RC Time Constant Measurement Lab at UT Pan American, Lab Reports of Microelectronic Circuits

Download RC Time Constant Measurement Lab at UT Pan American and more Lab Reports Microelectronic Circuits in PDF only on Docsity! University of Texas – Pan American Electrical Engineering ELEE 2120 spring 2009 1 of 5 LAB 8: RC Time Constant Measurements A. OBJECTIVES  Setup square wave pulse waveform and RC network to measure RC time constant.  Simulate RC network transient solution using PSPICE. B. EQUIPMENT REQUIRED  Oscilloscope (preferably basic type dual trace analog)  Digital Multimeter  Function Generator  Breadboard  Power Supply (Dual or Triple Output)  Miscellaneous Cables C. PARTS REQUIRED  0.01 uF or 0.015 uF capacitor, 20kohm resistor,200 ohm resistor.  Hook-up wire (#20 or #22 solid conductor) D. PRIOR TO LAB D.1 RC Time Constant Measurement Using Square Wave Pulse Generator. A resistor discharging through a capacitor as shown in Figure 1 satisfies the KCL node equation voltage differential equation:     0 R tV dt tdV C CC Setting the coefficient of the first derivative to one by dividing by the capacitance results in:     0 1  tV RCdt tdV C C The solution of a first order differential equation with no or constant sources is           t ExpKKtVC 21 University of Texas – Pan American Electrical Engineering ELEE 2120 spring 2009 2 of 5 Figure 1 Figure 2 Substituting the solution into the equation results in: 0 1 212 2121                                       t Exp RC K RC Kt Exp Kt ExpKK RC t ExpKK dt d Requiring that the coefficient of the Exp term be zero results in the equation: 022  RC KK  The equation can only be satisfied if  = RC and is satisfied for any K2. The non-exponential part of the differential equation requires that K1/RC = 0 or K1 = 0. At t = 0 then VC(t) = VC(0) = K1+K2 = K2 since K1 = 0 which is just the initial voltage on the capacitor. The resulting solution is then:            RC t ExpVtV CC 0 If the resistor is connected to voltage source VDC at t=0 instead of ground as shown in Figure 2 the resulting differential equation is:     DC CC V R tV dt tdV C  Repeating for this equation the same steps as for the first equation results in K1=Vc()=VDC and now K1+K2=VC(0)=0 so K2= - K1 = - VDC and the solution for VC(t) is:                          RC t ExpVVV RC t ExpVVVctV DCCDCCCC 00 If the capacitor is initially uncharged VC(0)=0 the solution is then