Chapter 4 Solar Energy ppsx

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 shad

Solar Energy II

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)

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

ISC = Id+ Ip+ I (4.1) Substitute (3.23) into (4.1) and re-arrange the terms, we obtain

I = ISC−I0(eqVd /kT −1)−Vd

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 Vdto 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 × 10−10A 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

kT = 38.9 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

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

Load V

6 I SC V d

I d

-I +

−

R p

R s

x

I p

−

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

cell

PV

cell

PV

cell

PV

cell

PV

cell

PV

cell

+

− V

6 I 1 6 I 2

+

−

V a

+

−

V a

+

−

V a

Current

Voltage

I

I1, I2

Figure 4.3: PV cells arranged in an array to increase voltage and current capability

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 ISCincreases at a rate of 0.065%/◦C and VOCdecreases 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

Tcell= Tamb+ (NOCT − 20◦C

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

Tcell = 27 + (47 − 20

0.8 ) · 1 = 60.75

◦C 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 max = 10W[1 − 0.005(60.75 − 25)]W = 8.21W

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

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

Vn−1= (n − 1

where V is the output voltage when n cells receive sunlight and no shading happens Substituting (4.8) into (4.7) gives

VSH = (n − 1

n )V − I(Rs+ Rp) (4.9)

Figure 4.4: I-V characteristics under various cell temperatures for the BP Solar SX5 and SX10 PV modules

6

(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 − VSH = V − (1 − n1)V + I(Rs+ Rp)

= Vn + I(Rs+ Rp)

(4.10)

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

n + I(Rs+ Rp) =

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 would be

Pmodule = V I = 5.23V × 2.25A = 11.77W

6

R s

R p

I SC I d ?

I1

I

D bypass



I2

6

I Rp

I

6 6

I 1

6

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 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

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

I1 = 0.6 5.6 + 0.005 = 0.11A 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

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

V +

−

-I

PV

Figure 4.7: A PV module powers a variable resistor

Figure 4.8: Different operating points by varying the resistance

Figure 4.9: Different operating points by varying the resistance.

e = kω +

-I

PV Panel

− M +

− V

R a

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

I-V characteristics can be written as

where kω is the back emf related by angular speed ω and Rais 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+ RiI (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

I = 1

RiV −

1

Figure 4.11: I-V curve of a dc motor on a PV I-V characteristic Source: Masters 2004.

V B

+

-I

PV Panel

−

+

− V

R i

Figure 4.12: Equivalent circuit of a PV module connected to a battery

-

V B,old V B,new

slope = 1

Ri

Voltage Charging

6

-V B,old

V B,new

slope = −1

Ri

Voltage

Discharging

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

RiV +

1

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

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+ RiI = 11.7 + 0.03 × 6 = 11.88V

(b) While drawing 20A after VB has reached 12.7V, the output voltage of the

PWM DC/DC Switching

-I 1

V1

I 2

-R1

V 2

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

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