121B_1_bjt

121B_1_bjt - Bipolar (Junction) Transistor (BJT) E N+ poly...

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Unformatted text preview: Bipolar (Junction) Transistor (BJT) E N+ poly p+ poly P+ P Intrinsic base N+ N- Collector N+ B C While BJT is two-dimensional in nature, intrinsic behavior mainly one-dimensional between the dotted lines. Jason Woo 1 EE121B Spring 2010 EC EFn VBE EV EFp VCB EFn Under normal operation, the base-emitter junction is forward biased, while the base-collector junction is reverse biased Contact n+ E h h p e e n Contact n+ C B Jason Woo 2 EE121B Spring 2010 Similarly, for PNP transistor under normal mode of operation: EC EFp VCB VBE<0 VCB<0 EFp EV VBE EFn Contact p+ E e e n h h p Contact p+ C B Jason Woo 3 EE121B Spring 2010 There are many usages for BJT. For example, as a power amplifier: ie N P N RL ic vout v VBE VCB The AC resistance the small input voltage source "sees" is the AC resistance of a pn junction. i.e., kT rin = rE = qI E 2 vs input power = rE = ie2 rE Jason Woo 4 EE121B Spring 2010 But output power = 2 vout RL = ic2 Rout i.e., ic2 Rout Rout 2 power gain = 2 since ic2 ie (we will show this later) ie re re Jason Woo 5 EE121B Spring 2010 We will now mainly study NPN transistor. PNP characteristics can be derived similarly. First, let us look at the DC characteristics. Contact IE E n+ e h IB B p e n+ n Contact IC C I E = I En + I Ep I C = I Cn + I CB 0 I B = I E - I C (Kirkoff's Law) Note that, Jason Woo I Cn I En if the base is small few electrons recombine in the base 6 EE121B Spring 2010 The key action is therefore in the quasi-neutral base, consider a uniformly doped NPN BJT, the equation that governs the electron flow under low level injection is: J n = qDB n x This is very similar to deriving the diffusion currents in PN junction under forward bias. Now in steady state, the continuality equation is: J n = qR = q n n B x where Combined the two equations, we have 2 n n = 2 2 LB x Jason Woo LB = DB B LB is the diffusion length of the minority carriers (electrons) in the base. 7 EE121B Spring 2010 Similarly, in the quasi-neutral emitter region: p J p = -qDE x and J p p = -qR p = q E x 2 p p = 2 2 LE x LE = DE E Combined the two equations, we have where LE is the diffusion length of the minority carriers (holes) in the emitter. Jason Woo 8 EE121B Spring 2010 To solve the equations, we need the boundary conditions: E B NDE -xE-WE -xE 0 NAB WB C NDC xC WB is neutral base width, WE is neutral emitter width Recall that when a bias VJ is applied to a PN junction, the minority carrier concentrations at the depletion edges are: ni2 qVJ n ( xP ) = exp NA kT Jason Woo ni2 qVJ p ( xn ) = exp ND kT 9 EE121B Spring 2010 Therefore, in the BJT case: In the quasi-neutral base: ni2 qVBE n ( 0 ) = exp kT N AB - 1 ni2 qVBC n ( W B ) exp N AB kT -1 ( 0) In the quasi-neutral emitter: ni2 p ( - x E ) = N DE qVBE exp kT - 1 p ( -W E - x E ) 0 Also in the quasi-neutral collector: ni2 p ( xC ) = N DC Jason Woo qVBC exp kT 10 - 1 EE121B Spring 2010 The general solution for the electron equation in the base: 2 n n = 2 2 LB x is given by: x n ( x ) = A exp LB -x + B exp LB With the B.C. at x=0 and x=W, we have: n ( x ) = and, n ( 0 ) sinh ( (WB - x ) / LB ) sinh (WB / LB ) n (WB ) sinh ( x / LB ) + sinh (WB / LB ) qDn n ( 0 ) cosh ( (WB - x ) / LB ) n (WB ) cosh ( x / LB ) - Jn ( x ) = + LB sinh (WB / LB ) sinh (WB / LB ) 11 EE121B Spring 2010 Jason Woo n n(0) Decreasing WB/Lp n(W) 0 0.2 0.4 0.6 0.8 1 x/WB We see that, if WB LB I Cn I En Jason Woo 12 EE121B Spring 2010 Similarly, in the quasi-neutral emitter, 2 p p = 2 2 x LE with B.C. p ( x ) = p ( i.e., sinh x / L E - xE sinh W E / LE ) ( ( ) ) with x = - ( x + x E ) + WE ( x + xE ) p ( x ) p ( - x E ) exp LE if W E >> LE WE + ( x + x E ) p ( x ) p ( - x E ) WE Jason Woo if W E << LE EE121B Spring 2010 13 For the collector current, just like the case for PN junctions, it is given by the sum of diffusion currents and recombination currents. * J C = - J n ( x = WB ) - J p ( x = xC ) - J rec Diffusion Currents n p * J rec = - qDB + qDC - x x =W x x = xC 0 except when B breakdown Note the sign! Assuming the collector is long, we have (c.f. PN junction under reversed bias) qDC pC 0 p qDC (e =- LC x x = xC qVBC kT - 1) 0 (since VBC < 0 typically) - 1 i.e., qDB nB 0 W W qV qV JC = csch B exp BE - 1 - coth B exp BC LB kT kT LB LB 0 D p qV * - q C C 0 exp BC - 1 - J rec LC kT 0 Jason Woo 14 EE121B Spring 2010 Similarly for JE - J E = J n ( x = 0 ) + J p ( x = - x E ) + J rec n p = qDB - qDE x n x = 0 + J rec x = - xE Note the sign! W W qDB nB 0 qVBE qV coth B exp = - 1 - csch B exp BC LB kT kT LB LB 0 WE qDE pE 0 qVBE coth + exp - 1 + J rec LE kT LE - 1 qDE pE 0 WE << W = W E if W E * E Jason Woo LE 15 * WE = LE if W E >> LE EE121B Spring 2010 Also, J B = - JC + J E (Kirkoff's law) For practical BJTs, WB LB and VBC < 0 VBE > 0 under nomal mode operation, expanding the hyperbolic functions using Taylor series, 3 1 WB - qDB N B 0 LB / qVBE -J E = 1 + exp - 1 LB 3 LB / kT WB qDE pE 0 qVBE - exp kT - 1 + J rec * WE 2 qDB nB 0 LB qV 1 WB 1- exp BE - 1 JC = kT LB WB 6 LB / DC pC 0 * +q - J rec LC Jason Woo 16 EE121B Spring 2010 From PN junctions' recombination currents under forward bias and the generation currents under reversed bias, we have: qVBE qni exp nkT W J rec rec DBE with 1 < n < 2 J * rec = - J gen - qni g W DBC with g n + p Jason Woo 17 EE121B Spring 2010 Current Gain Common emitter current gain: I C I B = Common base current gain I C 0 = I E 1 I B Since I B = - I C + I E = = -1 + I C 0 1 i.e., Jason Woo 0 = 1 - 0 18 EE121B Spring 2010 Now, using chain rule I C I nE I nC I C 0 = = I E I E I nE I nC We shall look at these terms one by one First I nE I nE = = I E I nE + I pE + I rec ( ) = 1+ I p ( - x E ) I n ( 0 ) 1 + I n ( 0 ) I rec Jason Woo 19 EE121B Spring 2010 Now I p ( - x E ) I n ( 0 ) DEWB N AB DBW E N DE Since typically WB<<LB and WE<<LE and I rec = I n ( 0 ) WDBEWB N AB exp DB ni rec - qVBE kT n - 1 n is small for large V BE But can be much large than 0 for small VBE Jason Woo 20 EE121B Spring 2010 log I IC InE IBIPE+Irec IPE dominates Irec dominates VBC<0 VBE Jason Woo 21 EE121B Spring 2010 Next, we consider I nC I n (WB ) = I nE I n ( 0 ) This is called the base transport factor, T T Now I n ( W B ) I n ( 0 ) T = I n ( W B ) I n ( 0 ) = 1- ( I n ( 0 ) - I n (WB ) ) I n ( 0 ) i.e., T is the percentage of injected electrons in the quasi-neutral base region that do not recombine with the majority carriers. Jason Woo 22 EE121B Spring 2010 Using the equations for In(WB) and I0(W0) , we therefore have: WB csech WB LB T = sech WB LB coth LB 2 WB 1- 2 2 LB In general, since J n - q n = - qRn = n x J n (W B ) - J n ( 0 ) = -q WE 0 ndx n if WB LB , T 1 Jason Woo 23 EE121B Spring 2010 Next, let us consider Recall I C I nC I C = I nC + I pC - I rec = I nC + I pC + I gen Again, under normal mode operation qDC N DC I p ( xC ) + A LC * - I rec is small I n ( 0 ) ( ) * * is also small since I rec = - I gen is small * Of course, I gen is large when the PN junction is close to or at breakdown. Jason Woo 24 EE121B Spring 2010 We can write I C =M I nC I C = M 1 I nC i.e. , if VBC is not too large (much less than BVBC) Together, we have 0 = T M For advanced BJTs, 1 - >> 1 - T 1 1 i.e., >> 1 - T 1- Therefore, Typically 0 = T M Jason Woo 25 EE121B Spring 2010 That is, J n ( 0) 0 = = 1 - 0 1 - J p ( - x E ) + J rec J n ( 0) DEWB N AB J p ( - x E ) DBWE N DE if Jrec<< Jp(-xE), i.e., JE is not too small Therefore, to maximize , we need to make WE NDE>>WB NAB (i.e., heavily doped the emitter). Note that in the low current regime when Jrec Jp(-xE), J n ( 0) 1- J rec = D B ni WB N AB W DBE rec e qVBE n -1 n kT can be rather small Jason Woo 26 EE121B Spring 2010 At moderate current = At high current, DBW E N DE constant DEWB N AB - and I C n I n -1 n C - IC n Irec dominates IB VBC<0 - IC 1 *In Jason Woo IC if the base reaches high level injection condition. 27 EE121B Spring 2010 reality, drops with large IC due to base-push-out effect. 0.8 0.7 0.6 5A 4A 3A 2A IB=1A 1 2 3 4 5 6 IC (mA) 0.5 0.4 0.3 0.2 0.1 0 0 VCE Jason Woo 28 EE121B Spring 2010 Output conductance (Early effect) At fixed VBE, IC increaes with VCB since WB = WBM - W DBE - W DCB and WDCB as VCB WB I C Now, ignoring the recombination in the base AqDB ni2 qVBE IC exp N BWB kT Jason Woo 29 EE121B Spring 2010 In this case I C VCE But VBE I C - I C W B = = VCB W B VCB W B = W BM - W DBEp - W DCBp W B VCB W DCBp =- VCB Now, recall from pn junctions: Vbi + VCB = N ABW DCBp Jason Woo q 2 2 N ABWDCBp + N DCW DCBn 2 = N DCWDCBn 30 ( ) EE121B Spring 2010 Solving for WDBC, we have: W DBCp = - 2 (Vbi + VCB ) N qN AB 1 + AB N DC That is: W B VCB W DCBp =- VCB =- 2qN * (Vbi + VCB ) -1/ 2 where Jason Woo N N = N AB 1 + AB N DC * EE121B Spring 2010 31 Recall C DBC =W = DBC = 2 (Vbi + VCB ) N N q AB DC N AB + N DC = 2 (Vbi + VCB ) 2 N AB q N 2 (Vbi + VCB ) N 2 qN AB (Notes: Capacitance/A) with W DBC = W DBCn + W DBCp W B VCB W DCBp =- VCB C DBC =- qN AB 32 EE121B Spring 2010 Jason Woo or I C VCE VBE - I C W B = W B VCB IC C DBC IC C DBC = = W B qN AB qWB N AB Define Early voltage, VA , via IC I C = VCE VA we have VA = Jason Woo qWB N AB C DBC 33 EE121B Spring 2010 IC IB6 IB5 IB4 IB3 IB2 IB1 -VA ~0.7V VCE As the equation implies, VA is independent of IC Jason Woo 34 EE121B Spring 2010 In advanced bipolar transistors, the base is doped reasonably high and the base does not reaches high level injection. However, the collector is only moderately doped. Base push out can occur --- Kirk's effect. In the base-collector depletion region, the E-field is large i.e., the Poisson's eqn is: e sat J 2 - = = N ( x) - n = N ( x) - 2 q sat x where N ( x) = = ND ( x) in the collector in the base EE121B Spring 2010 = - N A ( x) Jason Woo If J / q sat is comparable to ND(x) in the collector, then the collector no longer has a depletion region with the +ve charge. E-field lines are terminated in the e-. log N e- ND NA n ND ND New Base Region Jason Woo EE121B Spring 2010 B E N + D e p NA WB (pre Kirk's Effect) e ND + ND C WB (post Kirk's Effect) For good gain at high IC, collector must be doped high enough to avoid Kirk's effect In other words, we need to have: J < N Dcollector q sat As the device size reduces, I constant, and J N Dcollector has to increase. Note that as a result, CBC and becomes very important in determining the transient behavior of the BJT. Jason Woo EE121B Spring 2010 Bipolar transistor Breakdown Start with J C = M T J En We are interested in the case when M>1. Again M is caused by impact ionization in the base-collector depletion region. i.e.: M= J n (W B + X BC ) J n (W B ) JC # e - leaving depregion = - # e entering depregion J n (W B ) and I C o = (- I E ) Jason Woo = MT J C J n (W B ) J En = J n (W B ) J En (- J E ) 38 EE121B Spring 2010 Including the reversed junction currents, we can write I E = I EBo (exp(qVBE / kT ) - 1) + oR IC IC = - ICBo (exp(qVBC / kT ) - 1) + oF I E where ICB0 and IEB0 are the saturation currents of the collectorbase and the emitter-to base junctions when the other terminals (emitter and collector respectively) are open. In the common emitter case under normal mode operation, when IB=0 IC = I E ICBO = 1 - OF Jason Woo 39 EE121B Spring 2010 Let ICEO ICBO /(1 - OF ) i.e., if IB = 0 I E = IC = ICEO Jason Woo 40 EE121B Spring 2010 BVCB0 BVCB0 is the (collector) breakdown voltage when the emitter is open and the base is grounded. It is therefore just the junction breakdown of the base-collector junction. That is, when M= J n (W B + X BC ) J n (W B ) Experimentally, it was found that: 1 M= 1 - (V / BV )m i.e., for the case of BVCB0 with 3 M 6 1 M (VCB ) = 1 - (VCB / BVCBO )m Jason Woo 41 EE121B Spring 2010 BVCE0 BVCE0 is the breakdown voltage of the collector when the base is open and the emitter is grounded. In this case, there are interaction between the 2 junctions and BVCE0 happens when: ICEO i.e., when M (VCB )T = M ( BVCEO )T = 1 OF = 1 With the expression for M, we have M ( BVCEO )T Jason Woo 1 = m T 1 - ( BVCEO / BVCBO ) 42 =1 EE121B Spring 2010 We therefore have: BVCEO = BVCBO (1 - T )1/ m as T = o 1 BVCEO < BVCBO Since o 1 o = 1 - o 1 - o BVCEO 1 BVCBO o 1/ m we have Jason Woo 43 EE121B Spring 2010 Jason Woo 44 EE121B Spring 2010 Dynamic Behavior of Bipolar Transistors AC model of a BJT: hybrid- model RL C B v rbb B E b rbb rbe C b c b c C b e ib rbc gm be ro e Jason Woo 45 EE121B Spring 2010 and I BE kT kT rbe = = = qI BE qI C VBE C b e Q B = VBE -1 with QB = QBdep + QBf + QBimg QBdep is the base-emitter depletion charge. QBf = Aq WB 0 ndx is the charges stored in the quasi-neutral base region 2 i Since Jason Woo qVBE qn DB exp kT J n = qn ( x ) ( x ) = N BEWB 46 EE121B Spring 2010 QBf = AJ n QBf VBE WB 0 1 dx ( x) and qI C = B kT where B 0 WB 1 dx ( x) is the base transit time. For uniformly doped base n ( 0 ) ( W B - x ) n ( x ) = WB n ( 0 ) J n = qDn WB Jason Woo 47 EE121B Spring 2010 i.e., QBf Jn = Dn (W WB 0 B - x )dx 2 J nWB = 2 Dn WB2 B = 2 Dn For non-uniformly doped base, since the built-in E-field actually increases the electron velocity across the base 2 WB B = Dn with > 2 Jason Woo 48 EE121B Spring 2010 QBing is the "hole charge" stored in the base, due to the electrons in the base-collector depletion region. First, the electrons stored in the depletion region is Qelecdep = AJ n WB + xc WB 1 dx ( x) AJ n xc sat The hole charges induced in the base are the electrons in the depletion region, the other half being imaged in the collector. i.e., QBimg I c xc 2 sat Jason Woo QBimg VBE qI C xc = kT 2 sat EE121B Spring 2010 49 Similarly, C b c = Q B is the depletion capacitance of the base-collector junction VBC i rb c = bc vbc is due to the reversed biased base-collector junction current Also gm vb e = ic ic 1 gm = * ib rbe * with vbe = ib rbe Recall ic = * ib VA ro = IC VCB is the Early effect factor: = 1 + VA Also: Jason Woo and VA is the Early voltage 50 EE121B Spring 2010 Finally, if we ac ground the collector terminal (tie to the ac emitter). ic gm rbe h fe = = ib qI c qI c xc 1 + j rbe C bedep + + C b c B + kT kT 2 sat That is, the cut-off frequency for current gain, fT, ic (T ) T =1 fT = with ib (T ) 2 can be derived to be: rbe xc 1 C bedep + C b c + B + EC = 2 fT gm rb e 2 sat ( ) In the case where there is other parasitic capacitance, Cp, between the base-collector and the base-emitter junctions xc 1 kT C b edep + C bc + C p + B + = 2 fT qI C 2 sat ( ) C Jason Woo 51 EE121B Spring 2010 Thus, to have good high-frequency AC performance, we need to have 1 2 3 High IC Small B short base and large "built in" field Small C higher doped collector. Note that (3) is in contradiction to small CBC Jason Woo 52 EE121B Spring 2010 Large signal switching performance of a BJT In addition to the EC mentioned above, the RC delay (due to RB'B) is also important when BJT is used as a switch (such as in ECL circuit). Since switches are typically VB (not IB) controlled, the delay is characterized by RB B ( C bc + C b e ) That is, to improve BJT performance, we need to eliminate any parasitic CP and R B'B. Jason Woo 53 EE121B Spring 2010 Using the large signal equivlanet model : E Re Ce + CD E IB B C2 Rbx C1 Rbi B IB Cc CS C S Rc C S it can be shown that the delay can be expressed as Td k1 R1C c + k2 R1C e + k3 R1C s + k4 R1C1 + k5 + k6 Rbi C c + k7 Rbi C d + k8 Rbx C d That is, fast switching speed requires small parasitic, and a thin base. Jason Woo 54 EE121B Spring 2010 ...
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This note was uploaded on 06/01/2010 for the course EE EE161 taught by Professor Wang during the Spring '09 term at UCLA.

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