6.1 What should the load be to obtain maximum power transfer from the solar cell to the load at 1 Solar cell driving a load A Si solar cell of area 4 cm2 is connected to drive a load R as in Figure 6.8 (a). It has the I-V Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 23 April 2001 6.3 a characteristics in Figure 6.8 (b) under an illumination of 600 W m-2. Suppose that the load is 20 Ω and it is used under a light intensity of 1 kW m-2. What are the current and voltage in the circuit? What is the power delivered to the load? What is the efficiency of the solar cell in this circuit? b kW m-2 illumination. What is this load at 600 W m-2? c Consider using a number of such a solar cells to drive a calculator that needs a minimum of 3V and draws 3.0 mA at 3 - 4V. It is to be used indoors at a light intensity of about 400 W m-2. How many solar cells would you need and how would you connect them? At what light intensity would the calculator stop working? Solution a up by 1000/600 = 1.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 1.67 along the current axis. The load line for R = 20 W and its intersection with the solar cell I-V characteristics at P which is the operating point P. Thus, The power delivered to the load is This is not the maximum power available from the solar cell. The input sun-light power is The efficiency is Pout Pin The solar cell is used under an illumination of 1 kW m-2. The short circuit current has to be scale = (1000 W m-2)(4 cm2 ´ 10-4 m2/cm2) = 0.4 W Pin = (Light Intensity)(Surface Area) Pout = I¢V¢ = (22.5×10-3)(0.45V) = 0.0101W, or 10.1 mW. which is poor. b V¢ ≈ 0.44 V. The load should be R = 18.7 W, close to the 20 W load. At 600 W m-2 illumination, the load has to be about 30 W as in Figure 6.8 (b). Thus, the maximum efficiency requires the load R to be decreased as the light intensity is increased. The fill factor is Point M on Figure 6Q3-2 is probably close to the maximum efficiency point, I¢ ≈ 23.5 mA and I¢ ≈ 22.5 mA and V¢ ≈ 0.45 V = 2.5 0 0 η= 100 = 100 0.010 0.4 FF = ImVm IscVoc = (23.5 mA )(0.44 V) (27 mA)(0.50 V) ≈ 0.78 The solar cell is used under an illumination of 400 W m-2. The short circuit current has to be c scale up by 400/600 = 0.67. Figure 6Q3-2 shows the solar cell characteristics scaled by a factor 0.67 along the current axis. Suppose we have N identical cells in series, and the voltage across the calculator is Vcalculator. The current taken by the calculator is 3 mA in the voltage range 3 to 4 V and the calculator stops working when Vcalculator < 3 V. The cells are in series so each has the same current and equal to 3 mA, marked as I¢ in Figure 6Q-2. The voltage across one cell will be V¢ = Vcalculator/N. which is marked in Figure 6Q3-2. V¢ = 0.46 V. Minimum number of solar cells in series = N = 3 / 0.46 = 6.5 or 7 cells , since you must choose the nearest higher integer. cells in series until we reach about 4 V; N = 4 V / 0.46 V = 8.7 or 9 cells in series. will stop working when the light intensity cannot provide energy for the solar cell to deliver the 3 mA calculator current. The short circuit current at 400 W m-2 is 11 mA in Figure 6Q3-2. Thus If we want the calculator continue to work under low intensity levels, then we can connect more The easiest estimate for the minimum required light intensity is the following: The calculator 

Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 23 April 2001 6.2 I (mA) 0 –10 –20 –30 Figure 6Q3-1 I (mA) 0 I′ 3 mA –10 –20 Minimum light intensity = 3 mA 11 mA 400 W m−2 =109 W m-2 V′ V 0.2 0.4 0.6 I-V for a solar cell under an illumination of 1000 Wm-2. I′ P M The load line for R = 20 ž (I-V for the load) I−V for a solar cell under an illumination of 400 W m-2. 0.2 0.4 V′ 0.6 V P Figure 6Q3-2 6.4 Open circuit voltage A solar cell under an illumination of 100 W m-2 has a short circuit current Isc of 50 mA and an open circuit output voltage Voc, of 0.55V. What are the short circuit current and open circuit voltages when the light intensity is halved? Solution The short circuit current is the photocurrent so that at   50 W m−2   = 50 mA )  100 W m −2    (  I  2  I 1 Isc2 = Isc1 Assuming n = 1, the new open circuit voltage is = 25 mA )ln 0.5( ) = 0.508 V Voc2 = Voc1 + nk BT e ln  I  2  I 1   = 0.55 + 1 0.0259  ( Assuming n = 2, the new open circuit voltage is 

Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 23 April 2001 6.3 Voc2 = Voc1 + nk BT e ln  I  2  I 1   = 0.55 + 2 0.0259  ( )ln 0.5( ) = 0.467 V Series connected solar cells Consider two odd solar cells. Cell 1 has Io1 = 25´10-6 mA, n1 = 1.5, 6.7 Rs1 = 10 W and cell 2 has Io2 = 1´10-7 mA, n2 = 1, Rs2 = 50 W. The illumination is such that Iph1 = 10 mA and Iph2 = 15 ma. Plot the individual I-V characteristics and the I-V characteristics of the two cells in series. Find the maximum power that can be delivered by each cell and two cells in series. Find the corresponding voltages and currents at the maximum power point. What is your conclusions? Solution Iph2 Rs2 A I Id1 V1 Iph2 Rs1 RL V Id2 I V2 Iph1 Iph2 Figure 6Q7-1 the same. Thus, for cell 1 The equivalent circuit is shown in Figure 6Q7-1. The current through both the devices has to be Two different solar cells in series = −I ph1 + Io1 exp     V1 − IRs1   n1VT    − 1    ∴ ∴ For cell 2 ∴ But ∴ I = − I ph1 + Io1 exp V1 − IRs1 n1VT = ln V1 = n1VT ln Vd1 n1VT        I + I ph1   Io1    − 1      − 1  I + I ph1      + IRs1 − 1     − 1       Vd 2 n2VT Io1          V2 − IRs2 n2VT    − 1    I = − I ph1 + Io2 exp = −I ph 2 + Io2 exp V2 = n2VT ln    I + I ph 2 Io2   + IRs2 + 1  V = V1 + V2    V = n1VT ln I + Iph1 Io1   + IRs1 + n2VT ln + 1     I + I ph2 Io2   + IRs 2 + 1  

Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 23 April 2001 We can now substitute Io1 = 25´10-6 mA, n1 = 1.5, Rs1 = 10 W, Io2 = 1´10-7 mA, n2 = 1, Rs2 = 50 W and then plot V vs. I (rather I vs. V since we can calculate V from I) for each cell and the two cells in series as in Figure 6Q7-2. Notice that the short circuit is determined by the smallest Isc cell. The total open circuit voltage is the sum of the two. 6.4 Cell 1 in series with 2 Voltage Cell 2 Cell 1 Current (mA) Figure 6Q7-2 6.9 Solar cell efficiency The fill factor FF of a solar cell is given by the empirical expression v oc − ln(voc + 0.72 ) FF ≈ voc + 2 where voc = Voc /(nkBT/e) is the normalized open circuit voltage (normalized with respect to the thermal temperature kBT/e). The maximum power output from a solar cell is P = FFIscVoc Taking Voc = 0.58 V and Isc = Iph = 35 mA cm-2, calculate the power available per unit area of solar cell at room temperature 20 °C, at -40 °C and at 40 °C Solution The open circuit voltage depends on the temperature whereas Isc has very little temperature dependence. Use ′ V oc = Voc   ′ T   T + Eg e  1 −   ′ T  T to calculate the Voc at different temperature given Voc at one temperature. Then calculate voc using then FF using voc = Voc /(nkBT/e) (1) (2) v oc − ln(voc + 0.72 ) voc + 2 FF ≈ and then P using P = FFIscVoc (3) (4) 

6.5 Voc; Eq. (1) 0.580 V 0.686 V 0.545 Isc mA/cm2 35 35 35 Solutions Manual for Optoelectronics and Photonics: Principles and Practices S.O. Kasap 23 April 2001 as summarized in Table 6Q9-1. Table 6Q9-1 n = 1 20 °C -40 °C 40 °C n = 2 20 °C -40 °C 40 °C Conclusions: n = 2 case has a lower FF and also lower power delivery. NOTE: The temperature dependence of the open circuit voltage Voc was derived in the text as Isc mA/cm2 35 35 35 Voc; Eq. (1) 0.580 V 0.686 V 0.545 FF; Eq. (3) 0.666 0.744 0.638 FF; Eq. (3) 0.793 0.847 0.773 voc; Eq. (2) 11.49 17.10 10.09 voc; Eq. (2) 22.97 34.19 20.19 P mW cm-2; Eq. (4) 13.52 17.90 12.15 P mW cm-2; Eq. (4) 16.03 20.34 14.73 ′ V oc = Voc   ′ T   T + Eg e  1 −   ′ T  T This expression is valid whether n is 1 or 2. Recall that n = 1 represents diffusion in the neutral regions 2 and in the n = 2 case Io µ and n = 2 is recombination in the space charge layer. In the n = 1 case Io µ ni ni, thus in general Consider the open circuit voltage,    At two different temperatures T1 and T2 but at the same illumination level, by subtraction, eVoc nk BT Voc = Io µ ni nkBT or          = ln 2/n KI KI I o ln Io e eVoc2 nk BT2 − eVoc1 nk BT1 = ln    Io1 Io2   ≈ ln  2 / n 2 / n    ni1 ni2    where the subscripts 1 and 2 refer to the temperatures T1 or T2 respectively. and Nv compared with the exponential part to obtain, We can substitute ni 2/n = (NcNv)1/2exp(-Eg/nkBT) and neglect the temperature dependences of Nc eVoc2 nk BT2 − eVoc1 nk BT1 = Eg nk B    1 T2 − 1 T1    Rearranging for Voc2 in terms of other parameters we find, Voc2 = Voc1  T2   T1   +  E g e   1 −  T2 T1   