^ s V M)""p,SA7V qNa Prob. 6.29 An n-channel MOSFET has a lam long channel with N = 1016 and JV, = 1 02 0^ in cm ' the source and drain. Find the VD which causes punch-through. V=kT-ln 'N.V a cm 1016 1 02 0^ (1.5-1010-)2 , v There are two depletion regions, one at the source end and one at the drain end of the channel. Nd Na so most of W is in the p-side (channel). At the (zero-bias) source end, = 0.0259V-In = 0.933V cnr cm V n. cm3 J J x Ps ^ 2e V q-N, a J S 0 2-11.8-8.85-10"14 ^ 0 . 9 93 V 1.609-10~19C-10 cm 0.35um i In the drain end, xpD = 2S(V0+VD) q-Na Punch-through occurs when xpD= L xpS = 0.65um 0.933V + VD = (0.65-10"4cm)2-1.609-10^0-1016^ 2-11.8-8.85-10-14^ VD = 2.3V Chapter 7 Solutions Prob. 7.1 Plot the doping profile. For the base diffusion, D-t = 3-10"13^-3600s = 10.8-10"10cm2 2A/ETT = 6.58-10"5cm Vn-D-t = 5.82-10'5cm Na(X): N Vn-D-t e UVm J = g _6.1 0i7 _j_. e" 4.3-io-W

For the emitter diffusion, D-t = 3-10~14^-900s = 27.0-10"12cm2 2 7 r >t = 1.04-10"5cm V3i-D-t=9.21-10"6cm Nd(x) = Ns-erfc ,2Vr>T = 5-1020^r-erfc f ~ 1.04-KPcm \ ioM . \W<(: 0 (ni ttr) 10* 1 Base : width sU $tm 10 "N. Wf fflle r . : . X m collect M m u 1.4 m) x (fan) Prob. 7.2 Sketch Ic versus -VCE for the BJT and find-VCE for h=0.1mA. ic(mA) 20 C ( V) VCE=5VforIB=0.1mA Prob. 7.3 Plot 8p across the base ofap-n-p with Wt/Lp=0.5. Mj = 1.58 and M2 = 0.58 from Equation 7-14 e* = 1.58 and e"* =0.606 These values may be filled in to obtain a plot such as Figure 7-7 with normalized axis. ^= M,e Lp M , e4 = 1.58e_I 0.58c1 = 0 for ^ = 0 .5 ApE 5P Ap ^ = Mx Lp MxLp = 1.58e 0.58e = 1.0 for ^ =0 "* Prob. 7.4 For a n+pnBJT, show the current contributions, and band diagram. :E^:h*; :etectrbri:-fl6w:: :e:^nn+ EQUILIBRIUM P V n+ NORMAL BIAS B Prob. 7.5 Find a and Bfor the given BJT characteristics.

5 -^ Wb=0.1-Lp xp=xn = ^ a = B-y= cosh 'w^ + ^ ^ s i nh rW^) cosh (0.1) + ^2-0.1-0.5 sinh (0.1) VLP7 vLPy a 82 = 0.988 Prob. 7.6 Calculate the desired values for each BJT modification. Let us assume a PNP BJT; the results are the same for NPN a) find the change in the collector current NE ; wB, wE ! NB L Nc I lc 4 ifw;= then Ic p Jrn _ H w 0.5-WB andNB = k 5 _ Ic 10-0.5 ^ ^ p i ii NBWB = 10-NB .P kT _^u I wB forward \U b) find yandB WB, WEL Since we assume that the base and the emitter are much shorter than the diffusion length, the carrier concentration profiles vary linearly with distance in the top figure. IfNE=100-NB andWE=0.1-WB then using IE a n; NE-WE e kT and IF a n: NB-WB 3 kT lK + \ NB-WB 1 NBWB 1 +NEWE 1+Mi

NE-WE NB-WB 100-NB-0.1-WB 1+1 1+0.1 Base carrier profile is linear so B = 1 c) find B and y Here, since we assume that the base and emitter regions are much longer than the diffusion lengths, the carrier concentrations decay exponentially with distance, as shown in the lower figure. Base carrier profile for long diode exponentially decays to 0; so B = 0 IfNE= 100-NB andLn=Lp=L then using IE a n; NE-L 1 'VBB e kT and IE a n; 'P NB-L q-vBE > kT ~P k + k NB-L -+NB-L NE-L 1 l 1 1 +^ 1 + ^100-Nc 1 1+0.01 0.99 0.91 Prob. 7.7 Identify which gives the best diode characteristics. ApE (a) IE = Ic , IB = 0Since V is large, the collector is strongly reverse biased, Ape = PnSince Is = Ic, APE = Ape = pn from Eq.(7-34). The area under Sp(xn) is zero. (b) VCB = 0, thus Ape = 0. Notice that this is the narrow-base diode distribution. S? + V fn:<& Sp 901 Aps ApE Ape ctApE 0 Wh Wh (c) Since Ic = 0, Ape = aAps from Eq.(7-34b).