20111ee121B_1_EE121B_HW4_soln

20111ee121B_1_EE121B - 2011 Winter EE121B HW4 SOLUTIONS 1(a From Slide 4-34 Q p(t = I B p(1 e Turn-on time=tS t p t>=0 Plug in the above equation Q

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2011 Winter EE121B HW4 SOLUTIONS 1. (a) From Slide 4-34 ) 1 ( ) ( / p t p B p e I t Q τ - - = t>=0 Turn-on time=t S CC t S p I t t Q = = ) ( , I CC =E CC /R L =10V/5Kohm=0.002A Plug in the above equation CC t t p B I e I p S = - - ) 1 ( / b ] ) 1 ln[( 1 - - = B p CC t p S I I t Plug in I B =50uA, τ t =1x10 -9 s, τ p =2.2x10 -5 s b t S =4x10 -8 s (b) From Slide 4-36 ) 1 2 ( ) ( / - = - p t p B p e I t Q To solve t sd CC t sd p I t t Q = = ) ( CC t t p B I e I p sd = - - ) 1 2 ( / b ) 1 2 ln( + = B p CC t p sd I I t The time interval t 1 is for the collector current to drop to 1/e of the initial value ) 1 2 ( / ' - = - p t t p B C e I I ( exponential decay beyond t sd point, set (t’=0)=(t=t sd )) e t I t t I C C 1 ) 0 ( ) ( ' 1 ' = = = e I e I t p B t t p B p 1 ) 1 2 ( / 1 = - - b ) 1 1 2 ln( 1 + = e t p So, the total turn off time=t sd +t 1 =2.35x10 -5 s =1.48x10 -5 s =8.35x10 -6 s Ic drops to 1/e t 1
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2. (a)(i) Punchthrough occurs when X 1 2 merge Dp E = kT/q x 140 = 3.626 cm 2 /s Dn B =kT/q x 230 = 5.957 cm 2 /s Lp E = (Dp E x τ p )^0.5 = 8.52 x 10 -4 cm Ln B = (Dn B x τ n )^0.5 = 1.14 x 10 -2 cm 2 , ln( ) 0.895 dE aB BE i N N kT Vbi q n × = = 2 , ln( ) 0.656 dc aB BC i N N kT Vbi q n × = = Wbase=2um W BE : depletion region in Emitter-Base junction W BC : depletion region in Base-Collector junction
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This note was uploaded on 03/28/2011 for the course EE 121b taught by Professor Bjt-gamma during the Winter '08 term at UCLA.

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20111ee121B_1_EE121B - 2011 Winter EE121B HW4 SOLUTIONS 1(a From Slide 4-34 Q p(t = I B p(1 e Turn-on time=tS t p t>=0 Plug in the above equation Q

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