Download 1. 1 cm solar cell consists of a p

1. 1 cm2 solar cell consists of a p-type region containing 1018 cm-3 acceptors and an n-type
region containing 1015 cm-3 donors. wp' = 0.1 μm and wn >> Lp. Use μn = 1000 cm2/V-s
and μp = 300 cm2/V-s. . The minority carrier lifetime is 10 μs. The diode is illuminated
with sunlight, yielding a photocurrent density of 30 mA/cm2.
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 = qAni2 (
Dp
Dn
+
) = 55.5 pA
'
wp N a Lp N d
The open circuit voltage under illumination equals
Voc = Vt ln
I ph + I s
Is
= 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:
Vm = Vt ln(
I ph + I s
Is
1
) = 0.445 V
Vm
1+
Vt
The current and maximum power are:
I m = I s (exp(
Vm
) − 1) − I ph = - 28.4 mA
Vt
Pm = Vm × I m = - 12.6 mW
c. Calculate the fill factor of the solar cell.
The fill factor equals:
Fill Factor =
Pm
= 0.81
Voc × I sc
d. Calculate the fill factor for the same cell when illuminated by a concentrator so
that the photocurrent density equals 300 A/cm2.
Repeating the same procedure as in part a), b) and c) one finds:
Voc = Vt ln
I ph + I s
Is
= 0.759 V
I sc = − I ph = - 300 A
Vm = Vt ln(
I ph + I s
I m = I s (exp(
Is
1
) = 0.673 V
Vm
1+
Vt
Vm
) − 1) − I ph = - 289 A
Vt
Pm = Vm × I m = - 195 W
Fill Factor =
Pm
= 0.855
Voc × I sc
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
) − 1)
Vt
Current (A)
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
Voltage (V)
The simulated I-V characteristic is plotted on a semi-logarithmic scale and fitted to a
straight line described by
I = I s (exp(
V
) − 1)
nVt
The line was obtained using the simulated points at 4 and 5 V, yielding the following
saturation current and ideality factor.
Is = 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 wp’. 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.