Chapter 4 Solar Energy ppsx

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 P V p anel, which is made up of an array of PV cells. Later in this chapter we will discuss the concept of maximum p ower 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 an d 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 P V panel I-V characteristics. We need some path to be added to the model to show the cur rent flow within the PV cells. Fig. 4.1 shows a more accurate PV cell equivalent circuit using one parallel resistor, R p , and one series resistor, R s , to realize the extra current paths. L et us first derive a mathematical expression for this equivalent circuit. By KCL, the cur rent enters node 65 4. Solar Energy II 66 x equal the currents leave this node: I SC = I d + I p + I (4.1) Substitute (3.23) into (4.1) and r e-arrange the terms, we obtain I = I SC − I 0 (e qV d /kT − 1)− V d R p (4.2) And V d and V are related by the following equation V d = I · R s + V (4.3) We cannot find a solution of (4.2) with close-loop form. However we may first use an assumed value of V d 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 th e PV cell has short-circuit current I SC = 4A and at 25 ◦ C its reverse saturation current is I 0 = 6 × 10 −10 A. Show the I-V curves under different resistance valu es (a) R p = ∞, R s =0; (b) R p = ∞, R s = 0.06Ω; (c) R p = 2.0Ω, R s = 0Ω; and (d) R p = 2.0Ω, R s = 0.06Ω. Solution: With the given temperature and reverse saturation current conditions, q kT = 38.9 Using the 4 set of values we can now plot the curves as sh own in Fig. 4.2. This figu re shows that both the series and parallel resistances decrease the m aximum 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 conn ected 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 Load V ✻ V d I SC I d ❄ ❄ ✲ I + − R p R s x I p PV + − 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 PV cell PV cell PV cell PV cell PV cell PV cell + − V ✻ ✻ ✲ I I 2 I 1 + − V a + − V a + − V a Current Voltage I I 1 , I 2 V a 2V a 3V a Figure 4.3: PV cells arr an ged 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/m 2 (1 sun) with special distribution of air mass ratio of 1.5 (AM 1.5). The cell temp erature 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-circu it current increases slightly. From the datasheet it is shown that I SC increases at a rate of 0.065%/ ◦ C and V OC 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/m 2 , and wind speed of 1m/s. For example, the BP SX5/10 has a NOCT of 47 ◦ C. To evaluate the cell temperature T cell in ◦ C at different ambient temperatures T amb in ◦ C, we may use the following expression T cell = T amb + ( NOCT − 20 ◦ C 0.8 ) · S (4.4) 4. Solar Energy II 69 where S is the solar insolation measured in kW/m 2 . 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 mod ule has a NOCT of 47 ◦ C, V OC at 16.8V, temperature co efficient of V OC -80mV/ ◦ C, and temperature coefficient of power -0.5%/ ◦ C. Solution: Using (4.4) with S = 1kW/m 2 , cell temperature will become T cell = 27 + ( 47 − 20 0.8 ) · 1 = 60.75 ◦ C The V OC drops by V OC -80mV/ ◦ C, the new value will be V OC = 16.8V − 0.08(60.75 − 25)V = 13.94V The maximum power output will drop according to -0.5%/ ◦ C, therefore we have P new max = 10W[1 − 0.005(60.75 − 25)]W = 8.21W 4.5 Shading Impacts on I-V Curves From the equivalent circuit showning n cells in Fig. 4.5, the cur rent I consists of three currents and is given by I = I SC − I d + I Rp (4.5) If the nth cell is completely shaded, I SC = I d = 0, then (4.5) reduces to I = I Rp (4.6) The output voltage V SH can be written as V SH = V n−1 − I(R s + R p ) (4.7) We may express the voltage of (n-1) cells as V n−1 = ( n − 1 n )V (4.8) where V is the output voltage when n cells receive sunlight and no s hading happens. Substituting (4.8) into (4.7) gives V SH = ( n − 1 n )V − I(R s + R p ) (4.9) 4. Solar Energy II 70 Figure 4.4: I-V characteristics under variou s cell temperatures for the BP Solar SX5 and SX10 PV mod ules. 4. Solar Energy II 71 ✻ ✻ (n-1) cells R s R p I SC ❄ I d ✛ I Rp I + ✛ − I + − V SH nth cell ✲ I V n−1 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 ∆V = V − V SH = V − (1 − 1 n )V + I(R s + R p ) = V n + I(R s + R p ) (4.10) Example: The 36-cell PV module has a parallel resistance per cell of R P = 5.6Ω and series resistance per cell of R S = 0.005Ω. In full sun and at current I = 2.25A the output voltage was f ou nd 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 = V n + I(R s + R p ) = 18.35 36 + 2.25(0.005 + 5.6) = 13.12V The new output voltage will be (18.35-13.12)V = 5.23V. Power delivered by the module with one cell shaded completely wou ld be P module = V I = 5.23V × 2.25A = 11.77W 4. Solar Energy II 72 ✻ ✻ R s R p I SC ❄ I d I 1 I D bypass ✛ I 2 ✻ I Rp I ✻ ✻ I 1 ✻ I 1 I 2 ✲ 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 cu rrent of the PV module will flow thr ough R P and R S . Therefore the drop across the shaded cell is V shaded cell = I(R s + R p ) = 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 th e cell. In order to p revent 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 s how how the current is bypassed. Example: A bypass diod e is added to one P V 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 D bypass is constant at 0.6V from 0-2.5A of diode current, find the values of I 1 and I 2 . Solution: By KCL we have I = I 1 + I 2 = 2.5A (4.11) Since the voltage across the bypass diode equals to that of series and parallel resistances, that is 0.6 = I 1 (R P + R S ) = I 1 (5.6 + 0.005) Therefore I 1 is calculated as I 1 = 0.6 5.6 + 0.005 = 0.11A Hence the current of I 2 can be calculated from (4.11) I 2 = I − I 1 = 2.5 − 0.11 = 2.39A We can see that now the cell resistance only condu cts 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 s imply the operating point of the PV modu le? In short, the operating point is defined by the resistance of the load and we may use graphical method to fi nd this point. Let us examine the following different types of loads and how we can find the op erating 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 ✼ V + − ✲ I PV Panel R Figure 4.7: A PV module powers a variable resistor. Figure 4.8: Different operating points by varying the resistance. [...]... and Ri is the internal resistance of the battery Rearranging the terms in (4. 13), we get I= 1 1 V − VB Ri Ri (4. 14) 4 Solar Energy II 77 Figure 4. 11: I-V curve of a dc motor on a PV I-V characteristic Source: Masters 20 04 I + PV Panel Ri + V VB − − Figure 4. 12: Equivalent circuit of a PV module connected to a battery 4 Solar Energy II 78 Current 6 Current Charging I slope = VB,old VB,new... V2 = V1 Tperiod = V1 D (4. 18) 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 I2 = I1 D (4. 20) 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 R2 = V2 I2 (4. 21) 4 Solar Energy II 81 On the other... = −1 Ri 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 I= 1 −1 V + VB Ri Ri (4. 15) where the slope becomes −1/Ri To plot (4. 14) and (4. 15), we need to find the... 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... 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 V1 PWM DC/DC Switching Power Converter − - + Vint I2 Filter − + V2 − - R1 Figure 4. 14: A conceptual diagram showing a buck converter serving as a MPPT battery... is to use a switching (or switch-mode) power 4 Solar Energy II Voltage 80 D=  Ton - Ton Tperiod Vint  Tof f 6 V1 - 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-widthmodulated... 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 : D= Ton Ton = Ton + Tof f Tperiod (4. 17) The output filter in Fig 4. 14 is.. .4 Solar Energy II 75 Figure 4. 9: Different operating points by varying the resistance I + PV Panel V Ra + M − e = kω − 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... 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 R1 = V1 I1 (4. 22) Substitute (4. 18), (4. 20) and (4. 22) into (4. 21), we get R1 = R2 D2 (4. 23) 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:... I= 1 −1 V + VB Ri Ri (4. 15) 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 . happens. Substituting (4. 8) into (4. 7) gives V SH = ( n − 1 n )V − I(R s + R p ) (4. 9) 4. Solar Energy II 70 Figure 4. 4: I-V characteristics under variou s cell temperatures for the BP Solar SX5 and SX10. Rear- ranging the terms in (4. 13), we get I = 1 R i V − 1 R i V B (4. 14) 4. Solar Energy II 77 Figure 4. 11: I-V curve of a dc motor on a PV I-V characteristic. Source: Masters 20 04. V B + ✲ I PV Panel − + − V R i Figure. to I = −1 R i V + 1 R i V B (4. 15) where the slope becomes −1/R i . 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