EECS 105 Microelectronic Devices and Circuits Spring 2001 Prof A R Neureuther Dept EECS 510 Cory 642 4590 UC Berkeley Office Hours M11 Tu2 W2 Th2 F11 Course Web Site http www inst EECS Berkeley EDU ee105 First Midterm Exam February 23 2001 53 Minutes Print Your Name Solutions see the following pages Sign Your Name Use the following parameters for MOS devices W I Dn nCOXp VGS VTn VDS 2 1 nVDS VDS L p I Dn 1 W n COXn VGS VTn 2 1 nVDS 2 L n W I Dp pCOXp VSG VTp VSD 2 1 pVSD VSD L p ID p 1 W p COXp VSG VTp 2 L p 2 1 pVSD NMOS PMOS nCox 50 A V2 pCox 25 A V2 VT0n 1 0V VT0p 1 0V n 0 6V1 2 p 0 6V1 2 1 n 0 1 L V L in m p 0 1 L V 1 L in m p 0 42 n 0 42 Assume L 2 m Here are a few fundamental constants and room temperature values o 8 85x10 14 F cm Vth 0 026eV r ox 3 9 ni 2 2x1020 r si 11 7 K 1 38x10 23 J K mobility n 1015 1400 1016 1200 1017 750 units cm2 V s p 500 400 350 cm2 V s q 1 6 x10 19 C Problem Possible I 20 II 30 III 20 IV 30 Total 100 Score EE 105 Sp 2001 A R Neureuther Solutions First Midterm Exam February 23 2001 53 Minutes Version 2 25 01 I 20 Points Threshold Voltage An NMOS process produces a device with a threshold voltage VTn 1V and p 0 42 Complete the table below that describes the percentage change in the electrical properties listed that is produced by the change shown in the columns for the oxide thickness and doping You must briefly explain your listed percentages with footnotes 1 6 as indicated Electrical Parameter Oxide thickness tOX Substrate doping Na doubles doubles 2 p 1 0 4 4 3 COX 2 50 5 0 QB MAX COX 3 100 6 44 1 No effect 2 COX is inverse with tOX 3 COX decreases causes inverse Q COX increase 4 2 26mV ln 2 36mV or 4 3 of 840 mv 5 No effect 6 Sqrt 2 1 043 1 44 or 44 II 30 Points Electrostatic and Mobile Carrier Analysis An NMOS device is made in a p type substrate doped Na 10 16 and an oxide thickness of 0 2 m A region 0 25 m deep under the gate is depleted a 15 Points Sketch the electric field versus position from the gate electrode Give the quantitative value of the maximum electric field qN a xd 1 6x10 19 1x1016 0 25x10 4 1 16x105V cm 14 OX 3 9 8 85x10 E MAX 0 2 O 25 b 15 Points Find the local potential and the density of mobile electrons at the siliconoxide interface x A x x p E x 2 2A x x p x E x 0 2 Ax p x 0 E x 0 x xp 2xp Ex 0 x p 2 2 1 16 x105 1016 483mV 11 7 3 9 2 2 x10 20 483 n 0 exp 2 x10 4 1 16 x108 2 3 x1012 16 10 26 III 20 Points MOS Circuits VDD a 10 Points Assuming the device is in saturation and n 0 find the large signal quantity VOUT 2 mA W L 50 VIN 2V vin I D 0 5 50 50 2 1 2 1 25mA VOUT 2 MA 1 25mA 5k 3 75V b 10 Points Find the small signal gain vOUT vIN for the circuit in part a when n 0 vin gmvin vout RL g m 50 50 2 1 2 5mS VOUT g m R L 2 5mS 5k 12 5 vout RL 5k VDD IV 30 Points Advanced MOS Circuits 2 mA a 7 Points A resistor RS is inserted as shown and has a value such that VS 0 5V Find the large signal quantity VOUT when n 0 W L 50 vout RL 5k VIN 2V vin VT 1 0 6 0 5 0 84 0 84 1 145V RS I D 0 5 50 50 2 0 5 1 145 2 0 158 mA VOUT 2mA 0 158mA 5k 9 21V b 14 Points Draw the small signal equivalent circuit of the circuit in part a and give quantitative values for each element in the circuit when n 0 vin vin g m 50 50 2 0 5 1 145 0 89mS vv gsgs ggmmvvgsgsggmb mbvvbs bs RLL R vout v out g mb 0 6 0 89 mS 0 23mS 2 0 84 0 5 bs vvbs RS 0 5V 3 16 K 0 158 mA c 9 Points Write a sufficient set of equations to determine the small signal gain vOUT vIN when n 0 Solve for the gain if you have completed all other questions on this exam Vin V gs Vbs g m v gs g mb vbs vbs 0 Rs VOUT g m v gs gbs vbs RL VOUT R L R S Vin g mb 1 1 g m g m RS Use the second equation to find vgs in terms of vbs and then substitutes in the first equation to get vbs in terms of vin Now go backwards and get vgs and then substitute both into the third equation
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