Two Channels of Oscilloscope in Circuit Analysis II Laboratory | EE 304, Lab Reports of Microelectronic Circuits

Download Two Channels of Oscilloscope in Circuit Analysis II Laboratory | EE 304 and more Lab Reports Microelectronic Circuits in PDF only on Docsity! Name: ___________________________ Lab Instructor: _________________ Date Performed: ________________ Date Due: _______________________ Lab Partner(s): ______________________________________________________ © 2008 Simon J. Tritschler. All Rights Reserved. EE 304 Laboratory II Kirchoff & Thévenin In Laboratory I, we measured voltage phasors using a two-channel oscilloscope with unbalanced inputs. In cases where a circuit element was floating with respect to the common ground node required by the oscilloscope, we were able to determine the voltage across the element by measuring the node voltage phasor at each end of the element and calculating their difference. Now we turn our attention to finding current phasors and using them to verify Kirchoff’s Current Law. Finally, we will use these techniques to determine a Thévenin equivalent circuit. Consider the following circuit: + Kirchoff’s Current Law effectively states that the sum of all currents flowing away from a node must equal zero. If we consider the principal node at the top of the above circuit, then the KCL equation for that node, acknowledging the directionality of the currents indicated, would be as follows: I1 = I2 + I3 Verifying Kirchoff’s Current Law experimentally requires that we determine the three currents I1, I2, and I3. However, since these currents are all phasors due to presence of reactance in the circuit, and the only equipment available to us in the lab for measuring wide-band AC phasors is a two- channel unbalanced oscilloscope that measures voltage rather than current, we must use “guerilla” tactics to determine these currents based on what we can measure. I1 3.3 kΩ 10 VPK-PK @ 10 kHz SINE + V6.8k − + V3.3k − + V4.7n − 6.8 kΩ4.7 nF 1.0 nF I3 I2 © 2008 Simon J. Tritschler. All Rights Reserved. 1. Build the circuit as shown. Generate a 10-VPK-PK sine wave with a frequency of 10 kHz. Make sure your function generator is properly terminated to avoid voltage errors! Connect channel two of the oscilloscope in parallel with the function generator to monitor the input signal. Connecting channel one to the appropriate signal node, measure the following voltage phasors using the techniques and conversions developed in Laboratory I. Leave your voltages in peak-to-peak form for convenience. V4.7n = _____ ∟ _____ VPK-PK V6.8k = _____ ∟ _____ VPK-PK V3.3k, the voltage across the 3.3-kΩ resistor, is determined by calculating the difference between the node voltage at its right-hand terminal, V4.7n, and that at its left-hand terminal, the input voltage source (10 ∟ 0° VPK-PK ): V3.3k = VSOURCE − V4.7n = _____ ∟ _____ − 10 ∟ 0° VPK-PK V3.3k = _____ ∟ _____ VPK-PK Now we have the necessary information to determine I1 and I3 through the simple application of Ohm’s Law: I1 = V3.3k / 3.3 kΩ = _____ ∟ _____ mAPK-PK I3 = V6.8k / 6.8 kΩ = _____ ∟ _____ mAPK-PK Now we must determine I2, the AC current flowing through our 4.7-nF capacitor. Rather than simple division by a resistance, we must divide our voltage phasor, V4.7n, by the complex impedance of the capacitor under given circuit conditions; i.e., the operating frequency of 10 kHz.