1 UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences Midterm 1 EECS 105B. E. BOSER September 28, 2004 FALL 2004 Show derivations and mark results with box around them. Erase or cross out erroneous attempts. Simplify algebraic results as much as possible! Mark your name and SID at the top of the exam and all extra sheets. For Office Use Only: Points Problem 1 25Problem 2 25Problem 3 24Problem 4 25TOTAL 100 Name [1 point]: ___________AMIN______________ SID: ______________________________2Problem 1 [25 points]: Sheet resistance Shown below is the layout of a p-type resistor with NA = 1017 cm-3, Vscm2502=pµ and thickness t = 1 µm. Electronic charge C106.119−×=q . (a) [10 pts] Calculate the sheet resistance shR in Ω/. (b) [10 pts] Assuming Ω= k1shR / (not the correct answer for part a), calculate the resistance between terminals V1 and V2 for µm1=s . (c) [5 pts] Repeat part (b) with µm2=s . ANSWERS (a) squareRcmVscmcmCxtNqtpqRshpApshΩ=∴⋅⋅⋅=⋅⋅⋅=⋅⋅⋅=−−−2500)10()250()10()106.1(1114231719µµ (b) The number of squares in the segment marked by red is 515=ss. The number of squares in the segment marked by green is 339=ss. Hence, the total number of squares equals 835 =+. As a result, Ω=⋅Ω=⋅= ksquaressquareksquaresofnumberRRsh8)8()1()(. (c) From part (b), the value of s does not enter into the expression for R. Hence, the value for R is unaffected from part (b) and is equal to Ωk8. ssV 1 V2ssV 1 V23Problem 2 [25 points]: Carrier transport In this problem, you are to design an over-current protection device. Assume that the slab of silicon shown below is doped with an acceptor concentration of Na (with Na >> ni). 316cm 10−=aN , Vscm 2502=pµ, scm106,=psatv, µm10=t , µm 50=W , µm 100=L , electronic charge C106.119−×=q . (a) [15 pts] Derive an analytical expression and calculate the numerical value for the maximum current Imax. (b) [10 pts] Derive an analytical expression and calculate the numerical value for Vc (see the I-V curve below for the definition of Vc). Wt LVI V VcImax04ANSWERS (a) mAcmxcmxscmcmCxItWvNqAvpqIpsatapsat8)1010()1050()10()10()106.1(44631619max,,max=⋅⋅⋅⋅=∴⋅⋅⋅⋅=⋅⋅⋅=−−−− (b) VVscmcmxscmVLvVLVEvcppsatccpcppsat40250)10100()10(246,,=⋅=∴⋅=⇒⋅=⋅=−µµµ5Problem 3 [24 points]: Region of operation Shown below are both NMOS and PMOS transistors with various terminal voltages referred to ground. Identify the source terminal (V1 or V2) and the region of operation (cutoff or triode) of each transistor by circling the correct answer in the table provided. V1= 2 VV2= 0 VV4= 0 VV3= 4 VExampleV1= 2 VV2= 0 VV4= 0 VV3= 4 VExample V1= 5 VV2= 0 VV4= 5 VV3= 5 V(a) (b) (c)V1= 3 VV2= 0 VV4= 0 VV3= 0.5 VV1= 5 VV2= 2 VV4= 5 VV3= 0 VV1= 5 VV2= 0 VV4= 5 VV3= 5 VV1= 5 VV2= 0 VV4= 5 VV3= 5 V(a) (b) (c)V1= 3 VV2= 0 VV4= 0 VV3= 0.5 VV1= 3 VV2= 0 VV4= 0 VV3= 0.5 VV1= 5 VV2= 2 VV4= 5 VV3= 0 VV1= 5 VV2= 2 VV4= 5 VV3= 0 V V1= 3.1 VV2= 3 VV4= 0 VV3= 4.5 V(d) (e) (f)V1= 102 VV2= 100 VV4= 100 VV3= 104 VV1= 5 VV2= 4 VV4= 13 VV3= 0 VV1= 3.1 VV2= 3 VV4= 0 VV3= 4.5 VV1= 3.1 VV2= 3 VV4= 0 VV3= 4.5 V(d) (e) (f)V1= 102 VV2= 100 VV4= 100 VV3= 104 VV1= 102 VV2= 100 VV4= 100 VV3= 104 VV1= 5 VV2= 4 VV4= 13 VV3= 0 VV1= 5 VV2= 4 VV4= 13 VV3= 0 V Circuit Source Terminal Region Example V1 V2 Cutoff Triode (a) V1 V2 Cutoff Triode (b) V1 V2 Cutoff Triode (c) V1 V2 Cutoff Triode (d) V1 V2 Cutoff Triode (e) V1 V2 Cutoff Triode (f) V1 V2 Cutoff Triode NMOS PMOS VT0n = 1 V VT0p = -1 V n = 1 V1/2 p = -1 V1/2 p = -0.5 V n = 0.5 V ()pBSpnnTTnVVVϕϕγ220−−−−+=()nSBnppTTpVVVϕϕγ220−−+=6ANSWERS (a) 2V is the source terminal. Also, VVVTnTn10== because the source (2V ) and bulk (4V ) are at the same potential. The NMOS transistor is cutoff because VVVVVVVVTnGS15.005.023=<=−=−= . (b) 1V is the source terminal. Also, VVVTpTp10−== because the source (1V ) and bulk (4V ) are at the same potential. The PMOS transistor is in triode because VVVVVVVVTpSG150531=>=−=−= and VVVVVVVVVVVVVVVTpTpSGSD41053253121=−−=−−=−<=−=−= . (c) 1V is the source terminal. Also, VVVTpTp10−== because the source (1V ) and bulk (4V ) are at the same potential. The PMOS transistor is cutoff because VVVVVVVVTpSG105531=<=−=−= . (d) 2V is the source terminal. Also, 4V is the bulk terminal. As a result, VVVVVVVVVVVVVTnppnTnpBSpnTnTn2)1)3(1(11)2)(2()22(2400=−−−⋅+=∴−−−−−+=−−−−+=φφγφφγ The NMOS transistor is cutoff because VVVVVVVVTnGS25.135.423=<=−=−= . (e) 1V is the source terminal. Also, 4V is the bulk terminal. As a result, VVVVVVVVVVVVVTpnnpTpnSBnpTpTp3)1)8(1()1(1)2)(2()22(4100−=−−−⋅−+−=∴−−−+=−−+=φφγφφγ The PMOS transistor is in triode because VVVVVVVVTpSG350531=>=−=−= and VVVVVVVVVVVVVVVTpTpSGSD23051453121=−−=−−=−<=−=−= .7(f) 2V is the source terminal. Also, VVVTnTn10== because the source (2V ) and bulk (4V ) are at the same potential. The NMOS transistor is in triode because VVVVVVVVTnGS1410010423=>=−=−= and VVVVVVVVVVVVVVVTnTnGSDS3110010421001022321=−−=−−=−<=−=−= .8Problem 4 [25 points]: CMOS switch For this problem, ignore the backgate effect; that is, let n = p = 0. 5 VVi0 VVoLn= 4 µmWn= ? µmLp= 4 µmWp= ? µm5 VVi0 VVoLn= 4 µmWn= ? µmLp= 4 µmWp= ? µm Figure 1: CMOS switch for part (a). (a) [15 pts] Find Wn and Wp such that Rio = 1 k for both Vi = Vo = 0 V and Vi = Vo = 5 V (see Figure 1 above). Rio is the resistance between the input and output terminals, Vi and Vo. (b) [10 pts] Now, let Wn = 10 µm and Wp = 40 µm (see Figure 2 below). NOTE: these are not the answers you found for part (a). Calculate Rio for Vi = Vo = 2 V. 5 VVi0 VVoLn= 4 µmWn= 10 µmLp= 4 µmWp= 40 µm5 VVi0 VVoLn= 4 µmWn= 10 µmLp= 4 µmWp= 40 µm Figure 2: CMOS switch for part (b). NMOS PMOS VT0n = 1 V VT0p = -1 V µnCox = 200 µA/V2 µpCox = 100 µA/V29ANSWERS For 0=DSV , )()(10TnGSnoxnNMOSVVLWCR−⋅⋅=µ (1). Also, for 0=SDV , )()(10TpSGpoxpPMOSVVLWCR−⋅⋅=µ (2). (a) For VVVoi0== , the PMOS transistor is off. Hence, Ω=−−⋅⋅==
View Full Document
Unlocking...