Transcript Lesson 24: Photocell Electrical Characteristic and Circuit Model

Lesson 24: Photocell Electrical
Characteristic and Circuit Model
ET 332a
Dc Motors, Generators and Energy Conversion Devices
1
Lesson 24 332a.pptx
Learning Objectives
After this presentation you will be able to:
Identify and interpret a photocell electrical
characteristic
 Find the maximum power output from a photocell
 Calculate a photocell’s efficiency
 Determine circuit model parameters for a photocell
given its characteristic curve
 Perform a calculation using the circuit model of a
photocell.

2
Lesson 24 332a.pptx
Photocell Characteristic Curve
Pm = Maximum cell power
IC
Max Power Pt.
(Vm, Im)
ISC
Fill Factor = FF
PM  IM VM
Pm
FF 
ISC VOC
0
0
Cell Efficiency
3
VOC
V
PC
C 
PI
0.7 <FF<0.85 Typical FF range
PI = incident solar power
Lesson 24 332a.pptx
Solar Cell Characteristics Example (1)
Example: A photocell has a saturation current of 2.5 x 10-12 A
and a short circuit current of 35 mA. It has an area of 1.5 cm2.
The incident solar power is 1000 W/m2. Assume that the cell
operates at room temperature. Find Voc, Pm, Fill Factor and
conversion efficiency.
Is  2.5  10
 12
A
VT  0.026 V

VOC  VT  ln  1 

4
Isc  0.035
A
IL  Isc
IL 

Is

VOC  0.607
Lesson 24 332a.pptx
V
Solar Cell Characteristics Example (2)
Find Pm graphically
Create function for plotting
Define plot range
Calculate power as
function of V
5
 V

 VT

Ic ( V)  IL  Is   e
 1
V  0.0  0.02  0.62
P ( V)  Ic ( V)  V W
Lesson 24 332a.pptx
V
Solar Cell Characteristics Example (3)
Cell Characteristic
Cell Current (A), Cell Power (W)
0.04
I sc
0.03
Vm  0.5360 V
0.02
Im  Ic Vm
 
Pm  0.0177 W
Im  0.0328 A
Find the Fill Factor (FF)
0.01
0
Pm
FF 
VOC  Isc
0
0.1
0.2
0.3
0.4
Cell Voltage (V)
Cell Current
Cell Power
6
0.5
0.6
0.7
V
OC
Lesson 24 332a.pptx
FF  0.8
Solar Cell Characteristics Example (4)
Find the cell efficiency at maximum power output
Find the incident power, PI
W
I  1000 
2
m
2
A  1.5  cm
2
Am  A 
1 m
10000  cm2
PI  Am I
Pm
 C 
PI
7
Am  1.5  10
PI  0.2 W
 C  0.1
4
m
2
Pm  0.0177  W
Cell efficiency is 10%
Lesson 24 332a.pptx
Circuit Model of Solar Cell
+
D
IL
+
Rs
VOC Rsh
RL
Vc
-
Cell Characteristic
0.04
Rs slope of characteristic near Voc
Rsh slope of characteristic near Isc
1 /R
sh
0.02
1/Rs
Values determined by cell construction
Cell Current (A)
0.03
0.01
0
0
0.1
0.2
0.3
0.4
Cell Voltage (V)
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Lesson 24 332a.pptx
0.5
0.6
0.7
Solar Cell Circuit Model Parameters
Finding Model Parameters, Rsh and Rs
Cell Characteristic
Pick points on characteristic plot and compute slope.
For Rs use points 1 and 2.
1/R
0.04
sh
1.)
V1s  0.5781 V
I1s  0.02395 A
0.03
2.)
V2s  0.6058 V
I2s  3.9  10
0.02
Gs 
4
2
1/Rs
Cell Current (A)
3
I2s  I1s
V2s  V1s
1
Rs 
Gs
0.01
4
A
Conductance is 1/Rs
Gs  0.8505
S
Rs  1.176 
For Rsh use points 3 and 4
0
1
0
0.1
0.2
0.3
0.4
Cell Voltage (V)
0.5
0.6
0.7
3.)
V1sh  0.512
4.)
V2sh  0.550
Gsh 
I2sh  I1sh
V2sh  V1sh
1
Rsh 
Gsh
9
V
V
A
I2sh  0.0311
A
Conductance is 1/Rsh
Gsh  0.0842
Lesson 24 332a.pptx
I1sh  0.0343
S
Rsh  11.875 
Solar Cell Circuit Model Example (1)
IRL
+
ID
VOC Rsh
RL
Ish
D
-
Vc
-
I L  100  mA
R sh  11.875  
+
R eq 
D
VOC Req
 R s  R L   R sh
 R s  R L   R sh
Parallel circuit so.....
-
I D  I L  I eq
10
R s  1.176  
R eq  8.599 
I D  17.4 mA
Lesson 24 332a.pptx
R L  30  
0
I eq 
Ieq
100 mA
IL
VOC  0.71  volt
I L  I D  I eq
By KCL
ID
IL
+
Rs
Example: Find the power delivered to a 30 ohm resistive load
by the solar cell with a light current of 100 mA and model
parameters of R s =1.176  and R sh of 11.875 . Determine
the cell load voltage for this load resistance.
VOC
R eq
I eq  82.6 mA
Solar Cell Model Example (2)
IRL
Ish
+
-
Ish
Rsh
VOC
+
Rs
RL
Vc
-
Use Current Divider rule to find IRL
Rsh
IRL  Ieq 
Rsh  RL  Rs

2
PL  IRL  RL
Find VC from KVL

IRL  22.77 mA
PL  15.6 mW
VC  VOC  IRL  Rs
VC  0.683 V
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Lesson 24 332a.pptx
Solar Cell Efficiency

AM1.5 Solar Intensity (Incident power density) 1000 W/m2 or
100 W/cm2
 Losses





Photon Energy -47% of photons have eV<1.1, 30% goes to heat
Voltage factor – ratio of energy given to energy required to produce
electron 0.65
Recombination – electron/holes that recombine 10%
Reflection – reduced to 4%
Overall Efficiency c = (0.47)(0.65)(.10)(.96)=.26

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26% Maximum efficiency using current technologies
Lesson 24 332a.pptx
End Lesson 24
ET 332a
Dc Motors, Generators and Energy Conversion Devices
13
Lesson 24 332a.pptx