UNIVERSITY OF CALIFORNIA AT BERKELEY

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1 Homework #10 Soluions EECS 40, Fall 2006 Prof. Chang-Hasnain Due a 6 pm in 240 Cory on Wednesday, 04/18/07 oal Poins: 100 Pu (1) your name and (2) discussion secion number on your homework. You need o pu down all he derivaion seps o obain full credis of he problems. Numerical answers alone will a bes receive low percenage parial credis. No lae submission will be acceped expec hose wih prior approval from Prof. Chang-Hasnain. 1. Hambley, P10.46 (a) he inegral of V m = sin(! ) over one cycle is zero, so he dc volmeer reads zero. [2ps] (b) = / 2 / 2 1 ( % 1 ( Vm % Vm V avg = & Vm " d d " = * sin( ) + * 0 # = & ) cos( ) " # '! 0 / 2 $ ' $ = 0 [4ps] (c) V avg / 2 1 ( % 2V m = & Vm sin( " ) d Vm sin( " ) d * + * ) # = [4ps] '! 0 / 2 $ 2. Hambley, P10.48 As in Problem P10.47, he peak volage mus be 10 V. For a full-wave recifier, he capaciance is given by Equaion (10.12) in he ex: I L 0.1(1/ 60) C = = = 417µ F [4ps] 2V r 2(2) he circui diagram is: [6ps] Page 1 of 10

2 3. Hambley, P10.52 (a) he curren pulse sars and ends a he imes for which [5ps] v ( ) = V s B 20sin(200! ) = 12 Solving we find ha sin! 1 (0.6) = = 1.024ms and =! 3.976ms = sar end sar " Beween hese wo imes he curren is 20sin(200" )! 12 i( ) = 80 A skech of he curren o scale versus ime is Page 2 of 10

3 (b) he charge following hrough he baery in one period is end sar end Q = ( i( ) d = ( sar 20sin(200) ) ' 12 d 80 & 1 12 # = $ ' cos(200) ) ' % 800) 80! " end sar [5ps] Finally, he average curren is he charge divided by he period! 6 Q 194x10 I avg = = = 19.4mA! 3 10x10 4. Hambley, P10.55 [10ps] Refer o Figure P10.55 in he book. When he source volage is negaive, diode D 3 is on and he oupu v o () is zero. For source volage beween 0 and 10 V, none of he diodes conducs and v o () = v s (). Finially when he source volage exceeds 10 V, D 1 in on and D 2 is in he breakdown region so he oupu volage is 10 V. he waveform is: Page 3 of 10

4 5. Hambley, P10.57 [10ps] Page 4 of 10

5 6. Hambley, P10.60 (EE40 Only) [10ps] Refer o he circui shown in Figure P10.60 in he book. If he oupu volage aemps o become less han -5 V, he Zener diode breaks down and curren flows, charging he capaciance. hus he negaive peak is clamped o -5 V. he inpu and oupu waveforms are: 6. Hambley, P10.62 [EE100 Only] [10ps] A suiable circui is: Page 5 of 10

6 7. Consider he following circui: [10ps] Use he 0.7 model for all he diodes, and le R 2 =2R 1. Skech he volage ransfer characerisic (v o vs v in ). he inpu volage is beween -3V and +3V. Using he volage divider formula, he volage across R1 is 1/3Vin. herefore, if he volage across R1 is beween -0.7 and 1.4, all he diodes are off. Noice ha D3 has a reverse polariy han D1 and D2. his means ha if -2.1 < Vin < 4.2, hen all he diodes are off. Given he max Vin=+3, D1 and D2 are always off. As a resul, Vo=2/3Vin or -1.4 < Vo < 2. If -2.1 > Vin > -3, hen he volage across D3 is greaer han 0.7 and he diode urns on. As a resul, all he curren goes hrough he diode (since i s shor) and he volage across D3 is 0.7 Vol, -2.3 < Vo < -1.4 Page 6 of 10

7 Vo Vin Diode Logic: In his problem, we will compare 2 implemenaions in diode logic of he same logic funcion. Vcc Vcc R R D2 3 4 A D1 D1 A D2 B D3 C F (a) (b) a) he logic funcion performed is he AND of he inpus A, B, C. [2ps] Page 7 of 10

8 b) In Figure (a), when all he inpus are low, D1 is off and he oher diodes are on so Vou = 0.7V. In figure (b), all he diodes will be on, so Vou = 0.7[3ps] c) he power dissipaed in circui (a) is: P1=(3-0.7) 2 /100k and P2=(3-1.4) 2 /100k and oal power is P=P1+P2= 7.85 *10 5 Wa. he power dissipaion in circui (b) is; P= 5.29 *10 5 Wa [3ps] d) For he circui in Figure (b), when inpus A and B are low and C is high, D1 and D2 are on and D3 is off, so he oupu F is 0.7 since he diodes ha are on and carry a 0.7V across hem. However, in Figure (a), when he inpus A and B are all low and C is high, D1, D2, and D3 are on and D4 is off. herefore, he oupu F is 1.4, which is very close o being logic high and could resul in an error. [2ps] 9. Hambley, P10.64 [EE 100 Only] (a) A suiable circui is: [5ps] We choose he resisors R 1 and R 2 o achieve he desired slope. Page 8 of 10

9 Slope = 1 3 = R2 R + R 1 3 hus, choose R 1 = 2R 2. For example, R 1 = 2 kω and R 2 = 1 kω (b) A suiable circui is: [5ps] Oher resisor values will work, bu we mus make sure ha D 2 remains forward biased for all values of v in, including v in = -10V. o achieve he desired slope (i.e., he slope is 0.5) for he ransfer characerisic, we mus have R 1 = R Diode + Op-Amp: Consider he following circui: [EE40 Only] Use he 0.7V model for he diode. Page 9 of 10

10 a) When he diode is on, he volage across i is 0.7V. Using he Summing poin consrain, he volage a he invering inpu of he op-amp is 2. Now we use KCL a he (2 # ( d Vc invering inpu node: Vin # 0.7)) ( = 20µ F! ) where Vc= 2! Vou (). Subsiuing 5k" d Vc in he KCL equaion and inegraing gives Vou in erms ofvin. 1 2 (! vin + 2.7) d RC " [3 ps] 1 b) When he diode is off, here is no curren in he resisor. herefore, Vc is consan and Vou is also consan. [2ps] c) When he diode is on, he volage across i has o be greaer han 0.7 (wih he correc polariy applied). Given he volage a he invering inpu of he op-amp is 2V (Summing poin consrain), he inpu volage Vin has o be greaer han 2.7V in order for he diode o urn on. [2ps] d) he periods is =1/50=0.02 sec. Vin is a riangle wave wih a peak-o-peak value of 20 V, so he heigh of he riangle is Vpp/2=10 V. In order o find he oupu volage Vou, we have o inegrae he inpu wave beween 1=0 and 2=0.06 sec. Given he period is 0.02, we can inegrae he inpu wave over one period and muliply he resuls by 3. Vou=3*0.1=0.3 V [3ps] 10. Doping [10ps] a) he majoriy carrier for Silicon doped wih Phosphorus is elecron, and he resuling maerial is n-ype. Concenraion of holes=10 15 and concenraion of elecrons is /10 15 [3ps] b) he majoriy carrier for Silicon doped wih Arsenic and Boron is holes, and he resuling maerial is p-ype since he concenraion of Boron is more han Arsenic. Concenraion of holes= 5*10 16 and concenraion of elecrons=10 20 /5*10 16 [3ps] c) he majoriy carrier for Silicon doped wih all hree is elecron, and he resuling maerial is n-ype. Concenraion of elecrons=10 15 and concenraion of holes is /10 15 [4ps] Page 10 of 10