10/4/0

2 Eergy E c E E E F E v E c E F E -tye -tye E v < W > Jucto Posto We kow that ths shft the trsc eergy level must corresod to a electrc feld. Further, we kow that the chage desty of the charge carrers wll result dffuso across the boudary. Thus, we fd that the artcles wll tally move aroud the jucto ad ths wll result the formato of a electrc feld across the boudary. We ca look at what each of the movemets do Partcle flux desty, Γ Mechasm Partcle curret desty, J < > > < Hole drft (E) ffuso Electro drft (E) ffuso < > < > Exermetally, we usually do ot have a system whch we smly lace a ece of -tye materal o to of a ece of -tye. (Such a system ca be made but t s very rarely used because of the cost.) Rather what s tycally doe s that the semcoductor s doed, va o mlatato, oe area to be -tye ad the t s doed to be -tye a adjacet locato. (For S techology, B s the most commo -tye doat whle s s the most commo -tye doat.) Io mlatato s a -exact rocess, for a umber of hyscal reasos, ad thus we caot get erfectly abrut (ste) doat chage. (These reasos are dscussed other classes ad are ot artcularly mortat here.) The best ossble s kow as ultra shallow juctos ad s the result of very low eergy o mlatato ad very secal actvato stes. O the other had, we ca certaly get the doat to chage slowly or graded the area of the jucto, through hgh eergy mlats ad/or log (tme) actvato stes. The ushot of the above dscusso s that we ca have ste juctos as well as graded juctos. Both are mortat the roducto of electroc devces. For the uroses of the class, we wll oly cosder ste juctos. Let us go back ad look at our cture of the jucto. UT EE3301 otes art Page 81 of (49+79) Last udate :3 PM 10/4/0

3 E Eergy qv b E c E E F E v E c E F E Metallurgcal jucto (ot always where E F E but where a d ) -tye -tye < W > Jucto Posto (ote that the jucto s also kow as the eleto Rego ad the Trasto Rego.) E v Lookg at the fgure, begs the questo, what are W, V b ad E? To get at ths, lets us look at the areas outsde of the jucto. Frst, we kow the locato of the Ferm eergy wth resect to the trsc eergy o both sdes of the jucto. ( EF E) 0 ex kt EF E 0 ex kt Thus, x x EF ( EF E) ktl 0 x x 0 EF ( E EF) ktl Where we have defed x 0 ad -x 0 as the edge of the jucto o the sde ad sde resectvely. Whe the Ferm eergy o each sde shfts such that EFF E the the trsc eergy must shft o each sde. Ths gves rse to the voltage shft across the jucto. UT EE3301 otes art Page 8 of (49+79) Last udate :3 PM 10/4/0

4 qv E + E b F F x ktl ( ) x + ktl 0 0 ( ) x 0 x 0 ktl ow what are x ( 0) ad ( x ) 0? They are smly ( x0) (really o the sde) x ( 0) (really o the sde) so qvb ktl ow takg to accout x ( 0) x ( 0 ) ad x ( 0) ( x0) so x ( 0) x 0 qvb ktl ( ) 0 ( 0) x ktl x x ktl x or! x0 x ( x 0) x 0 ( ) 0 0 ( 0) e qvb / kt We ow eed to fd W ad E. To get them requres that we kow the currets the jucto. To get at the currets, we eed to make some aroxmatos, kow as the deleto aroxmato. UT EE3301 otes art Page 83 of (49+79) Last udate :3 PM 10/4/0

5 To uderstad why solvg these s tough to do exactly, we eed to cosder the equato descrbg the electrc feld across the boudary. E E x ρ x ε 1 ( ( x) ( x)+ ( x) ( x) ) ε (Remember that whle our charge carrers are ad, we stll have the boud charges, ad! The door stes are ostvely charged whle the accetor stes are egatvely charged.) However the hole ad electro destes are set art by the electrc feld ad we have o clue as to what the fuctoal form s for those destes. Our aroxmato s related to those destes. Frst, outsde of the jucto, the charge carrers cover the boud charges ad thus the charge desty s zero. If we assume that the hole ad electro destes are zero sde the jucto, the we ca come u wth a aroxmato that we ca solve. (Ths s ot a bad aroxmato, as the electrc feld should ush all of the charge carrers out of the jucto rego.) Ths s because we ca assume that the umbers of door ad accetor stes are costat o each sde of the metallurgcal jucto. Thus E E x ρ x ε 1 ( x x ) x x x 0 0 ε 0 elsewhere where ( x ) ad x are grahcally gve by: ρ -tye -tye x 0 -x 0 Posto (Ths s actually very close to realty most ste - juctos. lso ote that we have dected a ste jucto. graded jucto wll have ( x ) ad ( x ) vary the rego.) We ca tegrate o each sde of the jucto to get: UT EE3301 otes art Page 84 of (49+79) Last udate :3 PM 10/4/0

6 E E x x x q x dx 0 ε q x + x 0 ε x q 0 x dx ε q ( x 0 x ) ε Grahcally ths looks lke: ρ -tye -tye x 0 -x 0 Posto E(x) ote that the maxmum electrc feld s at x 0 ad that t goes to zero at the edges of the jucto. We ca ow get the otetal at all ots sde the jucto (whch wll lead us to our jucto wdth!) E V x V Vx Vx x x1 x x0 Edx Edx q x + x 0 V x 0 x 0 x ε x0 x ( ) Edx q Vx ( 0) ( x 0 x ) 0 x x ε By lookg at our cture, we see that we ca set Vx ( 0) 0 Vx ( 0) Vb Thus, 0 0 UT EE3301 otes art Page 85 of (49+79) Last udate :3 PM 10/4/0

7 q Vx x + x 0 x 0 x 0 ε q Vx V b ( x 0 x ) 0 x x 0 ε t the metallurgcal, x 0, these must match. q q Vb x ( 0) + ( x 0) ε ε t ths ot, we ote that the total charge sde the jucto must be zero. (I electromagetsm, oe lears that f the total chage a rego s ot zero, the there s a electrc feld outsde of that rego. We have setu our defto of the jucto such that all of the electrc feld s sde.) Thus, Q qx0 Q qx 0 Q + Q 0 x0 x0 We ca ow lug ths to our above equato to get q V q x x b ( 0) + 0 ε ε q q + x ( 0) ε ε q ( + ) ( x0 ) ε x 0 εv b q + 1 / 1 / x x 0 εvb 0 q( + ) So the wdth s the sum of x 0 ad x 0. Or, UT EE3301 otes art Page 86 of (49+79) Last udate :3 PM 10/4/0

8 w x0 + x0 1 / 1 / εvb εvb q( + + ) q( + ) εv b q + 1 / 1 / + 1 / / / εv b / q + + / ( ) 1 / εvb q ( / ) + 1 / 1 / εv + + b ( + q ) 1 / εvb( + ) q We ca rearrage these to show x w 0 ( + ) x w 0 ( + ) 1 q V b w ε ( + ) kt q Examle: x x 0 0 kt l l q -tye Gas layer s grow o -tye Gas: cm -3, X10 15 cm -3, X10 6 cm -3, e r 13. > e e r e 0. Fd V b, w, E max, x 0, x 0, ad Q/ a) UT EE3301 otes art Page 87 of (49+79) Last udate :3 PM 10/4/0

9 V b kt q l X V l 6 ( X10 ) V b-d) 1 / εvb( + ) w q x w 0 ( + ) 15 X µ m X / X X X10 X µ m ( larg e!) 37. µ m x0 w µ m e) q EMX ( x 0 ) 4π ε 811. kv / m f) Total charge o each sde Q qx 0 qx X10 C/ cm ow, we make our lfe eve more comlex by lacg a voltage across t. UT EE3301 otes art Page 88 of (49+79) Last udate :3 PM 10/4/0

10 PPLIE VOLTGE CROSS P- JUCTIO How we aly the voltage across the - jucto makes a sgfcat dfferece how the devce resods. To uderstad why, we frst look at the devce a qualtatve maer. (We wll come back ad do ths quattatvely later.) Bas ad Jucto To get at the bas ad jucto arameters we eed to uderstad that most of a aled bas wll fall across the jucto. Ths s because both the -sde ad -sde have charge carrers that are moble ad thus wll move so as to get rd of ay electrc feld outsde of the jucto. Isde of the jucto, the umber of moble charge carrers s very small ad thus the electrc feld ca be mataed. Ths meas that to get the jucto arameters we smly eed to relace V b the above dervato wth V b - V. (We are takg V to be ostve the forward drecto. [Reverse ad forward have to do wth curret flow.] Ths sg fl makes lfe easer whe we look at dodes as a whole crcuts. For ow, t seems backwards.) Curret flow Qualtatve aalyss Wthout ay aled bas, ad hece o curret, we have our orgal eergy bad structure. Eergy -tye E E c -tye E E F E v E c E F E qv b < W > Jucto Posto E v Let us ut some electros ad holes o ths dagram ad exame where they go. Majorty carrers Eergy -tye E E c E E c E F E v qv b E F E E v -tye Posto UT EE3301 otes art Page 89 of (49+79) Last udate :3 PM 10/4/0

11 The most eergetc electros ad holes ca overcome the electrc feld barrer ad move to the other sde of the jucto. (Ths s the dffuso rocess at work.) Morty carrers Eergy -tye E E c E E c E F E v qv b E F E E v -tye Posto y morty carrer that waders to the jucto rego, gets draw by the electrc feld to the other sde of the jucto. I our ubased system, these currets balace ad thus the total curret desty s zero. Reverse Bas Reverse bas mles that we are addg a bas the same drecto as jucto bas. It s kow as reverse bas as because ths s, as we wll see, the low curret drecto. ow our eergy dagram looks lke: Eergy -tye E E c E E F E c E v q(v +V b ) E F E E v -tye Posto ow the morty carrers ca stll move across the barrer but the majorty carrers do ot have eough eergy to over come the barrer. ddtoally, we see that the Ferm eerges o the two sdes are dfferet, by qv, where -V s the aled bas. Thus we would exect that the total curret s set by ay morty carrers that just hae to ru to the jucto. Thus we would exect a low level curret that s effectvely deedet of the aled bas. UT EE3301 otes art Page 90 of (49+79) Last udate :3 PM 10/4/0

12 Forward Bas Forward bas mles that we are addg a bas the ooste drecto to the jucto bas. It s kow as forward bas as because ths s, as we wll see, the hgh curret drecto. ow our eergy dagram looks lke: Eergy -tye E E c E E c E F E v q(v b -V ) E F E -tye E v Posto ow the curret s set by the umber of majorty carrers that have suffcet eergy to overcome the reduced barrer. We kow that the umber of electros/holes s ( E ) de f( E ) c( E ) de Electros the Coducto bad * 3 / 1 m E EcdE ( E EF) + ex π h 1 kt * 3 / ( E EF) m ex E EcdE kt π h E EF ex kt de ( E ) de 1 f( E ) ( E ) de v 1 1 E 1 + ex E ex E ex E kt E kt F F E kt F * m π h * m π h de Holes the Valace bad 3 / E 3 / v E E de v E de Thus we mght exect that forward bas, the curret creases exoetally wth voltage. UT EE3301 otes art Page 91 of (49+79) Last udate :3 PM 10/4/0

13 Total curret flow Before we go o, we wat to look at the total ath of the curret. For ths, we wll use the reverse bas codto. The other codtos look smlar. Morty carrers Eergy -tye E E c E E F E c E v q(v +V b ) E F E E v V -tye Posto Majorty carrers Eergy -tye E E c E E F E v q(v +V b ) E c E F E E v V -tye Posto There are two thgs to ote: 1) For both the majorty ad morty carrers, the electros flow the ooste drecto to the holes ad thus, because of the dfferece charge, roduce curret flow the same drecto. ) The majorty carrers flow the ooste drecto as the morty carrers. Thus swtchg the sg of V wll chage the drecto of the curret flow. Quattatve alyss of curret flow. To make ths aalyss, we eed to make a few smlfyg assumtos. 1) The dode s beg oerated uder steady state codtos. UT EE3301 otes art Page 9 of (49+79) Last udate :3 PM 10/4/0

14 ) The jucto ca be modeled as a o-degeerately doed ste jucto. (More comlex juctos ca be modeled at the rce of creased dffculty.) 3) The dode s oe-dmesoal. (ga, more comlex dodes ca be modeled at the rce of creased dffculty.) 4) Low-level jecto revals the quas-eutral rego. (Ths s the tycal mode of oerato.) 5) The oly rocesses occurrg the system are drft, dffuso, thermal geerato ad recombato. (o lght emsso or absorto etc.) Each of these assumtos are reasoable ad are smlar to what we assumed for electrostatcs a ubased - jucto. I fact, the oly dstcto that we have s that the bas across the jucto s ow V +V b ad there s a et curret flow. ow what s our curret? I J J J( x)+ JP( x) costat J q x x q x µ E+ x J q x x q x P µ E x where s the cross secto of the dode. ote that because we do ot have source/sk of charge ad we are equlbrum, the the total curret flowg across the dode must be costat. ow how does quas-eutralty fluece our results? Uder steady state codto, the drft of the morty carrer must be balaced by the dffuso. Thus, Reeat of dervato otcal thermal exctato exctato recombato 1 } J + go t + αr αrtt () () t q otcal thermal exctato exctato recombato 1 } J ot + g + αr αrtt () () t q ow, however, we have show that the last two terms ca be combed to leave a smler sgle term UT EE3301 otes art Page 93 of (49+79) Last udate :3 PM 10/4/0

15 1 + + J go t αr αr ( 0 + )( 0 + ) t q 1 J + go t αr( 0 + 0) q 1 J + go t q τ 1 ot + g + J αr αr ( 0 + ) 0 t q + 1 J ot + g αr( 0 + 0) q 1 J ot + g q τ t ths ot, we ca ow ut what we have derved for our currets J q µ E+ q J q µ E q Pluggg these gves 1 ( µ + )+ q E q go t t q τ µ E + + go t τ 1 ( q µ E q g )+ ot t q τ µ ot E + + g τ fally, we eed to look at the left-had sde of the equato ad ote that the desty s made u of the equlbrum art ad the jected art ad further ote that the equlbrum art s a costat; thus µ E + + go t t τ µ ot E + + g t τ Thus for the morty carrers o the -sde (x -x 0 ) UT EE3301 otes art Page 94 of (49+79) Last udate :3 PM 10/4/0

16 t E } 0 µ E + ot + g τ τ ad o the sde (x x 0 ) 67 4E 084 } 0 0 µ E + + got t τ τ Lkewse, we ca exame the morty curret o each sde of the jucto. ga, E 0, so J q ( x ) ( x) q x> x0 x x J q x q τ ad J q ( x ) P q x JP q x q τ x x x< x s we move away from the jucto, we should exect the morty carrer desty to dro toward the equlbrum destes. Thus at a log dstace from the jucto, e.g. the cotact locato, the morty carrer desty dros to the equlbrum desty. ( x ) 0 ( x+ ) 0 Because the total curret each sde s costat, these equatos mly that the curret desty shfts from the morty carrer to the majorty carrer as oe moves away from the jucto toward the cotact. We wll retur to ths cocet later. Jucto rego ow we eed to look at what haes the jucto rego. Here, we caot set the electrc feld to zero. Thus 0 UT EE3301 otes art Page 95 of (49+79) Last udate :3 PM 10/4/0

17 t t 1 J q } 0 0 µ E + + got τ 1 J P q } 0 0 µ E + + got τ If the thermal recombato s small sde the jucto the the currets must be costat across the jucto. Physcally, the recombato mechasm requres both hole ad electros to be letful for the rate to be hgh. I the jucto, the umber of holes ad electros s relatvely low, so the recombato rate s relatvely slow. Ths make the last term the above equatos aroxmately zero. Thus 1 J 0 J( x0 x x0) J( x0) q 1 JP 0 JP( x0 x x0) JP( x0) q J J x JP x + + J x0 JP x0 costat ow we eed to match the boudary codtos. (We caot have a dscotuous soluto.) To do ths, we wll make a assumto for whch we do ot have a good a-ror reaso other the t works. That assumto s that the quas-ferm eerges are costat through the jucto ad equal to the Ferm eergy o the arorate sde. Thus F EF FF E. ow the quas-ferm eergy s related to the carrer destes by F E kt e ( ) * e F E kt / / F F / kt e Ths holds everywhere. If we assume that the quas-ferm eerges are costat sde the jucto the qv e / kt x x x t the -edge 0 0 UT EE3301 otes art Page 96 of (49+79) Last udate :3 PM 10/4/0

18 F F kt ( x0) ( x0) e e ( 0) x / qv / kt x e ( 0) F F / kt qv / kt + x x ( 0) x ( 0) e F F / kt e qv / kt 1 1 t the -edge ( ) ( x ) F F / kt qv / kt 0 0 x x e e x e F F kt ( 0) + ( x ) x ( 0) 0 / qv / kt x ( 0) e F F kt / qv / kt e e 1 e 1 ow, we ca solve our equato for the morty carrers subject to the boudary codtos. Frst, we wll start wth x -x 0, (The -sde). τ ( x ) 0 ( x 0) 1 e qv / kt tegratg UT EE3301 otes art Page 97 of (49+79) Last udate :3 PM 10/4/0

19 + ( x+ x 0)/ L ( x+ x 0)/ L e + Be where L τ Pluggg our boudary codtos, we fd that qv / kt e 1 ad B 0 Thus e 1 e + + qv / kt ( x x0)/ L Lkewse for the -sde ( a reeat of the above dervato) τ x ( 0) ( 1) e qv / kt ( x+ ) 0 tegratg + ( x x L x x L e )/ ( )/ Be where L τ Pluggg, we fd B e qv / kt ( 1) ad 0 Thus e qv / kt e x x L ( ) ( 1 0)/. ow, we ca get the currets UT EE3301 otes art Page 98 of (49+79) Last udate :3 PM 10/4/0

20 J x q x q x ( e 1) e qv kt x x L / + ( + 0)/ qv kt x x L / + ( + 0)/ q e 1 x e q ( x) L ad x J q P x q x ( e 1) e q e 1 x e q ( x) L qv kt x x L / ( 0)/ qv kt x x L / ( 0)/ t ths ot, we have assumed that the currets of electros ad holes are costat through the jucto. Thus, the total curret s smly the sum of the morty currets at the resectve jucto edges Jtotal J( x)+ JP( x) J( x x 0)+ JP( x x0) q L e qv kt q / L e qv/ kt ( 1)+ ( 1) L L q e qv / kt + ( 1) qv kt J e / 0 1 ( I J) qv / I I0 e kt 1 I0 q + L L O a sde ote, we ca show a addtoal relatosh betwee the curret ad the total excess charge o each sde. From above: UT EE3301 otes art Page 99 of (49+79) Last udate :3 PM 10/4/0

21 I x q P ( x) m or L q L e qv kt e x x L / ( )/ ( ) 0 1 but the total charge s Q q ( x) dx x0 q e qv / kt e x x L ( 0)/ ( 1 ) x dx 0 q L qv kt x x L e ( ) / ( 0)/ 1 e x0 q L qv kt e / ( 1) luggg ths to our total hole curret, e.g. the morty curret at the jucto edge, we fd I I x x q P P ( ) ( x x ) total m or 0 0 L q L e qv / kt ( 1) q L qv kt e / ( 1 ) L ( Q) τ Q τ Lkewse, we ca show Q I total τ Ths meas that we have a coule of equvalet methods of calculatg the curret, frst from the sloe of the morty carrer cocetrato, the method we used frst, or the total charge ad the carrer lfetmes, our secod method, show just ow. So what does ths look lke ad what does ths mea? basc lot of the I-V character looks lke: UT EE3301 otes art Page 100 of (49+79) Last udate :3 PM 10/4/0

22 Curret () I Voltage (V) O ths scale, we see that the dode has very lttle curret below V 0.5 V. bove that, the curret skes. We ca blow ths cture u to get a better vew of the extremes. 0.0E E-07 I Curret () -4.0E E E E-06-1.E Voltage (V) Here we have blow the u the area of egatve curret. From ths, we ca see that the dode begs to tur o about 0 V ot 0.5. (It s just that the frst cture s ot o the correct scale.) UT EE3301 otes art Page 101 of (49+79) Last udate :3 PM 10/4/0

23 Curret () 3.0E+0.5E+0.0E+0 1.5E+0 1.0E+0 5.0E E E Voltage (V) ow we are blowg u the area ear 0.5 V. Ths gves us a reasoable cture of how the curret chages as a fucto of V; t s aroxmately logarthmc. oe of these results are surrsg. Ca we ga a uderstadg of what wll hae to smlar dodes wth dfferet materal characterstcs? Yes Frst a sgle dode s usually oly deedet o the materal characterstc of oe sde of the dode. How ca we tell ths? qv / kt 0 I I e 1 I q 0 + L L q / τ / τ + The mortat art s the equato for I 0. We see that equato that the cotrbuto to the curret deeds learly o the doat cocetrato ad weakly o the dffuso coeffcet ad the mea collso frequecy. (The deedece s a square root ad thus ot eve lear.) For ay gve sgle tye of materal, the dffuso coeffcets ad collso frequeces of the holes ad electros are very smlar though ot exactly the same. Thus our oly true deedece s the doat cocetratos. Thus our curret s hghly deedet o the doats. If oe of the cocetrato s sgfcatly hgher tha the other.e. by a order of magtude (factor of te) the the cotrbuto to the curret from that sde s sgfcat. Thus, I UT EE3301 otes art Page 10 of (49+79) Last udate :3 PM 10/4/0

24 q / τ + tye dode < I q / τ / τ 0 + tye dode q / τ + < tye dode Ofte, our dodes are ether + - tye or - + tye. Examle: Gve two detcal S based + - dodes, excet for the door cocetrato o the -sde, lot the currets of the two dode. Use 1 1E15 cm -3 ad 1E16 cm -3. swer: q 1 / τ1 I 01 1 I0 q / τ 1 / τ1 / τ 1 ow the dffuso ad collso frequeces wll be almost detcal the two devces. Thus, I01 I 0 1 UT EE3301 otes art Page 103 of (49+79) Last udate :3 PM 10/4/0

25 Curret () 5.0E E E-06.0E E E E-06 I I1 -.0E-06 Voltage (V) Secod, f we chage the tye of base materal, ths ca have a major mact o the curret from the dode. For examle, f we have equally doed ad szed Gas ad S based dodes, ca we tell whch wll have the hgher currets? Whle ths deeds weakly o the dffuso ad collso rates, t s rmarly set by the trsc destes of the materals. For S 1.5 E10 cm -3, whle for Gas E6 cm -3. Thus the curret the S dode wll have sgfcatly hgher currets by ~8 orders of magtude! Carrer currets We ca also use our equatos to look at the carrer currets. We kow from before that the morty carrer curret s gve by: ( x) J q x q ( x) L x JP q x q ( x) L Further we kow that the jected morty carrer cocetratos dro off as: UT EE3301 otes art Page 104 of (49+79) Last udate :3 PM 10/4/0

26 e 1 e qv / kt + ( x+ x0)/ L e qv / kt e ( x x0)/ L 1 Thus we would exect a morty carrer curret to look lke: I total Morty curret I (x) I (x) -x o 0 x o Because the total curret must be costat the we ca come u wth the majorty carrer currets I total Morty curret I (x) Majorty curret I (x) -x o 0 x o Ths of course assumes a forward bas: If we stead look at reverse bas, what would we exect? ow, the morty carrer cocetrato must be greatly reduced as t s oly due to the dffuso across the jucto. Further the drecto of the curret must be ooste to that whch t had before. The ertet equatos are: UT EE3301 otes art Page 105 of (49+79) Last udate :3 PM 10/4/0

27 e qv / kt e + ( x+ x0)/ L 1 e + ( x+ x0)/ L e qv kt e x x L / ( 0)/ ( 1) e ( x x0)/ L Ths meas that outsde of the chage sze of the curret (by a magtude of ~e qv / kt ) we have a sg fl. Thus, 0 Morty curret Majorty curret I (x) I (x) I total 0 -x o x o ZEER BREKOW If we reverse bas the dode suffcetly, t wll go to a breakdow mode. Ths breakdow mode does ot mea that we have ecessarly damaged the dode whch s ossble but rather we are causg aother mortat rocess to occur. That rocess s kow as tuelg. Zeer dodes are created by heavly dog both sdes of the jucto. (If the dode s moderately or weakly doed the other thgs occur.) I such a stuato we get a eergy dagram that looks lke: UT EE3301 otes art Page 106 of (49+79) Last udate :3 PM 10/4/0

28 Eergy -tye E E c -tye E E F E v E c E F E qv b < W > Jucto Posto ow we ca aly a moderate reverse bas: Eergy -tye E E v E c -tye E E F E v qv b E c E F E < W > Jucto E v Posto ow we have the stuato whch the to valace bad the -sde s at the same or smlar eergy as the coducto bad the -sde. I such a case, we ca have drect tuelg from the valace bad of the -sde to the coducto bad of the -sde. secod way to thk about ths s that the electrc feld the jucto s so strog that t feld ozes a atom the -sde. Ths requres a electrc feld o the order of 10 6 V/cm hece the jucto legth eeds to be very short. valache Breakdow If we have ot heavly doed the two sdes the the jucto, the t ca be farly wde. Remember: w x0 + x0 1 / ε( Vb V )( + ) q Ths meas that the dstace that the electro eeds to travel to get across the jucto ca be sgfcat. The loger the dstace the hgher the lkelhood that t wll collde wth lattce atoms. If the electro colldes wth suffcet eergy, t ca oze a lattce atom, e.g. create ad EHP. The old electro ad hole are ow accelerated by the electrc feld to hgh eerges utl they collde wth addtoal atoms, roducg more EHPs. Ths ca result a sgfcat ozato of the atoms that make u the jucto ot ulke a lghtg strke through ar. Ths s kow as valache Breakdow. UT EE3301 otes art Page 107 of (49+79) Last udate :3 PM 10/4/0

29 odes uder C ad traset codtos exact aalyss of traset behavor across a dode s a o-trval edeavor. It would take much more tme tha we have ad requre alcatos of o-trval mathematcal fuctos. Further, t would ot hel the studets develo a hyscal uderstadg of the traset behavor of dodes. Thus, we wll make a umber of smlfyg aroxmatos that wll lead us to aroxmately the correct solutos. We kow from our aalyss of the steady state soluto that the jucto wdth s deedet o the aled bas. w x0 + x0 1 / ε( Vb V )( + ) q Further we kow that the jected charges deed o the aled bas: e qv / kt e + ( x+ x0)/ L ( 1) e qv / kt e ( x x0)/ L ( 1) Lkewse the total ject charge deeds o the aled bas: Q q ( x) dx x0 q L qv kt e / ( 1) qv kt qle / 0( 1) Q q ( x) dx x 0 q L qv e / kt 1 qv kt ql e / 0 1 Let us ow defe the curret. We wll break t u to the tme-deedet ad tme-deedet arts Total curret through the dode I C curret through the dode d C curret through the dode () t I + d() t We wll do the same thg for the aled voltages UT EE3301 otes art Page 108 of (49+79) Last udate :3 PM 10/4/0

30 v V v v t V v t a Total voltage across the dode C voltage across the dode C voltage across the dode () + () a ow let us thk about the curret. The ac art of the curret s smly the tme dervatve of the total charge the dode. The dc art ca be wrtte a umber of ways but oe smle way s to use the charge dvded by the recombato tme. (Ths s derved above.) From above: I x q P ( x) m or L q L e qv kt e x x L / ( )/ ( ) 0 1 but the total charge s Q q ( x) dx x0 q e qv / kt e x x L ( 0)/ ( 1 ) x dx 0 q L qv kt x x L e ( ) / ( 0)/ 1 e x0 q L qv kt e / ( 1) luggg ths to our total hole curret, e.g. the morty curret at the jucto edge, we fd I I x x q P P ( ) x x total m or 0 0 L q L e qv / kt ( 1) q L qv kt e / ( 1 ) L ( Q) τ Q τ Lkewse, we ca show Q I total τ UT EE3301 otes art Page 109 of (49+79) Last udate :3 PM 10/4/0

31 otg of course that the above ales to the C art of our curret. (Ths s also kow as the recombato curret.) Thus, we fd that: Q t dq t ( t) () + τ dt () Examle ssume that we have a + - jucto ( >> ). I ths case all of the acto s domated by what haes o the -sde of the jucto. The excess hole desty wll look lke: (x) Q (x) Q > chage V > x x Thus, () () t t () + () Q t dq t τ dt ow let us suose that we start wth a forward based dode, (I I F ) but at t 0 we clam the curret to zero. IF t < 0 () t 0 t 0 For t < 0, we are steady state. Thus UT EE3301 otes art Page 110 of (49+79) Last udate :3 PM 10/4/0

32 () t IF Q t dq t τ dt () + () Q τ Q τif For t 0, the total curret s zero. Thus () t 0 Q() t dq() t + τ dt () () dq t Q t dt τ We ca ow tegrate to get dq() t Q() t dt dt [ ] ex/τ t We ow match the solutos at the boudary (t 0) to fd IFτ t < 0 Q() t IFτex/ [ t τ] t 0 lot of ths shows Q (t) (x) > > t t x Whle I have draw the satal dstrbuto of the holes as a exoetal decay (e -ax ) t does ot look qute lke that a real devce. The loss rate at a gve ot deeds o both recombato ad dffuso. (Our smle model oly accouts for recombato.) The jucto sde of the dstrbuto has greater dffuso tha the cotact sde. (Remember, you had a much hgher dffuso rate at the UT EE3301 otes art Page 111 of (49+79) Last udate :3 PM 10/4/0

33 jucto before but you also had the electrc feld to balace that lose rate. That electrc feld ad hece balacg curret s ot there ay more.) Thus oe would fd a o-exoetal rofle to the charge dstrbuto. (I fact, you would fd that the sloe of the satal dstrbuto goes to zero rght at the jucto edge.) What we have foud s that eve though we ca force a stataeous chage the dode curret, we ca ot force a stataeous chage the charge dstrbuto. (Ths should come as o surrse, as wll take some tme to move the charges to ther ew locatos.) Because we ca ot move the charges stataeously, the voltage across the jucto ca ot chage stataeously ether. Thus the tme varyg jucto voltage deeds o the recombato rate of the charge carrers o both sdes. We ca calculate a aroxmate value for the temoral voltage rofle from the total chage. IFτ t < 0 Q() t IFτex/ [ t τ] t 0 ql x ( x0) If we assume that the satal rofle of the holes follow a exoetal rofle the we get Q( t > 0) IFτex/ [ t τ] ql x ( x0) qv kt qle / 0( 1) [The same aroxmato wll hold for the electros.] Thus, kt IFτ v l [ t ]+ q ex/ τ 1 ql 0 [Remember that ths s a rough aroxmato!] What we have foud s that the voltage lags behd the curret. Ths s dcatve of a caactve tye of devce! To try to uderstad ths, we wll frst look at swtchg dodes. I artcular, we wll look at tur off ad tur o. (These are kow as tur-off traset ad storage decay tme, for reasos that should be obvous soo.) Let us assume that we are sulyg our dode wth a fucto geerator. The bas suled looks lke UT EE3301 otes art Page 11 of (49+79) Last udate :3 PM 10/4/0

34 + Fucto geerator R +V F -V R t We wll further assume that the aled bases are much larger tha the self bas (V b ). Toward the ed of the forward bas hase of the voltage, most of the bas s droed across the resstor. Ths s because t takes very lttle forward voltage to allow us to draw a curret. Further the curret through the resstor must equal that through the dode. The VF VF dode + VF resstor VF resstor V I F F R Further, the jected charge dstrbutos wll be ear ther steady state values, ad thus satally fall of as a exoetal. Whe the voltage source swtches sg, a umber thgs occur: 1) The excess holes/electros do ot go away stataeously ) The voltage across the jucto remas ~ small utl the excess hole cocetrato falls to the ew equlbrum codto. [.e. t acts as f t where stll forward based utl the excess holes/electros are used u.] 3) The curret remas large o the order of what t was before we swtched the voltage UT EE3301 otes art Page 113 of (49+79) Last udate :3 PM 10/4/0

35 4) The excess holes decay by both recombato ad by beg swet back to the + sde. +V F -V R t +V F /R t RR -V R /R t t sd t R Here t sd s the storage delay tme, t r s the recovery tme ad t RR s the total recovery tme. The book calculates the storage delay tme as I I t f + r sd τ l Ir where I r s the curret the reverse drecto ad I f s the forward curret. more rgorous calculato shows that the storage delay tme s UT EE3301 otes art Page 114 of (49+79) Last udate :3 PM 10/4/0

36 1 I tsd erf r τ If + I r where erf s kow as the error fucto ad erf -1 s the verse error fucto. These are rather ugly tegrals that caot be solved aalytcally. For your formato (do ot memorze ths!) 1 x t erf e dt π 0 To reduce the storage delay tme oe ca do a umber thgs 1) Reduce the recombato tme by addg tras etc. ) Make the currets smaller. What ths really does s: a. Make the -rego short - thereby cuttg off Q b. o t forward bas too hard, thus reducg the curret ratos Tur o traset For ths, we assume that we are drvg a fxed curret through the devce, I F after t 0. (Before t 0 the devce mght be ubased or reversed bas. Ether way, that curret level s so small that t does ot lay a major role the result.) Here, after t 0 we are ot oly drvg our curret through but we are also jectg holes, from the + sde to the -sde. Therefore dq() t Q() t IF dt τ Ths equato comes drectly from Q() t dq() t () t () t + τ dt, where s the total curret through the devce. We wll assume that I F s a costat ad s the fal curret. (I our system, as show above, IF VF R.) Thus, UT EE3301 otes art Page 115 of (49+79) Last udate :3 PM 10/4/0

37 dq() t Q() t IF dt τ t t 0 t τ Q 0 dq 1 Q t τ I t τ l 1 Q () t τ I F F () Q Q t dt IF 0 τ 1 () τ e 1 Q () t τ I Q() t τi F 1 e 1 F t τ dq ( τif) ow let us assume that the jucto voltage ca be determed from the total jected charge. (Ths s kow as the quas-statc steady state aroxmato.) Thus t Q( t > ) IF τ 0 τ 1 e ql x ( x0) qv kt qle / 0( 1) qv kt I e / 0τ ( 1) Rewrtg ths, we fd t τ qv I e I ( e / kt F 1 0 1) qv I e / kt F t τ 1 e I kt I + v F t τ l 1 e 1 q I 0 (Ths haes to be very fast.) UT EE3301 otes art Page 116 of (49+79) Last udate :3 PM 10/4/0

38 ode caactace/resstace ow that we kow how the voltage chages wth tme, or wth curret, or wth charge, we ca beg to develo a caactace/resstace model for dodes. We wll have to do ths for both forward ad reverse bases. 1) Reverse bas: a. eleto layer caactace due to chages the wdth of the jucto, w, wth the aled voltage, v..e. the chage the charge wth v. b. Here coducto s effectvely zero, realze that very lttle curret flows reverse bas, so coducto s very low. ) Forward bas: a. Charge storage caactace domates, due to chage morty charge vs v at low frequeces. (ote that our model oly real works at low frequeces t assumes quaseutralty.) b. Coductace s rmarly due to morty carrer jecto. Let us frst look at Krchoff s Law. V LQ«+ RQ«1 + C Q LQ«1 + G Q«1 + C Q where R s the resstace, G s the coducto ad C s the caactace. Thus, f we oly have caactace, C dq dv Ths what haes reverse bas, so let us start wth that: Reverse Bas Caactace We kow from several weeks ago that the total charge the jucto s zero ad thus, Q qx0 q w ( + ) Q qx0 qw where w x0 + x0 + ( + ) 1 / ε Vb v q ow let us calculate the caactace UT EE3301 otes art Page 117 of (49+79) Last udate :3 PM 10/4/0

39 C dq dv d dv qw ( + ) d q [ + dv w ] q ε ε + q + d dv ( + ) ε V v q 1 / q + V v b b 1 / [( b ) ] d dv V v ( ) 1 / 1 / 1 / q ε ε( + ) ( Vb v ) ε w C j (C j s the jucto caactace.) Here we see that the caactace vares weakly o the aled (reverse) voltage but ths s ot sgfcat. ow, let us look at the coductace: dq«di G0 dv dv d dv I e qv kt [ 0( 1) ] q kt I e qv kt 0 q ( kt I + I 0) 0 reverse bas (Ths of course assumes a erfect or deal dode usually ths s a reasoable estmate.) Forward Bas (aga low frequecy) ow what s the charge? It s o loger just what we had before fact we ow eed to look at the charge outsde of the jucto. To make our lfe easer, let us assume that we have a + - jucto. For forward bas, the charge s UT EE3301 otes art Page 118 of (49+79) Last udate :3 PM 10/4/0

40 Q Iτ qv kt τ I e 0 1 [ ] ql0 qv kt τ e 1 τ qv kt qle 0( 1) qv kt qle 0 ow the charge storage caactace s C dq dv d qv kt ql e dv [ 0( 1) ] q qv kt qle 0 kt q kt Q q kt I q ( kt I I τ ) 0 τ - f doe wthout aroxmatos CS The coducto s the same as above, as we have the same equato for the curret. Thus we see that CS G 0 τ For hgh frequecy basg, the equatos become aroxmately G GFB 0 1 / ( 1+ ωτ + 1) G CS 0 1 / ( 1+ ωτ 1) ω Fally, for small sgal resoses, G v C dv a FB a + S dt a Metal-Semcoductor Juctos Thus far, we have oly cosdered the sdes of our devce. However, to use t we must coect t to the outsde world. Ths meas that we have addtoal juctos, otably the metal cotact wth each ed of the devce. What haes at these juctos ad do they affect our results? (If they dd t, would I be askg the questo?) UT EE3301 otes art Page 119 of (49+79) Last udate :3 PM 10/4/0

41 Let us frst look at what the eergy dagram mght look lke for a metal. Metals Eergy level ga states s states searato Ths mles that the Ferm level s eergy level at whch all of the states would be full at T 0K. (s wth the other materals, a hgher temerature results some of the electros movg to hgher eergy levels.) To remove a electro from the metal, thus requres that we move a electro from the Ferm level to the vacuum otetal. (It s commo to set the vacuum otetal to zero but that s just a referece ot. We wll ot do that here.) We kow from very early exermets, that metals have a work fucto for removg a electro ths s what was foud from shg a lght o metals. Thus, we ca draw the eergy dagram for metal as: E Vac qφ M E FM Electros fll most states Eergy Posto where φ M s the metal work fucto. ow let us ush our metal ext to our semcoductor materal. UT EE3301 otes art Page 10 of (49+79) Last udate :3 PM 10/4/0

42 Eergy E Vac qφ M qφ S qχ E c E F E FM Electros fll most states -tye E E v Posto s wth - juctos, we eed the Ferm level to be costat f there s o curret flow. Ths meas that we have a ew jucto at the cotact ot. There are two dfferet tyes of juctos: a) Shottky dode (rectfcato) (Shottky dodes are majorty carrer devces) ( φm > φs, ty e) ( φm < φs, tye ) b) Ohmc cotact (small resstace) ( φm < φs, ty e) ( φm > φs, tye ) Let us ow ut them cotact Eergy E Vac qφ M qφ S qχ qφ B E c E F E FM Electros fll most states qv b -tye E E v Posto There are two otetals that we eed to cosder. Frst, the bas betwee the two materals s gve by the dfferece the trsc eergy or the shft the vacuum bas. Ths s smly qvb qφm qφ S. The other eergy s the amout of eergy that a electro must ga (or lose) order to move from the metal to the semcoductor. That eergy s qφb q( φm χ). The wdth of the jucto s smly the wdth of the -sde there s o -sde ths case. Thus UT EE3301 otes art Page 11 of (49+79) Last udate :3 PM 10/4/0

43 } 0 w x0 + x0 1 / ε Vb v q For metal--tye juctos: } 0 w x0 + x0 1 / ε Vb v q ow let us forward bas the jucto Eergy E Vac qφ M qφ S qχ E c E FM Electros fll most states qφ B q(v b -V F ) -tye E F E E v ad ow reverse bas t Eergy Posto E Vac qφ M qφ S qχ qφ B E c E F E FM Electros fll most states q(v b +V R ) -tye E E v Posto ote that the edge of the coducto ad valace bad must follow the trsc eergy level. ow let us look at what the morty ad majorty carrers do,.e. how do they move from sde to sde. The majorty carrers the metal are the electros. 1) Electro drft from the metal to the semcoductor s slow because of the voltage betwee the Ferm level the metal ad the coducto bad eergy the semcoductor, φ B. ) Hole dffuso from the metal to the semcoductor s very small because there are very few holes the metal. UT EE3301 otes art Page 1 of (49+79) Last udate :3 PM 10/4/0

44 The majorty carrers the Semcoductor are the electros (But they could be holes.e. -tye materal) 3) For our examle, -tye, the hole drft s very small because they are the morty carrer. 4) Electro dffuso ca be sgfcat. We have a majorty carrer devce ow dog the same thg for -tye Eergy E Vac qφ M qφ S qχ E c E FM Electros fll most states qφ B qv b -tye E F E E v Posto 5) For our examle, -tye, the electro drft s very small because they are the morty carrer. 6) Hole dffuso ca be sgfcat. ga, we have a majorty carrer devce If ths does ot make sese, draw stacks of charge carrers ad look at t aga. Lookg at the exected curret flow we fd Flow Equlbrum Reverse Forward Set by S -> M q(v b -V ) M -> S q(φ B ) Ths very smlar to what we saw for stadard - juctos. For the same reasos, we would exect to see currets that are costat for reverse bas ad exoetally creasg for forward bas. Thus, we would exect Itotal Iforward + Ireverse qv kt Ie f + Ir Wth o aled bas we would exect o curret. Thus If Ir I0 qv kt I I e total 0 1 Pctures of all of the ossble cotacts are UT EE3301 otes art Page 13 of (49+79) Last udate :3 PM 10/4/0

45 ( φm > φs, ty e) 1) ( φm < φs, tye ) - Shottky Eergy E Vac qφ M qφ S qχ qφ B E c E F E FM Electros fll most states qv b -tye E E v Posto Eergy E Vac qφ M qφ S qχ E c E FM Electros fll most states qφ B qv b -tye E F E E v ) Eergy Posto ( φm < φs, ty e) ( φm > φs, tye ) - Ohmc E Vac qφ M qφ S qχ qφ B E c E F E FM Electros fll most states qv b -tye E E v Posto ot a barrer to electro movemet just some small resstace -> Ohmc cotact UT EE3301 otes art Page 14 of (49+79) Last udate :3 PM 10/4/0

46 Eergy E Vac qφ M qφ S qχ E c E FM Electros fll most states qφ B qv b -tye E F E E v Posto o real barrer to hole movemet just some small resstace -> Ohmc cotact. Fally, there are two ractcal ssues: 1) I ca make ay jucto act ohmc. Ths s ecessary as some materals make t tough to obta ohmc cotact otherwse. The trck s very heavy dog. The the jucto wdth s very arrow ad tuelg occurs. Thus all juctos are tycally heavly doed. ) The surface effects ofte E F, whch ca alter the work fucto of the metal or χ for the semcoductor. So geeral the values are emrcally determed from the devce characterstcs. Examle S dode: -sde 3E18 cm -3 τ 0.1 µs τ µ 100 cm /V-s µ 00 cm /V-s -sde 4E15 cm -3 τ 1 µs τ µ 450 cm /V-s µ 1300 cm /V-s Cross sectoal area 10-4 cm 1.5E10 cm -3. UT EE3301 otes art Page 15 of (49+79) Last udate :3 PM 10/4/0

47 1) Fd V b. kt Vb l q V ) Fd the jucto wdth at equlbrum w x0 + x0 1 / ε( Vb v) ( + ) q ε εrε ( 8, 85E 14F/ cm) w 517. E 5cm 3) Fd I f the devce s forward based by 0.38 V qv kt I I e 0( 1) I q L L o + 0 o q o + τ o τ q kt µ ktµ o + o q q τ τ 9E 16 I 1. E 9 ( or ) ote that at 0.9 V the dode releases 1 of curret! UT EE3301 otes art Page 16 of (49+79) Last udate :3 PM 10/4/0

48 Useful equatos debrogle mometum h/λ h k. Heseberg ucertaty rcle x h E t h Photo eergy Bohr Model E hoto hν, r K me 0. 59Å E h Bohr 4 1 me K h eV / h h Schrödger s equato + V Ψ r, t tψ r, t m j Maxwella dstrbuto 3 / m mv fv ex πkt kt 1 E f( E ) ex kt kt. Equato of moto * F ee m a * F + ee m a (electro) (hole) Ferm-rac fucto. STTE ESITY f( E ) c v 1 E 1+ ex E kt * F 3 / m ( E ) de E EcdE π h * 3 / m ( E ) de Ev E de π h Coducto bad Valace bad UT EE3301 otes art Page 17 of (49+79) Last udate :3 PM 10/4/0

49 state dstrbuto fucto. 1 ( E ) de f( E ) c( E ) de E EF 1+ ex kt ( E ) de 1 f( E ) v( E ) de 1 total umber of electros ad holes Itrsc Eergy Moblty * 3 / m π h E EcdE Electros the Coducto bad 1 * 3 / m ( E EF) π Ev E de Holes the Valace bad h 1 + ex kt ( Ec EF) ( EF E) 0 c exex kt kt 0 v E c v Ev EF EF E exex kt kt * mkt πh * mkt πh 3 / 3 / ( E + E ) kt m + l m v c * * v τ q µ E * m v τ q µ E * m 3 / Coducto ffuso J q v ( 0 0 ) q µ + µ σe E / ρ kt µ q kt µ q E UT EE3301 otes art Page 18 of (49+79) Last udate :3 PM 10/4/0