Download lecture 4:pv system

LECTURE 4
PHOTOVOLTAIC
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
1
I   V
• For the load curve,
 R
• 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:
Vm
Rm 
Im
• Where Vm and Im are voltage and current at maximum
power point (MPP).
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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 permanentmagnet dc motor.
• As motor spins, it develop a back emf proportional to the
speed of motor. The V-I relationship for dc motor is:
V  IRa  k
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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.
V  VB  Ri I
• 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).
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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
Refrigerator
Steady state
(Watts)
300
Surge
(Watts)
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:
500 Ah / day  2days
Battery storage 
 1667 Ah
0.60
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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:
1667
Battery storage( rated condition) 
 2083 Ah
0.8
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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 amphour 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:
E in  VC I C TC
• Suppose that the battery is discharged at ID and VD over
a period of time ΔTD, delivering energy:
E out  V D I D TD
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• The energy efficiency of battery:
Energy efficiency 
E out V D I D TD

E in
VC I C TC
• The current (A) x time (h) is Coulomb charge expressed
as Ah, then the above equation becomes:
 VD
Energy efficiency  
 VC
 I D TD

 I C TC
  VD
  
  VC
 coulombs out , Ahout

 coulombs in, Ahin



• The ratio of discharge voltage to charge voltage –
voltage efficiency of the battery and the ratio of Ahout to
Ahin – Coulomb efficiency.
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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:
No min al (C / 20,25 0 C ) battery capacity 
Usable battery capacity
( MDOD )(T , DR)
• Where MDOD stands for max depth of discharge
(default:0.8 for lead-acid) and T,DR stands for
temperature and discharge-rate factor.
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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:
I d  I 0 (e 38.9Vd  1) at 25o C
• The nighttime current from the battery through each cell
will be:
V
38.9V
I B  I d  I Rp  I 0 (e
d
 1) 
d
Rp
• 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.
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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:
I B  I d  I Rp  10 10 (e 38.90.347  1) 
0.347
 0.000073  0.043  43mA
8
• 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 12.5S.A.A
= 06/07
6.45 Wh
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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
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