Transcript Power Electronics Applications

Photovoltaics characteristics
Lecture Note 7
The photovoltaic effect is the basic physical process by which a solar cell
converts sunlight into electricity.
Sunlight contains photons or “packets” of energy sufficient to create electronhole pairs in the n and p regions.
Electrons accumulate in the n-region and holes accumulate in the p region,
producing a potential difference (voltage) across the cell.
When an external load is connected, the electrons flow through the semiconductor
material and provide current to the external load.
The Solar Cell Structure
Although there are other types of solar cells and continuing
research promises new developments in the future, the
crystalline silicon solar cell is by far the most widely used.
n region is much
thinner than the p
region to permit
light penetration
A grid-work of very thin
conductive contact strips
are deposited on top of
the wafer by methods
such as photoresist or
silk-screen
This allows the solar cell to
absorb as much of the sun’s
energy as possible by reducing
the amount of light energy
reflected away from the surface
of the cell.
The contact grid must maximize the surface area of the silicon wafer that be exposed
to the sunlight in order to collect as much light energy as possible.
The conductive grid across the top of the cell is necessary so that the electrons have a
shorter distance to travel through the silicon when an external load is connected.
completed solar cell
a glass or transparent plastic layer is attached to the top of the cell with transparent
adhesive to protect it from the weather.
Operation of a Solar Cell

The sun produces an astounding amount of energy

The small fraction of the sun’s total energy that reaches the earth is enough
to meet all of our power needs many times over. There is sufficient solar
energy striking the earth each hour to meet worldwide demands for an entire
year.

The n-type layer is very thin compared to the p region to allow light
penetration into the p region.

The thickness of the entire cell is actually about the thickness of an eggshell.
the electron becomes a free electron and leaves a hole in the valence band
The amount of energy required to free an electron from the valence
band of a silicon atom is called the band-gap energy and is 1.12 eV (electron volts)
photon penetrates either the n
region or the p-type region
In the p region, the free electron is swept across the depletion region by the
electric field into the n region. In the n region, the hole is swept across the
depletion region by the electric field into the p region. Electrons accumulate in
the n region, creating a negative charge; and holes accumulate in the p region,
creating a positive charge. A voltage is developed between the n region and p
region contacts
When a load is connected to a solar cell via the top and bottom contacts,
the free electrons low out of the n region to the grid contacts on the top
surface, through the negative contact, through the load and back into the
positive contact on the bottom surface, and into the p region where they
can recombine with holes.
Solar Cell Characteristics
Higher light intensity produces more current.
The load on the solar cell controls the maximum operating point (RL / VI).
The operating point for maximum power output for a given light intensity should be in
the “knee” area of the curve, as indicated by the dashed line.
A special method called maximum power point tracking will sense the operating
point and adjust the load resistance to keep it in the knee region.
Example
assume the solar cell is operating on the highest
intensity curve the voltage is 0.5 V and the current is 1.5 A
The output voltage and current of a solar cell is also
temperature dependent
for a constant light intensity the
output voltage decreases as the
temperature increases but the
current is affected only by a
small amount.
Solar Cell Panels
The Solar Power System
Single-Phase Transformer Tap Changers
37

Thyristors can be used as static switches for on-load tap changing of
transformers.

The static tap changers have a very fast switching action.

The circuit diagram of a single-phase transformer tap changer is shown
below.
0  vo  V1
0  vo  (V1  V2 )
V1  vo  (V1  V2 )
38
0  vo  V1
39
0  vo  (V1  V2 )
40
V1  vo  (V1  V2 )
41

The gating pulses of thyristors can be controlled to vary the load voltage.
0  vo  V1
0  vo  (V1  V2 )
V1  vo  (V1  V2 )
ForV1  vo  (V1  V2 )
Dr. Oday A. Ahmed
Cycloconverter
Frequency Converter
Dr. Oday A. Ahmed
This power electronics converter converts input power at one frequency to
output power at a different frequency through one-stage conversion.
Cycloconverter are of two types, namely:
(i) Step-down cycloconverter and
(ii) Step-up cycloconverter.
,
the applications of cycloconverter include the following
(i) Speed control of high-power AC drives
(ii) Induction heating
(iv) For converting variable-speed voltage to constant frequency
output voltage for use as power supply in aircraft or shipboards.
(v) Wind generators
Cycloconverter Configuration
T
α= 0
T
α= 0
α apply after 3 input pulses
T
3T
α=
600
α=
600
T
T
α apply after 3 input pulses
T
3T
Note that when one of the converters
operates the other one is disabled, so that
there is no current circulating between
the two rectifiers.
Constant α operation gives a crude output
waveform with rich harmonic content.
If the square wave can be modified to
look more like a sine wave, the harmonics
would be reduced. For this reason α is
modulated
α=
600
α= 00
T
T
α apply after 3 input pulses
T
3T
Cycloconverter Configuration
α= 0
T
α= 0
T
α apply after 4 input pulses
T
3T
The DC output voltage can be obtained as:
The Fundamental output voltage can be obtained as:
The RMS output voltage can be obtained as:
T
3T
Example:
single-phase bridge-type cycloconverter has input voltage of 230V and 50Hz and
load of R =10Ω. Output frequency is one-third of input frequency. For a firing angle
delay of 300, calculate (a) RMS value of output voltage (b) RMS current of load(c)
RMS current of each thyristor and (d) input power factor.
Solution:
Here Vs =230V, 𝑉𝑚 = 2 × 230, R =10Ω, 𝛼 = 300 =
a
b
𝜋
3
c
Each thyristor handles RMS current for π
radians with a periodicity of 2π rad.
d
Input pf
Input VA
Po = 22.67 2 × 10 = 5.14 𝑘𝑊
Input VA = 230× 22.67 VA
pf
For a Single-phase Bridge Cycloconverter the output voltage waveform of
it is look like the figure below:
2𝑉𝑚
𝜋
3𝑉𝑚
2𝜋
𝑉𝑚
𝜋
T
Find the firing angles of each step in the waveform above and determine the
output frequency
2𝑉𝑚
𝜋
3𝑉𝑚
2𝜋
T
Uninterruptible Power Supply
(UPS)
There are several applications where even a temporary power
failure can cause a great deal of public inconvenience
leading to
large economic
losses. Examples of such
applications are major computer installations, process control
in chemical plants, safety monitors, general communication
systems, hospital intensive care units etc. For such critical
loads, it is of paramount importance to provide an
uninterruptible power supply (UPS) system so as to
maintain the continuity of supply in case of power outages.
Earlier UPS systems were based on rotating machines
arrangement as shown below:
At present, however, static UPS systems are becoming popular
up to a few kVA ratings.
Hence
What is a UPS?
a UPS is a device that Provides backup power when utility
power fails, either long enough for critical equipment to shut
down gracefully so that no data is lost, or long enough to keep
required loads operational until a generator comes online.
Static UPS systems are of two types; namely short-break UPS
and no-break UPS.
In situations where short interruption (4 to 5msec) in supply can be tolerated, the shortbreak UPS shown below can be used.
Short-break static UPS configuration
When a no-break supply is required, the static UPS system
shown below can be used.
No-break UPS configuration
In this system, main ac supply is rectified and the rectifier
delivers power to maintain ,required charge on ,the batteries.
Rectifier also supplies power to inverter continuously which is
then given to ac-type load through filter and normally-on
switch. In case of main-supply failure, batteries at once take
over with no-break of supply to the critical load.
Figure below shows one of no break UPS configuration with
different power electronics Converters.
Switching Regulated DC Power
supplies or Switch-Mode DC Power
Supply (SMPS)
The rising cost of electricity and continuous reductions in the
size of many electronic devices have stimulated recent
developments in switching regulated power supplies. These
supplies are smaller, lighter, and dissipate less power than
equivalent linear supplies.
SMPS is based on the chopper principle. The output DC
voltage is controlled by varying the duty cycle of chopper
by PWM techniques .
SMPS refer to transformer-isolated DC-DC converters. The
typical block diagram of SMPS is shown below:
By employing high switching frequencies, the sizes of the
power transformer and associated filtering components in the
SMPS are dramatically reduced in comparison to the linear
RDPS.
In the uninterruptible power supply system as shown in the Figure
below:
The desktop Personal computer PC consumes a power of 500W supplied by the inverter at a sinusoidal voltage of 220V
and at a frequency of 50Hz. The battery is required to provide a 15-minutes backup time when the utility power fails.
The efficiency of the charger is 80%, the efficiency of the inverter is 75%, and the battery must be restored within 4
hours after the utility power returns. Determine,
1. The average output voltage and current of the charger.
2. RMS values of output and input currents, and supply pf of the charger.
3. The RMS output current of the inverter if the input impedance of the PC power supply is 10Ω.
1. The input power to the inverter,
2. The average battery discharge current, with the assumption that the battery voltage is averaged at 150V
throughout the backup period,
3. The minimum battery capacity required in ampere-hours,
4. The maximum charger input power.
Solution:
1. The circuit diagram of the DC charger is a single-phase full
bridge diode rectifier as shown below
Thus, the average output voltage Vo is
2𝑉𝑚 2 2𝑉𝑠 2 2 × 220
𝑉𝑜 =
=
=
= 198𝑉
𝜋
𝜋
𝜋
1. The output current must be ripple free ⟹ RMS value of output current Ior = average value of output current Io ,
this current can be obtained as shown below:
Since, the output power of the inverter is Po =500W and the inverter’s efficiency 𝜂=75% ⟹ the input power
can be obtained as
𝑃𝑜 500
=
= 667𝑊
𝜂
0.75
The power input of the inverter is the power output of the charger ⟹
𝑃𝑖𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑣𝑒𝑟𝑡𝑒𝑟 = 𝑃𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑟𝑔𝑒𝑟 = 𝑉𝑜 × 𝐼𝑜
𝑃𝑖𝑛 =
⟹
𝑃𝑜 667
𝐼𝑜 = =
= 3.36𝐴 = 𝐼𝑜𝑟
𝑉𝑜 198
RMS value of source current Is
𝐼𝑠 =
𝜋𝐼𝑜2
= 𝐼𝑜 = 3.36𝐴
𝜋
Hence, Supply pf is
𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑟𝑔𝑒𝑟 𝑜𝑟 𝑟𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟
𝑐𝑜𝑠𝜑 =
𝑖𝑛𝑝𝑢𝑡 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐ℎ𝑎𝑟𝑔𝑒𝑟 𝑜𝑟 𝑟𝑒𝑐𝑡𝑖𝑓𝑖𝑒𝑟
𝑐𝑜𝑠𝜑 =
𝑃𝑜
667
=
= 0.9 𝑙𝑎𝑔
𝑉𝑠 × 𝐼𝑠 220 × 3.36
1. The topology of the inverter is a single-phase bridge DC-AC converter, as shown below
The RMS inverter’s output current Iorms is equal to:
𝐼𝑜𝑟𝑚𝑠 =
𝑉𝑜𝑟𝑚𝑠
𝑅
R is the load resistance and its equal to the input impedance of the PC power supply. Thus, R = 10Ω ⟹
220
𝐼𝑜𝑟𝑚𝑠 =
= 22𝐴
10
1.
2.
3.
4.
The input power to the inverter is found in step 2
For an average voltage of 150 V at 667 W within the back-up period, the battery current = 667/150 = 4.45 A
Minimum battery capacity = 4.45 x15/60 = 1.11 Ah (ampere-hour)
At 198V output from the charger,
Charger current to the inverter = 3.36A,
Charger current to the battery = 1.11/4=0.278A
Total charger current = 3.36A+0.278A= 3.638A
Maximum output power of the charger =3.638×198=720W
⟹ For a charger efficiency of 80%, the maximum input power = 720/0.8=900W