Lecture 02 CSE 40547/60547 Computing at the Nanoscale

Transcription

1 PN Junctin Ntes: Lecture 02 CSE 40547/60547 Cmputing at the Nanscale Letʼs start with a (very) shrt review f semi-cnducting materials: - N-type material: Obtained by adding impurity with 5 valence elements t a valence 4 semicnductr This will increase the amunt f negative charge carriers - P-type material: Obtained by adding impurity t increase the number f psitive charge carriers Accepts nly weakly bnded, uter electrns frm ther semi-cnductr atms Semi-cnductr atms that lse electrns are hles Purpse f p-type dping is t create an abundance f hles Brn r aluminum substituted int a silicn crystal lattice Result: 1 electrn missing frm 1 f 4 cvalent bnds Dpant can accept electrn frm neighbring atm t cmplete the 4 th - Bringing P-N type material tgether creates a PN junctin and dide Picture: (1), (2) tw dide symbls, (3) dide crss-sectin, (4) Lewis structure - Bringing P, N-type material tgether causes a large cncentratin at the gradient bundary Thereʼs a higher electrn cncentratin in N material, lwer electrn cncentratin in P material Gradient causes: Electrns t diffuse frm N P Hles t diffuse frm P N When a hle leaves the P material, it leaves behind an immbile acceptr in (with a negative charge) Therefre, a negative charge will exist in the P material near the vicinity f the PNbundary Similarly, psitive charge builds up n the N side as a diffusing electrn leaves a psitive charge behind Picture: PN-junctin with depletin regin

2 - At the depletin regin, there is an electric field acrss the bundary directed frm N P Picture: Depletin regin; field will cunteract diffusin, and eventually there is 0 net current flw Therefre, under 0 bias, a vltage φ 0 exists which serves as a built-in ptential. - If we raise the ptential f the P regin with respect t the N regin, we lwer the ptential barrier I.e. given a frward vltage V d applied t the junctin The flw f mbile barriers acrss the junctin increases as diffusin current is dminated by drift current The net result is current flwing in the dide frm P N in this frward biased mde - In reverse bias mde, the depletin regin increases, and current flw is shut ff Dides fr Blean Lgic: Questins t cnsider: (1) Hw can we make lgic gates frm dides? (2) Are dides a gd switch fr making lgic gates? (Why r why nt?) Answer t Questin (1): Picture: (a) dide based OR gate, (b) dide based AND gate - OR gate peratin: If 1 input (A r B) is high, current flws thrugh the assciated, (frward biased) dide and brings the utput nde up. - AND gate peratin: If 1 input (A r B) is grunded, current flws thrugh the dide and utput X is at a lw vltage (assumes that dides in the abve picture are frward biased) The nly way t bring / keep utput X high is fr bth inputs t be high (i.e. the dides need t be reverse biased) - Freshadwing: What if I tld yu, that yu culd build a dide in an area thatʼs apprximately 25 nm 2? Keep in mind, that the minimum feature size (which represents just 1 part f 1 device) fr the current technlgy nde is ~ 16 nm. This is an appealing thught, but dides have their issues i.e. hw is gain achieved?

3 Picture: Experimental results f dide-based, OR gate made frm nanwires Example setup: - Fr bth AND, OR gates, letʼs assume V high = 5V 1, V lw = 0V - Als, (fr nw) weʼll assume that dides dnʼt intrduce any errrs r lses int circuits - In actuality, this is nt really the case a silicn-based dide wuld have a drp f ~0.7V while cnducting - Maybe we culd vercme this prblem by saying: If vltage is >> 3.5V, we have a lgic 1 If vltage is << 1.5 V, we have a lgic 0 Example: - Hw wuld we use dides t implement the lgic functin: AB + CD Picture: (a) AB+CD based n dide lgic; (b) vltage divider Example discussin: - If all inputs are lgic 0 (a lw vltage) the utput will be held at lgic 0. - If bth inputs t ne f the AND gates are a lgic 1, what will the utput be? Dide in OR gate will becme frward biased fr the AND gate where bth inputs are at a high vltage Current flws thrugh the AND gate resistr and the OR gate dide Current flws thrugh the OR gate resistr - If all resistrs are the same value, we have a vltage divider and equally share a (5V) supply - After the OR gate dide inserts its lss, the utput vltage may be n the rder f ~2.1V 2.2 V Questin: - What lgic state is represented by 2.1V? 1 In reality, mre like 0.9V, 1V nw, but 5V is fine fr illustrative purpses.

4 Take Aways: - Fr dide lgic, itʼs difficult t create cascades f multiple lgic levels (The vltage drps add up) - This leads us t 3 terminal devices - Ging back t the nanwire example: Itʼs great t have junctins that are 10s f nm 2 But, as seen abve, dide-based lgic may be limited as an infrmatin prcessing technlgy Mrever, hw d yu d inversin? Fr a functinally cmplete lgic set, need AND, OR, + NOT. - Can engineering gain, inversin int nanwire-based circuits (as yu will see later), but verhead strips away ptential density and perfrmance gains Digressin: S, ideally, what features shuld a device have it is t be used t implement a digital lgic system? Cnsider 5 tenets f digital lgic: 1. a device shuld have nn-linear respnse characteristics 2. a device shuld enable a functinally cmplete lgic set 3. pwer amplificatin (r gain) is needed 4. the utput f ne device must drive anther (i.e. with n state variable change) 5. dataflw directinality must be well defined Use nanmagnet lgic as an example Transistr Basics: Disclaimer: This discussin is by n means meant t replace an electrnics r VLSI design curse. Rather, it is just meant t briefly review/intrduce material frm these curses, and set the stage fr a discussin f radblcks t transistr-based scaling and cmputatin. Devices at a glance: - At the mst superficial level, transistrs can be thught f as a switch - If a vltage is applied t a gate (that is greater than sme threshld vltage V t ), a cnducting channel is frmed between drain and surce - If the vltage difference between drain and surce, than current flws between them The greater the vltage difference between gate and surce, the smaller the resistance f the channel and the greater the current - If the gate vltage is lwer than the threshld, n channel exists and the switch is pen (Assuming an NMOS device, current is carried by electrns mving thrugh an N-type channel between the transistrʼs surce and drain.)

5 Threshld Vltage: - Cnsider V gs = 0 the drain, surce, and bulk are all cnnected t grund - The drain and surce are effectively cnnected by back-t-back PN junctins (substrate-surce and substrate-drain) - With the cnditins abve, bth junctins have 0 bias and the device can be cnsidered ff (i.e. there is a high resistance between surce and drain indicative f an pen switch) See Slides Nw, assume that a psitive vltage is applied t the gate (with respect t the surce) The gate and substrate frm plates f a capacitr where the gate xide frms the dielectric A psitive gate vltage causes +/- charge t accumulate n gate/substrate side respectively See Slide 6 - When the gate vltage becmes sufficiently high, the ptential at the silicn surface reaches a critical value, and the semi-cnductr surface inverts t N-type (strng inversin) Further increases t the gate vltage prduce n mre changes in the depletin layer width, but results in a thin inversin layer under the xide Drawn int inversin layer frm heavily dped N surce regin N-type channel frmed between surce and drain, mdulated by V gs Value f V gs where strng inversin ccurs = threshld vltage V t 5 See Slide 6 Resistive / Linear Operatin: - Nw, assume V gs > V t and a small vltage is applied between drain and surce - Vltage difference causes current i D t flw frm drain t surce - At a pint x alng the channel, the vltage V(x) at gate-t-channel vltage at that pint = V gs V(x) - If vltage exceeds V t at all pints alng the channel, current can be calculated See Slides 9-13 Saturatin Regin: - As the value f the drain-surce vltage is further increased, the assumptin that the channel vltage is greater than the threshld vltage at all pints ceases t hld Occurs if V gs V(x) < V t - Here, the cnducting channel is pinched ff - Cnditins must be met at the drain regin therefre V gs V ds <= V t E.g. if V gs = 5V, V ds = 5V, and V t = 0.7V 5V 5V < 0.7V - Nw, the transistr is in the saturatin regin, and current is n lnger a functin f V ds (it acts as a current surce) See Slides 14-16

6 Hw scaling impacts transistr parameters, perfrmance: See Slide 17 - Letʼs talk mre abut the vltage scaling factr U. Assumes all vltages scale by the same factr U Prduct cmpatibility might be ne reasn A bit mre n 3 mdels in the chart: Full scaling: - Vltages and dimensins are reduced by the same factr S; leads t: Greater density Better perfrmance (intrinsic delay gets better) Pwer cnsumptin reduced t (and chip-level pwer density stays cnstant as mre devices placed in same area) - Fr a lng time, this mdel held - Cnsider impact n smething like Pwer: P = I sat x V I sat scales as 1/S V scales as 1/S Therefre P scales as 1/S 2 seemingly great! What abut t x? Affects ther parameters which influence delay and pwer What if it desnʼt scale in lckstep (as it must) See Slide 18 fr t x scaling trends Fixed Vltage Scaling: - T keep new parts cmpatible with existing cmpnents, V cannt be scaled arbitrarily - Can be hard, expensive t supprt multiple supplies (e.g. 5V, 3V) - What if vltage des nt scale? Lk at example: P density P density = P / Area Scales as 1 / (1/S 2 ) increases as S 2! General Scaling: - Supply vltage scales, but nt at the same rate as technlgy - Sme fundamental limits cme int play here Yu canʼt make V t arbitrarily lw as yuʼll see, yu culdnʼt turn the transistr ff There are intrinsic device vltages Material parameters canʼt be changed Need new materials