Masters G.M. Renewable and Efficient Electric Power Systems

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568 PHOTOVOLTAIC SYSTEMS

c. At C/20, the current is 100 Ah/20 h = 5 A . The input voltage needed to

drive 5 A through the 0.035- resistance is

= V

+ IR = 12.5 + 5 × 0.035 = 12.68 V

So at the C/20 = 5-A rate, the losses are now only

Power lost in R

Input power

(5)

× 0.035

12.68 × 5

= 0.014 = 1.4%

The simple Thevenin model of a battery is complicated by the fact that both

and R

depend on the battery’s state of charge (SOC), its temperature, and its

history. But since the model is so simple, it does help provide an intuitive under-

standing of the actual charging and discharging curves for a battery. Figure 9.45

shows representative values of battery voltage for differing charging and dis-

charging currents as a function of SOC.

During charging, notice the sudden rise in cell voltage around the 14-V level as

the battery nears full charge. This is when charging is most inefﬁcient and when

gassing occurs. The release of generated hydrogen and oxygen gases removes

water from the battery, which has to be replaced or the plates may be damaged. It

also creates a potentially dangerous situation since hydrogen gas is so explosive.

In addition, very small quantities of the poisonous gases arsine (AsH

) and stibine

(SbH

) can be released when hydrogen comes in contact with the lead alloys

arsenic and antimony. Proper ventilation is clearly an important consideration in

the design of a safe battery storage system.

To reduce the need for water replacement, some lead-acid batteries now use

pressure-relief valves that allow most of the oxygen formed during overcharging

to recombine with lead rather than being released. Hydrogen gas releases are

also suppressed. While they do open as necessary to relieve the pressure built up

by hydrogen and oxygen gases, these valve-regulated lead-acid batteries (VRLA)

can reduce gas emissions by over 95% (Linden, 1995).

Gassing and charging losses can be minimized, by using a c harge controller

that has been designed to slow the charging rate as the battery approaches its fully

charged condition. Charge controllers also protect batteries from overcharging by

completely disconnecting the PV array at some predetermined battery voltage,

usually around 14 V (for a 12-V battery). They also keep the batteries from being

overly discharged by disconnecting the load when battery voltage drops below

another set point, usually around 11.5 V.

9.5.7 Battery Sizing

If good weather could be counted on, battery sizing might mean simply providing

enough storage to carry the load through the night and into the next day until

STAND-ALONE PV SYSTEMS 569

120100806040200

11.0

11.5

12.0

12.5

13.0

13.5

14.0

14.5

15.0

15.5

16.0

Battery State of Charge (%)

Battery Voltage (V)

/10

/20

/100

/10

/20

/40

DISCHARGING

CHARGING

Figure 9.45 Terminal voltage and state of charge for 12-V lead-acid batteries for various

rates of charging and discharging. Based on Sandia National Laboratories (1991).

the sun picks up the load once again. The usual case, of course, is one in which

there are periods of time when little or no sunlight is available and the batteries

might have to be relied on to carry the load for some number of days. During

those periods, there may be some ﬂexibility in the strategy to be taken. Some

noncritical loads, for example, might be reduced or eliminated; and if a generator

is part of the system, a trade-off between battery storage and generator run times

will be part of the design.

Given the statistical nature of weather and the variability of responses to

inclement conditions, there are no set rules about how best to size battery storage.

The key trade-off will be cost. Sizing a storage system to meet the demand

99% of the time can easily cost triple that of one that meets demand only 95%

570 PHOTOVOLTAIC SYSTEMS

8765432

Peak Sun Hours

Days of Usable Storage

99% Availability

95% Availability

Figure 9.46 Days of battery storage needed for a stand-alone system with 95% and 99%

system availability. Peak sun hours are for the worst month of the year and availability

is on an annual basis. Based on Sandia National Laboratories (1995).

of the time. As a starting point for estimating the number of days of storage

to be provided, consider Fig. 9.46, which is based on the excellent guidebook

Stand-Alone Photovoltaic Systems Handbook of Recommended Design Practices

(Sandia National Laboratories, 1995). The graph gives an estimate for days of

battery storage needed to supply a load as a function of the peak sun hours

per day in the design month, which is the month with the worst combination of

insolation and load. To account for a range of load criticality, two curves are

given: one for loads that must be satisﬁed during 99% of the 8760 h in a year

and one for less critical loads, for which a 95% system availability is satisfactory.

When doing designs with a computer, it is handy to have equations for the

curves given in Fig. 9.46. The following match pretty well.

Storage days (99%) ≈ 24.0 − 4.73 (Peak sun hours)

+ 0.3 (Peak sun hours)

(9.30)

Storage days (95%) ≈ 9.43 − 1.9 (Peak sun hours)

+ 0.11 (Peak sun hours)

(9.31)

Figure 9.46 refers to days of “usable storage,” which means after accounting for

impacts associated with maximum allowable battery discharge, Coulomb efﬁ-

ciency, battery temperature, and discharge rate. The relationship between usable

storage and nominal, rated storage (at C/20, 25

◦

C) is given by

Nominal (C/20, 25

◦

C) battery capacity =

Usable battery capacity

(MDOD)(T, DR)

(9.32)

STAND-ALONE PV SYSTEMS 571

where MDOD stands for maximum depth of discharge (default: 0.8 for lead-acid,

deep-discharge batteries, 0.25 for auto SLI; subject to freeze constraints given in

Fig. 9.39) and T, DR stands for temperature and discharge-rate factor (Fig. 9.42).

The following example illustrates the process.

Example 9.18 Battery Sizing for an Off-Grid Cabin. A cabin near Salt Lake

City, Utah, has an ac demand of 3000 Wh/day in the winter months. A decision

has been made to size the batteries such that a 95% system availability will be

provided, and a back-up generator will be kept in reserve to cover the other 5%.

The batteries will be kept in a ventilated shed whose temperature may reach as

low as −10

◦

C. The system voltage is to be 24 V, and an inverter with overall

efﬁciency of 85% will be used.

Solution. With an 85% efﬁcient inverter, the dc load is

DC load =

AC load

Inverter efﬁciency

3000 Wh/day

0.85

= 3529 Wh/day

With a 24-V system voltage the batteries need to supply

Load =

3529 Wh/day

24 V

= 147 Ah/day @ 24 V

From Appendix E the following monthly insolation data are found for Salt

Lake City:

Tilt Jan Feb Mar Apr May Jun July Aug Sept Oct Nov Dec Year

Lat − 15 2.9 4.0 5.0 5.9 6.6 7.2 7.3 7.0 6.3 5.0 3.3 2.5 5.2

Lat 3.2 4.3 5.2 5.8 6.2 6.6 6.7 6.7 6.4 5.4 3.7 2.9 5.3

Lat + 15 3.4 4.4 5.1 5.4 5.5 5.6 5.8 6.1 6.1 5.5 3.9 3.1 5.0

Even though annual insolation is highest for a tilt angle equal to the local latitude,

we’ll use an L + 15 tilt to help meet winter needs. December has the lowest

insolation so that will be our design month. From Fig. 9.46 at 95% availability

and 3.1 peak sun hours in December, it looks like we need about 4.6 days

of storage.

Usable storage = 147 Ah/day × 4.6day= 676 Ah

We’ll pick deep discharge lead-acid batteries that can be routinely discharged

by 80%. But, we need to check to see whether that depth of discharge will

expose the batteries to a potential freeze problem. From Fig. 9.39, at −10

◦

Cthe

572 PHOTOVOLTAIC SYSTEMS

batteries could be discharged to over 95% without freezing the electrolyte, so an

80% discharge is acceptable.

Batteries that are nominally rated at C/20 and 25

◦

C will be operated in much

colder conditions which degrades storage capacity, but they will be discharged

at a much slower rate, which increases capacity. Figure 9.42 suggests that at

−10

◦

CandaC/72 rate (the 5-day rate for the cabin is even slower, so this is

conservative), storage capacity would be about 0.97 times the nominal capacity.

Applying the 0.80 factor for maximum discharge and the 0.97 factor for dis-

charge rate and temperature in (9.32) gives

Nominal (C/20, 25

◦

C) battery capacity =

676 Ah

0.80 × 0.97

= 871 Ah (at 24 V)

Table 9.15 can help us pick a battery. None of those batteries comes close to

providing that many Ah, so we’re going to have to have parallel strings.

Partly to keep the weight of each battery low enough to be able to handle,

and partly to keep the number of batteries under control, suppose we choose the

Trojan 6 V, 225 Ah, T-105. Three parallel strings would give us 675 Ah, four

would give 900 Ah, and our goal is 871 Ah. Suppose we slightly oversize with

four strings.

To get 24 V, we need each string to have four batteries, so the total battery

bank will have four parallel strings of four batteries each as shown in Fig. 9.47.

9.5.8 Blocking Diodes

The simplest PV–battery system consists of just a single module connected to

a battery and a load, with no charge controller, inverter, or anything else to

−

24 V

900 Ah

6 V, 225 Ah ea.

Figure 9.47 The 24-V, 900-Ah battery bank for the cabin in Example 9.18.

STAND-ALONE PV SYSTEMS 573

−

Load

Ah/d to

battery

Ah/d to

load

−

Blocking

diode

(a)

(b)

Figure 9.48 Simplest PV–battery system (a). Adding a blocking diode to prevent losses

from the battery through the PV at night (b).

complicate things (Fig. 9.48a). Such a system might provide someone with a bit

of light at night and maybe a few other simple amenities. As long as the user is

careful about not letting the batteries discharge too deeply, or be overcharged, the

system will perform well. There is another concern, however. The system shown

in Fig. 9.48a allows the battery to leak current back through the PV module

at night, which raises the question of whether it might be worthwhile to add a

blocking diode as shown in Fig. 9.48b to prevent that nightly discharge.

The equivalent circuit of a single PV cell shown in Fig. 9.49a will help us

analyze the potential nighttime battery loss problem. Remember that the black-

ened symbol for a diode means that it is a real diode, rather than an ideal one.

Ignoring the insigniﬁcant impact of the very small series resistance and elimi-

nating the ideal current source I

because the cell is in the dark at night leaves

the simple circuit of Fig. 9.49b.

Current through the diode in the equivalent circuit for a cell (at 25

◦

C) is given

by the Shockley equation introduced in Chapter 8:

= I

38.9V

− 1)(9.33)

(a)

−

n-cells

(b)

Figure 9.49 Nighttime leakage from a battery back through a PV module with n cells.

(a) Equivalent circuit of one PV cell. (b) A simpliﬁed equivalent circuit at night for one

cell having V

/n volts from the battery across it.

574 PHOTOVOLTAIC SYSTEMS

The nighttime current from the battery through each cell will be

= I

+ I

= I

38.9V

− 1) +

(9.34)

where the voltage V

across the diode will be equal to the battery voltage V

divided by the number of cells n in the PV module.

With this simple nighttime equivalent circuit, we can decide how much leakage

will occur from the battery through the PVs. The following example shows how

to evaluate the potential advantage of using a blocking diode to prevent that

current from ﬂowing.

Example 9.19 Impact of a Blocking Diode to Control Nighttime Battery

Leakage. A PV module is made up of 36 cells, each having a reverse saturation

current I

of 1 × 10

−10

A and a parallel resistance of 8 . The PVs provide the

equivalent of 5 A for 6 h each day. The module is connected without a blocking

diode to a battery with voltage 12.5 V.

a. How many Ah will be discharged from the battery over a 15-h night?

b. How much energy will be lost due to this discharge?

c. If a blocking diode is added, how much energy will be dissipated through

the diode during the daytime. Assume the diode while conducting has a

voltage drop of 0.6 V.

Solution. The voltage across each PV cell will be about 12.5 V/36 cells =

0.347 V. From (9.34) the current discharged from the battery while the PV is in

the dark will be

= 10

−10

38.9×0.347

− 1) +

0.347

= 0.000073 + 0.043 = 0.043 A = 43 mA

a. Over a 15-h nighttime period, the loss in Ah from the battery will be

Nighttime loss = 0.043 A ×15 h = 0.65 Ah

b. At a nominal 12.5 V, the energy loss at night will be

Nighttime loss = 0.65 Ah ×12.5V= 8.1Wh

c. During the day the PVs will deliver

PV output = 6h× 5A= 30 Ah

STAND-ALONE PV SYSTEMS 575

The nighttime loss without a blocking diode is 0.65 Ah/30 Ah = 0.02; that

is, 2% of the daytime gains.

With the blocking diode dropping 0.6 V, the daytime loss caused by that

diode is

Blocking diode loss = 30 Ah × 0.6V= 18 Wh

In the example just presented, the blocking diode loses more energy during the

day while it is conducting (18 Wh) than it saves overnight (8.1 Wh). Another

way to look at it is that without a blocking diode, only about 2% of the Ah

of daytime solar gains are lost overnight. In spite of these arguments, blocking

diodes still make some sense. Since the operating point of the battery–P V system

is generally some distance to the left of the knee of the I –V curve, shifting

the battery I –V curve approximately 0.6 V to the right to cover the diode drop

barely changes current delivered by the PVs. That is, in terms of amp-hours, the

diode doesn’t lose any during battery charging, but does stop nighttime battery

discharge, so in terms of amp-hours it does offer a modest net beneﬁt.

9.5.9 Sizing the PV Array

Designing stand-alone PV–battery systems is clearly much more demanding

than sizing grid-connected systems. Month-by-month load estimates and solar

resource evaluations, making trade-offs between ac and dc loads, choosing a sys-

tem voltage, and determining battery storage with or without a back-up generator

are things that simply don’t apply to grid-connected systems. Having addressed

those topics, we can now deal with the most important part of the system: the

PV array itself.

In Fig. 9.50 a 1-sun, PV I –V curve has been drawn along with a vertical

I –V line for a battery. As shown, during battery charging, the operating point of

the PVs is almost always above the knee of the PV I –V curve, which means that

charging current will exceed the rated current of the PVs. It is a fairly conservative

estimate therefore to simply use the rated current of the PVs as an indication of the

battery charging current at 1-sun insolation. There are circumstances in which

this assumption should be checked—as, for example, when a 12-V battery is

charged in a high-temperature environment with a “self-regulating” PV module

having fewer than the usual 36 cells in series. Fewer cells and higher temperatures

move the maximum power point (MPP) toward the battery I –V curve, and the

conservatism of the assumption decreases.

Our simple sizing procedure will be based on the same “peak hours” approach

used for grid-connected systems, except that it will be applied to current rather

than power. So, for example, an area with 6 kWh/m

-day of insolation is treated

as if it has 6 h/day of 1-sun, 1-kW/m

radiation. Then using the rated current

at 1-sun, times peak hours of sun, gives us amp-hours of current provided to

the batteries.

576 PHOTOVOLTAIC SYSTEMS

Voltage

Current

Rated current

Maximum

Power Point

Battery current

Battery

−

Curve

"1-sun"

−

Curve

Figure 9.50 Estimating battery charging at 1-sun to be the rated current of the PVs is a

fairly conservative assumption.

Notice that the operating point for battery charging is usually some distance

away from the MPP. This means that a considerable fraction of power that the

PVs could provide based on the rated power P

of the module is not being

delivered to the batteries. It would not be appropriate, therefore, to multiply the

peak hours of sunlight times the rated power of the module to estimate energy to

the batteries.

The product of rated current I

times peak hours of insolation provides a good

starting-point estimate for Ah delivered to the batteries. It is common practice

to apply a de-rating of about 10% to account for dirt and gradual aging of the

modules. The temperature and module-mismatch factors that were quite important

for grid-connected systems with MPP trackers are usually ignored for PV–battery

systems. This is because the operating point for battery systems is far enough

away from the knee of the I –V curve that those variations are minimum and are

to some extent offset by the conservatism associated with assuming that charging

current is only I

Here is another important feature of stand-alone PV sizing. It will be based

on (a) amp-hours from the PVs into the batteries and (b) amp-hours from the

batteries to the load. That is, the appropriate battery efﬁciency measure will be

the Coulomb efﬁciency. The current delivered to the batteries needs to multiplied

by the Coulomb efﬁciency (Ah

out

/Ah

) to give Ah delivered from the batteries

to the load:

Ah to load = I

× Peak sun hours × Coulomb efﬁciency × De-rating factor

(9.35)

The only other thing to keep track of is the system voltage. For a 12-V system

voltage and 12-V modules, modules are added in parallel until sufﬁcient Ah are

provided for the load. For a 24-V system voltage and 12-V modules, two modules

in series are needed to provide the 24 V, and then parallel strings of two series

STAND-ALONE PV SYSTEMS 577

modules each are added to deliver the Ah needed by the load. Or, if 24 V is

the system voltage, it may make more sense to simply choose 24-V PV modules

rather than using two 12-V versions.

Example 9.20 PVs for the Cabin in Salt Lake City. The cabin from Example

9.18 needs 3000 Wh/day of ac delivered from an 85%-efﬁcient inverter. For a

24-V system voltage, a 90% Coulomb efﬁciency, and 10% de-rating (de-rating

factor = 0.90), size a PV array using Kyocera KC120 modules.

Solution. From Table 9.3, the Kyocera KC120 is a 120-W module with its

maximum power point at a current of 7.1 A and a voltage of 16.9 V. From

Example 9.18, the worst solar month is December, with 3.1 peak hours of sun-

light at a tilt of L + 15. Using (9.35), one string of modules will therefore deliver

in December about

Ah to inverter = 7.1A× 3.1h/day× 0.90 × 0.90 = 17.83 Ah/day per string

For an 85% efﬁcient inverter to deliver 3000 Wh/day of 120 V ac, it needs a

24-V dc input of

Inverter dc input =

3000 Wh/day

0.85 × 24 V

= 147 Ah/day @ 24 V

Since these modules have a rated voltage of 16.9 V, they are nominally “12-V

modules.” Therefore two modules are needed in series to provide a single 24-

V string.

The number of parallel strings of modules needed is

Number of parallel strings =

147 Ah/day

17.83 Ah/day-string

= 8.25 strings

Suppose that we undersize it slightly and use eight parallel strings with two

modules per string, for a total of 16 modules.

Including the 0.90 de-rating factor, the PVs will deliver

PV output = 8 strings × 7.1 A/string × 3.1h/day× 0.90

= 158 Ah/day @ 24 Vdc

The batteries with 0.90 Coulomb efﬁciency will deliver

Battery output = 158 Ah/day × 0.90 = 142 Ah/day @ 24 Vdc