1 PSU-Physics PH-315 Andrés La Rosa RLC SERIES CIRCUIT RESONANCE (Complex impedance) _______________________________________________________________________________ PURPOSE To observe the frequency-dependence of impedance in an alternating current (AC) circuit. We will measure the resonance frequency and use its value to determine the inductance of a coil, assuming the values of the capacitance and the resistor. The oscilloscope will be used to measure the phase lags between voltage and the current across a resistor, a capacitor and an inductor. The analysis is undertaken using complex variable. THEORETICAL CONSIDERATIONS I. Electrical components Resistance Capacitor complex impedance Inductor complex impedance II. Kirchhoff law for the RLC circuit III. How to calculate and measure current amplitude Io() and phase lag () ? Complex impedance of the RLC-series circuit Calculation of Io and IV. Resonance Resonance condition Relative orientation of the phasors input voltage vA and current response i at different values of Example: Analysis at resonance and out of resonance conditions EXPERIMENTAL CONSIDERATIONS Experimental setup Measurements a) Finding L b) Plot o I as a function of . c) Plot the experimental values of | z | = Zo = o oA I V as a function of d) Plot the experimental values of the phase vs frequency, = () e) Influence of the resistance (Repeat the experiments using at least two different values) f) Measurement of the complex voltages across R, C and L.
PURPOSE To observe the frequency-dependence of impedance in an alternating current (AC) circuit. We will measure the resonance frequency and use its value to determine the inductance of a coil, assuming the values of the capacitance and the resistor. The oscilloscope will be used to measure the phase lags between voltage and the current across a resistor, a capacitor and an inductor. The analysis is undertaken using complex variable.
The peculiar characteristic is in this expression is that oI and depend on the frequency ;
that is, )( oo II and )(
Given C, L and R, as well as VoA, we have to figure out the values of Io and .
Resistance
Voltage across the resistor: i R Rv (3)
Capacitor complex impedance Since the current i is the same across each element, we will express their drop of voltage in terms of that current.
Voltage across the capacitor:
On one hand: C
q Cv , which implies
C
i
dt
d
Cv (i)
On the other hand: The voltage across the capacitor should be changing also
harmonically at the frequency ; like t o )(
jeCvvC .
This implies,
e j dt
d j t o)(
CC v
v
j dt
dC
C vv
(ii)
Equating the two expressions (i) and (ii), one obtains,
C
i j Cv ;
Or, equivalently,
i Cj
1
Cv (4)
Since )
1 /2j-e
j
i eC
/2j-
)
1
Cv Voltage across the capacitor is
lagging 90o with respect to the current.
4
Exercise: Assuming )(
-tjeIoi , draw the phasors Cv and i ( together in the
same diagram) for arbitrary fixed values of , , and t.
In (4), the expression
Cj
zC
1
(5)
is called the capacitance complex impedance. Thus, one re-writes expression (4) as,
i z CCv (6)
Notice in (5) that C
z changes with the frequency .
Inductor complex impedance Voltage induced across the inductor:
On one hand: dt
i dL Lv
On the other hand: Since the current is assumed to be changing according to
)( -tjeIoi
Then
)( -tjeI j L
dt
i dL o Lv
i j L
Thus,
i jL Lv (7)
Since )/2jej
i eL /2j ][ )
Lv Voltage across the inductor
leading 90o with respect to the current.
Exercise: Assuming )(
-tjeIoi , draw the phasors Lv and i (together in the
same diagram) for arbitrary fixed values of , , and t.
The expression Lj zL (8)
is called the inductance complex impedance.
5
i z L
Lv (9)
Notice L
z changes with the frequency .
II. Kirchhoff law for the RLC circuit
Consider the circuit shown in Fig. 1 Kirchhoff law states that, at any instant of time, the addition
of the three voltages Rv , Cv , and Lv should be equal to vA..
Av Rv + Lv + Cv
i R + i jL + i Cj
1
iCj
LjR ]
1 [
Av . (10)
Experimental assignment: Place the sinusoidal driving voltage in channel-1 of the oscilloscope. In channel-2 monitor, one at a time, the voltage across R, C, and L, respectively. Plot all the waveforms in a single graph, for comparison.
One can write expression (10) more explicitly,
)(
]
1 [ V
-tt jj
eICj
Lj ReA oo
Cancelling the factor t je ,
]
1 [ V
j-eI
Cj Lj RA oo , (11)
where oI and are still unknown.
Equivalently,
]
1[ ][ ][ V
j-j-j-eI
Cj eI Lj eI RA oooo
(12)
2/2/
]
1[ ][ ][ V
jj-jj-j-
eIC
eI L eI R oooAo
where )( oo II and )(
Expression (12) is very convenient for answering the questions that appear in the experimental section below [in particular see the requirement f ) in that section].
6
For a given , you have to measure both Io and , and subsequently verify that those experimental values fulfill expression (12).
Notice, the left side in expression (12) is a real quantity; therefore the right side has to be real as well (i.e. when entering the experimental values the right side has to be a real number).
III. How to calculate and measure current amplitude Io() and phase lag () ? We have to calculate Io and in terms of R, L, C, and VoA . It is convenient to calculate first the total complex impedance of the circuit.
Complex impedance of the RLC-series circuit In expression (10),
iCj
LjR ]
1 [
Av ,
we identify the total impedance of the circuit iz / Av as,
) (
ωC
1L jR
Cj
1LjRz
(13)
Total impedance RLC series circuit.
Exercise. Show that z can alternatively be written in the form,
jeZz o (14)
where
1
arctanR
CL
and
22 )
1 (
CLRZ
o
7
R 1/(C)
Fig. 3 Representation in the complex plane of the total
complex impedance z, given by expression(14).
L L– 1/(C)
Im
Real
Calculation of Io and
From expression (11)
]
1 [ V
j-eI
Cj Lj RA oo
Using (13) and (14),
][V-j
eIz A oo
] [V-jj
eIeZA ooo
Since VoA is a real number, the right side of the last expression has to be real as well. This requires that:
has to be equal to . (15)
The latter also implies that IoA = VoA / Zo.
That is,
o
oA
ZI
V)( o =
22 )1
(
V
CLR
oA
and (16)
)
1 (
arctan)(R
LC
8
In summary, for the RLC-series circuit in Fig. 1 above,
t j
eV oAAv , )
( - tjeoIi (17)
where )(
)1
(
V
2
2
oo I
CLR
I
oA
)
1 (
arctan)(R
CL
IV. Resonance
Resonance condition Notice in (14) and (16) that,
when )/(1 CL ,
the impedance oZ is minimum, and the current oI is maximum
Hence,
LCo
1 (18)
is called the resonance frequency.
Fig. 6 Resonance curve (from expression (16).
Relative orientation of the phasors vA and i at different values of
At very low frequencies: 0
-
)
)( /2 (
tjeoIi (current i leads the voltage vA)
9
at LC
o
1 : =
)
)( ( tjeoIi
o (current i in phase with the voltage vA)
at very high frequencies
)
)( /2 (
tjeoIi (current i lags the voltage vA)
vA
vA
vA
i i
~ 0 ~ 0 ∞
v
i
i
Real
Im Im Im
Real
Fig. 4 Phasor vA and phasor i rotate with the same frequency. Their phase difference
remains constant at fixed frequency, but changes as the frequency varies. Here t j
eV oAA v
A similar diagram is shown in Fig. 5.
Fig.3 Notice, the phase is measured relative to
the voltage phasor
vA
i
Case: <o Case: >o < 0 > 0
vA
i
Real Real
Im Im
Fig. 5 The phasor voltage and the phasor current rotating counterclockwise. Left: The current leads the voltage. Right: The current lags the voltage.
Example: Analysis at resonance and out of resonace conditions (Taken from James Brophy, “Basic Electronics for Scientists,” McGraw Hill).
10
Calculate the current in the circuit and the voltages across each component for the circuit in Fig. 7. Consider a) the case of resonance, and b) a case out of resonance.
L= 250 mH
R= 100
C= 0.1 F
10V
i
Fig. 7 RLC series circuit.
A
Fig. 8 Resonance curve of circuit 7.
Io (mA)
The input voltage is Av 10 Volts t je
After substituting the values for R, C and L into Eq. (17),
)1
(
V )(
22
CLR
Io
oA , it is
found that the current in the circuit changes with frequency as illustrated in Fig. 8. The maximum current is I0.max = 10/100 = 0.1 Amps (which occurs at the resonance frequency).
Hz1006 101.01025028.6
1
2
1
6-3-
LCfo
a) Case of resonance
The voltage drop across each element at resonance illustrates an important feature of alternating currents. Let’s see.
Current at resonance: I = 0.1 Amp
Voltage amplitude across the resistor: VR = R I = 100 0.1 A = 10 Volts
Voltage amplitude across the capacitor: VC = C
1
I =
F1 7-10 Hz006 2
1
0.1 A = 158 Volts
Voltage amplitude across the inductor: VL = L I = Hz006 2 1 0.25 H 0.1 A = 158 Volts
It is evident the amplitude voltage drops around the circuit do not add up to zero.
Notice however that,
11
According to expression (4), the phase angle at the capacitor voltage Cvis -90o with respect to the current.
According to expression (7), the phase angle of the voltage across the inductor is, +90o with respect to the current.
According to expression (16), the phase difference between the applied
voltage and the current is zero; = 0.
From (17), the current at resonance (f=1006 Hz) is: i = 0.1 Amp t j e
Voltage across the resistor: vR = R i = 100 i = 10 Volts t j e
Voltage across the capacitor: vC = Cj
1
i = j 158 Volts t j
e
Voltage across the inductor: vL = Lj i = j 158 Volts t j e
Notice, the instantaneous voltage drop across the two reactances (capacitor and inductor) cancel each other out.
Thus , adding the three voltages vR + vC + vL one obtains 10 Volts t j e ,
which equals the input voltage Av 10 Volts t je .
This confirms the Kirchhoff law.
Kirchhoff’s voltage rule is valid when both the magnitude and phases of the currents and voltages are taken into account.
b) Case: Selecting a frequency different than the resonance frequency Of course Kirchhoff’ law is equally valid at frequencies away from resonance. Consider the voltage drop around the circuit in Fig. 7 at a frequency of f= 1200 Hz.
According to expression (17), at f=1200 Hz the phase angle is,
)0 12002
1 ) 25.012002(
arctan7-
100
1
3261 8851 arctan
100 = .58655 arctan = 79.85o or 1.39 radians.
According to expression (17), at f=1200 Hz the current is,
)
)( (
- tjeoIi =
)(
tj -j eeoI
= 22 )
1 (
V
CLR
oA -j e
tj e
12
= 22 )13261885(00
10
1
.391 -j e
tj e
= 9.567
10 .391 -j e
tj e
= 1.76 10-2 .391 -j e
tj e
= 1.76 10-2 [ cos(1.39) - j sin (1.39) ]
tj e
Voltage across the resistor:
vR = R i = 100 i = 1.76 Volts [ cos(1.39) - j sin (1.39) ] t j e
= [ 0.31 - j 1.73 ] t j e
Voltage across the capacitor:
vC = Cj
1
i = j
7-0 12002
1
1 i =
/2-j e
1326 i
= /2-j
e 1326 1.76 10-2 .391 -j e t j
e
= 23.3 2.96-j e
tj e
= 23.3 [ cos(2.96) - j sin (2.96) ]
tj e
= [ -22.91 - j 4.2 ] t j e
Voltage across the inductor:
vL = Lj i = j ) 25.012002( i = /2j e
1885 i
= /2j e
1885 1.76 10-2 .391 -j e t j e
= 33.17 .18j e
0
tj e
= 33.17 [ cos(0.18) + j sin (cos(0.18) ]
tj e
= [ 32.63 + j 5.9 ] t j e
Rv + Lv + Cv = { [ 0.31 - j 1.73 ] +
[ 32.63 + j 5.9 ] +
[ -22.91 - j 4.2 ] }Volts t j e
13
= { 10.03 - j 0.03 } t j e
10 Volts t j e
which is equal to Av 10 Volts t je , thus verifying Kirchhoff law
In summary, for two different driving frequencies, f = 1006 Hz and 1200 Hz, we have verified that the Kirchhoff’s voltage rule is valid for the circuit in Fig. 7.
EXPERIMENTAL CONSIDERATIONS
Experimental setup
Build the circuit shown in Fig. 9. A two-channel oscilloscope is our basic measuring tool in alternating current (AC) circuits.
The resistor Ro is a known resistor (whose value is on the order of 100 ).
The ground of the oscilloscope will be connected to G.
Channel-1 will monitor the input voltage at A.
Channel-2 will monitor the voltage at B.
B
G
C= 10 nF
L
Ro
A
Ground
t A
jeVoAv VA(t) = Real { Av }
= Real { t
jeVoA }
=VoA cos (t )
VoA is a constant real-value. (It is the amplitude of your sinusoidal input voltage).
)(
-tj
eI oi I(t) = Real { i }
= Real { )(
-tjeIo }
= Io cos (t-)
Io = Io() is a real-value.
Fig. 9 RLC series circuit.
Suggestion: use C~ 10 nF , 100 nF (or higher) for this experiment.
Note: If a capacitor is, for example, labeled 473G, it means C = 47 103 picoFarads.
Pico means 10-12.
14
Measure the frequency of the input voltage using the oscilloscope. (Do not trust the reading from the signal generator knob.)
In Fig. 9 the voltage at B is the voltage across the resistor VR = VR (t). This is the trace you see in your oscilloscope. Since VR(t) = R0 I (t), then the current I= I (t) can be obtained.
R
tVtI
)()( R (18)
The voltage at A is the driving voltage VA = VA(t). It is the trace you see in your oscilloscope. Measurements
a) In order to compare your results to the theory, determine first the value of the inductance
L. This is best done early on by locating the resonance frequency =o, the one that
makes i) the impedance Zo minimum, ii) the phase between VA and VR equal to zero, and iii) the current I0 maximum. (See expressions 14 and 15 above.)
Select values of around o (that is the frequency region where a majority of your data need to be taken).
b) Plot oI as a function of .
(At the bottom of this file see the suggestion about how to overcome the unwanted variability of the input voltage amplitude as the frequency changes).
c) Plot the experimental values of | z | = Zo as a function of .
Here Zo is the magnitude of the complex impedance z of the RLC circuit (see expression 14).
Zo, is determined experimentally from the ratio of amplitudes of the two signals VA(t), and I(t) (see expression (15):
| z |= Zo = )(I
)(VA
t
t
signal the of Amplitude
signal the of Amplitude=
o
oA
I
V
Compare your experimental results with the ones obtained using expression (14) above.
Plot the theoretical (curve) and experimental values (discrete points) of Zo as a function of
(both in the same graph, for comparison).
d) The phase can be measured on the oscilloscope as a distance between the points where the two traces cross the horizontal axis, and converted to degrees by comparing the half (or full) wavelength as shown on the oscilloscope. See figure 10 below.
Pay attention during the measurements to verify if is positive or negative. That is, whether the VR is lagging or ahead of VA.
Plot the experimental values of the phase vs frequency, = ().
Plot also the theoretical values for the phase predicted by expression (17).
15
vA
vR
180o
VA
time
Phasors vA and vR rotating with
angular velocity vR lags vA by .
vR
t
VA =VoA Cos (t); VR = VoR Cos (t - )
The traces show VR lagging VA by .
Real voltages Complex analysis measured by the oscilloscope Phasors
Fig. 10 Methodology to measure the phase using the oscilloscope.
e) During the course of measurements you take enough data to make a graph of both
impedance Zo and phase as a function of frequency . (as requested above).
How do the graphs change when using a higher or lower value of the resistance Ro? Repeat the experiments using at least two different values
f) Measurement of the complex volages across R, C and L. By swapping C with R, and L with R, it is possible to measure the voltage across C and
across L respectively.
f1) Measurement at resonance condition:
Measure the amplitude and phase (relative to vA) of the current i. Express the
experimental value of i in phasor format.
Measure the amplitude and phase (relative to vA) of the individual voltages vc, vL, vR. Express their experimental values in phasor format. Verify if Kirchhoff law is fulfilled
Verify if i Cj
1
Cv .
Verify if i jL Lv
f2) Measurement at out of resonance condition
Choose a frequency * at which Io() is ~50% of the current obtained at resonance.
Measure the amplitude and phase (relative to vA) of the current i. Express the
experimental value of i in phasor format.
Measure the amplitude and phase (relative to vA) of the individual voltages vc, vL, vR. Express their experimental values in phasor format.
16
Verify if Kirchhoff law is fulfilled.
Verify if i Cj
1
Cv .
Verify if i jL Lv
_______________________________
OVERCOMING some SHORTCOMINGS ENCOUNTERED in the EXPERIMENT Problem: Variability of the input voltage amplitude
Solution: By the normalization method.
While sweeping the frequency of the driving voltage VA around the resonance frequency, it is observed that the amplitude VoA also changes. Ideally, it would be desirable that this amplitude remains constant.
The instability of the VoA makes somewhat inaccurate the procedure of locating the resonance frequency by monitoring the frequency at which the current is maximum. (When you do this, you may find that at the frequency where the current is maximum, VA(t) and I(t) are not in phase.) Still this is still a good preliminary step, since it helps to identify the frequency range where we have to take a closer look.
Next, sweep the frequency a bit, until you find that the phase between VA and I is zero.
Now we suggest to manually normalize the input voltage VA. By this we mean to choose a fixed amplitude value for VA, let’s call it VA, fixed. Then:
For a given frequency, record the current amplitude and phase.
Move to another frequency (you may notice that the amplitude of VA has changed.) Turn the knob that controls the amplitude of VA until you obtain back the predetermined fixed value VA, fixed. Then record the current amplitude and phase.
And so on, repeat the procedure for each frequency around the resonance frequency.
