Capacitors and Resistors: Capacitor Time Constants

Transcription

1 Not the Flux Kind A capacitor is comprised of a pair of conductors surrounding some non-conducting region (either empty space or a dielectric). The point of these things is to store energy in the electric field between conductors. Capacitors are exploited for other purposes in electronics (time travel not among them), but in this lab, we ll only concern ourselves with their defining property: the capacitance. So how do you get a field between two conductors anyway? Suppose one of the conductors, say a metal plate, is connected to a battery. Assume the other conductor is oriented parallel to the first and is connected to the other battery terminal. Furthermore, assume a resistor is in series with the capacitor so that we have a well defined current (see Figure below). As current flows to one of the conductors, it acquires some charge. This charge situates itself on the surface of the conductor and, via its electric field, begins to repel a like-signed charge on the other conductor (this is how current flows through a conductor). As the process continues, more charge builds up on the conductor, so the field strength between the conductors grows, as does the energy stored. And there you have it. + C C (a) Charging circuit (b) Discharging circuit Capacitance is defined by Figure : Schematics for charging and discharging C circuits. C Q. () To make sense of this, conceptualize it as a measure of how much charge a capacitor can hold given some applied voltage. A large capacitance either means that a large charge is accommodated for some given voltage, or that only a small potential is required to build up a given charge. In terms of the parallel plate capacitor, you can increase the capacitance by making the plates larger (more room for charges) or brining them closer together (so that the field between the plates is kept stronger). The capacitance equation is a little deceiving. For ideal capacitors, C does not depend on Q or. BS! you say, there s a Q and in the definition. It turns out that they end up canceling, leaving only geometric terms. As such, the capacitance depends only on the geometry of the system. Since budgetary constraints only allow for finite systems, capacitances are also finite. So for a given voltage, Equation () gives that a finite charge will be stored on the capacitors, to wit, Q = C. What eventually stops the charge from accumulating? How does the capacitor know when it s full? Is the process instantaneous? Let s consider the charging state more carefully. When a constant voltage is applied to a capacitor, charges accumulate on the plates. As the charge builds, they repel one another more insistently, discouraging any more accumulation. This repulsion manifests itself as a potential that counteracts the battery. After a certain amount of time, the charges already on the capacitor succeed in halting charge buildup. No current flow implies no net voltage across the capacitor. Indeed, the voltage across a full capacitor is the same a the battery s,

2 but drives current in the opposite direction, impeding any charge flow. The charge time is related to the capacitance and resistance in the charging circuit. Let s think about the situation more mathematically. Say a constant voltage 0 is applied on the capacitor (capacitance C) in a charging circuit where it s in series with a resistance (refer to Figure again). In what follows, lower case variables indicate time dependent quantities. Adding up the voltage drops for the circuit elements should result in 0 (Kirchoff s ule): 0 = v + v C = i + q C. (2) Taking a time derivative, remembering that the rate of change of charge is current, we get 0 = di dt + C dt = di dt + C i (3) di dt = C i (4) So we re looking for a function whose derivative is proportional to itself. Of course, says you, it s an exponential! And you re right: i = I 0 e t/c = 0 e t/τ. (5) Here the initial current is I 0 (determined by the resistor) and τ is called the time constant (τ C), which sets the capacitor s (dis)charge rate. So we have the current through the circuit (the same for the resistor and capacitor because they are in series), but we d like to get the voltage across the capacitor, since it s easier to measure directly (with an oscilloscope, say). We could use the definition of capacitance, but that involves integrals. Let s be clever and exploit Kirchoff s ules again: 0 = v + v C v C = 0 v (6) Essentially, all we need to do is find the voltage across the resistor and subtract this from the total, and we have our answer. Our old, dead German friend Ohm gave us the prescription for the voltage drop across a resistor: ( ) 0 v = i = e t/τ = 0 e t/τ (7) Now we just subtract this from the 0 and we have our answer: v C = 0 0 e t/τ = 0 ( e t/τ ). (8) The behavior we described earlier is encapsulated in potential s plot (left half of Figure 2). Notice that the potential is initially null and rises rapidly at first, slowing down when the repulsion increases until finally leveling off to the match the battery s voltage. Those of you keen on mathematical details might notice that the potential never actually reaches 0, and so the capacitor never saturates. This is true for this model, so the convention is that we consider the capacitor full after 5 time constants have elapsed. How about a discharging capacitor? A discharging circuit involves only the capacitor and a resistor (no battery or power supply), so they share the same current and potential. Turning again to Kirchoff s ules: v = v C i = Q C di dt = C i (9) The tacit assumption is that for a charging capacitor /dt = i. You get a minus sign in the discharging state: /dt = i. 2

3 0 0 (t) (t) 0 /2 0 /2 T /2 t T /2 t (a) Charging (b) Discharging Figure 2: Plot of the potential across a capacitor as it charges and discharges. where the minus sign denotes a discharging capacitor. The solution for the current is the same as for the charging case (Equation (5)). Again we can take an integration route, but remember that the potential across the resistor and capacitor is the same. Ohm s law yields v, which immediately gives our answer: v C = v = i = 0 e t/τ (0) v C = 0 e t/τ () Finally, suppose we have a network of capacitors that we d like to replace by a single equivalent capacitor leaving any external variable unchanged. Such a capacitor should satisfy 0 Q/C eq where 0 is the voltage applied on the network. We ll only consider parallel or series arrangements of capacitors. Capacitors in series all share the same charge 2. Kirchoff s ules yield for N capacitors gives 0 = N = Q + Q + + Q ( = Q + ) + (2) C C 2 C N C C 2 That last term reveals the form for C eq. To see it explicitly, compare with the definition: ( Q = Q ) C eq C C 2 C N (3) = C eq C C 2 C N (4) When capacitors are connected in parallel, they share the same voltage. This uniform voltage across different capacitors induces different charges in all, but their total charge should be equal to the total charge for an equivalent capacitor: Q = 0 C eq. Let s see if we can find an expression similar to this: Q = Q + Q Q N = 0 C + 0 C C N = 0 (C + C C N ). (5) 2 A charge Q on one plate of the capacitor induces a charge Q on the other plate. One plate from the next conductor is connected to this plate, and thus will have a charge Q on it to maintain neutrality. This propagates ad infinitum and thus all share the same charge. C N 3

4 This is indeed what is sought. Compare with the C eq equation: 0C eq = 0 (C + C C N ) (6) C eq = C + C C N. (7) Half Time As stated earlier, the time constant determines rate of (dis)charge. Examine the equations for the potential across a capacitor. You ll notice that if we evaluate it at t = τ, the voltage changes by a factor of e. Measuring such changes is cumbersome, but factors like 2 are convenient, so we define the quantity T /2, which is the time it takes for the capacitor to either attain half it s maximum value (rising or falling). In either state, if you set the appropriate equation equal to 0 /2 with t = T /2, you ll get that T /2 = τ ln 2 = C ln 2.693C (8) This little gem is what we use to determine most of the capacitances. The procedure is simple. Once your scope is set to measure the potential across the capacitor in your C circuit (with a known ), simply read the T /2 off the scope. The procedure is the same for networks of capacitors. Treat them as a whole unit and perform the same measurement. The other method we use to arrive at a capacitance is to plot the potential of the capacitor as a function of time. The data acquisition consists of reading points off of the scope. To extract a τ from the plot, linearize and fit a line. Minding p s and q s A plot of the predicted Equation (8) (the form on the far right) would prove difficult to compare quantitatively with data without tedious calculations or non-trivial coding. Instead, we turn again to our linearization trick. Whatever it is we end up plotting, we d like the slope to be related to C. You ll notice that only the exponent has anything to do with C, so let s isolate it: = 0 ( ) e t/τ e t/τ = ) ln ( 0 = t (9) 0 τ Plotting ln ( / 0 ) vs t, then, yields a line with a slope of /C. Because is known, this results in a value for C. As always, the uncertainty in your data complicates matters as these propagate into the uncertainty of the slope via the error bars. Because we re not plotting the raw data ( vs. t), we need to play the propagation game to get the error bars. First, take our linearized function from the last equation, and call the potential side p and the temporal side q (so p vs. q gives the coveted linear plot). Function p depends on two quantities with intrinsic uncertainty, the variable potential and saturation potential 0, and q depends on the time t: ) p(, 0 ) = ln ( 0, q(t) = t. (20) Assuming uncertainties δ, δ 0, and δt, we can calculate the uncertainties δp and δq (i.e., the error 4

5 bars): ( ) p 2 ( ) p 2 δp = δ + δ 0 (2) 0 p = p ( ), = ) (22) ( 2 0 ( ) 2 ( ) δp = 0 δ ( 0 ) δ 0 (23) ( ) 2 δq = dt δt = dt δt (24) dt = dt δt = δt (25) δq = δt. (26) Having these, you can estimate the uncertainty in the slope by the range/2 method. Let s call the slope you find k and the uncertainty δk. emember that the slope is equal to /C, and thus C = /k. Because both and k have an uncertainty, the uncertainty in C follows from (δk ) 2 δc = C + k ( ) δ 2. (27) Now cross your fingers and hope that this range overlaps with your other ranges for C. Well, this wasn t very brief, was it? 5