Take Home Exam 001
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
DC Circuit Design Challenge
The following figure represents a system used to power certain elements in a
caravan with a Solar Panel.
Circuit Element
Solar panel shunt resistance
Solar panel current delivery
Refrigerator current
consumption
Battery internal series
resistance
Battery internal voltage
LED light internal series
resistance
LED light internal voltage
drop
Symbo
l
aΩ
bA
cA
Value from student
#9408363
27 Ω
7.8 A
1.2 A
dΩ
0.18 Ω
eV
fΩ
12.8 V
7.6 Ω
gV
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;
By simplifying some of the circuit elements, the following circuit was
produced to be used for analysis
1
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
(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;
1.2=I 2−I 3 (Equation 1)
Current Source:
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=27.2 I 2+ 0.18 I 3−0.18 I 4 (Equation 2)
Mesh 4:
0=−12.8+ 0.18 I 4 −0.18 I 3 +8.2 I 4 +6.8
6=−0.18 I 3 +8.38 I 4 (Equation 3)
Solving on a Graphics Calculator for simultaneous equations, the following values
were produced;
I1
7.8
I2
7.2377
I3
6.0377
2
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
I4
0.845679
To 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
Solar Panel
P=VI
2
P=( I 2 ) × R
P=VI
P=((I 1−I 2) ×27)× I
P=15.1821 ×7.8
P=118.42
P=( 7.2377 )2 ×0.1
P=5.23843
27Ω
LED
0.18Ω
Battery
8.2Ω
Refrigerator
P=VI
2
P=( I 1−I 2 ) × R
2
P=( 0.5623 ) ×27
P=8.53689
P=VI
P=6.8 × I
P=6.8 × 0.845679
P=5.7062
P=VI
P=( I 3−I 4 ) 2 × R
P=26.9571 ×0.18
P=4.85227
P=VI
P=12.8 ×(I 3 −I 4)
P=66.4579
P=VI
2
P=I 4 × R
P=0.715173 × 8.2
P=5.86442
P=VI
P=(12.8+0.18 × ( I 3−I 4 ) )× I
P=13.7346 ×1.2
P=16.4815
∑ Powe r ¿=∑ Powe r out
PSolar Panel =PLED + PBattery + PRefrigerator +2× P0.1 Ω + P27 Ω + P0.18 Ω + P 8.2Ω
PSolar Panel =5.7062+66.4579+16.4815+ ( 2× 5.23843 ) +8.53689+ 4.85227+5.86442
118.42calculated 118.376
The power transfer from the solar panel to the refrigerator is 13.9178%
((
) )
P Refrigerator
×100
PSolar Panel
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.
The program EveryCircuit (http://everycircuit.com/app) was used to verify these
values found in part b). It’s important to mention that in this simulation and the
ones that follow, the LED is represented by a voltage difference rather than a
diode. This is done in order to reflect theoretical conditions but in real life there
would be some fluctuation in the voltage drop across the diode, and therefore
the system.
3
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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 I 3 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 Amphours 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;
Amp-hours for the Battery;
Amps × Hours
7.8 ×8=62.4
(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 shown more clearly as;
4
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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:
Mesh 3:
I
(¿ ¿ 2−I 3)+12.8
0=27.2 I 1+ 0.18 ¿
−12.8=27.2 I 1 +0.18 I 2 −0.18 I 3
(Equation 2)
0=−12.8+ 0.18 ( I 3−I 2 ) +8.2 I 3 +6.8
6=−0.18 I 3 +8.38 I 4 (Equation 3)
Using a Graphics Calculator, the following values were found;
-0.455132
I1
-1.65513
I2
0.680439
I3
I1 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;
I 3 −I 2 =0.680439−−1.65513=2.33557
These values can be verified below where current over the battery is 2.34;
5
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
When Solar Panel inactive for 16 hours, (from diagram) 2.33557 Amps run
through the battery
2.33557 ×16=37.3691
Because this value is less than what is built up during the daytime (41.5362), the
system works.
6
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
(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 3
Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4
Equation 3: 6=−0.18 I 3 +8.38 I 4
A 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 0.6I 4 and fI4 making the new equations;
Equation 1: 1.2=I 1−I 3
Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4
Equation 3: 6=−0.18 I 3 +0.78 I 4+ f I 4
Letting I4 equal 0.5 Amps
Equation 1: 1.2=I 1−I 3
Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 × 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;
It can be seen that the new voltage over the diode (represented as a voltage
difference) is 0.5A
7
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
(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.
http://www.futureelectronics.com/en/technologies/semiconductors/discretes/diod
es/scottky-rectifiers/Pages/7024714-80SQ045NG.aspx?IM=0
8
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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.
2
P=I R
P=( 0.455 )2 ×(27+0.1+0.1)
P=5.63108 W
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
=0.188003=18.8 ℑ proved efficiency
29.952
The simulation below shows the use of this diode in the system during the
daytime when the solar panel is active;
9
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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
6.03 ×8=48.24
10
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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).
If Olive decided to use the diode described in part f, this would further benefit
the system for using two lights, and this is shown in the diagrams below in the
daytime and night-time respectively.
11
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
During the night-time it the modified system would last for 19.6 hours (
48.24
16.63−19.6
=hours ). This is a 17.8593% increase (
× 100 ) which is quite
2.46
16.63
substantial and Olive should consider this option.
12
EGB120 DC Circuit Design Challenge
DC Circuit Design Challenge
The following figure represents a system used to power certain elements in a
caravan with a Solar Panel.
Circuit Element
Solar panel shunt resistance
Solar panel current delivery
Refrigerator current
consumption
Battery internal series
resistance
Battery internal voltage
LED light internal series
resistance
LED light internal voltage
drop
Symbo
l
aΩ
bA
cA
Value from student
#9408363
27 Ω
7.8 A
1.2 A
dΩ
0.18 Ω
eV
fΩ
12.8 V
7.6 Ω
gV
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;
By simplifying some of the circuit elements, the following circuit was
produced to be used for analysis
1
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
(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;
1.2=I 2−I 3 (Equation 1)
Current Source:
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=27.2 I 2+ 0.18 I 3−0.18 I 4 (Equation 2)
Mesh 4:
0=−12.8+ 0.18 I 4 −0.18 I 3 +8.2 I 4 +6.8
6=−0.18 I 3 +8.38 I 4 (Equation 3)
Solving on a Graphics Calculator for simultaneous equations, the following values
were produced;
I1
7.8
I2
7.2377
I3
6.0377
2
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
I4
0.845679
To 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
Solar Panel
P=VI
2
P=( I 2 ) × R
P=VI
P=((I 1−I 2) ×27)× I
P=15.1821 ×7.8
P=118.42
P=( 7.2377 )2 ×0.1
P=5.23843
27Ω
LED
0.18Ω
Battery
8.2Ω
Refrigerator
P=VI
2
P=( I 1−I 2 ) × R
2
P=( 0.5623 ) ×27
P=8.53689
P=VI
P=6.8 × I
P=6.8 × 0.845679
P=5.7062
P=VI
P=( I 3−I 4 ) 2 × R
P=26.9571 ×0.18
P=4.85227
P=VI
P=12.8 ×(I 3 −I 4)
P=66.4579
P=VI
2
P=I 4 × R
P=0.715173 × 8.2
P=5.86442
P=VI
P=(12.8+0.18 × ( I 3−I 4 ) )× I
P=13.7346 ×1.2
P=16.4815
∑ Powe r ¿=∑ Powe r out
PSolar Panel =PLED + PBattery + PRefrigerator +2× P0.1 Ω + P27 Ω + P0.18 Ω + P 8.2Ω
PSolar Panel =5.7062+66.4579+16.4815+ ( 2× 5.23843 ) +8.53689+ 4.85227+5.86442
118.42calculated 118.376
The power transfer from the solar panel to the refrigerator is 13.9178%
((
) )
P Refrigerator
×100
PSolar Panel
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.
The program EveryCircuit (http://everycircuit.com/app) was used to verify these
values found in part b). It’s important to mention that in this simulation and the
ones that follow, the LED is represented by a voltage difference rather than a
diode. This is done in order to reflect theoretical conditions but in real life there
would be some fluctuation in the voltage drop across the diode, and therefore
the system.
3
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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 I 3 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 Amphours 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;
Amp-hours for the Battery;
Amps × Hours
7.8 ×8=62.4
(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 shown more clearly as;
4
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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:
Mesh 3:
I
(¿ ¿ 2−I 3)+12.8
0=27.2 I 1+ 0.18 ¿
−12.8=27.2 I 1 +0.18 I 2 −0.18 I 3
(Equation 2)
0=−12.8+ 0.18 ( I 3−I 2 ) +8.2 I 3 +6.8
6=−0.18 I 3 +8.38 I 4 (Equation 3)
Using a Graphics Calculator, the following values were found;
-0.455132
I1
-1.65513
I2
0.680439
I3
I1 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;
I 3 −I 2 =0.680439−−1.65513=2.33557
These values can be verified below where current over the battery is 2.34;
5
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
When Solar Panel inactive for 16 hours, (from diagram) 2.33557 Amps run
through the battery
2.33557 ×16=37.3691
Because this value is less than what is built up during the daytime (41.5362), the
system works.
6
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
(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 3
Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4
Equation 3: 6=−0.18 I 3 +8.38 I 4
A 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 0.6I 4 and fI4 making the new equations;
Equation 1: 1.2=I 1−I 3
Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 I 4
Equation 3: 6=−0.18 I 3 +0.78 I 4+ f I 4
Letting I4 equal 0.5 Amps
Equation 1: 1.2=I 1−I 3
Equation 2: 197.8=27.2 I 2+ 0.18 I 3−0.18 × 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;
It can be seen that the new voltage over the diode (represented as a voltage
difference) is 0.5A
7
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
(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.
http://www.futureelectronics.com/en/technologies/semiconductors/discretes/diod
es/scottky-rectifiers/Pages/7024714-80SQ045NG.aspx?IM=0
8
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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.
2
P=I R
P=( 0.455 )2 ×(27+0.1+0.1)
P=5.63108 W
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
=0.188003=18.8 ℑ proved efficiency
29.952
The simulation below shows the use of this diode in the system during the
daytime when the solar panel is active;
9
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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
6.03 ×8=48.24
10
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
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).
If Olive decided to use the diode described in part f, this would further benefit
the system for using two lights, and this is shown in the diagrams below in the
daytime and night-time respectively.
11
Michael O’Brien n9408363
EGB120 DC Circuit Design Challenge
During the night-time it the modified system would last for 19.6 hours (
48.24
16.63−19.6
=hours ). This is a 17.8593% increase (
× 100 ) which is quite
2.46
16.63
substantial and Olive should consider this option.
12