EE 242 EXPERIMENT 1: ENERGY STORAGE AND TRANSFER CIRCUIT
1
PURPOSE:
• To demonstrate that inductors and capacitors can store energy.
• To determine the exchange of energy between a capacitors and an inductors
• To observe the exponential decay of stored energy in a capacitor or an inductor
• To investigate how ideally diodes behave in a circuit
LAB EQUIPMENT:
1
1
HP/Agilent E3640A DC Power Supply (or TPS-4000, Dual-Tracking DC Power Supply)
Agilent 34410A Digital Multimeter/Timer/Counter
1 HP 427A Analog Voltmeter
1 8H Inductor
1 Decade Resistance Box (min rating of 300 mA at 10 Ohms)
1 IN2615 Diode, ½ A, 600 V (or equivalent)
1 Decade capacitor box, 1 μ F/Step
1 Watch with 1 second resolution
11 Banana-to-Banana leads (to be checked out from senior project lab, 20-111)
DISCUSSION
A current i
(W
L
L
through an inductor L causes an energy W
L
to be stored in the induced magnetic field.
= ½ L i
L
2
). Similarly, a voltage v
C
across a capacitor C causes an energy W induced electric field (W
C
= ½ C v
C
2
).
C
to be stored in the
Simple Model
To illustrate the exchange of energy between a capacitor and an inductor, consider the circuit shown on the right. Before the switch is opened, the current through the inductor is at its steady state maximum value and the energy stored in
2 the inductor is W
L
= ½ L i
L
. There is no energy stored in the capacitor since the voltage across it is zero (same as the inductor which acts as a short in the steady state).
V
S
+
-
R = 10
Ω t = 0
Figure 1 i
L
L
Diode
1
Original experiment, amended and 01/15/09, John Saghri
1
C
10
μ
F

Once the switch is opened, the current in the inductor will be forced to flow to the capacitor-diode branch (diode will be in forward biased) and will charge the capacitor negatively. Thus the energy stored in the inductor transfers to the capacitor. Once the inductor’s energy completely transfers to the capacitor (i.e., the current through the inductor becomes zero and the voltage across the capacitor reaches the maximum), the capacitor starts to discharge very slowly through the large back resistance of the diode.
Detailed Analysis
V
C
For the circuit of Figure 1, assuming actual components, we will show that for t > 0, the natural response of the circuit will look like the waveform shown. The magnitude of the voltage t peak t across the capacitor will peak from zero very quickly and then drops back to zero very slowly.
V peak
Before opening the switch, the circuit is assumed to be in the steady state. That is the inductor is shorted and the capacitor has zero voltage across it.
At t = 0
-
and t = 0
+
, the current through the inductor will be
V
S
/R
. When the switch is opened, the current will flow through the capacitor and will negatively charge it very quickly (since the diode is in the forward direction with respect to initial direction of current). Once the capacitor is fully charged, the current changes direction but now faces the diode in the backward direction (which acts like a very large, e.g., >100 k Ω , resistor) and hence will take a relatively very long time to discharge.
To analyze the natural response theoretically, we separately look at the two time periods to 0 < t
t peak
(when the diode is in the forward direction with respect to current) and t > t peak
(when the diode is in the backward direction with respect to current). For 0 < t
t peak
, the equivalent circuit looks like the circuit shown on the right. Here
R
F
represents the total forward direction resistance of the diode and the internal resistances of the inductor and the wires (
R
F
is typically
<5 Ω ). Note that in the actual circuit that we construct i
L
L
R
F
C
10
μ
F
2

in the lab, there is an analog voltmeter connected across the capacitor. However, for simplicity, since its internal impedance is large ( ≈ 10 M Ω ) we do not include it in the circuit. Also note that at t =0
+
,
I
L
=
V
S
/R
and
V
C
= 0.
V
C
Natural response for period
0 < t
t peak
Since the value of
R
F
is very small, the natural response of the series RLC circuit will be under-damped. This is because the damping rate α
= R
F
/ 2 L will likely be less than the resonant frequency ω o
= 1 / ( L C )
½ (for the selected values of L and C) . Hence the voltage across t peak t t t the capacitor will start
to be a damped oscillatory waveform as shown. Since α
= R
F
/ 2 L the damped frequency of oscillation ω
is very small, d
= (
ω o
2
-
α 2
) ½ will tend to be very large. This means the time it takes for the capacitor to reach its peak, i.e., t peak the graph), will be very short (i.e., t peak
, (shown on
0 ).
Now consider the period t > t peak
(when the diode is in the backward direction with respect to current).
V peak
V
C Natural response for period t > t peak t
Here we have the same RLC circuit as before except we replace the
R
F with the backward resistance of the diode
R
R
. Unlike
R
F.
,
R
R
is very significant (e.g., >
100 k Ω ) . In this case the circuit natural response will likely be over-damped since α > ω o
. The natural response will be the sum of two exponential decays of t peak t t the form;
V peak
V
C
(t) = k
1
e
- t /
τ
1
+ k
2
e
- t /
τ
2
where s with respective time constants of
1
= -
α
+ (
α 2
-
ω o
2
) ½ and s
2
=
τ
1
=
-
1 / s
-
α
(
α
1
2 and τ
-
ω o
2
2
=
-
1 / s
2
,
) ½ .
The magnitude of V
C
will decay to zero very slowly (large time constant) as shown in figure on the right.
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PRELAB
In the circuit shown (constructed of ideal elements), the inductor current at t = 0
+
was
2.0 A. The switch was opened at t = 0 sec and the voltage across the capacitor, V
C
, quickly reached a maximum (negative) value of
-
10
V
S
V. After an elapse of 250 sec, V
C
dropped down to
-
5 V. Determine the value of L, the
+
-
R = 10
Ω t = 0 i
L
L
Diode
C
10
μ
F back resistance of the diode, and the time constant ( τ = RC ), of the capacitor discharge circuit.
For simplicity, since the capacitor charges very quickly after opening the switch, you can assume that it peaks instantly after opening the switch, i.e., t peak
0
0
. Note that the conservation of energy principle always holds. That is the energy stored in the inductor at t = 0
, i.e.,
W
L
= ½ L i equal to the energy stored in the capacitor at t
+ , i.e.,
W
C
= ½ C v
C
L
2 , is
2 , Also, since the capacitor discharges very slowly after reaching its peak, you can assume that the inductor is shorted at hence
t /
τ the natural response approximates that of a simple RC circuit i.e.,
V
C
= V i
e
where
V i
is the initial value of the capacitor voltage and τ is time constant RC. R represents the backward resistance of the diode.
EXPERIMENTAL PROCEDURE
Safety Caution!!
The inductor and capacitor used in this experiment are capable of storing and discharging dangerous amounts of energy. The hazard is negligible provided that you adhere to the circuit specifications and changes outlined in the procedure.
Part a.
Determining the inductance L via measurement of its transferred energy to a capacitor
1.
Construct the circuit shown on the right. Use the Agilent/HP DC
Digital
Voltmeter power supply for the source V
S
and initially set it to zero volts. Leave leads A and B open as shown.
+
V
R
R = 10
Ω
-
Lead A i
L
1
Lead B
Diode +
Momentarily place a short across the capacitor to discharge its initial charge, if any.
V
S
+
-
L
10
C
μ
F
-
V
C
Analog
Voltmeter
2
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2.
Connect leads A and B. Gradually increase the power supply voltage control clockwise to achieve 50 mA inductor current. You should be able to measure i
L
indirectly via measuring
V
R
(remember that i
L
= i
R
= V
R
/R).
3.
Quickly disconnect lead A and record maximum V
C
via the attached analog voltmeter (use the
-
300 or
-
100 V scale). Repeat this step three times and average your result.
4.
Repeat step 3 for i
L
= 50, 75 mA, and 100 mA. Enter your results in a table as shown below. i
L mA
50
75
V
C volts
(trial 1)
V
C volts
(trial 2)
V
C volts
(trial 3)
V
C volts
Average
W
C mJ
L
H
100
5.
Calculate the maximum energy stored in the capacitor for each of the selected current values.
Recall that
W
C
= ½ C v
C
2 . Enter the results in the table you have constructed.
6.
Assuming complete energy transfer, calculate the inductance of the inductor for each value of the current. Enter the results in the table.
7.
Compare the calculated values of L to the value marked on the inductor (nominal value by the manufacturer) and to the measured value with the impedance bridge
at 1000 (zero dc bias). State the reasons for differences between the calculated, nominal, and the bridge values of L.
Part b.
Determining the analog voltmeter resistance
1.
Discharge the capacitor by momentarily connecting a shorting lead across it.
2.
“Charge”, i.e., “energize” the inductor with a current that will produce a maximum value of
V
C
in the 110
-
130 V range.
3.
Quickly disconnect lead A from point 1 (TOUCH ONLY INSULATED PARTS of the circuit) and then disconnect lead B from point 1 to prevent charge leakage through the back resistance of the diode (now the discharge of C is through the voltmeter only). As the capacitor discharges, make several measurements of V
C
and corresponding values of time
(with the aid of a watch). You should have about ten equally-spaced measurements
(samples) spanning v
C
from peak its value to zero.
4.
Select three pair of time-voltage values, e.g., [v
C
(t
1
), t
1
] , [v
C
(t
2
), t
2
], and [v
C
(t
3
), t
3
] from your measurements in step 3 above. For each pair, calculate the time constant τ of the RC
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discharge circuit and the voltmeter’s internal resistance R. Note that for an RC discharge circuit, v
C
(t) = v
C max
e –t/
τ
. In this equation v
C max
= v
C
(0
+
) which you measured in step 2.
Using this equation, you should be able to find τ using any one of the three time-voltage pairs. The voltmeter resistance R can then be found via τ = RC. Enter your calculated values of τ and R in the table below.
τ ms
(trial 1)
τ ms
(trial 2)
τ ms
(trial 3)
τ ms
Average
R ohms
(trial 1)
R ohms
(trial 2)
R ohms
(trial 3)
R ohms
Average
5.
Using your measurements in steps 2 and 3, plot V
C
as a function of time. This curve represents the discharge curve for the capacitor through the voltmeter’s impedance
(resistance).
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