1.
1 cm
2
solar cell consists of a p-type region containing 10
18 region containing 10 and
μ
15
cm
-3
donors. w p
'
= 0.1
μ m and w n
>> L p
. Use with sunlight, yielding a photocurrent density of 30 mA/cm
2
cm
.
-3
acceptors and an n-type
μ n
= 1000 cm
2
/V-s p
= 300 cm
2
/V-s. . The minority carrier lifetime is 10
μ s. The diode is illuminated a.
Calculate the open circuit voltage and short-circuit current of the solar cell.
The saturation current of the diode is obtained by applying the “short” diode equation to the p -type region and the “long” diode equation to the n -type region.
I s
= qAn i
2
D
( w
' p n
N a
+
D p
L p
N d
) = 55.5 pA
The open circuit voltage under illumination equals
V oc
=
V t ln
I ph
I s
+
I s
= 0.52 V
And the short circuit current equals the photocurrent.
I sc
= −
I ph
= - 30 mA b.
Calculate the maximum power generated by the cell and the corresponding voltage and current.
The voltage corresponding to the maximum power point is obtained from:
V m
=
V t ln(
I ph
I s
+
I s
1
1
+
V m
V t
) = 0.445 V
The current and maximum power are:
I m
=
I s
(exp(
V m
V t
)
−
1 )
−
I ph
= - 28.4 mA
P m
=
V m
×
I m
= - 12.6 mW c.
Calculate the fill factor of the solar cell.
The fill factor equals:
Fill Factor
=
V oc
P m
×
I sc
= 0.81 d.
Calculate the fill factor for the same cell when illuminated by a concentrator so that the photocurrent density equals 300 A/cm
2
.
Repeating the same procedure as in part a), b) and c) one finds:

V oc
=
V t ln
I ph
I s
+
I s
= 0.759 V
I sc
= −
I ph
= - 300 A
V m
=
V t ln(
I ph
I s
+
I s
1
1
+
V m
V t
) = 0.673 V
I m
=
I s
(exp(
V m
V t
)
−
1 )
−
I ph
= - 289 A
P m
=
V m
×
I m
= - 195 W
Fill Factor
=
V oc
P m
×
I sc
= 0.855
2.
Use SimWindows to calculate the current of the diode described in problem 1 (without light) at 0.1, 0.2, 0.3, 0.4, 0.5, 0.6 and 0.7 Volt. Plot the current on a semi-logarithmic scale and extract the saturation current as well as the ideality factor of the diode.
Calculate the ideal diode current and plot the result on the same graph. Discuss the agreement or lack thereof between the simulation and the ideal diode current.
The current voltage obtained with SimWindows is compared to the ideal diode of the diode and shown in the figure below. The ideal diode current is calculated from:
I
=
I s
(exp(
V
)
−
V t
1 )

1.E+01
1.E+00
1.E-01
1.E-02
1.E-03
1.E-04
1.E-05
1.E-06
1.E-07
0.0
0.2
0.4
0.6
0.8
The simulated I-V characteristic is plotted on a semi-logarithmic scale and fitted to a straight line described by
I
=
I s
V
(exp( nV t
)
−
1 )
The line was obtained using the simulated points at 4 and 5 V, yielding the following saturation current and ideality factor.
I s
= 676 pA and n = 1.08
The differences between the simulated and calculated current include the current due to recombination in the depletion region at low bias voltage ( V < 0.3 V) and the influence of the series resistance at high bias ( V > 0.55 V).
The ideality factor of 1.08 is due to the modulation of the quasi-neutral region, while the ideal diode calculation is based on a constant value for w p
’
. Since the ideality factor is not equal to one, one expects to obtain a different saturation current. A difference in mobility, minority carrier lifetime and intrinsic carrier density would also result in a different saturation current.