Chapter 4 Solar Energy

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In this chapter, we will discuss a more accurate equivalent circuit for a PV cell. This circuit helps us to have better prediction and understanding of PV cell current-voltage characteristics as well as at diﬀerent types of load connected to the PV panel, which is made up of an array of PV cells. Later in this chapter we will discuss the concept of maximum power point tracker (MPPT).

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Nội dung Text: Chapter 4 Solar Energy

Chapter 4 Solar Energy II 4.1 Introduction In this chapter, we will discuss a more accurate equivalent circuit for a PV cell. This circuit helps us to have better prediction and understanding of PV cell current-voltage characteristics as well as at diﬀerent types of load connected to the PV panel, which is made up of an array of PV cells. Later in this chapter we will discuss the concept of maximum power point tracker (MPPT). 4.2 More Accurate Equivalent Circuit for a PV Cell If we connect two simple PV cell model as described in Fig. 3.7 in series and simulate the situation when one of the PV cell is shaded while the other is under strong sun rays, the current source is essentially of zero current. However, in the real PV panel with PV cells connected in series , such as the BP SX10 with 36 PV cells connected in series, measurable current is recorded when some of the cells are shaded. This implies that the simple PV cell is insuﬃcient to model the PV panel I-V characteristics. We need some path to be added to the model to show the current ﬂow within the PV cells. Fig. 4.1 shows a more accurate PV cell equivalent circuit using one parallel resistor, Rp , and one series resistor, Rs , to realize the extra current paths. Let us ﬁrst derive a mathematical expression for this equivalent circuit. By KCL, the current enters node 65
4. Solar Energy II 66 x equal the currents leave this node: ISC = Id + Ip + I (4.1) Substitute (3.23) into (4.1) and re-arrange the terms, we obtain Vd I = ISC − I0 (eqVd /kT − 1)− (4.2) Rp And Vd and V are related by the following equation Vd = I · Rs + V (4.3) We cannot ﬁnd a solution of (4.2) with close-loop form. However we may ﬁrst use an assumed value of Vd to evaluate I and use the value of I to calculate V from (4.3). An example as follows shows how to work that out. Example: Given the PV cell has short-circuit current ISC = 4A and at 25◦ C its reverse saturation current is I0 = 6 × 10−10 A. Show the I-V curves under diﬀerent resistance values (a) Rp = ∞, Rs =0; (b) Rp = ∞, Rs = 0.06Ω; (c) Rp = 2.0Ω, Rs = 0Ω; and (d) Rp = 2.0Ω, Rs = 0.06Ω. Solution: With the given temperature and reverse saturation current conditions, q = 38.9 kT Using the 4 set of values we can now plot the curves as shown in Fig. 4.2. This ﬁgure shows that both the series and parallel resistances decrease the maximum power the PV cell can be obtained. Compared to parallel resistance, a small series resistance will cause a dramatic decrease of voltage hence the power delivered by the PV cell. 4.3 PV Cells, Modules and Arrays As the maximum voltage of each PV cell is about 0.6V or below, there is no practical value for application using only 1 PV cell. However, PV cells can be connected in series and in parallel to increase the voltage and current respectively. Fig. 4.3 show how the cells are connected both in series then parallel to increase the total deliverable power of the PV array.
4. Solar Energy II 67 x - Rs I ?Id ?Ip + + PV 6 Rp ISC Vd Load V − − Figure 4.1: A more accurate PV cell equivalent circuit with parallel and series resis- tances. Figure 4.2: Plot of I-V curves of a PV cell with equivalent circuit in Fig. 4.1.
4. Solar Energy II 68 I - 61 62 I I + Current + PV PV Va cell cell − I + PV PV Va V cell cell − I1 , I2 + PV PV Va cell cell − − 3Va Voltage Va 2Va Figure 4.3: PV cells arranged in an array to increase voltage and current capability. 4.4 Impacts of Temperature and Insolation on I-V Curves Manufacturers will often provide I-V curves that show how the change of cell tem- perature and insolation would shift the curves. Fig. 4.4 shows two examples of two PV modules from BP Solar, SX5 and SX10. The measurement was taken at standard test conditions (STC) which include a solar irradiance of 1kW/m2 (1 sun) with special distribution of air mass ratio of 1.5 (AM 1.5). The cell temperature is at 25◦ C. As can be seen in Fig. 4.4, as cell temperature increases, the open-circuit voltage decreases rapidly while the short-circuit current increases slightly. From the datasheet it is shown that ISC increases at a rate of 0.065%/◦ C and VOC decreases 80mV/◦ C. This corresponds to a shift of maximum power point to the left hence decrease in maximum deliverable power (around -0.5%/◦ C). In general, as the PV module eﬃciency is low (around 12%), only a tiny portion of the insolation striking on the panel is converted to electricity. Most of the energy turns to heat. To work out the cell temperature of the module, manufacturers usually provide an indicator called the NOCT which stands for normal operating cell temperature. It is measured at ambient temperature of 20◦ C, solar irradiation of 0.8kW/m2 , and wind speed of 1m/s. For example, the BP SX5/10 has a NOCT of 47◦ C. To evaluate the cell temperature Tcell in ◦ C at diﬀerent ambient temperatures Tamb in ◦ C, we may use the following expression NOCT − 20◦ C Tcell = Tamb + ( )·S (4.4) 0.8
4. Solar Energy II 69 where S is the solar insolation measured in kW/m2 . Example: Estimate cell temperature, open-circuit voltage, and maximum power output for the 10W BP-SX10 module under conditions of 1-sun insolation and ambient temperature 27◦ C. The module has a NOCT of 47◦ C, VOC at 16.8V, temperature coeﬃcient of VOC -80mV/◦ C, and temperature coeﬃcient of power -0.5%/◦ C. Solution: Using (4.4) with S = 1kW/m2 , cell temperature will become 47 − 20 Tcell = 27 + ( ) · 1 = 60.75◦ C 0.8 The VOC drops by VOC -80mV/◦ C, the new value will be VOC = 16.8V − 0.08(60.75 − 25)V = 13.94V The maximum power output will drop according to -0.5%/◦ C, therefore we have Pnew = 10W[1 − 0.005(60.75 − 25)]W = 8.21W max 4.5 Shading Impacts on I-V Curves From the equivalent circuit showning n cells in Fig. 4.5, the current I consists of three currents and is given by I = ISC − Id + IRp (4.5) If the nth cell is completely shaded, ISC = Id = 0, then (4.5) reduces to I = IRp (4.6) The output voltage VSH can be written as VSH = Vn−1 − I (Rs + Rp ) (4.7) We may express the voltage of (n-1) cells as n−1 Vn−1 = ( )V (4.8) n where V is the output voltage when n cells receive sunlight and no shading happens. Substituting (4.8) into (4.7) gives n−1 VSH = ( )V − I (Rs + Rp ) (4.9) n
4. Solar Energy II 70 Figure 4.4: I-V characteristics under various cell temperatures for the BP Solar SX5 and SX10 PV modules.
4. Solar Energy II 71 - I + Rs nth cell IRp ? ISC Id 6 Rp VSH 6I + Vn−1 (n-1) cells − − I Figure 4.5: Equivalent circuit to investigate the eﬀect of shading when the nth cell is shaded while (n-1) cells received full sun. The decrease of voltage ∆V at any given I , caused by the shaded cell, is given by 1 ∆V = V − VSH = V − (1 − n )V + I (Rs + Rp ) (4.10) V = + I (Rs + Rp ) n Example: The 36-cell PV module has a parallel resistance per cell of RP = 5.6Ω and series resistance per cell of RS = 0.005Ω. In full sun and at current I = 2.25A the output voltage was found there to be V = 18.35V. If one cell is shaded and this current somehow stays the same, then: 1. What would be the new module output voltage and power? 2. What would be the voltage drop across the shaded cell? 3. How much power would be dissipated in the shaded cell? Solution: (1) From (4.10) the drop in module voltage will be V 18.35 ∆V = + I (Rs + Rp ) = + 2.25(0.005 + 5.6) = 13.12V n 36 The new output voltage will be (18.35-13.12)V = 5.23V. Power delivered by the module with one cell shaded completely would be Pmodule = V I = 5.23V × 2.25A = 11.77W
4. Solar Energy II 72 6I I2 6 I1 Rs IRp Dbypass 6 I1 ? ISC Id 6 Rp 6 - I1 I2 6I Figure 4.6: Equivalent circuit to investigate the eﬀect of shading with the presence of bypass diode. Comparing to the full power 18.35V×2.25A = 41.29W when no cell is shaded, there is a drop of 72% of power when one cell is shaded! (2) For the fully shaded cell, the short circuit current is zero, all the current of the PV module will ﬂow through RP and RS . Therefore the drop across the shaded cell is Vshaded cell = I (Rs + Rp ) = 2.25(0.005 + 5.6) = 12.61V (3) The power dissipated in the shaded cell is just the voltage drops times current, which is given by P = V I = 12.61V × 2.25A = 28.37W All of the power dissipated in the shaded cell will be converted to heat, which can cause overheat in that spot therefore permanently damage the plasitc laminates enclosing the cell. In order to prevent the cell from damaging due to shading, a bypass diode can be used to reduce the voltage drop across the cell and hence the power of the cell. The equivalent circuit is shown in Fig. 4.6. We will use a simple example to show how the current is bypassed. Example: A bypass diode is added to one PV cell shown in Fig. 4.6. The series and parallel resistances are 0.005Ω and 5.6Ω respectively. If the current I is 2.5A and
4. Solar Energy II 73 the forward voltage drop of diode Dbypass is constant at 0.6V from 0-2.5A of diode current, ﬁnd the values of I1 and I2 . Solution: By KCL we have I = I1 + I2 = 2.5A (4.11) Since the voltage across the bypass diode equals to that of series and parallel resistances, that is 0.6 = I1 (RP + RS ) = I1 (5.6 + 0.005) Therefore I1 is calculated as 0.6 I1 = = 0.11A 5.6 + 0.005 Hence the current of I2 can be calculated from (4.11) I2 = I − I1 = 2.5 − 0.11 = 2.39A We can see that now the cell resistance only conducts 4.4% of the total current and the rest is ﬂowing through the bypass diode. 4.6 I-V Curves on Diﬀerent Loads We have so far discussed the I-V characteristics of PV cells, but what happens when a load is connected to a PV module? What is the output voltage and current or simply the operating point of the PV module? In short, the operating point is deﬁned by the resistance of the load and we may use graphical method to ﬁnd this point. Let us examine the following diﬀerent types of loads and how we can ﬁnd the operating points of each of them. 4.6.1 Resistive Load When a resistor is connected to the output terminals of the PV module shown in Fig. 4.7, the voltage across the resistors equals the voltage of the terminals. Therefore we can plot the I-V curve of the resistor and the panel and we will ﬁnd a point where the two curves intercept. This is the operating point. An example is depicted in Fig. 4.8. Although it is easy to implement, the resistor can only achieve one point of maximum
4. Solar Energy II 74 - I + 7 PV Panel R V − Figure 4.7: A PV module powers a variable resistor. Figure 4.8: Diﬀerent operating points by varying the resistance.
4. Solar Energy II 75 Figure 4.9: Diﬀerent operating points by varying the resistance. - I Ra + + PV V M e = kω Panel − − Figure 4.10: Equivalent circuit of a dc motor connected to a PV module. power point (MPP), as shown in Fig. 4.9. When the insolation changes, the operating point is slipped away from the MPP. We will introduce later a Maximum Power Point Tracker (MPPT) which can keep the PV operating at the highest point out of its available power at a particular insolation condition. 4.6.2 DC Motor The dc motor is diﬀerent from a pure resistive load because it will generate back electro- motive force (emf) when the rotor start to rotate. It is by Faraday’s Law which states that a conductor cuts through a magnetic ﬁeld will induce a voltage at its terminal.
4. Solar Energy II 76 Fig. 4.10 shows a simple equivalent circuit of a PV panel connecting a dc motor. The I-V characteristics can be written as V = IRa + kω (4.12) where kω is the back emf related by angular speed ω and Ra is the armature resistance. Fig. 4.11 shows the I-V curves of how a dc motor works on a PV module. At startup, the e = 0 as the motor is stalled. Therefore V = IRa and large amount of current passes through the armature. Through the magnetic ﬁeld of the permanent magnet, a torque is applied on the rotor and it starts to turn. As it is turning, emf is building up and the current I reduces according to (4.12). This explains why there is a sharp increase of current against voltage in Fig. 4.11. As the voltage applied across the motor increases, the armature current increases for a while and it produces a larger torque to spin the rotor. The angular speed ω increases and reduces the armature current a bit. Therefore we can see that there is a roughly linear relationship between current and voltage of the motor after it starts up. In the case of Fig. 4.11, the PV module can provide suﬃcient start-up current when its insolation is at 400W/m2 . When the motor starts spinning, the required insolation to maintain its speed drops to 200W/m2 . This implies the motor won’t work until the insolation is higher than 400W/m2 . To remedy this problem, we may use a power converter to convert the PV module at high-voltage low-current output to low-voltage high-current output (power remains constant assuming 100% eﬃciency of power converter) for supplying suﬃcient current to start-up the motor. 4.6.3 Battery As solar energy is of intermittent nature, some form of storage is needed to provide power when solar energy is insuﬃcient to feed the load. Battery is a part of the key components in PV systems. Fig. 4.12 shows a simple equivalent circuit of a PV panel connecting a battery. The I-V characteristics can be written as V = VB + Ri I (4.13) where VB is the battery voltage and Ri is the internal resistance of the battery. Rear- ranging the terms in (4.13), we get 1 1 I= V− VB (4.14) Ri Ri
4. Solar Energy II 77 Figure 4.11: I-V curve of a dc motor on a PV I-V characteristic. Source: Masters 2004. - I Ri + + PV VB V Panel − − Figure 4.12: Equivalent circuit of a PV module connected to a battery.
4. Solar Energy II 78 Current Current 6 6. .. .. .. .. . .. . .. .. .. . . Discharging Y .. Charging I .. - .. .. . 1 −1 slope = slope = . . . .. Ri Ri .. .. . .. ... .. .. - - . . . VB,old VB,old VB,new VB,new Voltage Voltage Figure 4.13: I-V characteristics when charging and discharging a battery with internal resistance Ri . Equation (4.14) tells us the slope of the I-V of PV panel is 1/Ri when it charges the battery. When the battery is discharging or powering the load, the equation will be changed to −1 1 I= V+ VB (4.15) Ri Ri where the slope becomes −1/Ri . To plot (4.14) and (4.15), we need to ﬁnd the x-axis intercept. By putting I = 0 in (4.14) and (4.15), we get the same following result of x-axis intercepts V = VB (4.16) The curves are plotted in Fig. 4.13. We can see from the two plots that the I-V curve shears in both process. It is because the voltage of the battery VB is not a constant. When it is charged VB will increase and vice versa. Example: Suppose that a nearly depleted 12V lead-acid battery has an open- circuit voltage of 11.7V and an internal resistance of 0.03Ω. (a) What voltage would a PV module operate at if it is delivering 6A to the battery? (b) If 20A is drawn from a fully charged battery with open-circuit voltage 12.7V, what voltage would the PV module operate at? Solution: (a)Using (4.13), the PV voltage would be V = VB + Ri I = 11.7 + 0.03 × 6 = 11.88V (b) While drawing 20A after VB has reached 12.7V, the output voltage of the
4. Solar Energy II 79 V2 V1 - - - + + + I1 I2 PWM DC/DC Switching V1 Vint V2 Filter Power Converter − − − - R1 Figure 4.14: A conceptual diagram showing a buck converter serving as a MPPT. battery would be Vload = VB − IRi = 12.7 − 20 × 0.03 = 12.1V and since the voltage that the PVs operate at is determined by the battery voltage, they would also be at 12.1V. An added advantage of the I-V characteristics of PV panel on charging battery is that we can reduce the charging current of the battery when its voltage is reaching the point where the battery is fully charged. In this case, by the self-regulating PV module, we may prevent the battery from overcharging. The design of such control is by setting the PV panel MPP before the the fully-charged battery voltage point. Since the operating voltage when PV panel and battery are connected is determined by the battery voltage, we may arrange the PV open-circuit voltage close to the battery fully-charged voltage to achieve such charging current limiting function. 4.7 Maximum Power Point Tracker (MPPT) Since from Fig. 4.9 that the resistance value is ﬁxed therefore we can only obtain one maximum power point (MPP) out of a particular I-V combination of the PV panel. One would think that if it is possible to vary the resistance value so that we may still get the MPP whenever the insolation changes. The answer is yes. We may employ a maximu power point tracker (MPPT) which is able to track the insolation and change the internal resistance in order to obtain MPP at that particular insolation level. A common technique to implement a MPPT is to use a switching (or switch-mode) power
4. Solar Energy II 80 Ton D= Voltage Tperiod - Ton Vint 6 V1 - Tof f ............................................................... 6 V2 - Tperiod Time Figure 4.15: Deﬁnition of duty cycle: the ratio of on-time to the whole switching period. converter. A conceptual diagram of a voltage step-down switching power converter (or a buck converter) is shown in Fig. 4.14. The buck converter consists of a pulse-width- modulated dc/dc switching part or (a pulse-width-modulation PWM block) and an output ﬁlter. We can see from Fig. 4.14 that after the input voltage V1 has passed through the PWM block, the intermediate voltage Vint becomes a train of pulses with peak voltage equals V1 . A detailed inspection of Vint waveform is shown in Fig. 4.15. Here we deﬁne an important term, the duty cycle. It is the ratio of on-time Ton to one switching period Tperiod : Ton Ton D= = (4.17) Ton + Tof f Tperiod The output ﬁlter in Fig. 4.14 is used to smoothen the pulses to become a theoretically ripple-free voltage V2 . We can relate the two voltages by inspecting and equating the area of V1 during Ton and the area of V2 during Tperiod , that is V2 × Tperiod = V1 × Ton (4.18) Ton V2 = V1 Tperiod = V1 D Assume there is no loss in the buck converter, then the input power equals output power, that is P1 = P2 (4.19) V1 I1 = V2 I2 Substitute (4.18) into (4.19) we obtain I1 I2 = (4.20) D On one hand, if we connect a resistor R2 across the output terminals of the buck converter, I2 will ﬂow through R2 and V2 is established. V2 R2 = (4.21) I2
4. Solar Energy II 81 On the other hand, we can view the input of buck converter as a equivalent resistance of, say, R1 . Therefore we can write the following V1 R1 = (4.22) I1 Substitute (4.18), (4.20) and (4.22) into (4.21), we get R2 R1 = (4.23) D2 We can see that the buck converter is eﬀectively a resistance converter which varies the equivalent resistance of R1 by changing the value of duty cycle D. Example: Under certain ambient conditions, a PV module has its maximum power point at Vm = 15V and Im = 0.6A. What duty cycle should an MPPT (using buck converter) have if the module is delivering power to a 5ohm resistance? Solution: The maximum power delivered by the PV panel is P = 15V × 0.6A = 9W. Using (4.22) and (4.23) we can ﬁnd the duty cycle directly R2 · I1 5Ω × 0.6A D= = = 0.447 V1 15V
Bibliography [1] G. M. Masters, Renewable and Eﬃcient Electric Power Systems, John Wiley & Sons, 2004. 82