6.013 Recitation 13: Introduction to Electric Forces and Generators

Transcription

1 6.013 Recitation 13: Introduction to Electric Forces and Generators A. Overview This section introduces electrostatic orces and generators and explores their application to micro-electromechanical systems (MEMS). One o the major current revolutions in electrical engineering involves the extension o integrated circuit technology to abrication o MEMS as independent components and as elements integrated on the same substrate as the integrated circuits with which they interoperate. Signiicant advances in cost and/or unctionality have already been achieved with MEMS incorporating optical switches, radio-requency switches, microphones, accelerometers, thermometers, pressure sensors, chemical sensors, micro-luidic systems, electrostatic and magnetic motors, biological sensors, and others. MEMS devices have been used in everything rom video projectors to automobile air bag triggers and mechanical digital memories or hot environments. The size o MEMS devices ranges rom the micron or sub-micron scale up to one or more millimeters, although the basic electromagnetic principles apply to devices o any scale. Recent advances in micro-abrication techniques, such as new lithography and etching techniques, precision micro-molds, and laser cutting and chipping tools, have simpliied exploiting MEMS technology. Eicient motors and actuator conigurations also generally can work as sensors. The irst example explored below is that o a parallel-plate capacitor where the charged plates are electrically attracted and can do work as they squeeze together. Conversely, the same plates can be mechanically orced apart so as to charge a battery. Two dierent ways to calculate these orces are explored below: use o the Lorentz orce equation and use o system-energy derivatives with respect to distance. Lorentz Forces The Lorentz orce law expresses the orce vector [Newtons] acting on a charge q [Coulombs]. The orce is a unction o the local electric ield E, magnetic ield H, and the charge velocity vector v [ms -1 ]: = q( E + v µ o H) [Newtons] (1) A cathode-ray tube (CRT) provides a simple example o this law, as illustrated in Figure -1, where electrons thermally excited by a heated cathode escape at low energy into a vacuum. There the electric ield E between anode and cathode draw it toward the anode at acceleration a [ms -2 ], as governed by Newton's law: - 1 -

2 = m a (2) where m is the mass o the accelerating electron. Cathode -V Heated ilament Figure -1. Cathode ray tube - + Anode, phosphors Cathode ray tube CRT Electron D The kinetic energy o the electron w k equals the accumulated work done on it by the electric ield E. That is, the increased kinetic energy o the electron is the product o the orce acting on it and the distance D the electron moved while experiencing that orce. I D is the separation between anode and cathode, then: w k = D = (ev/d)d = ev [J] (3) Thus the kinetic energy acquired by the electron in moving through the potential dierence V is ev Joules. I V = 1 volt, then w k is one "electron volt", or "e" Joules, where e Coulombs. Typical values or V in television CRT's are 25,000 volts or more. This voltage is limited in part by the desire not to generate excessive x- rays as the electrons impact the phosphors on the CRT aceplate, which is oten leaded glass designed to reduce x-ray release. Figure -1 illustrates how lateral electric ields can delect electron beams in CRT's so as to scan the beam across the aceplate, "painting" the image to be displayed. At higher tube voltages the electrons move so quickly that they do not suiciently experience the lateral orces, and magnetic delection is used instead because the lateral magnetic orces increase with v and are then greater. Static electric orces on capacitor plates can also be calculated using the Lorentz orce equation (1), which becomes = q E. To compute the total attractive orce on each plate we must sum the orces acting on all the charges. I we assume the surace charge on the plates is distributed over some ininitesimal depth δ, as illustrated in Figure -2, then the electric ield E diminishes to zero at that depth δ. E must go to zero when ρ = 0 because there can be no E in a perect conductor unless ε E = ρ 0. It is easy to see that i the charge density is uniorm over the depth δ, then E declines linearly with depth to zero. The average electric ield acting on each charge then has hal the maximum ield strength E, and the total orce pulling on the plate is thereore QE/2, where Q is the total charge on the plate. 1 1 This can be shown or any charge distribution ρ(z) varying with depth z

3 +Q +V d E(z) +Q ε E E σ = E -Q - z Figure.2. Electric charge within skin depth δ in capacitor plates δ > 0 The orce on each capacitor plate, = QE/2 [N] (4) can be expressed in terms o E alone, using the boundary condition at a perect conductor: εe = ρ s [Coulombs m -2 ] (5) which ollows rom ε E = ρ. Multiplying (5) by A we obtain εea = Q, and rom (4) we obtain = εe 2 A/2, and a orce density F Newtons per square meter: F = εe 2 /2 [Nm -2 ] (6) Thus the orce density (pressure, Nm -2 ) pulling on a charged conductor is the same as the energy density [Jm -3 ]. These dimensions are actually the same, because J = Nm. The maximum achievable E thus limits the maximum achievable pressure. In MEMS devices with micron-level plate separations, E can approach [Vm -1 ] beore electric breakdown occurs, so F = εe 2 /2 40 N/cm 2, or ~10 lb/cm 2. For plate separations larger than a ew mean-ree paths or ree electrons, E max is approximately two orders o magnitude smaller. Larger gaps are more vulnerable to breakdown because any ree electrons accelerated to suicient energies beore a collision can produce additional ree electrons that repeat the process several times in a cascade (breakdown) beore they all strike the wall. I the two plates are each charged with the same Q, instead o oppositely, the surace charges repel each other and reside on the outer suraces o the two plates, away rom each other. Since there is now no E between the plates, it can apply no orce. However, the charges Q on the outside are associated with E = Q/εA, which pulls the plates apart with the same orce density F = εe 2 /2. Calculating Forces using Energy Derivatives Mechanics teaches that a orce on object pushing an object a distance dz expends energy dw = dz [J], so: - 3 -

4 on object = dw/dz (7) The orce required to separate two capacitor plates oppositely charged with Q (thus doing work and increasing the stored energy w) is thereore: = dw/dz = d(q 2 D/2εA)/dD = Q 2 /2εA [N] (8) where the plate separation here is D and the energy stored in a capacitor C is CV 2 /2 = Q 2 /2C = Q 2 D/2εA (we recall Q = CV and C = εa/d). We can put (8) in a more amiliar orm or orce density F by noting F = /A and Q = εea: F = Q 2 /2εA 2 = εe 2 /2 [Nm -2 ] (9) which is the same expression as (6). Note that the derivative in (8) was easily evaluated because Q remains constant as the plates separate. I (8) had been expressed instead as d(cv 2 /2)/dD, the derivative would have been more diicult to evaluate because both C and V depend on D. Thereore we always seek to express energy in terms o parameters that remain constant as dz changes. This static attractive orce or V volts remains the same i the plates are connected to a battery o voltage V. A more awkward way to calculate the same orce density (9) is to assume (unnecessarily) that a battery is connected and that V remains constant as D changes. In this case Q must vary with dz, and dq lows into the battery, increasing its energy by VdQ. Since dw in the orce expression (7) is the change in total system energy, the changes in both battery and electric ield energy must be calculated to yield the correct energy when we use (8). As noted above, this complexity can be avoided by careully restating the problem without the source. The power o the energy method (8) is much more evident i we wish to calculate the orce needed to pull two capacitor plates apart laterally, as illustrated in Figure -3. The Lorentz orce law requires knowledge o E, the lateral components o which are responsible or the orce o interest and are not readily determined. Since energy derivatives can be computed accurately and easily, it is the preerred method in this case. Figure -3. Capacitor plates separated laterally d V+ W L - z In this case, or purposes o illustration, we shall solve the problem the more diicult way by including the change in battery energy. The capacitance C = εa/d = εlw/d, where the plate overlap is L meters and the width is W. The orce pulling the plates apart is: - 4 -

5 = dw T /dz = -dw T /dl = -(d/dl)(εwlv 2 /2D - VQ) (10) where w T is the total energy and the two terms relect the energy changes in the capacitor and battery respectively. In (10) only L and Q vary with L, where Q = CV = εwlv/d. Thus (10) becomes: = -(d/dl)(εwlv 2 /2D - εwlv 2 /D) = -(d/dl)(-εwlv 2 /2D) = εwv 2 /2D [Ν] (11) Had we noted that the static orce at voltage V is the same with or without the battery, we could more easily have set Q = constant and taken the derivative only with respect to the energy in the electric ields, yielding the same answer (11). A simple example using (11) is the ollowing. Let W = 10 cm, δ = 10 microns in air, and V = 10 volts. Then = ε o WV 2 /2D = / Newtons, or ~one micro-pound o orce. Even i we decrease D to one micron and increase V to 100 volts, the orce still is only ~one milli-pound. This orce does not depend on L but can be boosted by increasing W, or example, by increasing the number o teeth in the electrodes, as illustrated in Figure -4. W Figure -4. Multi-electrode electrostatic motor Rotary Electrostatic Motors One way to boost power levels o a motor with limited orce or torque is to increase the velocity at which it moves since power P = v, where v is velocity. Consider the ideal 4-segment rotary electrostatic motor illustrated in Figure.5. It has radius R, plate separation D, total plate overlap A = R 2 θ [m 2 ], and operating voltage V. Stator plates occupy two quadrants o the motor and a rotating pair o quadrant plates (the rotor) can rotate to yield an overlap with the stator varying rom zero to perect as θ increases rom zero to π/2. I the voltage V is applied across the plates, a torque T is produced 2, where: T = -dw T /dθ [Nm] (12) and dw T is the increment by which the total system energy (ields plus battery) is increased as a result o the motion θ. This dierence in sign between (10) and (12) is due 2 Torque [Nm] equals the orce on a lever times its length. Thereore Tθ is work perormed by the torque, where θ is the angle (radians) through which the lever rotates about its pivot at one end. Power is Tdθ/dt = Tω [W]

6 to the act that the orce in (10) is applied to the system, and that in (12) the orce is applied by the system to its environment. Figure -5. Rotary electrostatic motor The total energy increase is: θ stator + - rotor T R dw T = d(cv 2 /2) VdQ [J] = d(aε o V 2 /2D) V 2 dc = R 2 dθ ε o V 2 /2D V 2 R 2 dθε o /D = V 2 R 2 dθε o /2D (13) Thereore the torque T rom (12) is: T = -dw T /dθ = V 2 R 2 ε o /2D [Nm] (14) I we assume R = 10-3, D = 10-6, V = 300 volts; then T = / A single such ideal motor can then deliver Tω watts, where we might assume the tip velocity v o the rotor is 300 ms -1, slightly less than the speed o sound. The corresponding angular velocity ω is v/r = 300/10-3 = radians s -1, and the available power is ~ = 0.12 watts i we neglect all losses. In principle one might pack ~2500 motors into one cubic centimeter i each motor were 10 microns thick, yielding ~300 W/cm 3. This can be compared to a 300-hp automobile engine that delivers watts 3, and would thereore occupy 746 cm 3, or roughly the volume o a sotball. Such densities are impractical here because o the diiculties in removing heat and torque rom such a package containing more than a million tiny motors. They have great potential, however, or extremely low power applications where torque extraction can be eicient. For example, the torque could drive a gas turbine at high speeds. The ield o MEMS motors is still young, so its ull potential remains unknown. MEMS Sensors A common type o MEMS sensor electrostatically measures small displacements o lever arms due to temperature, pressure, acceleration, chemistry, or other changes. Figure -6 portrays a standard coniguration that illustrates the basic principles, where the capacitor plates o area A are separated by the distance d, and the voltage V is determined in part by the voltage divider ormed by the source resistance R s and the sensor output resistance R. V s is the source voltage. 3 There are 746 watts per horsepower

7 Figure -3. Capacitive MEMS sensor + - V R s V+ sensor C A d R Cantilevered capacitor C, stressed at V The circuit response to an increase δ in the plate separation d is to increase the capacitor voltage V above its normal equilibrium value V e determined by the voltage divider, where V e = V s (R/[R+R s ]). The capacitor then discharges exponentially toward V e with a time constant τ = (R//R s )C. I R s >> R then τ RC. I R s >> R and R represents a sensor similar to the best o those used or communications systems, then that sensor can detect as little as ~ joules per "bit" o inormation. This can be compared to the incremental increase in capacitor energy dw due to the displacement δ << d as C decreases to C'. dw = (C C')V 2 /2 = V 2 ε o A(d -1 [d+δ] -1 )/2 V 2 ε o Aδ/2d 2 [J] (15) A simple example illustrates the extreme potential sensitivity o such a sensor. Assume the plate separation d is one micron, A is 1-mm square (10-6 ), and V = 300. Then the minimum detectable δ given by (15), assuming dw = , is: δ min = dw 2d 2 /V 2 ε o A (10-6 ) 2 /( ) meters (16) At this level o sensitivity we are limited instead by thermal and mechanical noise due to the Brownian motion o air molecules and conduction electrons. A more practical set o parameters might involve a less sensitive detector (dw ), and lower voltages (V 10); then δ min 1 angstrom (very roughly an atomic diameter). The dynamic range o such a sensor would be enormously greater, o course