Designing Information Devices and Systems I Spring 2018 Lecture Notes Note 16
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1 EECS 16A Designing Information Devices an Systems I Spring 218 Lecture Notes Note Touchscreen Revisite We ve seen how a resistive touchscreen works by using the concept of voltage iviers. Essentially, for a resistive touchscreen, we map the value of analog voltag to the position touche. Another way to esign a touchscreen is to break the screen own into a bunch of pixels. At each one of the pixels, we can etect whether the finger is touching it or not. This allows the touchscreen to sense multiple touch points at a time. In orer to o this efficiently, we nee a new element: capacitors. One of the nice parts about the capacitor is that you on t have to ben anything like we i for the resistive touchscreen the presence or absence of the finger can moify the capacitance irectly! 16.2 Capacitor Capacitance represents how much charge can be store for a given amount of voltage. We usually enote capacitance with C. Let s represent voltage across the capacitor as V an the charge store on the capacitor as Q. This relation is often written as: Q = CV. (1) When we raw circuits, the symbol for a capacitor is: C The unit of capacitance is Fara (F). A capacitor with capacitance of 1F stores 1C of charge across is when it has 1V of potential energy across its electroes. Let s examine the efining relationship of a capacitor: Q = CV C. (2) Differentiating both sies with respect to t an assuming a constant capacitance, we have Q = CV C = C V C. (3) EECS 16A, Spring 218, Note 16 1
2 However, we know that current I = Q. Hence, I = C V C. (4) Thus, the current through a capacitor is just the prouct of the capacitance an the rate of change of the voltage across the capacitor. This yiels the following graph, with slope equal to C: I V An important implication of this is in orer for current to flow through the capacitor, the voltage across that capacitor must be changing with time. If the voltage is no longer increasing, then the current through the capacitor will equal. What about solving for the voltage across the capacitor; how o we fin V elem (t)? There are many ways to solve this equation, but we will use separation of variables with our equation for current, an then integrating both sies. Note that it s okay if you haven t seen this technique before, it is not require for this class. t I = I = C V C. (5) I = CV C (6) VC CV C (7) It = C(V C (t) V C ()) (8) Don t forget about the initial voltage across the capacitor! Our limits of integration in the erivation set the point of our initial conition- in this case, at t =. This yiels the following graph, with slope equal to I C. V C (t) = I C t +V C(). (9) EECS 16A, Spring 218, Note 16 2
3 + V elem (,V ()) t Let s see what happens when we connect a capacitor with a voltage source. I V S + - C V C We know that initially, charges are going to buil up on the two plates positive charges on the top plate an negative charges on the bottom plate. Once enough charge has been store, the voltage across the capacitor becomes V C = V S ; because the voltage can no longer increase past this point, the current I flowing through the circuit becomes zero Capacitor Equivalence Note that when we initially talke about equivalence it was in terms of I V. But what about for capacitors? Now we will specify equivalence in terms of I V. That means we nee to apply I test an measure V test or apply V test an measure I test. Just like resistors, we can connect capacitors in series an in parallel. We ll first focus on capacitor-only circuits, which means that we on t nee to fin Thevenin/Norton equivalent circuits; it will just be a capacitor. Remember that a capacitor has an I V for equivalence we will use a test V or measure V measure a V test Capacitors in parallel Given the following circuit: curve, not an IV curve. To be clear, this means that when solving. For example, we may apply a test current I test an then EECS 16A, Spring 218, Note 16 3
4 C 2 C eq = I total Vt What is the equivalent capacitance? By analogy with our resistor case, we will try an apply a test voltage to this circuit: i 1 i 2 i test C 2 V test i 1 = V T i 2 = C 2 V T i T = i 1 + i 2 = ( +C 2 ) V T C eq = +C 2 We can thus reuce our original circuit to: +C 2 We fin that in general, the equivalent capacitance of capacitors in parallel is the sum of their capacitance Capacitors in series Given the circuit: C 2 EECS 16A, Spring 218, Note 16 4
5 By symmetry with the resistor case, let s apply a test current an measure the resulting voltage: u 2 C 2 i C2 + u 1 V T I T i C1 i C1 = i C2 = I T i C1 = I test = u 1 i C2 = I test = C 2 (u 2 u 1 ) V test = u 2 = C 2 ( u2 u ) 1 u 1 = I T u 2 I test = C 2 C 2 I T I T = C 2 u C 2 I T = C 2 V test 1 + C 2 C eq = C C 2 = C 2 +C 2 Note that this is in fact the same as saying C 2. Remember that the operator is mathematical notation; in this case, the capacitors are actually in series, but mathematically their equivalent circuit is foun via the parallel resistor operation. We can thus reuce our original circuit to: C 2 +C 2 EECS 16A, Spring 218, Note 16 5
6 16.4 A Little Bit of Physics Remember that this is 16A physics - esigne not to be exact or fully etaile, but close enough to get us the unerstaning we nee to create circuit moels from physical structures. Any two pieces of conuctive material, typically metal, that are separate by some other material that is not in general conuctive (aka an insulator), forms a capacitor. Note: What oes this mean about you in the case of a capacitive touch screen? That s right, you re a conuctor! If there is voltage across the two pieces of conuctors, charges will buil up on the surface of the capacitor. As epicte below, when we apply voltage V across the two plates, positive charges buil up on the (bottom) surface of the plate connecte to the positive terminal an negative charges buil up on the (top) surface of the plate connecte to the negative terminal. + V Let s try to get some intuition on capacitors by rawing an analogy between capacitors an water buckets simply put, we can think of a capacitor as a bucket. The more water (charge) you put in the bucket, the higher the water level (voltage) woul be. The water level in the bucket is etermine by its imensions. Similarly, the capacitance is set by the imensions of the conuctors an some properties of the material separating them. Please note that the water analogy is imperfect for many cases in electrical engineering- be careful not to try an exten or use it arbitrarily. In particular, the capacitance of a capacitor is: C = ε A (1) where A is the area of the surface of the plates facing each other, is the separation between the two plates (illustrate below), an ε is the "permitivity" of the material between the plates. We can euce that ε has unit F/m. For example, if the material between the capacitors is air (vacuum), then ε = F/m. EECS 16A, Spring 218, Note 16 6
7 A When a capacitor is charge, there is voltage across it. What oes this mean? This means that there is energy store in the capacitor. Why is there energy store? Recall that like charges repel each other. So for example if we take a positive charge an move it closer to another positive charge, we nee to exert force an thus supply energy to o so. Let s now ask the question: when the capacitor is fully charge, how much energy is store in it? We know from previous lectures that the energy require to store an aitional q amount of charge when the voltage across the capacitor is V C is We also know that q = CV C. With this in min, we have Integrating both sies, we have E E = V C q. (11) E = V C (CV C ) = CV C V C. (12) VS VS E = CV C V C = C V C V C (13) E = 1 2 CV 2 S. (14) Hence, the energy store in the capacitor after it s fully charge is 1 2 CV S 2. The charges across the capacitor cannot just isappear for no reason, an therefore the energy continues to be store - that means we can use this element as a tool to esign interesting circuit which measure this store charge/energy. Let s return to our capacitor equivalence moels an try to gain some more physical intuition: Suppose we combine two capacitors in parallel as follows A 1 A 2 C 2 If we look at the combine capacitor above, its capacitance is equal to C eq = ε A 1 + A 2, (15) which is equal to the sum of the capacitance of the two capacitors C eq = ε A 1 + A 2 = ε A 1 + ε A 2 = +C 2, (16) If we take the following capacitor an raw a horizontal line in between, then essentially we can view it as two capacitors connecte in series EECS 16A, Spring 218, Note 16 7
8 1 2 C 2 Assume that the surface area is A. Intuitively, we know that the equivalent capacitance shoul be smaller than both an C 2 since the plates are farther from each other. We know that = ε A 1 an C 2 = ε A 2. Let the capacitance of the combine capacitor be C eq. It satisfies Hence, the equivalent capacitance is equal to 1 = = 1 1 C eq ε A ε A ε A = (17) C 2 1 C eq = = C 2. (18) C +C 2 2 The capacitor on the left has capacitance = ε A 1 an the capacitor on the right has capacitance C 2 = ε A 2. Intuitively, the equivalent capacitance for capacitors in parallel woul just be the sum of the capacitance of each of the capacitors since connecting them in parallel is analogous to summing the surface area of the two plates. EECS 16A, Spring 218, Note 16 8
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