PHYS 211L Lab Report, Spring 2008

PHYSICS 211L Exp. 1: Basic Oscilloscope Operations Date: _08/03/2007_ NAME: _ PARTNER’S NAME:_ SECTION NUMBER & NAME OF INSTRUCTOR: _-1- MS. SARA NAJM _ 1. Data Table 1: Amplitude Measurements Amplitude control Function Generator CH1 VOLTS/DIV (13) Number of divisions Peak to Peak Voltage Fully CW 2 V/div 6.8 div 13.4 V ½- Way CW 2 V/div 3 div 6 V ¼- way CW 2 V/div 2.6 div 5.2 V Fully CCW 0.5 V/div 1.6 div 0.8 V Fully CW Square wave: Vp-p = 27 V Fully CW Triangular wave: Vp-p =28 V Table 2: Frequency Measurements Set the frequency using the frequency counter (counters are accurate to within about 10 parts per million) Frequency of Function Generator Time/DIV (23) Number of divisions Period Frequency (from scope) % Difference in Frequency 100 Hz 2 ms/div 5 div 10*10-2 s 100 Hz 0 % 2.00 kHz 0.1 ms/div 5 div 0.5 * 10-3 s 2000 Hz 0.75 % 40.0 kHz 5 μs/div 4.2 div 21*10-6 s 47619 Hz 18 % 150 kHz 2 μs/div 2.8 div 5.6 * 10-6 s 178571 Hz 18 % 1.30 MHz 0.2 μs/div 3.3 div 0.66 * 10-6 s 1515151 Hz 17 % Comment on the accuracy of the scope measurement. The scope measurements are more or less close to the frequency given by the generator but not very close especially in the case of high frequencies where we noticed a 17-18% difference. Describe what happens when you vary the trigger and when you press the +/-trigger control. What is the use of the trigger on the oscilloscope? When we vary the trigger and we press the +/- trigger control, we notice that the screen becomes blurry and the signal not very clear. The trigger is used to stabilize the signal and to make it start at the same point with a positive or negative slope. Grade: 1 Table 3. Wave form Unit: Sketch and Explanation Note that the dotted line represents the center horizontal graticule (a) K1 and K2 open Sketch of Wave Form Explanation When K2 open, the current does not pass through the lower most branch. K1 is open; the current passes through the diode in part A of the circuit. First, the current in the positive sense in the upper diode when we watch a sinusoidal signal on the oscilloscope’s signal. But, it is blocked when it flips direction. We observe the following graph. In part B, K2 is open; no current passes. So, the observed graph is as follows: The super position of both graphs gives the signal of the current. (b) K1 open and K2 closed Sketch of Wave Form Explanation In part A, K1 is open; the current passes through the diode only in one sense. No negative current can pass. We observe the following graph. 2 Part A of the circuit is the upper part of the circuit. Part B of the circuit is the lower part of the circuit. In part B, K2 is closed; the current will pass through the diode only in the positive sense. No current passes in the opposite sense. We observe the following graph. The super position of the two signals gives the signal shown in the left box. (c) K1 closed and K2 open Sketch of Wave Form Explanation In part A, K1 is closed; this means that we have short circuited the diode. The current passes normally giving the form of a sinusoidal wave form. In part B, K2 is open; no current passes. The wave form is as follows: The super position of the signals gives the form of the current as shown on the left. (d) K1 and K2 closed Sketch of Wave Form Explanation In part A, K1 is closed; the diode is short circuited and the current won’t pass through it. The current passes normally in both directions. The signal observed is as follows: In part B, K2 closed; the current passes only in the positive direction. So, there will be no negative voltage. 3 The super position of the two signals gives the signal on the left. The negative part of the first will cancel with the positive part of the second resulting in the signal shown on the left. P.S.: The transformer/rectifier unit changes the DC current into AC current. The form of signal of an AC current is as follows: Table 3. X-Y Operation Phase Measurements: 5 kHz Phase shift unit 2A 2D Phase angle  (Measured) A A=1.8 2A=3.6 D=1 2D=2 33.75° B A=1.2 2A=2.4 D=0.5 2D=1 24.62° Table 4. X-Y Operation Phase Measurements: 10 kHz Phase shift unit 2A 2D Phase angle  (Measured) A A=2 2A=4 D=0.8 2D=1.6 23.578° B A=2.2 2A=4.4 D=0.4 2D=0.8 10.47° Work out the derivation steps between eq (1) and eq (2). What happens when =0 and =90? 4  *=0 From the equation (1) we have x/B=y/A => y= (A/B)x The signals are in phase, then the graph that we get is a straight line passing by the origin.  *=90 x/B=sin (wt) y/A=sin (wt+90) = - cos(wt) (x/B) 2 + (y/A) 2 = sin2 (wt) + cos2 (wt) = 1 The y-signal leads the x-signal by π/2. The signals are π/2 out of phase. x2/B2 + y2/A2 = 1 The graph that we obtain is an ellipse. If B=A, then we get x2+y2=A2, The graph that we get is a perfect circle of radius A 5