First Midterm Exam, February 23, 2001 (53 Minutes) Use the following parameters for MOS devices. Here are a few fundamental constants and room temperature values. Problem Possible Score I 20 II 30 III 20 IV 30 Total 100 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/ Print Your Name:_____Solutions (see the following pages) Sign Your Name:________________________ ( )[ ]( )DSDSnDSTnGSOXpnpDnVVVVVCLWI λµ +−−= 12/( ) ( )DSnTnGSOXnnnDnVVVCLWI λµ +−= 1212( )[ ]( )SDSDpSDTpSGOXpppDpVVVVVCLWI λµ +−+=− 12/( ) ( )SDpTpSGOXppppDVVVCLWI λµ ++=− 1212cmFxo/1085.814−=ε9.3,=oxrε7.11,=sirεKJxK /1038.123−=NMOS PMOS ìnCox=50ìA/V2 ìpCox=25ìA/V2 VT0n=1.0V VT0p=-1.0V γn=0.6V1/2 γp=0.6V1/2 ën=(0.1/L)V-1 L in ìm ëp=(0.1/L)V-1 L in ìm φp=-0.42 φn=0.42 202102xni=eVVth026.0=cm2/V-s 350 400 500 µp cm2/V-s 750 1200 1400 µn units 1017 1016 1015 mobility Cxq19106.1−=Assume L = 2 µmI. (20 Points) Threshold Voltage An NMOS process produces a device with a threshold voltage VTn = 1V and φp = - 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 2φφp 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% EE 105 Sp 2001 A. R. Neureuther Solutions First Midterm Exam, February 23, 2001 (53 Minutes) Version 2/25/01II. (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 µm. A region 0.25 µm 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. cmVxxxxxxqNEOXdaMAX/1016.11085.89.31025.0101106.151441619=⋅⋅⋅==−−−ε b) (15 Points) Find the local potential and the density of mobile electrons at the silicon-oxide interface. ()()2pxxAx −=φ ( )( )pxxAxxE −=∂∂−= 2φ O.25 -0.2 pxAxE 20==()∂φ( )( )mVxxEpx48329.3/7.11101016.1201650=⋅⋅===φ ( )12841620103.21016.110226483exp101020 xxxxn =⋅== ( )()202ppxxxxEx −==φIII. (20 Points) MOS Circuits a) (10 Points) Assuming the device is in saturation and λn = 0, find the large-signal quantity VOUT. b) (10 Points) Find the small-signal gain vOUT/vIN for the circuit in part a) when λn = 0. voutvingmvinRL vout2 mAVDDRL=5kΩΩW/L = 50VIN = 2V +vin ()mAID25.11250505.02=−⋅⋅⋅=()VkmAMAVOUT75.3525.12 =Ω⋅−= ()mSgm5.2125050 =−⋅⋅= 5.1255.2 −=Ω⋅−=−= kmSRgVLmOUTIV. (30 Points) Advanced MOS Circuits 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 λn = 0. 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 λn = 0. voutvingmvgsRLgmbvbsvbsvgs c) (9 Points) Write a sufficient set of equations to determine the small-signal gain vOUT/vIN when λn = 0. Solve for the gain if you have completed all other questions on this exam. 0=++sbsbsmbgsmRvvgvg ()LbsbsgsmOUTRvgvgV +−= ()++−=SmmmbSLOUTRgggRRVinV11/ vout2 mAVDDVIN = 2V +vinRSRL=5kΩΩW/L = 50 ()VVT145.184.084.05.06.01 =−++= ()mAID158.0145.15.0250505.02=−−⋅⋅⋅= VkmAmAVOUT21.95)158.02( =Ω⋅−=()mSgm89.0145.15.025050 =−−⋅= mSmSgmb23.05.084.0289.06.0=+⋅⋅= Ω== KmAVRS16.3158.05.0 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. bsgsinVVV−=
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