OBJECTIVES:

1. To use a square wave like a step function.
2. To become familiar with the transient response of R-C and R-L circuits.
3. To learn the voltage and current response due to a step change in voltage applied to an R-C or R-L circuit.
4. To gain further experience with use of oscilloscope.

BACKGROUND & THEORY:

Review theory given in Exp. 8 for transient response of a RC circuit. Note that components used in Exp. 8 resulted in a relatively long time constant where t(tau) > 15 seconds. In this experiment, the time constant, t, will be much shorter or approximately 1.0 ms.

In a R-C circuit, the response curve for the voltage across the capacitor (V_c) will be

and the current flow to the capacitor i_c will be

and the time constant, t = RC.

 In a R-L circuit, the response curve for the current through the inductor will be

and the voltage across the inductor will be

and the time constant, t =L/R.

* Note the formulas above are for Ideal capacitors and inductors.  The internal resistance in a real capacitor or a real inductor could have some effect on the time constants in the circuits.

EQUIPMENT AND PARTS LIST

Signal Generator
Breadboard
Fixed Resistors 100 kW, 1 kW
Fixed Capacitor 0.01 µF
Fixed Inductor 10 mH
Oscilloscope
Digital Multimeter
Digital LCR Meter

PROCEDURE

1 a. Construct the following circuit using the equipment and parts provided. However, first measure the value of R and C on the Digital multimeter.

b.     Display one complete cycle (approx. 10 div) of a 100 Hz square wave on Channel 2 of the oscilloscope by following these steps:

1) Turn selector to square wave on the signal generator.

2) Set oscillator to 100 Hz .

3) Set horizontal sweep time on oscilloscope to 1 ms/div.

4) Set vertical gain to 1 volt/div.

5) Change the output of signal generator to get 6 volts peak to peak (6 divisions peak to peak).  Use the measure mode on the oscilloscope to do this.  Check to see that Averaging is turned on in the Acquire menu to eliminate variations due to noise in the system.

6) Center Channel 1 and Channel 2 displays vertically by adjusting the position controls until the position display at the bottom of the screen reads zero. Then set both channels to "DC" coupling.

7) Set the triggering source to channel 2 and adjust the trigger level to zero volts on the positive slope.  The trigger position should be set to zero using the horizontal position control.

8) Observe and print one full cycle of the square wave.  Draw in the  x and y axis, respectively.  Then add the correct voltage and time scales on each axis.  The Y axis will be in volts for this step, with 0 at the center and increasing or decreasing by 1 volt for each division up or down from the center.  Place this scale to the left of the graph.  Place the X axis scale below the graph, with 0 at the center, trigger point, and increasing or decreasing by 1 ms per division to the right or left. 

Note: The square wave serves as a switch to first apply 3 volts to the circuit in the first half cycle, and then it switches to -3 volts during the second half cycle.  This provides a step change in voltage of -6 or +6 volts each time it changes.

d. Now switch the positions of R and C in the circuit leaving all other connection the same.  Note: This is equivalent to switching the positions of the reference node and node 1 in the circuit.  This results in displaying the voltage across the resistor (V_R) on  Channel 1. Note that when you observe the waveform for V_R, you are also observing the waveform for the current, because I = V_R/R. This is a standard way of observing current on an oscilloscope in an A.C. Circuit. You will note that the current waveform goes off scale.  Change the vertical sensitivity to 2 volts/div to bring the waveform back on scale (both channels to keep the same scale for both signals).  Now record one full cycle of square wave voltage and the response current, I_C, (or V_R) on the same graph.  Since R = 100 kW,  the current will be 10 µA per volt.  Since you are using the 2 V/div scale, it can be read as 20 µA/div on the graph.  That is 2 V/100 kW = 20 µA per division.  Make sure to label this scale on your printed graph.

e. Use the measured values of R and C and to calculate time constant t(tau) = RC.

2 a. Construct the following circuit using the equipment and parts provided.  Use the DMM to measure the resistor and the DC resistance of the inductor.  Then use the Digital LCR Meter provided to measure the inductance, L, and dissipation factor, D, of the inductor.  Make these measurements with the LCR meter frequency set to 10 kHz, since that is the frequency you will be using in this experiment.  The dissipation factor can be used to calculate the equivalent ac series resistance of the inductor.  It should be greater than the DC resistance.  R_{seried ac} = 2pfLD =2(pi)fLD, where f is the frequency at which the measurement was made (10 kHz).

b. Display one complete cycle of 10 KHz square wave on Channel 2 of the oscilloscope by following these steps:

1) Selector should still be on square wave.

2) Set the signal generator to 10 KHz.

3) Set sweep time/div on the oscilloscope to 10 µs/div.

4) Set the vertical gain to 2 volts/div.

5) Change the output level of signal generator to get 6 volts peak to peak (3 divisions peak to peak) if necessary.

d. Now switch the positions of R and L (or the positions of the reference node and node 1) so that the current waveform can be observed along with one full cycle of the voltage output of the signal generator.  Record both waveforms on the same graph using the correct scales for voltage and time. Adjust the vertical scale for maximum on screen display (1 volt/div).  What is the current scale at 1 Volt /Div for this circuit?

e. The signal generator has a Thévenin equivalent resistance or source resistance R_s = 50 Ohms.  This impedance is in series with the 1 K resistor.  The inductor also has an internal series resistance, R_{seried ac} calculated above.  Calculate the equivalent resistance, R_eq, of the 1 k resistor (using the actual value) in series with the 50 source impedance and the internal series resistance, R_{seried ac}, of the inductor.  Then calculate the time constant, t(tau) = L/R_eq using the, R_eq,  just determined and the measured value of the inductor.

QUESTIONS, COMPARISONS, & GRAPHS

1a. From the waveforms recorded in items 1c and 1d (of the PROCEDURE), estimate the value of the time constant in milliseconds by reading the time off the graph at which, V_C changed by 0.632 of  V_{C max} (Figure 1), and at which V_R  = R*I_R drops to 0.368 of  V_{R max}   = R*I_{R max} (Figure 2), respectively.  Show the construction lines on the graph.

Figure 1

Figure 2

b. How do these time constants compare with that calculated in item 1e?

2a. From the waveforms printed in items 2c and 2d (of the PROCEDURE), estimate the value of the time constant in micro-seconds by reading the time off the graph at which, V_L drops to 0.368 of  V_{L max} (Figure 2), and at which V_R  changes by 0.632 of  V_{R max}, (Figure 1) respectively.

b. How do these time constants compare with that calculated in item 2e?

3. Why might the calculated values of be different from that estimated from your graphs?

4. How are the voltage and current responses of the R-C and the R-L circuits related to each other?

CONCLUSIONS

Based on the results of this experiment.