1RL Circuits

Equipment DataStudio, RLC circuit board, 2 voltage sensors (no alligator clips), 2 leads(35 in)

Reading Review operation of oscilloscope, signal generator, and power amplifier II

1 Introduction

The three basic linear circuit elements are the resistor, the capacitor, and the inductor. Thislab is concerned with the characteristics of inductors and circuits consisting of a resistor andan inductor in series (RL circuits). The primary focus will be on the response of an RLcircuit to a step voltage and a voltage square wave.

2 Theory

2.1 Inductors

An inductor is a 2-terminal circuit element that stores energy in its magnetic field. Inductorsare usually constructed by winding a coil with wire. To increase the magnetic field inductorsused for low frequencies often have the inside of the coil filled with magnetic material. (Athigh frequencies such coils can be too lossy.) Inductors are the least perfect of the basiccircuit elements due to the resistance of the wire they are made from. Often this resistanceis not negligible, which will become apparent when the voltages and currents in an actualcircuit are measured.

If a current I is flowing through an inductor, the voltage VL across the inductor is pro-portional to the time rate of change of I, or dI

dt. We may write

VL = LdI

dt, (1)

where L is the inductance in henries (H). The inductance depends on the number of turns ofthe coil, the configuration of the coil, and the material that fills the coil. A henry is a largeunit of inductance. More common units are the mH and the H. A steady current througha perfect inductor (no resistance) will not produce a voltage across the inductor. The signof the voltage across an inductor depends on the sign of dI/dt and not on the sign of thecurrent. Figure 1 shows the relationship between the current and voltage for a resistor and ininductor. The arrow indicates the positive direction of the current through the two devices(resistor and inductor). That is, if the current is in the direction of the arrow, then I > 0.[Note that just because the arrows point down in the Fig. 1 doesnt mean that the current isflowing in the downward direction. The current can also flow in the upwards direction, butin that case I < 0.] VR and VL are the potential differences across the resistor and inductor,respectively, defined such that if V(R,L) > 0, then the potential at the +

end of the deviceis greater than the potential at the end. [According to this convention, if V(R,L) < 0then the potential at the bottom is greater than the potential at the top.]

According to these conventions, for the case of the resistor, V = IR. If the current I > 0(current flowing downward) the potential is greater at the top than at the bottom in Fig. 1.

2Figure 1: Relationship between current and voltage in a resistor and inductor.

In the case of the inductor, if dI/dt > 0 then the potential is greater at the top of theinductor than at the bottom. In the inductor, one can have the situation where the currentis flowing downward (I > 0), but decreasing in magnitude so that dI/dt < 0. In this caseVL = dI/dt < 0, so the potential at the top is smaller than at the bottom. The potentialis always lower on the downstream end of a resistor, but for an inductor, itcan be larger or smaller, depending on the rate of change of the current.

As mentioned above, the coil of wire making up an inductor has some resistance RL. Thisresistance acts like a resistor RL that is in series with an ideal inductor. (An ideal inductoris one with no resistance.) In this situation, the voltage across the inductor is

VL = LdI

dt+RL I. (2)

2.2 RL Circuits

A series RL circuit with a voltage source V (t) connected across it is shown in Fig. 2. The

Figure 2: RL circuit: The loopy arrow indicates the positive direction of the current. The + and signsindicate the positive values of the potential differences across the components.

voltage across the resistor and inductor are designated by VR and VL, and the current aroundthe loop by I. The signs are chosen in the conventional way; I is positive if it is in thedirection of the arrow. Also, the positive signs of the voltages across the various componentsare indicated by the + and symbols. Kirchoffs law, which says that sum of the voltage

3changes around the loop is zero, may be written

VL + VR = V, (3)

where it is assumed that the voltages V (t), VR(t), and VL(t) can all vary with time t.Assuming RL = 0 (any resistance in the inductor can just be added to that of the resistor)and letting VL = LdI/dt and VR = IR, Equation (3) becomes

LdI

dt+RI = V. (4)

The solution to the homogeneous equation [V (t) = 0] is I(t) = I0e tL/R , where I0 is the

current through the circuit at time t = 0. This solution leads immediately to VR = I0Re tL/R

and VL = I0Ret

L/R . The homogeneous solution decays exponentially with a time constant = L/R.

Of interest is the response of an RL circuit to an abrupt change in the voltage V of thesource from V = 0 to a constant voltage V = V0. Before the change, we assume that the

voltage of the source was at V = 0 for a long time (much longer than ). The function et

L/R

describes the time dependence of the circuit. To find the behavior as a function of time, itis only necessary to use physical intuition and put the appropriate two constants into thesolution. One guiding principle is that the current I, and therefore VR does not change theinstant the voltage across an RL circuit is changed. That is, VR and I are continuous. Thismeans initially, that the entire voltage change must appear across the inductor.

QUESTION: What is the basis for the argument that I doesnt change discontinuouslywhen the source voltage is suddenly changed from V = 0 to V = V0?

A solution to the inhomogeneous equation

LdI

dt+RI = V0 (5)

is I = V0/R. The general solution in this case is a linear combination of the homogenousand inhomogeneous solutions that satisfy the initial and final conditions. This is

I =V0R

(1 e tL/R

). (6)

This says that after the constant voltage V0 is applied, the current in the circuit will expo-nentially approach V0/R if the inductor has no resistance, or V0/(R + RL) if the inductorhas a resistance RL. These considerations apply whatever the previous history of the RLcircuit. Consider an RL circuit where V = 0 and there is no current. We assume that RL =0. If at t = 0 a constant voltage V0 is put across the circuit, for t 0, VL and VR are givenby,

VL = V0 e tL/R and VR = V0

(1 e tL/R

). (7)

This behavior is illustrated in Fig.3. The current is not plotted, but remember that thecurrent is proportional to VR. Initially all of V0 appears across L because the current justbefore and after the application of V0 is zero. As the current exponentially builds up thevoltage across the resistor increases and the voltage across the inductor decreases. If we wait

4VR Voltage VR across the resistor

0

V0

time

VL Voltage VL across the inductor

0

0

V0

Figure 3: Voltages VR across the resistor R and VL across the inductor L as a function of time after thesource voltage is switched from V = 0 to V = V0.

many time constants = L/R, the voltages reach steady state, and we have VR = V0 andVL = 0. If now V is set equal to 0 (this is equivalent to shorting the circuit) VL and VR willbe given by

VL = V et

L/R and VR = V e tL/R . (8)

The response of an RL circuit to an alternating pair of constant voltages, first at V0, and then0, can be observed by applying a square wave to the circuit, alternating between V = V0 andV = 0. Figure 3 shows half the period of such an oscillation for the case where the period Tis much larger than the time constant = L/R.

QUESTION: How would Fig. 3 be modified if the inductor had some resistance?

QUESTION: If T L/R, what is the response of an RL circuit to a symmetric squarewave that oscillates between +V and V (assume RL = 0)?

If a high frequency square wave, such that T L/R, is applied to an RL circuit, thechanges in current and VR are minimal. There is not enough time for the current through theinductor to change much before the voltage is reversed. If the square wave is not symmetricwith respect to ground the average VR will be the average voltage of the square wave,assuming RL = 0. Fig. 6 shows the voltages for a square wave that oscillates betweenthe constant voltage V and 0 (ground). The exponential dependences of VL and VR areapproximated as straight lines and it has been assumed that RL = 0 and T L/R.

53 Procedure

3.1 Measuring RL

In this section the DC value of RL will be measured. In the right experiment setup window,click analog plug channel C and choose power amplifier. Click the AUTO button in thesignal generator window that appears. Program the signal generator for 2 V DC. Drag thedigits display icon to Output Current under Data. This digits display will give the currentdelivered by the power amplifier. Program the digits display for 2 decimal places. Connectthe output of the power amplifier directly across the coil (see Fig. 4) and click Start andthen STOP. From your data calculate RL, the resistance of the coil. Click the FILE menu

Inductor

InductorCore

Resistors

Figure 4: Picture of the circuit board

button, NEW ACTIVITY, and DONT SAVE. Remove the wires from the coil.During your experiments compare the value of RL to the values of the resistances used in

the RL circuits.

3.2 Time-Dependent Voltages

In this section the response of an RL circuit will be examined experimentally using theDataStudio signal generator and power amplifier II, two voltage sensors, and the oscilloscopedisplay. Click analog plug channel C and choose power amplifier. Click the AUTO buttonin the signal generator window that appears. Drag the scope display icon to Output Voltageunder Data. The oscilloscope display will open with the top or green trace channel havingthe signal generator voltage (analog output) as the input. Click analog plug channel A andchoose voltage sensor. Do this again for channel B. Program the middle scope channel (redtrace) for channel A and the bottom scope channel (blue trace) for channel B.

In the experiment various resistors will be connected in series with an 8.2 mH inductor.Use the circuit board shown in Fig. 4 to assemble the circuit in Fig. 2. Connect the voltagesource with the polarity shown in Fig. 2. The voltage sensors should also be connected to the

6resistor and inductor with the polarities shown in Fig. 2. The polarities are very importantin this experiment. A positive voltage from a voltage sensor will correspond to a positivevoltage as given by the convention of Fig. 2. The input square wave to the RL circuit shouldbe in the top scope channel (green trace), VL should be on the middle scope channel (redtrace), and VR on the bottom scope channel (blue trace).

QUESTION: There are buttons on the far right of the scope display that add adjacentchannels. In the experiments try adding the middle (VL) and bottom (VR) scope channels.Do they match the output signal voltage? Why?

You may wonder about the particular choice of frequencies used in the experiments. Itturns out that the results are better if the signal generator does not operate at too high afrequency and the time constants and frequencies have been chosen accordingly.

Some remarks.

When you click STOP, the last traces are stored. The stored traces are better forexamination than the live traces as they are steady.

You can use the smart cursor on the stored traces. You can use many of the scope functions on the stored traces. You can print out the stored traces. Label your printed out traces immediately by V , VL, and VR as the printers are not

color printers.

Use the same vertical sensitivity for all 3 traces so that you can easily compare the tracevalues.

The default voltage amplitude of 5 V for the signal generator is NOT satisfactory forthe experiments. Use 2 V. (The current rating for the inductor is 0.8 A.)

For reliable triggering with a positive only square wave, make the trigger voltage positivebut less than 2 V.

Due to RL, your graphs will differ somewhat from Figs. 2-4 of this write-up, whichassumes RL = 0.

3.2.1 T L/R, Positive Only Square Wave

Hook up an RL circuit with the R = 33 resistor. Compute the expected value of the timeconstant L/Rtot (with uncertainty) for two cases: RL = 0 and RL given by your result fromthe previous section. How do these values compare with the period (T ) of a 300 Hz squarewave?

Examine VL, VR, and V using a 300 Hz square wave that oscillates between ground (0 V)and 2V. Be sure to set the sample rate to 5000 Hz. Are the traces what you expect?

Now to determine the time constant from the data. We want to plot one charging cyclein Python. In DataStudio create a table recording VL and V . Write down the sampling rate,and RECORD data for a very short time (less than one second). In order to record data, itis necessary to close the scope display, then click start and stop almost immediately. Click

7File, Export Data and choose your desired run from the voltage sensor which is across theconductor. Save the file. Repeat for the data of Output Voltage. Open the data file in atext editor and look for one chunk of data with V 2V. Copy these lines into a new file,this will be the data used for plotting.

We expect VL = V e tR/L . Using your data, make a plot of log(VL) versus t in Python,

and determine the slope and y-intercept (with uncertainties). What do you expect the slopeto be? The y-intercept? From your slope and intercept, compute the time constant and V .Do they agree with the expected values? Which expression for the time constant is best touse: L/R or L/(R +RL)?

3.2.2 T L/R, Positive Only Square Wave, Metal Core

Use the same setup as the previous section. With the DataStudio scope running, insert themetal rod into the inductor coil. How is the trace changed? What parameter is the metalrod changing?

With the rod inside the coil, redo the analysis from the previous section. After closingthe scope display, record a table of VL and V for a short time, export the table, and copyout one section of data with V 2V. Plot log(VL) versus t (the sampling rate should be thesame), and determine the slope and intercept. Are the numbers the same as in the previoussection?

Use the slope to calculate the new inductance L L, assuming the resistance Rtot isthe same as in the previous section. How significant is the effect of the metal core?

3.2.3 T L/R, Symmetric Square Wave

Use the same parameters as in section 3.2.1 except use a square wave that is symmetric withrespect to ground (varies from 2V to -2V).

The time constant = L/Rtot is defined to be the time it takes VL to reduce by a factor ofe (=2.718...). Knowing this, we can approximate the time constant visually with the SmartCursor. In DataStudio, click Start then click STOP to freeze the trace. Find the peak ofthe VL curve, and record the voltage Vpeak and the time tpeak. If we assume RL = 0, then VLshould decay to 0 with time. Using the Smart Cursor, find the point on the VL curve whereVL = Vmax/e, remember 1/e 0.37. Record the time of this point t1. Then = t1 tpeak,simple!

Compare your results to what you expect. Due to limitations in the equipment used andthe non-zero value of RL, the peak voltages will be less than calculated.

3.2.4 T L/R, Positive Only Square Wave

Apply a 3,000 Hz positive only square wave to an RL circuit with R = 10 . Change thex-axis interval to .5 ms/div and set the sample rate to 50,000 Hz. What is the expectedtime constant and how does it compare to T? Compare your results to Fig. 3. In Fig. 3,is approximating the exponentials by straight lines reasonable? Note the distortion in thesquare wave.

83.2.5 T L/R, Symmetric Square Wave

Use the same parameters as in section 3.2.4 except use a square wave that is symmetric withrespect to ground. Compare to Fig. 4. The effect of RL is quite small here as the averagecurrent is small.

3.2.6 T L/R, Symmetric Square Wave

Apply a 500 Hz square wave to an RL circuit with R = 10 . Are the results what youexpect? If you have time, try somewhat lower and higher frequencies.

3.2.7 RLC Circuit (Optional)

Make a series circuit with the 33 resistor, the inductor, and the 330 F capacitor. This iscalled an RLC circuit, for obvious reasons. Use the voltage sensors to monitor the voltageacross the inductor and the capacitor (we will ignore the resistor). Monitor the voltagesin the DataStudio oscilloscope, and apply a 300 Hz, 2V, positive-only square wave to thecircuit. Change the sample rate to 5000 Hz. Describe what you see.

4 Finishing Up

Please leave the lab bench as you found it. Thank you.

9Figure 5: The signal in the first row is the voltage across the inductor. The signal in the second row is thevoltage across the resistor and the bottom row is the voltage supply.

Figure 6: The signal in the first row is the voltage across the inductor. The signal in the second row is thevoltage across the resistor and the bottom row is the voltage supply.