Designing Information Devices and Systems I Spring 2019 Lecture Notes Note 17

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1 EES 16A Designing Informaion Devices and Sysems I Spring 019 Lecure Noes Noe apaciive ouchscreen In he las noe, we saw ha a capacior consiss of wo pieces on conducive maerial separaed by a nonconducive maerial. In a ypical off-he-shelf capacior, he conducive maerial is a meal. However, our skin is mosly waer and herefore also acs as a conducor. his means we can form a capacior by holding a finger over a shee of meal! Le s look a how we can use his concep o design a ouchscreen based on capaciance. Each pixel of our ouchscreen consiss of wo pieces of meal wih an insulaor (non-conducive maerial) beween hem. his forms a capacior wih capaciance 0. Anoher insulaor, such as glass, goes on op of he upper (red) piece of meal. We wan o be able o deec if here is a finger ouch on op of his insulaor. When he finger ouches he op insulaor, wo new capaciors are formed beween he finger and each of he wo elecrodes. F-E is he capaciance beween our finger (F) and elecrode E, and E1 - F is he capaciance beween our finger (F) and elecrode E1. Now, we can draw he equivalen circui model corresponding o he pixel we re looking a. Noe ha he nodes are labeled wih he elecrode or finger ha he node represens. EES 16A, Spring 019, Noe 17 1

2 F E1-F E 1 FE o E Our ouchscreen can only measure hings ha are beween elecrodes E 1 and E as i is impracical o physically connec a measuring device o our finger. So, we redraw our circui as follows: E 1 E1-F o F F-E E We can furher simplify he circui by replacing he series combinaion of E1-F and F-E by some capaciance. When he finger is presen, E1-F and F-E have posiive capaciance, so we know has some capaciance greaer han zero. E 1 o E Finally we consider he parallel combinaion of o and as some equivalen capaciance eq, E1-E = o. When he finger is presen, > 0, and he equivalen capaciance eq, E1-E will be greaer han EES 16A, Spring 019, Noe 17

3 o. When here is no finger presen, he capaciors forming do no exis, and herefore he equivalen capaciance eq, E1-E is equal o o. If we can measure his change in capaciance, we can deec he presence or absence of a finger. E 1 eq, E1-E E ouchscreen wih muliple pixels Our analysis from he previous secion only ells us if a figure is presen or absen. How can we deermine where he finger is? We can creae a grid of he individual pixels described above. Here s an example wih four pixels: he four pixels are labeled 1,, 3, 4. If we wan o know if a finger is presen a pixel 1, we can measure he capaciance beween elecrodes x 1 and y 1 if he capaciance is increased, we know here is a finger pressing a pixel 1. Similarly, o know he if here is a finger pressing on op of pixels, 3, 4, we can measure he capaciances beween (x 1,y ), (x,y 1 ), (x,y ). Since each pixel acs separaely, we can deec muliple presses simulaneously his enables many of he ouchscreen gesures we re familiar wih, such as pinching o zoom. 17. apaciance Measuremen How can we deec changes in capaciance? We can measure capaciance direcly, bu if we can ransform capaciance ino a volage, we can measure ha. here are muliple ways of doing his, bu in his noe we EES 16A, Spring 019, Noe 17 3

4 will cover only one specific mehod of measuring capaciance. Firs, le s review some basic physics for a capacior. From our las noe, I = dv d Based on his equaion, we hypohesize ha if we connec a known curren source I s o he capacior and measure he volage, we migh be able o solve for he capaciance. For our firs aemp, we build he following circui o measure : (1) I s V If he capacior sars off a = 0 wih no volage across i, we find ha V is: V = I s () V slope= I s where is he ime passed since we sared charging he capacior. If we measure he volage a a known ime, we can solve for he capaciance. However, as ime coninues o pass, he volage across he capacior (and also he charge sored on he capacior) will grow o infiniy. I is very challenging o build an ideal curren source ha works over his large range of volages, so our model quickly becomes unrealisic Measuring capaciance wih a periodic curren source o remedy his, insead of applying a consan curren source, we apply a periodic curren source EES 16A, Spring 019, Noe 17 4

5 I s () V () where he curren I s is a funcion of ime as follows: I s () 3 Wha does he volage V look like wih his curren source? Le s assume ha he capacior is iniially uncharged (ie. Q = 0). Since Q = V, his means ha a ime = 0 he volage V = 0. (We will revisi his assumpion laer.) When a consan curren source is applied o a capacior, we know ha he volage obeys he following equaion V () = I V (0). (3) Our periodic curren source I s is consan from = 0 o =, so we can apply equaion 3 over his ime period. We know he iniial volage is zero, so: V () = when 0 EES 16A, Spring 019, Noe 17 5

6 V () I s () slope = 3 In order o figure ou wha happens nex, le s consider a more generic version of equaion 3: V () = I ( 0) V ( 0 ). (4) Wih his equaion, we can consider an arbirary saring ime 0 insead of always saring a = 0. Plugging in 0 = 0 yields equaion 3. Like equaion 3, he above equaion is only rue when he curren is consan. he nex ime period wih consan curren is from = o =. Over his ime, he curren hrough he capacior is. Since we are saring a ime, we se 0 = and plug ino equaion 4. V () = I ( 1 ) ( ) V V () = I ( 1 ) ombining wih our previous relaionship yields: { I1 when 0 V () = ( ) when < V () I s () slope = slope = 3 EES 16A, Spring 019, Noe 17 6

7 o deermine he full behavior of V (), we could coninue o apply equaion 4 for each period of consan curren. However, we noice ha a =, he volage and curren are he same as hey were a = 0. Since he curren source is periodic (repeas every ), he volage paern will also repea. V () I s () slope = slope = Swiches In he model above, we made he assumpion ha he capacior was iniially uncharged. his assumpion ells us ha V (0) = 0, and our subsequen analysis relies on his fac. How can we ensure ha his assumpion is rue? Le s inroduce a new elemen, a swich, o help us. A swich has wo saes, a is on sae (or closed sae ), he swich becomes a perfecly conducing wire wih zero resisance, and when he swich is urned off (or he swich is opened ), i becomes an open circui. We can modify our design from above o include a swich in parallel wih he capacior: I s () V () When he swich is closed, i acs as an ideal wire, so he volage across he capacior is 0. Since Q = V, his means ha he charge is zero as well, so someimes we refer o his swich as discharging he capacior. EES 16A, Spring 019, Noe 17 7

8 o ensure ha V (0) = 0, we can keep he swich closed for < 0. A = 0, he swich is opened, urning i ino an open circui. Now he circui behaves he same way as i did before adding he swich, bu we ve ensured ha he volage over he capacior is 0 a = Effec of finger ouch on he measuremen Remember ha we d like o use our circui o deec if here is a finger pressing on a given pixel. Recall ha when he finger is absen, he equivalen capaciance beween he elecrodes is o. When he finger is presen, he equivalen capaciance is o where > 0. his difference in capaciance will change he slope of V () in our measuremen. We use he same circui design as before and measure he capaciance beween he wo ouchscreen elecrodes. I s () eq, E1-E V () V () I s () ouch No ouch slope = o slope = o 3 When here is no finger presen, he slope of he rising edge of V () is o. When here is a finger pressing, he slope is smaller:. his resuls in a smaller peak volage when here is a finger press. o his change in volage can be quie small since can be very small. Nex we will inroduce a comparaor ha will allow us o measure very small volage changes. EES 16A, Spring 019, Noe 17 8

9 17.3 omparaor & Op-amp Basics In his secion, we will inroduce a new circui componen called an op-amp. As we ll see shorly, an opamp can be used as a comparaor somehing ha compares wo volages and indicaes which is larger. However, op-amps have many oher uses oo, and we ll explore hese more in he nex noe. By definiion, an amplifier is somehing ha can ransform somehing small ino somehing much bigger. For example, a speaker is an audio amplifier if you connec your smar phone o a speaker, i can generae sounds much louder han your phone can. An op-amp (operaional amplifier) is a device ha ransforms a small volage difference ino a very large volage difference. he circui symbol for an op-amp is shown below: V DD U U U ou V SS An op-amp has wo inpu erminals marked () and () wih poenials U and U, wo power supply erminals called V DD and V SS, and one oupu erminal wih poenial U ou. he inernal circui design of an op-amp is beyond he scope of his course (if you are curious, you can search for schemaics online), bu we can model he op-amp wih he following circui: U ou U U A(U U ) V SS V DD V SS his circui includes wo volage-conrolled volage sources. (Noe ha he dependen volage sources in he schemaic above are referenced o V SS, no ground.) he boom volage source adds he volage V DDV SS so ha when U = U, he oupu volage is he midpoin beween V DD and V SS. he op source creaes a volage A(U U ), which amplifies he difference beween he wo inpu erminals. In a good op-amp, he consan A is very large approaching infiniy. If A is very large, does he op-amp produce infinie volage? I migh seem like his is he case from he equivalen circui schemaic, bu he op-amp s power mus come from somewhere i comes from he supply volages V SS and V DD. his means ha U ou canno be greaer han V DD or less han V SS. EES 16A, Spring 019, Noe 17 9

10 Following he schemaic given above, he behavior of an op-amp can be summarized wih he equaion U ou = A(U U ) V DD V SS, excep U ou canno be above V DD or below V SS. So U ou is acually given by he following equaion, where he firs and hird cases are he cuoffs: V SS A(U U ) V DDV SS < V SS U ou = A(U U ) V DDV SS V SS A(U U ) V DDV SS V DD (5) V DD V DD < A(U U ) V DDV SS For very large A (close o infiniy), he non-clipped oupu A(U U ) V DDV SS is infiniy if U U > 0, and negaive infiniy if U U < 0. hen, we are always in eiher he firs or hird case, which gives us he following plo: U ou (0,V DD ) U U (0,V SS ) Since A is very large, here is a very sharp ransiion in he cener of he plo. herefore, when U < U (or equivalenly, when U U < 0), hen U ou equals V SS. Similarly, when U > U, hen U ou equals V DD. In his case, he op-amp is acing as a comparaor since i indicaes which volage (U or U ) is larger, even if he difference beween hem is very small apaciive ouchscreen wih omparaor Now we can complee our ouchscreen by adding a comparaor o deec if a finger is pressing a a given pixel. EES 16A, Spring 019, Noe 17 10

11 3.3V I s () V c eq, E1-E V re f V ou Firs, noe ha he volage over he capacior V c is equal o he volage a he posiive erminal of he op-amp (U = V c ). he negaive erminal of he op-amp is conneced o V re f so U = V re f. Using he equaions for a comparaor (assuming he op-amp is ideal, A = ), we can wrie { 3.3V if V c > V re f V ou = 0V if V c < V re f How should we choose V re f? We would like he behavior of V ou o depend on he presence or absence of a finger ouch, so we should se V re f o be somewhere in beween he peak of V c wih a finger and he peak of V c wihou a finger. In fac, o be mos robus o noise, V re f should be exacly beween he wo peaks. V 3 V c ouch V c No V re f When here is a finger presen, V c is always less han V re f so he oupu volage is 0: ouch EES 16A, Spring 019, Noe 17 11

12 V 3 V c ouch V re f V ou When here is no finger presen, hen here will be an increase in V ou when V c is higher han V re f : V No ouch 3 V c No ouch V re f V ou Every period of he curren source cycle, we can measure if here is a finger ouching his pixel. Laer in his course, we will learn how o build a periodic curren source like he one needed in his design afer ha you will be ready o compleely build your own capaciive ouchscreen! EES 16A, Spring 019, Noe 17 1

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