EE-3352: Fundamentals of Electric Power Engineering
Prof. Gosney
Spring 2015
Homework #4
For Friday, January 30, 2015, read Chapters 6 (pages 99-133) Solar Energy
For Wednesday, February 4, 2015:
Chapter 6 Questions: 6.1, 6.6, 6.14, 6.15, 6.16, 6.17, 6.18, 6.19; (pages 209-210),
6.1 A solar panel consists of four parallel columns of PV cells. Each column has 10 PV cells in
series. Each cell produces 2 W at 0.5 volts. Compute the voltage and current of the panel?
Panel voltage = 10 x 0.5 volts = 5.0 volts.
Panel current = 4 x 2W/0.5 V = 16 Amps
6.6 An area located near the equator has the following parameters: αdt= 0.82; αp= 0.92;
βwa= 0.06. The solar power density measured at 11:00 AM is 890 W/m². Compute the solar
power density at 4:00 PM.
We know,
𝑃𝑚𝑎𝑥 = 𝑃0 cos 𝜀 (𝛼𝑑𝑡 − 𝛽𝑤𝑎 )𝛼𝑝
= 1353 cos(𝑜) (0.82 − 0.06)0.92
𝑊
= 946.02 2
𝑚
𝑃 = 𝑃𝑚𝑎𝑥 𝑒
−(
(𝑡−𝑡0 )2
)
2𝜎2
890 = 946.02 𝑒
𝜎 = 2.84
𝑃 = 𝑃𝑚𝑎𝑥 𝑒
−(
−
(11−12)2
2(𝜎)2
(𝑡−𝑡0 )2
)
2𝜎2
−
(16−12)2
2(2.84)2
= 946.02 𝑒
𝑊
= 350.86 2
𝑚
6.14 A solar cell with a reverse saturation current of 1nA is operating at 35o C. The solar
current at 35o C is 1.1 A. The cell is connected to a 5ohm resistive load. Compute output
power of the cell.
Solution:
We know, Reverse Saturation current (𝐼0 ) = 1 nA
Solar current I = 1.1 A
Resistance = 1 ohm
𝑉𝑇 =
𝐾𝑇 1.38 × 10−23 × (35 + 273)
=
= 0.0265 𝑉
𝑞
1.6 × 10−19
𝑉
𝑉 = 𝐼𝑆 𝑅 − 𝐼0 𝑅 (𝑒 𝑉𝑡 − 1)
𝑉
𝑉 = (1.1 × 5) − (5)(10−9 ) (𝑒 0.0265 − 1)
𝑉 = 0.549 𝑉
𝑃=
𝑉 2 (0.549)2
=
= 60.28 𝑚𝑊
𝑅
5
6.15 A solar cell of reverse saturation current of 1 nA has a solar current of 1.1 A. Compute
the maximum output power of the cell per unit of thermal voltage.
Solution:
We use,
(1 +
𝑉𝑎𝑚𝑝 𝑉𝑎𝑚𝑝 𝐼𝑠 + 𝐼0
𝑉𝑎𝑚𝑝
) 𝑒 𝑉𝑇 =
; 𝑤ℎ𝑒𝑟𝑒
=𝑋
𝑉𝑇
𝐼0
𝑉𝑇
(1 + 𝑋) × 𝑒 𝑋 = 1.1 × 109
𝑋 = 17.88
6.16 For the solar cell in the previous problem, compute the load resistance at the
maximum output power.
Calculate ID = (1x10-9)(e17.88 – 1) = 58.24 mA.
IL = IS – ID = 1.1 A – 58.24 mA = 1.042 A.
VmaxP = xVt = 17.88 kT/q
RL = VmaxP/ID = (17.88/1042)kT/q = 17.16 kT/q.
At 20oC, kT/q = 25.25 mV, hence RL = 0.433 ohms.
6.17 A PV module is composed of 100 ideal solar cells connected in series. At 25⁰C, the
solar current of each cell is 1.2 A, and reverse saturation current is 10 nA. Find the module
power when the module voltage is 45V.
𝐼𝑐𝑒𝑙𝑙 = 𝐼𝑠 − 𝐼𝑑
0.45
= 1.2 − 10−9 (𝑒 0.0257 − 1)
= 0.7978 𝐴
𝑃𝑐𝑒𝑙𝑙 = 𝑉𝑐𝑒𝑙𝑙 . 𝐼𝑐𝑒𝑙𝑙
= (0.45)(0.7978)
= 0.359
𝑃𝑚𝑜𝑑𝑢𝑙𝑒 = 𝑛 × 𝑃𝑐𝑒𝑙𝑙 = 100 × 0.359 = 35.9 𝑊
6.18 An 80 cm² solar cell is operating at 30⁰C where the output is 1A, the load voltage is
0.5V and the saturation current of the diode is 1nA. The series resistance of the cell is 10
mΩ and the parallel resistance is 500 Ω. At a given time, the solar power density is 300
W/m². Compute the irradiance efficiency.
Efficiency,
ɳ=
𝑉𝑑 𝐼𝑠
𝑃𝐴
𝐼𝑠 = 𝐼 + 𝐼𝑑 + 𝐼𝑃
0.51
𝐼𝑑 = 10−9 (𝑒 0.026 − 1) = 0.33 𝐴
𝐼𝑃 =
𝑉𝑑 0.51
=
= 0.00102 𝐴
𝑅𝑝 500
𝐼𝑠 = 1 + 0.33 + 0.00102 = 1.33102 𝐴
ɳ𝑖𝑟𝑟𝑎𝑑𝑖𝑎𝑛𝑐𝑒 =
(0.51)(1.33102)
(300)(0.008)
= 0.283 × 100% = 28.3 %
6.19 A solar cell is operating at 30⁰C. I= 1.1A, load voltage is 0.5V, Rs= 20 mΩ, Rp= 2 KΩ.
ɳirr= 22%.
We know that,
𝑃𝑙𝑜𝑠𝑠 = 𝐼 2 𝑅𝑠 + 𝐼𝑝 2 𝑅𝑝
𝑉𝑑 = 𝑉 + 𝐼𝑅𝑠
= 0.5 + (1.1 × 20 × 10−3 )
𝑉𝑑 = 0.522 𝑉
𝐼𝑃 =
𝑉𝑑 0.522
=
= 2.61 × 10−4 𝐴
𝑅𝑝 2000
𝑃𝑒−𝑙𝑜𝑠𝑠 = (1.12 )(20 × 10−3 ) + (2.61 × 10−4 )2 (2000) = 0.0243 𝑊
ɳ𝑒 = 𝑃
𝑃𝑜𝑢𝑡
𝑉𝐼
𝑜𝑢𝑡 +𝑃𝑙𝑜𝑠𝑠
= 𝑉𝐼+𝑃
𝑙𝑜𝑠𝑠
(0.5)(1.1)
= (0.5)(1.1)+(0.0243) = 95.77 %
ɳ = ɳ𝑒 . ɳ𝑖𝑟𝑟 = 0.22 × 0.9577 = 21.07 %
and:
1.
A p-n junction diode at 25◦C carries a current of 100 mA when the diode voltage is 0.5 V. What is the
reverse saturation current, I0?
1) A p-n junction diode at 25⁰ C carries a current of 100 mA when the diode voltage is
0.5 V. What is the reverse saturation current, Io?
Solution:
Current at 25o C = 100mA
Voltage on diode = 0.5 V
We know that,
𝑉
𝐼 = 𝐼0 (𝑒 𝑉𝑡 − 1)
𝐾𝑇
𝑞
1.38 𝑥10−23 𝑥 298
𝑉𝑡 =
1.6𝑥10−19
𝑉𝑡 = 0.0257 𝑉
𝑉𝑡 =
0.5
100𝑥10−3 = 𝐼0 (𝑒 0.0257 − 1)
𝐼0 = 3.59 × 10−10 𝐴
2.
For the simple equivalent circuit for a 0.005 m2 photovoltaic cell shown below, the reverse saturation
current is I0 = 10-9 A and at an insolation of 1-sun the short-circuit current is ISC = 1 A,. At 25◦C, find the
following:
a. The open-circuit voltage.
b. The load current when the output voltage is V = 0.5 V.
c. The power delivered to the load when the output voltage is 0.5 V.
d. The efficiency of the cell at V = 0.5 V.
2) For the simple equivalent circuit for a 0.005 m² photovoltaic cell shown below, the
reverse saturation current is Io = 10⁻⁹ A and at an isolation of 1-sun the shortcircuit current is Isc = 1 A. At 25⁰ C, find the following:
a. The open-circuit voltage.
b. The load current when the output voltage is 0.5 V.
c. The power delivered to the load when the output voltage is 0.5 V.
d. The efficiency of the cell at V = 0.5 V.
Solution:
Area of photovoltaic cell = 0.005m3
Reverse Saturation current = 10-9 A
Short circuit current at 25o C = 1A
𝑉
a) 𝐼𝑠𝑐 = 𝐼0 (𝑒 𝑉𝑡 − 1)
𝑉
1 = 10−9 (𝑒 0.0257 − 1)
𝑉 = 0.533 𝑣𝑜𝑙𝑡𝑠
b) 𝐼 = 𝐼𝑠𝑐 − 𝐼𝑑
𝑉
𝐼 = 𝐼𝑠𝑐 − 𝐼0 (𝑒 𝑉𝑡 − 1)
0.5
𝐼 = 1 − 10−9 (𝑒 0.0257 − 1)
𝐼 = 0.718 𝐴
c) Power delivered by the load,
𝑃 = 𝑉𝐼
𝑃 = 0.5 × 0.718
𝑃 = 0.359 𝑊𝑎𝑡𝑡𝑠
d) Efficiency of solar cell,
𝑉𝐼
ɳ=
𝑃𝐴
0.359 × 10−3
ɳ=
× 100%
1 × 0.005
ɳ = 7.18%
3.
The equivalent circuit for a PV cell includes a parallel resistance of RP =10 Ω. The cell has area 0.005 m2,
reverse saturation current of I0 = 10-9 A and at an insolation of 1-sun the short-circuit current is ISC = 1 A,
At 25◦C, with an output voltage of 0.5 V, find the following:
a. The load current.
b. The power delivered to the load.
c. The efficiency of the cell.
3) The equivalent circuit for a PV cell includes a parallel resistance of Rp = 10Ω. The
cell has area 0.005 m², reverse saturation current of Io= 10⁻⁹ A and at an isolation
of 1-sun the short-circuit current is Isc= 1 A. At 25⁰ C, with an output voltage of 0.5
V, find the following:
a. The load current.
b. The power delivered to the load.
c. The efficiency of the cell.
Solution:
Resistance= 10 Ω
Area of cell= 0.005m2
Output voltage=0.5 V
a) 𝐼 = 𝐼𝑠 − 𝐼𝑑 − 𝐼𝑟
0.5
𝐼 = 1 − 10−9 (𝑒 0.0257 − 1) −
𝐼 = 0.6686 𝐴
b) 𝑃 = 𝑉𝐼
𝑃 = 0.5 × 0.6686
𝑃 = 0.3343 𝑊𝑎𝑡𝑡𝑠
c) Efficiency of solar cell,
𝑉𝐼
ɳ=
𝑃𝐴
0.3343 × 10−3
1 × 0.005
ɳ = 6.68 %
ɳ=
0.5
10
4.
The following figure shows two I-V curves. One is for a PV cell with an equivalent circuit having an
infinite parallel resistance (and no series resistance).
What is the parallel resistance in the equivalent circuit of the other cell?
4) From the given graph,
Parallel Resistance,
𝑉
𝑅=
𝐼
(0.4 − 0)
𝑅=
(4 − 3.8)
𝑅 = 2Ω
5.
The following figure shows two I-V curves. One is for a PV cell with an equivalent circuit having no series
resistance (and infinite parallel resistance). What is the series resistance in the equivalent circuit of the
other cell?
5) From the given graph,
𝑆𝑒𝑟𝑖𝑒𝑠 𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑅𝑠 =
(0.62 − 0.58)
(1 − 0)
𝑅𝑠 = 0.04 Ω
𝑅𝑠 =
𝑉
𝐼