Phys 15b: Lab 3, Sprng 20071Physics 15b, Lab 3: The Capacitor...and a glimpse of DiodesREV01; March 14, 2007Due Friday, March 23, 2007.NOTE that this is the first of the labs that you are invited to do in your own room, if you like, although we will holdhelp labs on Wednesdays and Thursdays:• Wednesday: 6-9 p.m. (March 21)• Thursday: 3-9 p.m. (March. 22)There will be no lab sections, apart from these three help sessions.You can do this experiment alone, but we encourage you to do it as a partnership, taking data collaboratively.Contents1 Purpose 12 Background 23 Procedure 33.1 Dischargeofacapacitor.............................................. 33.2 Chargingofacapacitor .............................................. 54 Diodes: Background 54.1 Theory....................................................... 54.2 Rectifiers...................................................... 64.3 *OptionalTheory*................................................. 65 Diode Experiment: Silicon diode I − V curve 85.1 A Trick to fill in missing Current Ranges on your meter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 PurposeTo study the charging and discharging of a capacitor in an RC circuit, with emphasis on the time dependence of thevoltage and current. You should also think about how the capacitor stores energy in the electric field, and how thisenergy can be added and subtracted from the capacitor.You will also get a look at the way a diode can rectify an AC waveform, and with the help of a capacitor, can beginto produce a DC voltage from an AC sinusoid. This is a process you will look at more closely in Lab 4.1Revisions: add table of contents (3/06); ground added to figures, because questionsrefer to “voltages” not VCAP, etc.; cap % typo fixed,p. 3; most of this lab was called “Lab 4” till this term. This is old Lab 4 minus ripple, but adding in Lab 3’s introduction to diodes (old 3.1)(10/04).Phys 15b: Lab 3, Sprng 200722 BackgroundConsider the voltage and the current as a functionof time for a capacitor charged to voltage V0, discharging througha resistor:This problem is discussed in section 4.11 of Purcell. We want to find the voltage as a function of time, and we knowthree relationships:Q = CV I =VRI =dQdtWe can use these to write a differential equation for V , and integrate to find V as a function of time. Showing all thesteps:−dQdt= −CdVdt=VR(1)ZVV0dVV= −Zt0dt0RC(2)lnV (t)V0= −tRC(3)V (t)=V0e−tRC(4)In words: the voltage starts at V0, and decays toward zero.Since I =VR, we also have: I(t)=V0Re−tRCNotice that the variation of V and I with time depends on the product RC in the exponential. This quantity, whichis called the time constant (τ), is equal to the time for decay to 1/e times the initial value.A related problem is the charging of a capacitor through a resistor:Phys 15b: Lab 3, Sprng 20073Now we have the three relations:Q = CV I =V0− VRI =dQdt(5)Again we can write a differential equation for V and solve it. Without showing all the steps, which are similar to theexample above:CdVdt=V0− VR(6)ZV0dV0V0− V0= −Zt0dt0RC(7)V = V0(1 − e−tRC) (8)Since I =V0−VR, we have:I(t)=V0Re−tRC= I0e−tRCIn words: when you close the switch, the current starts at V0/R and decays toward zero.In this experiment we hope you will be able to demonstrate the rules stated above.3 Procedure3.1 Discharge of a capacitorYou can use the circuit below, momentarily closing and then opening SW1 (the “switches,” “SW1” and “SW2” canbe simply wires touched to the breadboard, then released) so as to charge the capacitor, with “SW2” open. Thenclose SW2 to begin the discharge, measuring t from the time you close SW2. More elegantly, you can close bothSWl and SW2, observing a steady current I0= V0/R through your meter, and then measure the decrease of thecurrent as the capacitance discharges after you open SW1.Question 1: Let C be your unused 1000 µF capacitor. If R were 1 kΩ, what would the time constant be? Pick aresistor out of your unused ones that will give a convenient time constant (of order one minute) for following thecurrent change as the capacitor discharges. Calculate the maximum current for voltages you can easily get with yourdry cells, and select a value of V0appropriate for use with your 50 µA meter scale.Question 2: What is the meter resistance on the 50 µA scale? On this scale (where the small fuse resistance isnegligible relative to the meter movement’s resistance), you can count on the voltage across the meter to be 250 mVPhys 15b: Lab 3, Sprng 20074full-scale, 125 mV half-scale, etc. What correction do you need to make?Record I as a function of time, and plot it. If you had semi-log paper, you could use it to verify exponentialdecay simply by looking for a straight line. If you use the linear paper of your notebook, instead (much more likely;few of you carry around semi-log paper!) you can achieve the same result, by plotting the log I vs t. (Do you wantto use the natural log (ln), or log base 10? Does it matter what units you use – i.e. ln( µA) vs ln(A)?)Estimate the time constant You can estimate the time-constant, tau (τ = RC), from the graph you have justdrawn. Since R is known (to ±5%) you can calculate a value for C. Recognize that the error in C comes from twosources:• error in R: 5%;• error in your measurement of τ (caused by uncertainties in your measurements of time and voltage).Combine these errors appropriately to calculate a resulting percentage uncertainty for C.Compare your calculated value of C against the capacitor’s nominal value. The C value may look wrong, at first.But electrolytic capacitors like the one you are using show wide and asymmetric variation from the nominal value:usually they are used in uncritical applications such as filters (The asymmetry may puzzle you: it reflects whatcustomers want in a filter cap: they are rate -20% to +100%. This scheme is not as crazy as it may seem, at firstglance: this kind of capacitor is used in uncritical applications such as filters; and the asymmetry reflects whatcustomers want in a filter cap: they’ll complain if C is undersized, because the cap’s smoothing effect will be lessthan expected; they won’t complain if the smoothing is a little more than expected. Cap manufacturers know this,and spec their parts accordingly.)Now record V as a function of time, on the 5 volt scale of your meter:and use this curve to calculate C. On the 5 volt scale the