Substrate - Microelectronic Devices and Circuits - Exam, Exams of Microelectronic Circuits

Download Substrate - Microelectronic Devices and Circuits - Exam and more Exams Microelectronic Circuits in PDF only on Docsity! UNIVERSITY OF CALIFORNIA College of Engineering Department of Electrical Engineering and Computer Sciences Professor Oldham Spring 1999 EECS105 — Midterm 1 Thursday, 25 February 1999 Your name: ______________ ____________________ Your discussion TA: ❒ Allan Chang ❒ Lily Tam • This is a closed book exam, but you may use your page of notes. • Please do all your work on the pages of this exam. Ask if you need extra paper. • Full credit will be given only when you indicate the source of your answer, such as a table, graph, or calcula- tion. • Please write your name in the above space • Special notes: 1. SOME GRAPHS AND FORMULAS ARE GIVEN AS APPENDICES TO THIS EXAM. BE SURE TO LOOK THESE OVER. 2. SOME PARTS OF THE EXAM ARE GRADED WITH NO PARTIAL CREDIT. They are noted. You may wish to double check your answers on those parts. 3. ONCE IN A WHILE SOME EXTRA CREDIT IS POSSIBLE FOR CLEVER INSIGHT. Again, these places are noted. But we will not answer questions about these problems. Just be very clear in your work. SCORE Problem 1 (20 pts.) Problem 2 (25 pts.) Problem 3 (30 pts.) Problem 4 (25 pts.) TOTAL (100 pts.) first last Yes, I have looked these over. (Check box) 1 of 8 Problem 1 (20 pts.) a. [No partial credit] In a certain process, a thick layer of n-type silicon (doping = is created over a p-type substrate. It is to be used for the purpose of making integrated circuit resistors. What is the sheet resistance of this layer? (Units must be .) b. Using the layers of (a), above, you need to make a resistor with value of . It is wide. What must its length be (ignoring contact effects)? c. Someone properly points out to you that the layer in part a), though it is physically thick, is electri- cally somewhat thinner, because there must be a depletion layer at the n-p interface. (You are to ignore this in part a.) Suppose the doping in the p region is also (but acceptors instead of donors). At zero applied voltage between the n and p regions (i.e., in thermal equilibrium), just what is the net electrical thickness of the n-region? (Thickness minus depleted portion.) 2µm 2 1015× cm3⁄ Ω square⁄ 200KΩ 5µm 2µm 2 1015× cm3⁄ µm( ) 4 of 8 Problem 3 (30 points) An MOS structure is made with the layout ................................................ 1... ........ ........... ................................................................................................................... ................ :::::::::::j::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: ::::::: 2 pm................................................................................................................................. ... ................................................................................................................................ .. ... : : :. ........ .................................................................................................................... ....................................... t T The cross-section is the same as the left-hand part of the figure in Prob. 2. The substrate is doped 2 x 1015/cm3 with acceptors and the gate oxide is 40nm thick. a. [No partial credit] What is the flat-band voltage V,,? b. [No partial credit] What is the capacitance of the gate to substrate for V,, = 0 and VGs < V,,? (Ignore overlap and other second order effects.) (pF) C. What is the Threshold voltage V, ? V 5 of 8 d. Now we adjust the flat-band voltage (and thus the threshold voltage) with an ion implant just at the bottom of the gate oxide. We set the implant value to get a final threshold of 0.5 V. In testing the device we short source to body, i.e., . d.1) [No partial credit] What is of this device if we set ? d.2) Neatly sketch the I-V characteristics on the linear axes below for three cases: , and . Cover the range . Assume the electron mobility in the channel is . You must put a scale on the current axis. (Note that partial credit will only be possible if you very carefully show your work, including giving any formulas you are using before evaluation.) VTn VSB 0= VDSAT VGS 2 V= VGS 0.5 V VGS, 1 V= = VGS 2 V= VDS 0 to 5 V= 500 cm 2 Vsec⁄ 0 1 2 3 4 5 D ra in C ur re nt Drain-Source Voltage Drain Current vs VDS6 of 8 Appendix A: Appendix B: 1013 10151014 1019 10201016 1017 1018 1400 1200 1000 800 600 400 200 0 holes Nd + Na total dopant concentration (cm −3) electrons m ob ili ty ( cm 2 / V s) Log-linear plot of electron and hole mobilities at room temperature, as functions of the total doping concentration . Note that electron and hole diffusivi- ties are given by: . Nd Na+ D kT qµ⁄= φ (mV) φ (mV) intrinsicp-type po, equilibrium hole concentration (cm −3) no, equilibrium electron concentration (cm −3) n-type p-type n-typeintrinsic 1019 101 102 104 106 108 1010 1012 1014 1016 1018 1019 1018 1016 1014 1012 1010 108 106 104 102 101 −550 −550 −480 −480 −360 −360 −240 −240 −120 −120 0 0 120 120 240 240 360 360 480 480 550 550A.1 Appendix C: Formulas (EQ 2.7) (EQ 2.34) (EQ 2.55) (EQ 2.56) (EQ 2.67) (EQ 3.1) (EQ 3.4) (EQ 3.56) (EQ 3.88) (EQ 3.89) (EQ 3.95) (EQ 4.18) (EQ 4.19) (EQ 4.59) (EQ 4.60) (EQ 4.67) no po⋅ ni2 T( )= vdn µnE–= Jn Jn dr Jn diff+ qnµnE qDn dn dx -----+= = Jp Jp dr Jp diff+ qpµpE qDp dp dx -----–= = R 1 qNdµnt -----------------    L W ----    R L W ----    R N= = = dE dx ------ ρ ε --= E x( ) dφ x( ) dx -------------–= Xdo xno xpo+ 2εsφB q -------------    1 Na ----- 1 Nd -----+   = = VFB φn+ φp–( )–= QG VGB VFB=( ) 0= VTn VFB 2φp– 1 Cox -------- 2qεsNa 2φp–( )+= VDSSAT VGS VTn–= IDSAT W 2L -----    µnCox VGS VTn–( )2 W 2L -----    µnCoxVDSSAT 2= = ID 0 A= VGS VTn≤( ) ID W L⁄( )µnCox VGS VTn VDS 2⁄( )––[ ] 1 λnVDS+( )VDS VGS VTn VDS VGS VTn–≤,≥( )= IDSAT W 2L⁄( )µnCox VGS VTn–( ) 2 1 λnVDS+( ) VGS VTn VDS VGS VTn–≥,≥( )= VTn VTOn γn VBS– 2φp– 2φp––( )+= gm W L ----    µnCox VGS VTn–( )≅ 2 W L ----    µnCoxID=A.2 (EQ 6.10) (EQ 6.22) (EQ 6.30) (EQ 6.31) (EQ 6.48) Appendix D: Values φB Vthln ppo pno -------    and φB Vthln nno npo -------   = = pn xn( ) pno e VD Vth⁄ and np xp–( )⋅ npoe VD Vth⁄= = J qni 2 Dp NdWn ------------- Dn NaWp -------------+    eVD Vth⁄ 1–( )= ID qni 2A Dp NdWn ------------- Dn NaWp -------------+    eVD Vth⁄ 1–( ) Io e VD Vth⁄ 1–( )= = Cj Cjo 1 VD φB⁄– -----------------------------= q 1.6 10 19–× C= Vt kT q⁄ 0.026 V= = ε0 8.85 10 14–× F cm⁄= εoxide 3.9ε0= εSi 11.7ε0=A.3