R ircuits Equipment: apstone with 850 interface, RL circuit board, 2 voltage sensors (no alligator clips), 3 leads 1 Introduction The 3 basic linear circuits elements are the resistor, the capacitor, and the inductor. The voltage across a resistor is proportional to the current through the resistor. The voltage across a capacitor is proportional to the time integral of the current into the capacitor. The voltage across an inductor is proportional to the time derivative of the current through the inductor. This lab is concerned with the characteristics of capacitors and circuits consisting of a resistor and a capacitor in series (R circuits). The primary focus will be on the response of an R circuit to a step voltage and a voltage square wave. (A step voltage is a constant voltage which instantaneously changes to a different constant voltage.) 2 apacitors A capacitor is a 2 terminal circuit element that stores energy in its electric field. In its simplest form, a capacitor consists of 2 parallel plates separated by a small gap which is usually filled with a non-conducting dielectric material. In practice, many capacitors are made by sandwiching a dielectric sheet, such as mylar, between 2 pieces of metal foil and rolling this assembly up into a cylinder. The most important specification for a capacitor of a given value is its maximum voltage rating. If a positive current I enters one plate of a capacitor and exits the other plate, a positive charge Q builds up on the first plate and a negative charge -Q builds up on the second plate. A voltage V then develops across the capacitor. The charge Q and voltage V are linearly related by the equation Q = V, (1) where is a constant called the capacitance. Eq. 1 can also be written as V = 1 Idt. (2) The unit of capacitance in S.I. units is the farad (F). One farad is a huge capacitance and more usual units are the micro farad (µf) and the pico farad (pf). The constant depends on the geometry of the capacitor and the dielectric properties of the insulator between the plates. 1
3 R ircuits A series R circuit, with a voltage source V(t) connected across it, is shown in Fig. 1. The voltage across the resistor and capacitor are designated by V R and V, and the current around the loop by I. The signs are chosen in the conventional way, I is positive if it is in the direction of the arrow. Kirchoff s law, which says that the voltage changes around the loop are zero, may be written V R + V = V. (3) Letting V R = IR = ( d Q)R and V dt = 1 Q Eq.(3) may be written as R d dt Q + 1 Q = V. (4) Of particular interest in this lab is the solution of Eq.(4) when V is a constant voltage. The solution is Q = (Q 0 V )e t R + V, (V = constant) (5) where Q 0 is the charge on the capacitor at time t=0. The quantity R is the time constant of the circuit. The time dependence of the circuit is exponential when V=constant. Beginning at time t = 0, the parameters of this circuit exponentially approach constant values. V approaches V, and I approaches 0. In a time 3R, those parameters are within 95% of the final values. As a function of time, the voltage across the capacitor is 1 Q, the current in the circuit is I = dq, and the voltage across the resistor is V dt R = IR. Exercise: Substitute Eq.(5) into Eq.(4) and show that it is a solution. We can use this result to understand what happens when a constant voltage is applied to a R series circuit, or if a constant voltage is instantaneously changed from one constant value. A key point is that when the voltage is switched, the charge on and the voltage across the capacitor do not change instantaneously. V and Q are continous, but not differentiable, at the switching point. The voltage across the resistor and the current in the circuit do change instantaneously. V R and I are neither continuous nor differentiable at the switching point. onsider the situation shown in Fig.(2). For times t < 0, the applied voltage is 0, and we assume that the applied voltage has been zero long enough so that V = Q = V R = I = 0. At t=0, a constant voltage V is applied to the circuit for a time T/2 R. For the time interval 0 < t < T/2, the voltage across the capacitor rises exponentially from 0 toward V, and the voltage across the resistor decays exponentially for V toward zero. Exercise: For the time interval 0 < t < T/2, obtain Q(t) by integrating Eq. (4) with the appropriate limits of integration. Your answer should be in terms of Q 0, R,, and V. Using your result, obtain expressions for V (t), I(t), and V R (t)? As shown in Fig.(2), at time T/2, the applied voltage is returned to 0. As T R at time T/2, the voltage across the capacitor is essentially V. For times t > T/2, V decays 2
exponentially to 0. Let ɛ be a very small increment of time. At time t = (T/2 ɛ) V R is almost 0. At time t = (T/2 + ɛ), V R has jumped to V. an you explain why? For t > T/2, V R goes exponentially to 0. The above response of an R circuit to a constant voltage which is first zero, then V, and then zero again, can be observed by applying a square wave to the circuit where half of the square wave has zero voltage and the other half voltage V. If T is the period of the square wave, we continue to assume that T R. Make a single plot that shows the square wave, V, and V R, but read the next paragraph first and check that your plot agrees with the statements made. Regardless of the relative magnitudes of T and R, if a periodic voltage is applied to a series R circuit, then V, V R, and I must obey the following conditions: 1. The time average value of V must equal the time average value of the applied voltage. You can prove this by integrating Eq.(3) over one period. 2. The time average value of I (and V R ) must be zero. (If they were not, what would happen to the voltage across the capacitor?) Exercise: If T R, graph the response of an R circuit to an applied symmetric square wave where the voltage oscillates between +V and -V. Your graph should show the applied square wave, V, and V R. Hint: At the instant the square wave switches the magnitude of the voltage across the resistor is either 2V or 2V. If a high frequency square wave, such that T R, is applied to an R circuit, the voltage changes across the capacitor are minimal. The capacitor does not have time to charge much before the voltage is reversed. Most of the voltage changes occur across the resistor. As mentioned previously, the average voltage of the capacitor will be the time average voltage of the square wave, and the average V R will be zero. Fig. 3 shows the voltages for a square wave that oscillates between the constant voltage V and 0 (ground), and Fig. 4 shows the voltages for a square wave that oscillates between +V and -V. In both these figures, the exponential dependences of V and V R are approximated as straight lines. Neither V nor V R approach steady state values. 4 Experiments In these sections, the response of an R circuit to a voltage square wave will be examined experimentally. The period of the square wave is T. Both the situations T R and T R will be examined. For the former case, due to limitations of the equipment, the peak values of V R will be less than the theoretical values. One equipment limitation that you can easily see on the scope is the finite rise time of the square wave. 4.1 Setting up and applying a T R, Positive Only Square wave Before you connect your circuit elements, check the polarities of your capacitors. If they are polarized, the current must flow in a particular direction as indicated by the capacitor. On 3
the R circuit boards that we are using for this lab, there is no polarity so the current can flow in either direction. First, attach a lead across the induction coil to short it out. Then, connect the resistor to the positive (red) output of apstone, and the capacitor to the ground (black) output. In this experiment, you will be using R = 10 Ω and = 100 µf. Now add the analog voltage sensors. Put the first voltage sensor across the capacitor and plug it into channel A of apstone. Put the other across the resistor and plug it into channel B. Now program apstone. In the hardware setup window, set up the voltage sensors for analog channels A and B. lick on the output terminals of the 850 interface and select Output Voltage - urrent Sensor. Then, click on the signal generator icon in the tools column. Select 850 Output 1. hange the waveform to Positive Square Wave. Set the frequency to 60Hz and the amplitude to 2.0 V, then click Auto. Open the scope display by dragging the scope icon to the center of the screen. lick on the orange tack to prevent overlapping. Next, click on Add a new y-axis twice. You should now have 3 y-axes. For each axis, click Select Measurements and choose one to be Output Voltage h 01 (V), the second to be Voltage h A (V), and the third to be Voltage h B (V). For assistance look at the image on the next page. The input to channel 1 of the scope should be the voltage output of the signal generator (Square wave output). Also, on the bottom of apstone switch the ontinuous Mode 4
to Fast Monitor Mode. alculate T and R to determine the parameters you are using. Is the assumption that T R valid? Before you start the experiment read the following remarks: You can use many of the scope functions on the stored trace, such as the Show data coordinates tool located on top of the scope. When measuring scope trace parameters, use the scroll wheel on the mouse to zoom in and out. Also, you can click on and scroll across the time axis to help aid in seeing the traces. If you use same vertical sensivity for all 3 scope traces, you can easily compare the trace values. The voltage of 2.0 V for the signal genertor is satisfactory for all the experiments. Higher values may saturate V R. For reliable triggering with a positive only square wave, make the trigger voltage a little positive. After you click Stop, the last traces the oscilloscope takes are stored. The stored trace is usually better for examination than the live trace, for the live trace may jitter. 5
You can print out the stored trace, but remember to label your printed out traces V, V, and V R immediately, as the printer is not a color printer. lick on Monitor. If the trace does not appear or is unstable, activate the trigger function on the scope and scroll the arrow head slightly up. Examine V, V R and V. Are the results what you expect? Use the Data oordinates Delta Tool to measure the time constant of the exponentials and compare your measurements to R (1/e = 0.37). Note that V is fairly similar in form and magnitude to the input square wave, and that V R looks a lot like the derivative of the square wave. Referring to the left hand side of Eq.(4), dq will be of the order of Q. With the assumption dt T that T R this term can be neglected compared to the term Q. We now have, to some approximation, that V = Q = V. The result of differentiating this expression is dv = I = V R. You have constructed an electronic differentiator! dt R For another R combination with T R, repeat the two measurements of R as described above. 4.2 T R, Symmetric Square Wave Use the same parameters as in section 4.1 except use a square wave that is symmetric with respect to ground Oscillates between (+V and V ). ompare your results to what you expect. 4.3 T R, Positive Only Square Wave Apply a 100 Hz positive only square to an R circuit with R=100 Ω and =330 µf. What is the time constant and how does it compare to T? ompare your results to Fig. 3. Note that V R is fairly similar in form to the input square wave and that V is similar in form to its integral. Using Eq.(4) can you show that this is plausible? In Fig. 3 is approximating the exponentials by straight lines reasonable? 6
4.4 T R, Symmetric Square Wave Use the same parameters as in Section 4.3 except use a square wave that is symmetric with respect to ground. ompare to Fig. 4. Is approximating the exponentials by straight lines reasonable? 4.5 T R, Symmetric Square Wave Use the same R components as in Section 4.3. Apply a symmetric square wave with a frequency of 30 Hz. After use 10 Hz and 5 Hz. Are the results what you expect? 5 Finishing Up Please leave your bench in a state of maximum order. Thank you. 7
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