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LECTURE 4PHOTOVOLTAIC SYSTEM

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Introduction..

• 3 common configuration of PV systems:– Grid-connected PV system– Stand-alone PV– Load connected directly to PV; PV water pumping system.

• Figure 1 shows a grid connected system in which PVs supplying power to a building. PV deliver dc power to power conditioning unit that convert dc to ac and send power to the building.

• Figure 2 shows PV water pumping where wires from array are connected directly to the motor running a pump. When the sun shines, water is pumped. No electric energy storage only potential energy may stored in a tank.

• Figure 3 shows stand-alone system with battery storage and a generator for back-up power. An inverter convert battery dc voltages to ac for household electricity.

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Grid-connected PV

PV water pumping

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Stand-alone PV

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Current-voltage curves for loads

• From figure below, the same voltage and current across both PVs and load. I-V curve for load is plotted onto the same graph that has I-V curve for PVs.

• The intersection point between these two curves is called operating point.

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Simple resistive load I-V curve

• For the load curve,• It is a straight line with negative slope. As R increases,

the operating point where PV and resistance I-V curves intersect moves along from left to right.

• Since power delivered is product of current and voltage, one value of resistance that result in maximum power:

• Where Vm and Im are voltage and current at maximum power point (MPP).

VR

I

1

m

mm I

VR

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• A module supplying power to a resistive load. As resistance changes, the operating point moves around on PV I-V curve.

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DC motor I-V curve

• For dc motor as a load, commonly used is permanent-magnet dc motor.

• As motor spins, it develop a back emf proportional to the speed of motor. The V-I relationship for dc motor is:

kIRV a

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• Electrical characteristic of a permanent dc motor is shown above.

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• At start up, while ω=0, the current rises rapidly with increasing voltage until current is sufficient to create enough starting torque to break the motor loose from static friction.

• Once the motor starts to spin, back emf drops the current and thereafter I rises gradually with increasing V.

• A dc motor I-V curve is superimposed on a set of PV I-V curves.

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• DC motor I-V curve on PV I-V curves for varying insolation.

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Battery I-V curve

• Since PVs only provide power during daylight hours and many application require energy when sun isn’t shining, some method of energy storage is needed.

• An ideal battery is one in which the voltage remains constant no matter how much current is drawn.

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• A real battery has some internal resistance and is modeled with equivalent cct consisting of ideal battery of VB I series with resistance, Ri.

• During the charge cycle, with +ve current flow into battery.

• Which plots as a slightly-tilted, straight line with slope equal to 1/Ri.

• During charging, the applied voltage needs to be greater than VB. As the process continues, VB itself increases so the I-V line slides to the right as shown in figure below (a).

IRVV iB

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• During discharge, the o/p voltage of the battery is less than VB, the slope of I-V line flips and move back to the left in (b).

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Stand-alone PV system

• A general stand-alone system includes a generator backup and a combination charger-inverter.

• As a charger, it converts ac from generator into dc to charge the battery. As an inverter, it converts dc from battery into ac needed by the load.

• The charger-inverter unit may include an automatic transfer switch that allow generator to supply ac loads directly whenever it is running.

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Inverter

• To figure out how much power the battery must supply, we need to take into account the losses in dc-to-ac inverter.

• Most inverter operate at around 90% efficiency over most of their range. For calculation, an overall inverter efficiency of about 85% is considered to be a default assumption.

• When no load is present, a good inverter will power down to less than 1W of standby power. When it senses a load, the inverter powers up and uses 5-20W of its standby power.

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• Typical efficiency of a stand-alone system inverter.

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System voltage

• Inverters are specified by their dc input voltage, ac output voltage, continuous power handling capability and the amount of surge power they can supply for brief periods of time.

• The inverter’s dc input voltage is the same as voltage of battery bank and PV array, is called system voltage.

• The system voltage is usually 12V, 24V or 48V.• Higher voltages need less current, making it easier to

minimize wire losses. Also means more batteries wired in series.

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• One guideline to pick the system voltage is based on keeping the maximum steady-state current drawn below 100A so that the available electrical hardware and wire sizes can be used.

Maximum ac power System dc voltage

<1200W 12V

1200 – 2400W 24V

2400 – 4800W 48V

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• The maximum ac power that inverter needs to deliver can be estimated by adding power demand of all the loads.

Load Watts

Refrigerator 300

Lights 180

TV, active mode 85

Cordless phone 4

Microwave 1000

Washing machine 250

Ceiling fan 100

Maximum ac power demand 1919

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• From table, the total ac power demand with everything tuned on at once is 1919W, which draw 160A if the system voltage only 12V. A 24V system should be chosen which allow more load demand without exceed 100A guideline.

• The important spec. of inverter are the amount of ac power that it can supply on a continuous basis and the ability of inverter to supply surges of current when electric motor is started. Until the rotor starts spinning, there is no back emf to limit motor current and surges can occur many times higher than steady state current.

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Steady-state and surge power requirements for example loads

Load Steady state

(Watts)

Surge

(Watts)

Refrigerator 300 1500

Dishwasher 700 1400

Washing machine 250 750

Dryer 500 1800

Air conditioner 1200 1500

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Batteries

• Used as energy storage.• Lead-acid battery is commonly used as energy storage

in PV system.• Also provide other important energy services for PV

system including ability to provide current surge that much higher than current from the array and automatic property of controlling the output voltage of array so that load receive voltages within their acceptability.

• Other type of batteries – nickel-cadmium, nickel-metal hydride, lithium-ion, nickel-zinc and fuel cell.

• Cheap, high efficiency but short lifetime.

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Battery storage capacity

• Energy storage in battery is given in Amp-hours (Ah) unit at some nominal voltage and at some specified discharge rate.

• A lead-acid battery has a nominal voltage of 2V per cell which means 6 cells for a 12-V battery. The amp-hour capacity at discharge rate that would drain the battery down to 1.75V over a specified period of time at 250C temperature.

• Example: A fully charged 12V battery is specified to have 10 hours, 200Ah capacity could deliver 20A for 10h. At any point, the battery would have a voltage of 10.5V (6x1.75=10.5) and is considered to be fully discharged.

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• Energy is volts x amps x hours but voltage varies throughout the discharge period. To avoid ambiguity, anything relate to battery storage capacity is specified in amp-hours rather than watt-hours.

• A 200Ah battery is delivering 20A is said to be discharging at a C/10 rate; where C refer to capacity of Ah and 10h it take to deplete.

• The amp-hour capacity depends on the rate at which current is withdrawn and also on temperature.

• Long discharge times result in higher Ah capacity.

• Battery capacity and output voltage decreases dramatically in colder condition.

• A rule-of –thumb estimate is battery life is shortened by 50% for every 100C above the optimum 250C operating temperature.

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• Lead-acid battery capacity depends on discharge rate and temperature. Ratio is based on a rated capacity at C/20 and 250C.

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Example 1

• Suppose that batteries located at a remote telecommunications site drop to -200C. If they must provide 2 days of storage for a load 500Ah/day at 12V, how many amp-hours of storage should be specified for the battery bank?

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Solution example 1

• From the figure above, to avoid freezing, the maximum depth of discharge at -200C is about 60%. For 2 days of storage, with discharge of no more than 60%, the batteries need to store:

AhdaysdayAh

storageBattery 166760.0

2/500

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• Since the rated capacity of batteries is likely to be specified at 250C at C/20 rate, we need to adjust the battery capacity to account for our different temperature and discharge period. From the rated figure above, the actual capacity of batteries at -200C discharged over 48h period is 80% of rated capacity. So, we need to specify batteries with rated capacity:

AhconditionratedstorageBattery 20838.0

1667)(

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• Most PV battery system are based on 6V or 12V batteries, which may wired in series and parallel combinations to achieve the needed Ah capacity and voltage rating.

• For batteries wired in series, the voltage add but since the same current flows through each battery, the amp-hour rating of the string is the same as it is for each battery.

• For batteries wired in parallel, the voltage across each battery is the same, currents add, so the amp-hour capacity is additive.

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• In below figure, there is no difference in energy stored in the two-battery series and parallel.

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Coulomb Efficiency

• Battery capacity C is given in amp-hours and charging and discharging are expressed in C/T rates, also amps.

• For example, a battery is charging with a constant current IC over a period of time ΔTC when applied voltage is VC. The input energy to the battery is:

• Suppose that the battery is discharged at ID and VD over a period of time ΔTD, delivering energy:

CCCin TIVE

DDDout TIVE

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• The energy efficiency of battery:

• The current (A) x time (h) is Coulomb charge expressed as Ah, then the above equation becomes:

• The ratio of discharge voltage to charge voltage – voltage efficiency of the battery and the ratio of Ahout to Ahin – Coulomb efficiency.

CCC

DDD

in

out

TIV

TIV

E

EefficiencyEnergy

in

out

C

D

CC

DD

C

D

Ahincoulombs

Ahoutcoulombs

V

V

TI

TI

V

VefficiencyEnergy

,

,

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Battery sizing

• Starting point is estimating the number of days the storage can provide. Days of battery storage needed for stand-alone system with 95% and 99% system availability.

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• 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 (month with the worst combination of insolation and load).

• To account for a range of load critically, 2 curves are given: one for loads must be satisfied during 99% and one for less critical loads, 95% system is satisfactory.

• Y-axis of the graph refer to days of usable storage which means after accounting for impacts relates with maximum allowable battery discharge, Coulomb efficiency, battery temperature and discharge rate.

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• The relationship between usable storage and rated storage (at C/20, 250C) is given by:

• Where MDOD stands for max depth of discharge (default:0.8 for lead-acid) and T,DR stands for temperature and discharge-rate factor.

),)(()25,20/(min 0

DRTMDOD

capacitybatteryUsablecapacitybatteryCCalNo

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Blocking diodes

• The simplest PV battery-system consists of a single module connected to battery and load. The system perform well if the user is careful not to let the battery discharge too deeply or be overcharged.

• One disadvantage is that the system allows the battery to leak current back through PV module at night.

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• The equivalent cct of a single PV cell shown below (a). Ignore impact of a very small series resistance and ideal source current because the cell is in the dark at night leaves a simple cct in (b).

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• Current through diode is given by:

• The nighttime current from the battery through each cell will be:

• Where Vd across the diode will be equal to the battery voltage VB divided by number of cells, n in PV module.

• With the simple nighttime equivalent cct, we can decide how much leakage will occur from battery through PVs.

CateII oVd

d 25)1( 9.380

p

dVRpdB R

VeIIII d )1( 9.38

0

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Example

Impact of a blocking diode to control nighttime battery leakage.

• A PV module is made up of 36 cells, each having I0=1x10-10 A and Rp=8Ω. The PV provide the equivalent of 5A for 6 hours each day. The module is connected without a blocking diode to a battery with voltage 12.5V.

a) How many Ah will be discharged from battery over a 12-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 conducting diode has a voltage drop of 0.6V.

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Solution

• The voltage across each PV cell is 12.5V / 36 cells = 0.347V.• The discharged current from battery while PV is in the dark will

be:

• a) Over a 12-h nighttime period, the loss in Ah from the battery will be:

Nighttime loss = .043 A x 12 h = 0.516 Ah

• b) At nominal 12.5V, the energy loss at night will be:

Nighttime loss = 0.516 x 12.5 = 6.45 Wh

mAeIII RpdB 43043.0000073.08

347.0)1(10 347.09.3810

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• c) During the day, PVs will deliver:

PV output = 6 h x 5 A = 30 Ah

The nighttime loss without blocking diode is 0.516 Ah / 30 Ah = 0.0172; that is 1.72% of daytime gains.

With blocking diode drop of 0.6V, the daytime loss caused by the diode is:

Blocking diode loss = 30Ah x 0.6 = 18Wh

The blocking diode loses more energy during the day while it is conducting (18Wh) than it saves overnight (6.45Wh) or without blocking diode, only about Ah=1.72% of daytime solar gains are lost overnight.

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Sizing PV array

• In figure below, PV I-V curve has been drawn along with a vertical I-V line for a battery.

• During battery charging, the operating point always above the knee of PV curve, which means that charging current will exceed the rated current of PVs.

• The operating point for battery charging is usually some distance away from MPP. This means that a fraction of power that PV could provide based on rated power PR of the module is not being delivered to the batteries.

• The product of IR x peak hours of insolation provides a good estimate for Ah delivered to the batteries.

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• Ah delivered from the batteries to the load is:

Ah to load = IR x peak sun hours x

Coulomb efficiency x derating factor