1 The behavior of AC circuits

The purpose of this experiment is to explore potential uses for simple AC ﬁlters, to map out the AC response curve, to use the oscilloscope to measure signal voltages and phases, and to gain some familiarity with the mathematical formulas for current amplitude and phase of an AC circuit.

The apparatus consists of the oscilloscope, a function generator, a breadboard, and commodity-grade electronics components (resistors, capacitors, inductors, and jumper wires).

1.1 AC circuits in a nutshell

We have seen in class that when an AC voltage (sinusoidal function generator)

V (t) = V_{0} sin ω t

is connected to a single-loop circuit containing a resistor $R$ , a capacitor $C$ and inductor $L$ is series, the current in the loop is

I (t) = \frac{V_{0}}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}} sin (ω t - {tan}^{- 1} (\frac{ω L - \frac{1}{ω C}}{R}))

This current is a function of the AC-input driving frequency $ω$ , and reaches a maximum value at the resonance frequency of the circuit

ω_{r e s} = \frac{1}{\sqrt{L C}}

of $I_{m a x} = \frac{V_{0}}{R}$ . This is also the frequency at which $φ = 0$ .
From this formula we can see that each linear device in the circuit reduced the magnitude of the current, as though it had a “resistance”, and we can get this result very simply; deﬁne the impedance $Z$ of each type of device according to

Z_{R} = R, Z_{C} = \frac{1}{i ω C}, Z_{L} = i ω L

which has units of resistance, and behaves like a resistor according to

Z_{s e r i e s} = Z_{1} + Z_{2} + \dots, \frac{1}{Z_{p a r a l l e l}} = \frac{1}{Z_{1}} + \frac{1}{Z_{2}} + \dots

Then for any sinusoidal voltage source

V_{i n} = V_{0} sin ω t = I m V_{0} e^{i ω t}

the current in the circuit will be

I (t) = I m \frac{V_{i n} (t)}{Z (ω)}

The phase between the input voltage signal $V_{i n} (t)$ and the current varies between $\frac{- π}{2}$ and $\frac{π}{2}$ , and could be read from an oscilloscope trace. Remember that a full wave has a period of $T$ , and this means a full phase cycle $2 π$ . The phase magnitude between $V_{i n}$ and $I$ can be gotten then by reading the time difference $Δ t$ between zero values of $I$ and $V_{i n}$ and using

\frac{Δ t}{T} = \frac{∣ φ ∣}{2 π}

The current magnitude and phase plotted against $ω$ are illustrated below. We refer to this as the response curve of the AC circuit.

Another very useful relation is

cos φ = cos ({tan}^{- 1} (\frac{ω L - \frac{1}{ω C}}{R})) = \frac{R}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}} = \frac{R}{∣ Z ∣}

and since the magnitude of the current signal is

∣ I_{m a x} ∣ = \frac{V_{i n, m a x}}{∣ Z ∣}

one can usually obtain $R$ and $∣ Z ∣$ from oscilloscope readings as well at $φ$ and $ω$ .

One of the goals of this experiment is to map out the response curve, or a signiﬁcant portion of it around the resonance frequency. This requires that one ﬁrst compute the resonance frequency.

1.2 Procedure

Obtain the following; an inductor and capacitor chosen so that $ω_{r e s} = \frac{1}{\sqrt{L C}} \approx 2000 π - 10, 000 π \frac{r a d}{s}$ , a resistor, and a breadboard. You want the resonance frequency to be high enough that you can use a resistor in the $1000 Ω$ range. A good choice of inductor would be a Radio Shack 273-1366A transformer primary coil (the two black leads). This coil will give a resonance frequency of around $6000 π \frac{r a d}{s}$ with a $2, 500 p f$ polystyrene capacitor. It is important that $\frac{R}{L}$ be around $10 %$ of $ω_{r e s}$ , so choose $R$ accordingly. Build the circuit below on the breadboard. The $V_{1}, V_{2}, V_{R}$ points are simply marking points on the circuit where voltages will be measured.

The generator pictured is our function generator. The function generator output and ground post are connected to the circuit with banana-plug wires and alligator clips. Follow the color conventions that wires passing to ground are black, wires passing to high voltage are red. Only cut and strip fresh jumpers if you cannot ﬁnd something of suitable length in the recycled wire cups.

Either measure $L$ and $C$ with the bridge, and compute the precise resonance frequency, or else ﬁnd the frequency of maximal gain with the oscilloscope and function generator. Derive formulas for the current amplitude and phase as functions of the generator frequency. Find similar formulas for the magnitudes of voltage drops across the resistor, inductor, and capacitor, and come to lab with these formulas in hand.

1.2.1 Resonance frequency measurement

Use the oscilloscope to search for the resonance frequency by setting the function generator on sine wave, $10.0 V$ amplitude. This stage of the experiment serves the purpose of giving you the chance to measure a frequency with the oscilloscope.
You will need to connect the function generator and oscilloscope to your circuit using coaxial cables. The coaxial cable has a ground conductor and high/low, and to attach this to the circuit, use one of the end-adapters pictured below, together with BNC cables and alligator clips.

To use the oscilloscope to measure the resonance frequency, connect the scope probes (channel 1) across the resistor. Loop a probe hook to the resistor leader between the resistor and the capacitor, and the ground clip of the probe to the grounded lead of the resistor. Adjust the frequency until the voltage drop across the resistor is a maximum. This occurs at resonance, since there the impedance is minimal

Z (ω_{r e s}) = \sqrt{R^{2} + (ω_{r e s} L - \frac{1}{ω_{r e s} C})^{2}} \to R

1.2.2 Mapping the response curve

After verifying the resonance frequency, measure the amplitude of the voltage across the resistor and the phase angle for ten frequencies on either side of the resonance value. What frequency range should you use? Hopefully your $R$ is such that $\frac{R}{L} \approx 0.1 \cdot ω_{r e s}$ .
Determine those frequencies in the band

\frac{1}{2} \leq \frac{R}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}} \leq 1

for which the circuit gain exceeds $50 %$ ;
solving this, we ﬁnd

ω^{2} L C \mp \sqrt{3} R C ω - 1 = 0

gives the limits of this band,

\sqrt{\frac{3}{4} \frac{R^{2}}{L^{2}} + ω_{r e s}^{2}} - \frac{\sqrt{3}}{2} \frac{R}{L} \leq ω \leq \sqrt{\frac{3}{4} \frac{R^{2}}{L^{2}} + ω_{r e s}^{2}} + \frac{\sqrt{3}}{2} \frac{R}{L}

If your resonance frequency is high, and $\frac{R}{L} < < ω_{r e s}$ , this range simpliﬁes to (approximately)

ω_{r e s} - \frac{\sqrt{3}}{2} \frac{R}{L} \leq ω \leq ω_{r e s} + \frac{\sqrt{3}}{2} \frac{R}{L}

Pick seven or eight frequencies, nearly even-spaced, between $ω_{r e s}$ and $ω_{r e s} + 3 \cdot \frac{\sqrt{3}}{2} \frac{R}{L}$ , and seven or eight in the range $ω_{r e s} - 3 \cdot \frac{\sqrt{3}}{2} \frac{R}{L}$ to $ω_{r e s}$ . Measure both the phase and gain

g a i n = \frac{∣ V_{R} ∣}{V_{R, r e s}} = \frac{R}{\sqrt{R^{2} + {(ω L - \frac{1}{ω C})}^{2}}}

for each of these frequencies. Note that $V_{R, r e s} \neq V_{0}$ . See the pre-lab notes.
The phase is measured by using the oscilloscope in dual-display mode, with the voltage across the resistor on channel $1$ and the generator voltage on channel $2$ . Measure the time-interval $Δ t$ between the points where the two signals reach zero and apply

\frac{Δ t}{T} = \frac{∣ φ ∣}{2 π}, ∣ φ ∣ = ω Δ t = 2 π f Δ t

Create a pair of graphs, one of $φ (ω)$ versus $ω$ , the other being the gain versus $ω$ . How do you measure gain? It is the ratio of the amplitude of the output signal (peak-to-peak voltage across the resistor) to the input voltage (peak-to-peak function generator voltage).
The graphs should be similar to those picture a few pages back. You should plot your actual data points (as circles) and the theoretical curves for gain and phase versus $ω$ as a solid curve, for the $L, R$ and $C$ values that you used. There is a software program responsecurve on the lab server to assist in this graph creation. A second program, circuitparams will assist in selecting components and frequencies.

1.3 Pre-lab exercises and notes

1. A circuit contains a generator

ℰ = ℰ_{0} sin ω t

and two devices. One of the devices is a resistor. The generator voltage and the voltage across the resistor in the circuit are shown in the ﬁgure, the generator being the full-scale curve (blue). The oscilloscope has vertical scale setting

0.2 V ∕ D I V

and time scale setting

0.05 m s ∕ D I V

. Determine

ω

and the phase angle

φ

, and the gain. From these determine

R C

\frac{L}{R}

, whichever is appropriate.

2. A circuit contains a generator

ℰ = ℰ_{0} sin ω t

0.2 V ∕ D I V

and time scale setting

0.05 m s ∕ D I V

. Determine

ω

and the phase angle

φ

, and the gain. From these determine

R C

\frac{L}{R}

, whichever is appropriate.

3. In a series $R L C$ circuit operating at $60.0 H z$ the maximum voltage across the inductor is $2.0$ times the maximum voltage across the resistor and $2.0$ times the maximum voltage across the capacitor. Determine the phase angle. If the maximum generator voltage is $30.0 V$ and maximum current is $300 m A$ , ﬁnd $R$ .

Note;
Due to hysteresis and radiative energy losses in the inductor, even at resonance you will ﬁnd that the voltage across the resistor is not equal to the generator voltage even though it should be theoretically. You will need to measure the voltage across the resistor at resonance and use this to ﬁnd the gain.
Keep in mind that $V_{0}$ is the magnitude of the generator voltage, and $V_{R, r e s}$ is the magnitude of the voltage dropped across the resistor at resonance, when the generator and current (and voltage across $R$ ) are in phase.

1.4 Lab report

In the lab report table, the ﬁrst column ( $T$ ) is for the period of the generator waveform, measured with the oscilloscope. The frequency of the generator signal is then $ω = \frac{2 π}{T}$ . You should adjust the function generator level to give you a full-scale waveform at resonance. At resonance the gain is one.

AC circuits


Experimenter 1		Experimenter 2

Experimenter 3		Experimenter 4

LRC response curve


$L = \underline{}$	$R = \underline{}$	$C = \underline{}$	$ω_{r e s} = \underline{}$


$V_{0} = \underline{}$	$V_{R, r e s} = \underline{}$

Gain and phase (LRC circuit)




$T (s)$	$ω (r a d ∕ s) = \frac{2 π}{T}$	$V_{R} (V o l t)$	gain	$Δ t (s)$	$φ (r a d) = ω Δ t$

Attach graphs of both the gain and phase angle $φ$ versus $ω$ here.