DC Circuit Design Challenge
The following figure represents a system used to power certain elements in a caravan with a Solar Panel.
Circuit Element Symbo
l Value from student #9408363 Solar panel shunt resistance a Ω 27 Ω
Solar panel current delivery b A 7.8 A Refrigerator current
consumption c A 1.2 A
Battery internal series
resistance d Ω 0.18 Ω
Battery internal voltage e V 12.8 V LED light internal series
resistance f Ω
7.6 Ω LED light internal voltage
drop g V 6.8 V
(a)Draw a circuit diagram of the system described above, with the LED light switch closed. Where possible, simplify the diagram by noting series and parallel resistances and sources.
From the original figure above a circuit diagram can be formulated;
(b)Using a circuit analysis technique of your choosing, calculate the power balance of the system. Show all working. Comment on the effectiveness of power transfer from the solar panels to the refrigerator, battery, and LED light.
A mesh analysis can be performed over four sections, current in all of which assumed to be moving clockwise.
Due to the current source, a supermesh was used (outlined with the dotted line)
Analysis of the each of the meshes produced 3 equations to be solved simultaneously;
Current Source:
1.2
=
I
2−
I
3 (Equation 1)Supermesh:
0
=
0.1
I
2+
27
I
2−(
27
× I
1)+
0.1
I
2+
0.18
I
3−
0.18
I
4+
12.8
197.8=27.2I2+0.18I3−0.18I4 (Equation 2) Mesh 4:
0
=−
12.8
+
0.18
I
4−
0.18
I
3+
8.2
I
4+
6.8
6=−0.18I3+8.38I4 (Equation 3)
Solving on a Graphics Calculator for simultaneous equations, the following values were produced;
I
4 0.845679To complete a power balance over the system, power over all parts needs to be found, Power across the Solar Panel will equal the sum of all other parts.
Resistors 0.1Ω
P=VI
P
=
(
I
2)2× R
P
=(
7.2377
)
2×
0.1
P=5.23843
Solar Panel P=VI
P
=((
I
1−
I
2)
×
27
)
× I
P
=
15.1821
×
7.8
P=118.42
27Ω
P
=
VI
P
=
(
I
1−
I
2)
2× R
P=(
0.5623)
2×27P
=
8.53689
LED
P
=
VI
P=6.8× IP
=
6.8
×
0.845679
P=5.7062
0.18Ω P=VI
P
=
(
I
3−
I
4)
2× R
P
=
26.9571
×
0.18
P=4.85227
Battery P=VI
P
=
12.8
×
(
I
3−
I
4)
P
=
66.4579
8.2Ω
P
=
VI
P=I42× RP
=
0.715173
×
8.2
P=5.86442
Refrigerator
P
=
VI
P=(12.8+0.18×
(
I3−I4)
)× IP
=
13.7346
×
1.2
P=16.4815
∑
Powe r
¿=
∑
Power
outPSolar Panel=PLED+PBattery+PRefrigerator+2× P0.1Ω+P27Ω+P0.18Ω+P8.2Ω
P
Solar Panel=
5.7062
+
66.4579
+
16.4815
+
(
2
×
5.23843
)+
8.53689
+
4.85227
+
5.86442
118.42
calculated118.376
The power transfer from the solar panel to the refrigerator is 13.9178%
(
(
P
RefrigeratorP
Solar Panel)
×
100
)
leaving 4.81861% to the LED and 56.1205% to the Battery. This means that the circuit and wire resistances are dissipating a quarter of the input energy through heat. Any reduction in these could improve the system(c)Use circuit simulation software to verify your calculations in (b). Show screen shots of the software that verify your results.
Over the 0.1Ω resistors (I2) 7.24 in the simulation ~ the calculated 7.2377
Over the 8.2Ω resistors (I4) 0.846 in the simulation ~ the calculated 0.845679
For, I3 the simulation wouldn’t show I3 alone on the wire but can be verified by
using the calculated values (I3-I4) to get current over the battery which gives
5.19202 ~ 5.19 on the simulation. Therefore I3 is verified.
(d)Olive mostly leaves her LED light turned on, day and night. During a sunny day, the solar panel delivers its rated current for 8 hours. How many Amp-hours are stored in the battery? (Initially assume the battery is discharged enough overnight to accept all of the next day’s charging). When the sun is not shining for the other 16 hours, the solar panel produces no current, but its shunt resistance remains connected. How many Amp-hours are drawn from the battery during this time? Comment on your result. Show all working. Use circuit simulation software to verify your calculations, supported by screen shots.
Amp-hours for the Solar Panel;
Amps × Hours
7.8×8=62.4
Amp-hours for the Battery;
(
I
3−
I
4)
×
8
5.192021
×
8
=
41.5362
During the night-time, the Solar Panel is inactive in the circuit and the current runs as shown below;
This can be simplified as follows, with 3 meshes used (all assumed clockwise)
From a mesh analysis, the following equations were determined; Current Source:
1.2
=
I
1−
I
2 (Equation 1)Supermesh:
I
(
¿¿
2
−
I
3)+
12.8
0
=
27.2
I
1+
0.18
¿
−12.8=27.2I1+0.18I2−0.18I3 (Equation 2)
Mesh 3:
0
=−
12.8
+
0.18
(
I
3−
I
2)+
8.2
I
3+
6.8
6=−0.18I3+8.38I4 (Equation 3)
Using a Graphics Calculator, the following values were found;
I
1 -0.455132I2 -1.65513
I
3 0.680439I1 and I2 are only negative because the current is flowing in the opposite direction
to what was assumed whilst doing the mesh analysis. Current through the battery;
I3−I2=0.680439−−1.65513=2.33557
When Solar Panel inactive for 16 hours, (from diagram) 2.33557 Amps run through the battery
2.33557×16=37.3691
(e)During the day the LED light is very hot and you know this will significantly shorten its lifetime. You suggest changing resistor f in the LED light to limit the current in the LEDs to 0.5 A during the day. Calculate the new value of resistor f. Show all working.
Using the simultaneous equations found from the initial mesh analysis Equation 1:
1.2
=
I
1−
I
3Equation 2: 197.8=27.2I2+0.18I3−0.18I4 Equation 3:
6
=−
0.18
I
3+
8.38
I
4A new value for resistor f (7.6) by first separating the three resistors around the diode (0.3, 0.3 and f) in equation 3 to 0.6I4 and fI4 making the new equations;
Equation 1: 1.2=I1−I3
Equation 2:
197.8
=
27.2
I
2+
0.18
I
3−
0.18
I
4 Equation 3: 6=−0.18I3+0.78I4+f I4 Letting I4 equal 0.5 AmpsEquation 1:
1.2
=
I
1−
I
3Equation 2: 197.8=27.2I2+0.18I3−0.18×0.5
Equation 3:
6
=−
0.18
I
3+
0.78
×
0.5
+
f ×
0.5
Using a Graphics Calculator, the following value was found;
f=13.3928
Therefore the new value of the resistor is 13.3928Ω
This resistor, when simplified and summed with both 0.3Ω resistors in line on the diode, giving 13.9928Ω, the near value for f can be verified below from the circuit simulator;
(f) Show Olive how a diode can be used to improve the efficiency of the system during the night. Quantify the efficiency gains made from your design using either circuit simulation software (supported by screen shots) or manual circuit calculations. Look online for an appropriate diode to use in this application. Ensure that your diode model in the circuit is equivalent to the real diode that you have chosen.
A diode may be used to stop current from flowing through the shunt resistor when the solar panel is not in use. Power is then not lost across a resistor that is serving no practical purpose when the sun is not shining. The diode would be placed between the shunt resistor and the refrigerator as shown below;
The diode below would be effective in increasing efficiency as it allows 8A
(greater than 7.2377A moving) to flow through it. With a voltage drop of 0.55 this diode will still allow the battery to charge up substantially during the day.
The efficiency gains that would result can be calculated by
Efficiency Gains
=
Power dissipated
∈
element
Total power
∈
the system
To find the current that is flowing through the shunt resistor without the diode, a circuit simulator was used as shown above.
P=I2R
P
=(
0.455
)
2×
(
27
+
0.1
+
0.1
)
P=5.63108W
When the Solar Panel is inactive in the system, in the same way as d) the current is moving anti-clockwise, anti-clockwise and clockwise respectively for the
meshes. The ‘Power in’ still equals ‘Power out’ at this state an ‘Power in’ is now coming from the battery. This is calculated below;
P
=
VI
P=12.8×2.34
P
=
5.63108
W
Using the efficiency gains calculation
Efficiency Gains
=
5.63108
29.952
=
0.188003
=
18.8
ℑ
proved efficiency
The diode would slightly reduce the current through the battery (this is considered negligible in the efficiency calculations).
The next simulation, shows the system at night-time when the Solar Panel is inactive;
It can be seen the current from the battery is now much less than without the diode, allowing the battery to last much longer.
(g)Olive asks if she could have a second identical LED light (both resistor and diodes) in parallel with the first if she bothers to turn them off during the day. Assume that the meaning of “day” here are the 8 hours during which the solar panel delivers its rated current. Calculate how many night-time hours Olive can run the two lights (with the original resistor value) and still run her refrigerator 24 hours a day.
The following circuit was created on the simulator. The 8.2Ω resistor was split up into two 0.3Ω resistors and one 7.6Ω resistor so that another could be added in parallel. Two switches were put in to turn off during the day.
It can be seen that the battery current is increased which is good for charging should Olive remember to turn the LEDs off. To calculate how long the lights will work for, the new daytime battery amp-hours can be calculated.
Amps × Hours
The following diagram shows the system at night without the Solar Panel
It can now be seen that the battery current is 2.9A, the number of hours this can be sustained for is calculated as;
Amps × Hours
2.9×hours=48.24 therefore; hours=16.63
Because this value is greater than the 16 hours of night-time, using the second light will be okay. However, it is important to consider that because the lights will only last a little bit longer than 16 hours if it was a cloudy day it would be
recommended that only one light should be turned on at night. This is why having a second switch would be useful (not to mention if one of the lights break).
During the night-time it the modified system would last for 19.6 hours (
48.24
2.46
=
hours
). This is a 17.8593% increase (16.63
−
19.6
