OBJECTIVES:

1. To understand phasor relationships of  V_R and V_C in a R-C circuit.
2. To understand phase relationships of two waves in the time domain.
3. To determine phase angle by four different methods, two of which include the use of the oscilloscope.
4. To gain further understanding of the use of the oscilloscope.

BACKGROUND & THEORY:

See the attached theory on voltage and current relationships in both the time domain and frequency (or polar) domain. Also note how to combine resistance (R) and reactance (X_L and X_C) into impedance (Z) for a circuit with an A.C. sinusoidal voltage supply.

Be aware that in an A.C. circuit, the vector addition of the voltages across the circuit elements must equal the supply voltage. Therefore, in a R-C series circuit--

The voltage across the resistance (V_R) will always be in phase with the current and the voltage across the capacitor (V_C) will lag the current by 90^o. (Conversely the voltage across an inductor (V_L) would lead the current by 90^o). These voltage relationships are represented using complex numbers and by the vector diagrams called phasor diagrams as shown below.  Note a complex number is equivalent to a two dimensional vector.  The real part of the complex number is the x displacement and the imaginary component is the y displacement.  In Electrical engineering we use, j, to represent the square root of (-1) to avoid confusion with current which we usually designate by the variable, i.

Let's look at the following example for a R-C series circuit in both frequency domain (using phasors) and the time domain (using cosine waves).

Given: AC Supply voltage V_s = 10 V_rms ,     R = 3 Ohms,   C = 125µF
At a frequency, f  = 2000/(2p) = 318.3 Hz   or omega = 2000 radians/sec  T = 1/f = 3.14 ms
The reactance of the capacitor will be

Therefore the total impedance of the R-C circuit will be

and

Which gives

First, observe these frequency domain relationships in the polar form plotted as two dimensional vectors in the complex plane.  Note: the vector sum of  V_R and V_C  is equal to V_S.

Second, observe these relationships for one period (or cycle) in the time domain. The current i(t) is not shown but it would be in phase will V_R.

Now let's look at the above example mathematically. Expressing the current in the time domain, we have

Using Ohm's Law, the voltage across the resistor is--

Also, look at the voltage across the capacitor, starting with math model--

 

Substituting (1) into (3)

Expressing the sine function in terms of a cosine, sin(x) = cos(x-90º), and substituting in values for (omega) and  (C)--
V_S(t) = V_R(t) +V_C(t) 

This shows that at any instant of time the sum of the voltage across the resistor and the voltage across the inductor is exactly equal to the source voltage.  Remember that omega is in radians per second so that the phase angle must be converted to radians before adding it to the (omega t) term inside the trig function.  

EQUIPMENT AND PARTS:

Breadboard
Resistor 1 kW
Capacitor 0.22 µF
Signal Generator (Oscillator)
Digital Multimeter (DMM)
Oscilloscope
LCR meter

PROCEDURE:

1. Construct the following circuit with the equipment and parts provided.  Set the signal generator to produce a sinusoidal waveform at 100 Hz at just slightly below 2 V_rms.

2. Using the value of the capacitor measured at 1 kHz.  Calculate the theoretical phase angle between the voltage and the current using--
Method 1

This angle will be considered as positive if measured from the source voltage to the current (or Resistor voltage) and negative if measured from the current to the source voltage.  That is the current leads the voltage in an R-C circuit. This is because the current must flow to charge the capacitor before there will be a voltage across the capacitor.  Calculate all phase angles to the nearest tenth of a degree.

3. Measure and record V_S, V_C and V_R using the DMM on the 2 V range. The phase is then given by - -
Method 2

Note: The inverse sine formula is very sensitive to measurement errors for angles near 90º  and the inverse cosine is very sensitive to errors near zero degrees.  Try all three formulas for the highest and lowest frequencies.  Use the inverse tangent for all frequencies.  Can you explain why the inverse cosine is sensitive to measurement errors near zero degrees and the inverse sine near 90º?

4. With connections indicated below, observe V_S on Channel 1 and V_R on Channel 2 of the oscilloscope. Note that V_R gives the wave shape of the current and its phase relation to the supply voltage.  Adjust the vertical position of both waveforms so that zero volts is at the center of the screen.  You will need to use 1.0 volts/div on vertical sensitivity and 1ms/div on the horizontal sweep for this first frequency.  Adjust the triggering to start the V_S waveform at a negative slope zero crossing at the center of the screen. This should give one complete cycle of V_S on the screen for 100 Hz.  Since 10 divisions now correspond to the full 360^o or one complete cycle, each division corresponds to 36^o.  You can make a rough estimate of the phase angle by multiplying the number of divisions between the zero of the V_S waveform and the same slope zero crossing of the V_R waveform by 36^o, but this only works when one full cycle corresponds to exactly 10 full divisions.  Try this estimate for the 100 Hz frequency only.  The time per division will have to be changed for other frequencies and the method used in the next step is more accurate.

In the picture above the waveform that crosses zero at the two edges of the screen and at the middle is the reference waveform.  That is it is the reference for zero phase angle.  The phase of the second waveform is to be measured relative to this waveform.  On the picture the difference in time (or phase) of the two negative slope zero crossings is labeled as, t.  The total width of the waveform, T,  has been adjusted to match the total width of the display to maximize the accuracy of the measurements.  The ten divisions of the horizontal scale correspond to 36^o each with this setting.  The second waveform shown in this illustration leads the reference waveform by 45^o.  It is leading because it crosses zero before the reference waveform in time.  In this case the angle is in the first quadrant since the zero crossing is between 0 and 90^o ahead of the reference.

5. A more accurate estimate of the phase angle can be obtained by first measuring the period, T, of the waveform.  Note: both waveforms are at the same frequency and therefore have the same period so it makes no difference which waveform is used to measure the period.  It is usually best to use the waveform with the largest voltage.  The period can be measured on the oscilloscope by using the measure mode.   You will need at least slightly more than one full cycle on the screen for the measure mode to be able to calculate the period.  Adjust the time/div control accordingly.  Next accurately measure the time, delta t, between the same slope zero crossings of the two waveforms.  The times where V_S and V_R cross zero volts on the oscilloscope with the same slope.  The picture above shows the time between the negative slopes of the waveforms.  This difference in time, t, can be more accurately measured using the cursor mode.  First center delta t on the screen then expand the time/div scale until the space between the two zero crossings is as wide as possible while still showing both on the screen.  This may require adjusting the horizontal position control to keep the space between the zero crossings centered on the screen.  Then again check to see that the zero level of both waveforms is exactly at the vertical center of the screen.  This is indicated by a 0 in the parentheses at the bottom of the screen as you adjust the vertical position.  Next adjust the volts/div controls until both lines cross zero at a steep angle so that it is easy to see where each one crosses zero.  The fact that the top of the waveform is off the screen will not effect the position of the zero crossing.  Then switch to the cursor mode select time measurement and adjust the cursors so that one is set on each of the two zero crossings.  Since the vertical position controls are used as the cursor controls, it is important that the zero levels of both channels be set before changing to cursor mode.  The delta t reading on the oscilloscope is the desired time, delta t.  Divide this delta t by the Period, T,  and multiply by 360^o to get the phase angle in degrees.  The voltage across the resistor (V_R) will lead the supply voltage (V_S), because the current leads the voltage in a series R-C circuit. Record the waveforms V_s(t) and V_R(t).
Method 3

The following graphs illustrate phase angles in other quadrants.

This waveform leads the reference by 135^o (second quadrant).  It is between 1/4 and 1/2 a period before the reference waveform.

The phase angle of this waveform is in the third quadrant.  It leads by 1/2 to 3/4 of a period or it lags by 1/4 to 1/2 of a period.  The actual value in this case is an angle of -135^o or 135^o lagging or 225^o leading.

In this picture the phase angle is -45^o or +315^o or 45^o lagging.  This angle is in the fourth quadrant between 0 and 1/4 period lagging behind the reference waveform.  In this experiment all angles will be in the first or fourth quadrant.

6. Another method for determining phase angle difference between two signals is as follows:
Method 4

a. Switch the display mode from Voltage-time mode, YT,  to X-Y mode, XY.  The display on the oscilloscope will show an elliptical pattern similar to the following.

This pattern corresponds to a phase difference of 45^o, or -45^o.

If the ellipse is sloping in the other direction as shown below it  corresponds to 135^o, or -135^o.

b. Then the phase angle is determined by the following:

 

Derivation

The total width of the ellipse will be twice the amplitude if the V₁(t)  waveform or the peak to peak value of that voltage.  The distance between the two zero crossings will be the difference between the V₁(t)  voltages at the times when the V₂(t)  voltage is zero.

Taking the ratio of these two horizontal measurements gives:

The advantage of this method over the previous method when a cursor mode is not available is that a 90^o range of angles spreads the measurement over the full width of the screen on the oscilloscope and therefore provides greater measurement accuracy.  The disadvantage is that the same elliptical pattern could represent equivalent angles in two of the four quadrants.  You must look at the voltage-time display to determine which quadrant the angle is in and then calculate the equivalent angle in that quadrant.  If the ellipse collapses to a straight line then the angle is 0^o if the line has a positive slope and 180^o if the line has a negative slope.  If the ellipse becomes exactly horizontal then the angle is 90^o or 270^o.  Any ellipse with a positive slope represents an angle in the first or fourth quadrant.  Any ellipse with a negative slope represents an angle in the second or third quadrant.

7. In summary, we have looked at four methods for determining phase angle in an R-C circuit:

Method 1: (Theoretical Calculation)

 

Method 2: (Voltage measurement)

Method 3: (Oscilloscope/time domain)

Method 4: (Oscilloscope/elliptical pattern)

9. Repeat items 2, 3, 5 and 6 for the following additional frequencies: 500 Hz, 1 kHz, 2 kHz, and 10 kHz.  Remember to check the source voltage each time you change the frequency.

QUESTIONS, COMPARISONS, AND GRAPHS:

1. From measurements recorded in items 3 and 4 of PROCEDURE draw a phasor diagram showing that

Do this for frequency equal to 100 Hz only.

2. For all five frequencies check to see that the three voltage vectors form a right triangle by:

a)  applying the Pythagorean Theorem to see how close V_S comes to matching the square root of the sum of the squares of the other two voltages.
b) applying the law of cosines to determine the angle between V_R and V_C to see how close it comes to 90°.

4. Plot a graph of phase angle versus frequency using Method 1 (calculations) with enough extra points to make a smooth curve. Plot this curve as a line with no symbols (Symbol size set to zero).  Note: the frequency axis must be a log scale or all the low frequency points will be jammed up next to the y-axis and the detail of the phase verses frequency response will not be visible on the graph.  To get evenly spaced points on the log frequency scale calculate the extra points by multiplying each frequency by a constant such as 1.05 to get the next frequency to be plotted.  This log step size should give evenly spaced points that are close enough together to give a smooth curve for the theoretical curve.  To avoid the problem of the last point being in the next decade on the graph change the first calculated frequency value that exceeds 10 kHz to exactly 10 kHz.  Then plot points only no line for Method 3 on the same graph.  Use the sample graphs below as a guide.  In general the scale increments for graphs should be steps of 1, 2, or 5 times some power of 10, as in the first example below.  However for graphs involving angles steps that match fractional parts of a quadrant such as 45, 30, or 15 can be used, as in the second example.  Note that two minor tick marks have been specified so that they fall at 5 degree increments.

CONCLUSIONS:

Based on the experimental results.