Metal Oxide Varistor (MOV)...Basics and Working ...

Metal Oxide Varistor (MOV)...Basics and Working ...

Basics of Varistor

A varistor/voltage dependent resistor (VDR) is a component which has a voltage – current characteristics that is very much similar to that of a diode. This component is used to protect electrical devices from high transient voltages. They are planted in the devices in such a manner that it will short itself when a high current is produced due to the high voltage. Thus the current dependent components in the device will remain safe from the sudden surge.

Metal Oxide Varistor – Basics

MOV is the most commonly used type of varistor. It is called so as the component is made from a mixture of zinc oxide and other metal oxides like cobalt, manganese and so on and is kept intact between two electrodes which are basically metal plates. MOV’s are the most used component to protect heavy devices from transient voltages. A diode junction is formed between each border of the grain and its immediate neighbour. Thus an MOV is basically a huge number of diodes that are connected parallel to each other.  They are designed to be in the parallel mode as it will have better energy handling ability. But, if the component is meant for providing better voltage rating, it is better to connect them in series.
A reverse leakage current appears across the diode junctions of each border when an external tiny voltage is applied across the electrodes. The current produced will also be very small. But, when a large voltage is applied across the electrodes, the diode border junction breaks down as a result of the combination of electron tunnelling and avalanche breakdown. Thus the device is said to show a high level of non-linear voltage – current characteristics. From the characteristics, it should also be noted that the component will have low amount of resistance at high voltages and high resistance at low voltages.
The only problem with this component is that they cannot withstand the transient voltage more than the exceeded rating. They tend to deteriorate after a certain level. If so, they will have to be replaced at times. When they absorb the transient voltage they tend to dissipate it as heat. When this process continues repetitively for some time, the device begins to wear out due to the excessive heat.
They can be connected in parallel for increased energy-handling capabilities. MOVs can also be connected in series to provide higher voltage ratings or to provide voltage rating between the standard increments.

MOV Specifications

  • Maximum working voltage is the maximum steady-state, DC voltage. In this case, the value of the typical leakage current will be lesser than a specified value.
  • Varistor voltage
  • Maximum clamping voltage is obtained when a certain pulse current is applied to the component to obtain a maximum peak voltage.
  • Surge current
  • Surge shift refers to the variation in voltage after a surge current is given.
  • Energy absorption refers to the maximum energy that is dissipated for a certain waveform without many problems.
  • Capacitance
  • Leakage current
  • Response time
  • Maximum AC RMS voltage refers to the maximum amount of RMS voltage that can be delivered to the component.

Working of Metal Oxide Varistor (MOV)

Working of Metal Oxide Varistor (MOV)
Working of Metal Oxide Varistor (MOV)
The working of a MOV is shown in the figure above.
The resistance of the MOV is very high. First, let us consider the component to have an open-circuit as shown in figure 1(a). The component starts conducting as soon as the voltage across it reaches the threshold voltage. When it exceeds the threshold voltage, the resistance in the MOV makes a huge drop and reaches zero. This is shown in the figure 1(b). As the device has very small impedance at this time due to the heavy voltage across it, all the current will pass through the metal oxide varistor itself. The component has to be connected in parallel to the load. The maximum voltage that will pass through the load will be the sum of the voltage that appears across the wiring and disconnect given for the device. The clamp voltage across the MOV will also be added. After the transient voltage passes through the component, the MOV will again wait for the next transient voltage. This is shown in the figure 1(c).

MOV Performance

The varistor is mainly used to perform as a line voltage surge suppressor. The device does not conduct when the voltage across it is below the clamping voltage. But, if a high surge (lighting) that is higher in rate that a varistor can handle is passed through it, the component will not perform. The resulting current will be so high that it will damage the MOV.
The performance of the varistor will slow down with time even if small surges pass through it. The life of a MOV will be explained through the manufacturers chart. The chart will have graphs and readings between the current, time and also the number of transient pulses that passes through the varistor.
Another main reason that affects the performance of a MOV is the energy rating. When there is an increase in the energy rating, there will be an exponential change in the life of the varistor. Thus, there will be a change in the transient pulses that the device can manage. This increases the clamping voltage when each transient breaks down.
The performance can be increased by connecting more varistors in parallel. An increase in rating will also help in the process.
One of the best features of the MOV is its response time. The spikes are shorted through the device within nanoseconds. But the response time can be affected by the mounting design method and inductance of component leads.

Surge Protection for Rectifier Circuits...

Surge Protection for Rectifier Circuits...

                        The rectifier circuits in power supplies are subject to a large variety of transient voltages from lightning or from inductive loads such as motor and coils that produce a spike when they are de-energized. When these transients occur, they may have voltage levels that are two to five times the original supply voltage. The fuses in a circuit are designed to protect a circuit against overcurrent, but they can't detect overvoltage.
                           A solid-state device called a metal oxide varistor (MOV) is designed to be connected in parallel across the input power supply. Fig. 1 shows three diagrams of MOVs connected in power supply circuits. In Fig. 1a, notice that this circuit uses two MOVs. MOV1 is connected in parallel across the incoming voltage lines immediately after the mainline fuse. The MOV exhibits high resistance until its voltage rating is exceeded. This means that if the incoming voltage to the power supply circuit that the MOV is protecting is normal, the MOV will have very high resistance and act like an open circuit. If the incoming voltage increases to a point that exceeds the MOV's voltage rating, the internal resistance of the MOV is quickly reduced to near zero, which will allow maximum current to flow through it. Since MOV1 is connected directly across the two wires of the incoming voltage immediately after the main circuit fuse F1, the excess current that is flowing through the MOV will cause fuse F1 to open.
Electronic diagram that shows two metal-oxide varistors (MOVs). MOV1 is for protection against overvoltage conditions from incoming voltage, & MOV2 provides surge protection against transient voltages that may be produced in an inductive load. (b) Electronic diagram that shows three MOVs connected in a three-phase power supply. (c) Electronic diagram that shows two MOVs connected in a single-phase power supply.
                            Above: Fig. 1 (a) Electronic diagram that shows two metal-oxide varistors (MOVs). MOV1 is for protection against overvoltage conditions from incoming voltage, and MOV2 provides surge protection against transient voltages that may be produced in an inductive load. (b) Electronic diagram that shows three MOVs connected in a three-phase power supply. (c) Electronic diagram that shows two MOVs connected in a single-phase power supply.
                              In some applications the F1 fuse must be sized as a time delay fuse so that inductive loads such as motors can draw locked rotor amperage (LRA) for several seconds when the motor is started. In these cases a second fuse (F2) that is faster acting is added in series with the MOV to protect it. When MOVI goes into conduction, the excessive current it draws will be sufficient to cause both fuses to open. Since fuse F2 is specifically designed to react fast enough to protect the MOV, the MOV will not be damaged by the overcurrent.
MOV2 is connected after the choke in this circuit so that it's parallel to the load. If excessive voltage occurs, the MOV will cause a short circuit which results in excessive current flowing through the choke. The choke will absorb some of this excessive current and protect the MOV until fuse F1 opens and clears the fault.
                            Fig. 1b shows a three-phase rectifier circuit with three MOVs. One MOV is connected across each incoming phase. If the voltage of any phase becomes excessive and exceeds the level of any MOV, the MOV will begin conducting current and cause sufficient current to flow so that the main fuses in the circuit will open.
                            Fig. 1c shows two MOVs connected across the single-phase lines for a fullwave bridge rectifier. If the voltage that is supplied to the bridge rectifier becomes excessive, the MOVs will go into conduction and cause excessive current to flow, which will cause the main fuses to open.

Source : Surge Protection for Rectifier Circuits

How to design a Transformer less Power Supply Circuit ...???

How to design a Transformer less Power Supply Circuit ...???
One of the major problems that is to be solved in an electronic circuit design is the production of low voltage DC power supply from Mains to power the circuit. The conventional method is the use of a step-down transformer to reduce the 230 V AC to a desired level of low voltage AC. The most simple, space saving and low cost method is the use of a Voltage Dropping Capacitor in series with the phase line.
Selection of the dropping capacitor and the circuit design requires some technical knowledge and practical experience to get the desired voltage and current. An ordinary capacitor will not do the job since the device will be destroyed by the rushing current from the mains. Mains spikes will create holes in the dielectric and the capacitor will fail to work. X-rated capacitor specified for the use in AC mains is required for reducing AC voltage.

Schematic of the Capacitor Power Supply Circuit

capacitor power supply circuit
capacitor power supply express pcb layout
X Rated capacitor 400 Volt
225k 400v capacitor
Before selecting the dropping capacitor, it is necessary to understand the working principle and the operation of the dropping capacitor. The X rated capacitor is designed for 250, 400, 600 VAC. Higher voltage versions are also available. The Effective Impedance (Z), Rectance (X) and the mains frequency (50 – 60 Hz) are the important parameters to be considered while selecting the capacitor. The reactance (X) of the capacitor (C) in the mains frequency (f) can be calculated using the formula:
X = 1 / (2 ¶ fC )
For example the reactance of a 0.22µF capacitor running in the mains frequency 50Hz will be:
X = 1 / {2 ¶ x 50 x 0.22 x( 1 / 1,000,000) } = 14475.976 Ohms 0r 14.4 Kilo ohms.
Rectance of the capacitor 0.22 uF is calculated as X = 1/2Pi*f*C
Where f is the 50 Hz frequency of mains and C is the value of capacitor in Farads. That is 1 microfarad is 1/1,000,000 farads. Hence 0.22 microfarad is 0.22 x 1/1,000,000 farads. Therefore the rectance of the capacitor appears as 14475.97 Ohms or 14.4 K Ohms.To get current I divide mains Volt by the rectance in kilo ohm.That is 230 / 14.4 = 15.9 mA.
Effective impedance (Z) of the capacitor is determined by taking the load resistance (R) as an important parameter. Impedance can be calculated using the formula:
Z = √ R + X
Suppose the current in the circuit is I and Mains voltage is V then the equation appears like:
I = V / X
The final equation thus becomes:
I = 230 V / 14. 4 = 15.9 mA.
Therefore if a 0.22 uF capacitor rated for 230 V is used, it can deliver around 15 mA current to the circuit. But this is not sufficient for many circuits. Therefore it is recommended to use a 470 nF capacitor rated for 400 V for such circuits to give required current.
X Rated AC capacitors – 250V, 400V, 680V AC
x rated ac capacitors
Table showing the X rated capacitor types and the output voltage and current without load
capacitor and current
Rectification
Diodes used for rectification should have sufficient Peak inverse voltage (PIV). The peak inverse voltage is the maximum voltage a diode can withstand when it is reverse biased. 1N4001 diode can withstand up to 50 Volts and 1N4007 has a toleration of 1000 Volts. The important characteristics of general purpose rectifier diodes are given in the table.
diode characteristic
diode symbol
So a suitable option is a rectifier diode 1N4007. Usually a silicon diode has a Forward voltage drop of 0.6 V. The current rating (Forward current) of rectifier diodes also vary. Most of the general purpose rectifier diodes in the 1N series have 1 ampere current rating.
DC Smoothing
A Smoothing Capacitor is used to generate ripple free DC. Smoothing capacitor is also called Filter capacitor and its function is to convert half wave / full wave output of the rectifier into smooth DC. The power rating and the capacitance are two important aspects to be considered while selecting the smoothing capacitor. The power rating must be greater than the off load output voltage of the power supply.
The capacitance value determines the amount of ripples that appear in the DC output when the load takes current. For example, a full wave rectified DC output obtained from 50Hz AC mains operating a circuit that is drawing 100 mA current will have a ripple of 700 mV peak-to-peak in the filter capacitor rated 1000 uF.
The ripple that appears in the capacitor is directly proportional to the load current and is inversely proportional to the capacitance value. It is better to keep the ripple below 1.5 V peak-to-peaks under full load condition. So a high value capacitor (1000 uF or 2200 uF) rated 25 volts or more must be used to get a ripple free DC output. If ripple is excess it will affect the functioning of the circuit especially RF and IR circuits.
Voltage Regulation
Zener diode is used to generate a regulated DC output. A Zener diode is designed to operate in the reverse breakdown region. If a silicon diode is reverse biased, a point reached where its reverse current suddenly increases. The voltage at which this occurs is known as “Avalanche or Zener“ value of the diode. Zener diodes are specially made to exploit the avalanche effect for use in ‘Reference voltage ‘regulators.
A Zener diode can be used to generate a fixed voltage by passing a limited current through it using the series resistor (R). The Zener output voltage is not seriously affected by R and the output remains as a stable reference voltage. But the limiting resistor R is important, without which the Zener diode will be destroyed. Even if the supply voltage varies, R will take up any excess voltage. The value of R can be calculated using the formula:
R = Vin – Vz / Iz
Where Vin is the input voltage, Vz output voltage and Iz current through the Zener
In most circuits, Iz is kept as low as 5mA. If the supply voltage is 18V, the voltage that is to be dropped across R to get 12V output is 6volts. If the maximum Zener current allowed is 100 mA, then R will pass the maximum desired output current plus 5 mA .
So the value of R appears as:
R = 18 – 12 / 105 mA = 6 / 105 x 1000 = 57 ohms
Power rating of the Zener is also an important factor to be considered while selecting the Zener diode. According to the formula P = IV. P is the power in watts, I current in Amps and V, the voltage. So the maximum power dissipation that can be allowed in a Zener is the Zener voltage multiplied by the current flowing through it. For example, if a 12V Zener passes 12 V DC and 100 mA current, its power dissipation will be 1.2 Watts. So a Zener diode rated 1.3W should be used.
LED Indicator
LED indicator is used as power on indicator. A significant voltage drop (about 2 volts) occurs across the LED when it passes forward current. The forward voltage drops of various LEDs are shown in Table.
led forward drop
A typical LED can pass 30 –40 mA current without destroying the device. Normal current that gives sufficient brightness to a standard Red LED is 20 mA. But this may be 40 mA for Blue and White LEDs. A current limiting resistor is necessary to protect LED from excess current that is flowing through it. The value of this series resistor should be carefully selected to prevent damage to LED and also to get sufficient brightness at 20 mA current. The current limiting resistor can be selected using the formula:
R = V / I
Where R is the value of resistor in ohms, V is the supply voltage and I is the allowable current in Amps. For a typical Red LED, the voltage drop is 1.8 volts. So if the supply voltage is 12 V (Vs), voltage drop across the LED is 1.8 V (Vf) and the allowable current is 20 mA (If) then the value of the series resistor will be
Vs – Vf / If = 12 – 1.8 / 20 mA = 10.2 / 0.02 A = 510 Ohms.
A suitable available value of resistor is 470 Ohms. But is advisable to use 1 K resistor to increase the life of the LED even though there will be a slight reduction in the brightness. Since the LED takes 1.8 volts, the output voltage will be 2 volts less than the value of Zener. So if the circuit requires 12 volts, it is necessary to increase the value of Zener to 15 volts. Table given below is a ready reckoner for selecting limiting resistor for various versions of LEDs at different voltages.
led resistor
Circuit Diagram
The diagram shown below is a simple transformer less power supply. Here 225 K(2.2uF) 400 volts X rated capacitor is used to drop 230 volt AC. Resistor R2 is the bleeder resistor that remove the stored current from the capacitor when the circuit is unplugged. Without R2, there is chance for fatal shock if the circuit is touched. Resistor R1 protects the circuit from inrush current at power on. A full wave rectifier comprising D1 through D4 is used to rectify the low voltage AC from the capacitor C1 and C2 removes ripples from the DC. With this design, around 24 volts at 100 mA current will be available at the output.This 24 volt DC can be regulated to required output voltage using a suitable 1 watt Zener. It is better to add a safety fuse in the phase line and an MOV across the phase and neutral lines as safety measure if there is voltage spike or short circuit in the mains.
Caution: Construction of this form of power supply is recommended only to those persons experienced or competent in handling AC mains. So do not try this circuit if you are not experienced in handling High voltages.
The drawback of the Capacitor power supply includes
  • No galvanic isolation from Mains.So if the power supply section fails, it can harm the gadget.
  • Low current output. With a Capacitor power supply. Maximum output current available will be 100 mA or less.So it is not ideal to run heavy current inductive loads.
  • Output voltage and current will not be stable if the AC input varies.
Caution
Great care must be taken while testing the power supply using a dropping resistor. Do not touch at any points in the PCB since some points are at mains potential. Even after switching off the circuit, avoid touching the points around the dropping capacitor to prevent electric shock. Extreme care should be taken to construct the circuit to avoid short circuits and fire. Sufficient spacing must be given between the components.
The high value smoothing capacitor will explode, if is connected in the reverse polarity. The dropping capacitor is non-polarized so that it can be connected either way round. The power supply unit must be isolated from the remaining part of the circuit using insulators. The circuit should be housed in metal case without touching any part of the PCB in the metal case. The metal case should be properly earthed.

Why engineers in electricity transport the electricity with 3-phase cables ...???

Why engineers in electricity transport the electricity with 3-phase cables ...???

                 

                            The history of electricity is a very interesting one indeed. In the early 1900s, Thomas Edison is said to have hated AC, simply because he didn't understand it. Edison made efforts to squash AC distribution, but ultimately failed due to the economics of power generation and distribution: he just couldn't send DC very far. Edisons rival, Nikola Tesla developed and patented much of AC power generation and distribution technology used today. George Westinghouse purchased Teslas patents and profited from them. Even with these patents, the company Edison founded, General Electric, is many times the size of Westinghouse. Telsa fell into relative obscurity, he is rarely mentioned in the history books. Nikola Tesla does not get the kind of recognition he truly deserves, even though he is the creator of polyphase transformers and machinery. Nikola Tesla is the real reason why we use 3-phase distribution.


  1. Why is alternating current (AC) used for power distrubution over direct current (DC)?
  2. Why is three-phase used rather than single phase?
                To answer the first part, AC generators cost less and can produce more power than an equivalent size DC generator. AC in general is easier to distribute because very efficient transformers can be used to step up and step down distribution voltage. The step up helps reduce distribution losses by substantially reducing the current carried by the long distance lines. The voltage on high tension transmission lines may range anywhere between 230kV to as much as 500kV with currents up to 450 Amperes. Transmitting that amount of power at low voltage (240 volts) would mean extremely high currents resulting in unacceptably high ohmic losses.

                  In some cases, very high voltage lines (upwards of 1MV) actually use DC! It turns out that very long AC runs (hundreds of miles or so) can actually introduce significant radiation losses. The wavelength of 60 Hertz (used in North America) is 5000 Kilometers or about 3000 miles. Even an antenna of 1/10 of a wavelength (300 miles) will radiate quite a bit. Some of the long haul lines on the US West coast between large metropolitan areas run DC.

There are several reasons why electricity is distributed in three phase:
  • Polyphase generators can generate more power and cost less to maintain than single phase generators
  • 3-phase distribution is more efficient than single phase AC
  • Many industries require 3-phase power
      Lets say that we have a generator with polyphase windings. For each phase winding there are two ends. For a 3-phase system there are six wire ends. At first, one would think that all six wire ends must be distributed, but it turns out that only three wires are needed. The reason is that at any point in time, the total current on all three conductors is exactly zero! This assumes, of course, that all three phases are exactly balanced and that current is equally distributed. In reality, this isn't always the case. Varying loads, broken segments, and system failures imbalance the system, but engineers still do a fairly good job of current balancing. The result is that a given amount of power can be distributed using much less wire with 3-phase than single phase.
                            A 450 Amp aluminum transmission wire is about an 1.5 inches thick. Over hundreds of miles, the cost of wire really adds up! Individual homes are usually wired for single phase 120/240 . Each neighboring house is fed from different phases in sequence to balance the load. Clearly, if one house uses significantly more power than another, the phases will be imbalanced. In reality, it more or less evens out when hundreds of houses are involved. 
                        

Torques in Electrical Induction Motors...

Torques in Electrical Induction Motors...

Torques in Electrical Induction Motors

There are some common torques used to describe and classify electrical motors

Torque is the turning force through a radius and the units is rated in - Nm - in the SI-system and in - lb ft - in the imperial system.
The torque developed by asynchronous induction motors varies with the speed of the motor when its accelerate from full stop or zero speed, to maximum operating speed.
electric motor current torque curves

Locked Rotor or Starting Torque

The Locked Rotor Torque or Starting Torque is the torque the electrical motor develop when its starts at rest or zero speed.
A high Starting Torque is more important for application or machines hard to start - as positive displacement pumps, cranes etc. A lower Starting Torque can be accepted in applications as centrifugal fans or pumps where the start load is low or close to zero.

Pull-up Torque

The Pull-up Torque is the minimum torque developed by the electrical motor when it runs from zero to full-load speed (before it reaches the break-down torque point)
When the motor starts and begins to accelerate the torque in general decrease until it reach a low point at a certain speed - the pull-up torque - before the torque increases until it reach the highest torque at a higher speed - the break-down torque - point.
The pull-up torque may be critical for applications that needs power to go through some temporary barriers achieving the working conditions.

Break-down Torque

The Break-down Torque is the highest torque available before the torque decreases when the machine continues to accelerate to the working conditions.

Full-load (Rated) Torque or Braking Torque

The Full-load Torque is the torque required to produce the rated power of the electrical motor at full-load speed.
In imperial units the Full-load Torque can be expressed as
T =  5252 Php / nr         (1)
where
T = full-load torque (lb ft)
Php = rated horsepower
nr = rated rotational speed (rev/min, rpm)
In metric units the rated torque can be expressed as
T = 9550 PkW / nr             (2)
where
T = rated torque (Nm)
PkW = rated power (kW)
nr = rated rotational speed (rpm)

Example - Electrical Motor and Braking Torque

The torque of a 60 hp motor rotating at 1725 rpm can be expressed as:
Tfl = (60 hp) 5,252 / (1725 rpm)
    = 182.7 lb ft

NEMA Design

NEMA (National Electrical Manufacturers Association) have classified electrical motors in four different NEMA designs where torques and starting-load inertia are important criterions.
IEC/NEMA Standard Torques (percent of full load torque)
Power (hp)2 Pole4 Pole
Locked Rotor TorquePull Up TorqueBreak Down TorqueLocked Rotor TorquePull Up TorqueBreak Down Torque
3170/160110/110200/230180/215120/150200/250
5160/150110/105200/215170/185120/130200/225
7.5150/140100/100200/200160/175110/120200/215
10150/135100/100200/200160/165110/115200/200
15 - 20140/130100/100200/200150/150110/105200/200

Accelerating Torque

Accelerating Torque = Available Motor Torque - Load Torque

Reduced Voltage Soft Starters

Reduced Voltage Soft Starters are used to limit the starting current and reducing the Locked Rotor Torque or Starting Torque and are common in applications which is hard to start or must be handled with care - as positive displacement pumps, cranes, elevators and similar.


How Asynchronous (induction) generators Works ...???

How Asynchronous (induction) generators Works ...???





Introduction
Asynchronous generator        Most wind turbines in the world use a so-called three phase asynchronous (cage wound) generator, also called an induction generator to generate alternating current. This type of generator is not widely used outside the wind turbine industry and in small hydropower units, but the world has a lot of experience in dealing with it anyway. The picture illustrates the basic principles in the asynchronous generator, much similar with the synchronous generator presented before. In reality, only the rotor part looks different.
        The curious thing about this type of generator is that it was really originally designed as an electric motor. In fact, one third of the world's electricity consumption is used for running induction motors driving machinery in factories, pumps, fans, compressors, elevators and other applications where you need to convert electrical energy to mechanical energy.
        One reason for choosing this type of generator is that it is very reliable and tends to be comparatively inexpensive. The generator also has some mechanical properties which are useful for wind turbines, like the generator slip and a certain overload capability.



The cage rotor
Cage rotor         The key component of the asynchronous generator is the cage rotor (it used to be called a squirrel cage rotor but after it became politically incorrect to exercise your domestic rodents in a treadmill, we only have this less captivating name).
        It is the rotor that makes the asynchronous generator different from the synchronous generator. The rotor consists of a number of copper or aluminium bars which are connected electrically by aluminium end rings, as you see in the picture. In the picture is shown how the rotor is provided with an "iron" core, using a stack of thin insulated steel laminations, with holes punched for the conducting aluminium bars. The rotor is placed in the middle of the stator, which in this case, once again, is a 4-pole stator which is directly connected to the three phases of the electrical grid.



Motor operation
The rotor of an asynchronous generator        When the current is connected, the machine will start turning like a motor at a speed which is just slightly below the synchronous speed of the rotating magnetic field from the stator.If we look at the rotor bars from the previous picture, there is a magnetic field which moves relative to the rotor. This induces a very strong current in the rotor bars which offer very little resistance to the current, since they are short circuited by the end rings. The rotor then develops its own magnetic poles, which in turn become dragged along by the electromagnetic force from the rotating magnetic field in the stator.


Generator operation
        If we manually crank this rotor around at exactly the synchronous speed of the generator, e.g. 1500 rpm, as we saw for the 4-pole synchronous generator on the previous page nothing will happen. Since the magnetic field rotates at exactly the same speed as the rotor, there will be no induction phenomena in the rotor and it will not interact with the stator.
        If speed is increased above 1500 rpm then the rotor moves faster than the rotating magnetic field from the stator, which means that once again the stator induces a strong current in the rotor. The harder is cranked the rotor, the more power will be transferred as an electromagnetic force to the stator, and in turn converted to electricity which is fed into the electrical grid.



Generator slip
        The speed of the asynchronous generator will vary with the turning force (moment, or torque) applied to it. In practice, the difference between the rotational speed at peak power and at idle is very small, about 1%. This difference in per cent of the synchronous speed, is called the generator's slip. Thus a 4-pole generator will run idle at 1500 rpm if it is attached to a grid with a 50 Hz current. If the generator is producing at its maximum power, it will be running at 1515 rpm.
        It is a very useful mechanical property that the generator will increase or decrease its speed slightly if the torque varies. This means that there will be less tear and wear on the gearbox, because of lower peak torque. This is one of the most important reasons for using an asynchronous generator rather than a synchronous generator on a wind turbine which is directly connected to the electrical grid.



Automatic pole adjustment of the rotor
        The clever thing about the cage rotor is that it adapts itself to the number of poles in the stator automatically. The same rotor can therefore be used with a wide variety of pole numbers.


Grid connection required
        On the page about the synchronous generator we showed that it could run as a generator without connection to the public grid. An asynchronous generator is different, because it requires the stator to be magnetised from the grid before it works.
        However, an asynchronous generator in a stand alone system can be used if it is provided with capacitors which supply the necessary magnetisation current. It also requires that there be some remanence in the rotor iron, i.e. some leftover magnetism to start the turbine. Otherwise a battery and power electronics will be needed, or a small diesel generator to start the system.

What is the Difference between synchronous generator and induction generator ...???

Difference between synchronous generator and induction generator

  • In a synchronous generator, the waveform of generated voltage is synchronized with (directly corresponds to) the rotor speed. The frequency of output can be given as f = N * P / 120 Hz. where N is speed of the rotor in rpm and P is number of poles.

    In case of inductions generators, the output voltage frequency is regulated by the power system to which the induction generator is connected. If induction generator is supplying a standalone load, the output frequency will be slightly lower (by 2 or 3%) that calculated from the formula f = N * P / 120.
  • Separate DC excitation system is required in an alternator (synchronous generator).

    Induction generator takes reactive power from the power system for field excitation. If an induction generator is meant to supply a standalone load, a capacitor bank needs to be connected to supply reactive power.
  • Construction of induction generator is less complicated as it does not require brushes and slip ring arrangement. Brushes are required in synchronous generator to supply DC voltage to the rotor for excitation.

Basic differences between induction generators and synchronous generators can be better understood from the figures shown below 

Induction generator

induction generator

Synchronous generator

synchronous generator

Synchronous motor and Generator operation...

Synchronous motor and Generator operation...
Synchronous generator (2 poles)

3-phase generator or motor principles

        All 3-phase generators, or motors, use a rotating magnetic field. In the picture three electromagnets have been installed around a circle. Each of the three magnets is connected to its own phase in the three phase electrical grid.
         As shown in the image, each of the three electromagnets alternate between producing a South pole and a North pole towards the centre. The letters are shown in black when the magnetism is strong and in light grey when the magnetism is weak. The fluctuation in magnetism corresponds exactly to the fluctuation in voltage of each phase. When one phase is at its peak, the other two have the current running in the opposite direction, at half the voltage. Since the timing of current in the three magnets is one third of a cycle apart, the magnetic field will make one complete revolution per cycle.


Synchronous motor operation

        The compass needle (with the North pole painted red) will follow the magnetic field exactly and make one revolution per cycle. With a 50 Hz grid, the needle will make 50 revolutions per second, i.e. 50 times 60 = 3000 rpm.
        In the picture above, we have in fact managed to build what is called a 2-pole permanent magnet synchronous motor. The reason why it is called a synchronous motor, is that the magnet in the centre will rotate at a constant speed which is synchronous with (running exactly like the cycle in) the rotation of the magnetic field.
        The reason why it is called a 2-pole motor is that it has one North and one South pole. It may look like three poles, but in fact the compass needle feels the pull from the sum of the magnetic fields around its own magnetic field. So, if the magnet at the top is a strong South pole, the two magnets at the bottom will add up to a strong North pole.
        The reason why it is called a permanent magnet motor is that the compass needle in the centre is a permanent magnet, not an electromagnet. You could make a real motor by replacing the compass needle by a powerful permanent magnet, or an electromagnet which maintains its magnetism through a coil, wound around an iron core, which is fed with direct current.
        The setup with the three electromagnets is called the stator in the motor, because this part of the motor remains static (in the same place). The compass needle in the centre is called the rotor, obviously because it rotates.



Synchronous generator operation

        If you start forcing the magnet around instead of letting the current from the grid move it, you will discover that it works like a generator, sending alternating current back into the grid (you should have a more powerful magnet to produce much electricity). The more force, or torque, you apply, the more electricity you generate, but the generator will still run at the same speed dictated by the frequency of the electrical grid.
        You may disconnect the generator completely from the grid and start your own private 3-phase electricity grid, hooking your lamps up to the three coils around the electromagnets. If you disconnect the generator from the main grid, however, you will have to crank it at a constant rotational speed in order to produce alternating current with a constant frequency. Consequently, with this type of generator you will normally want to use an indirect grid connection of the generator.
        In practice, permanent magnet synchronous generators are not used very much. There are several reasons for this. One reason is that permanent magnets tend to become demagnetised by working in the powerful magnetic fields inside a generator. Another reason is that powerful magnets, made of rare earth metals like Neodynium, are quite expensive, even if prices have dropped lately.



Wind turbines with synchronous generators

        Wind turbines which use synchronous generators normally use electromagnets in the rotor which are fed by direct current from the electrical grid. Since the grid supplies alternating current, they first have to convert alternating current to direct current before sending it into the coil windings around the electromagnets in the rotor.
        The rotor electromagnets are connected to the current by using brushes and slip rings on the axle (shaft) of the generator.


A (50/60 Hz) Transformer. Which one will give more Output? (When operates on 50 or 60 Hz frequency)...???

A (50/60 Hz) Transformer. Which one will give more Output? (When operates on 50 or 60 Hz frequency)...???

 A (50/60 Hz) Transformer. Which one will give more Output? 

A Transformer is designed to be operated on both 50 & 60 Hz frequency. For the Same rating, which one will give more out put; when,
  1.  Operates on 50 Hz
  2.   Operates on 60 Hz
A (50/60 Hz) Transformer. Which one will give more Output? (When operates on 50 or 60 Hz frequency)Obviously! It will give more out put when we operate a transformer (of same rating) on 50 Hz instead of 60 Hz.
Because in previous posts, we proved that, in inductive circuit, when frequency increases, the circuit power factor decreases. Consequently, the transformer out put decreases.
Let’s consider the following example.
Suppose,
When Transformer operates on 50 Hz Frequency
Transformer = 100kVA, R=700, L=1.2 H, f= 50 Hz.
XL = 2πfL = 2 x 3.1415 x 50 x 1.2 = 377 Ω
Impedance Z = √ (R2+XL2) = √ (7002+ 3772) = 795 Ω
Power factor Cos θ = R/Z = 700/795 =0.88
Transformer Output (Real Power)
kVA x Cos θ
100kVA x 0.88
88000 W = 88kW
Now,
When Transformer operates on 60 Hz Frequency
Transformer =100kVA, R=700, L=1.2 H, f= 60 Hz.
XL = 2πfL = 2 x 3.1415 x 60 x1.2 = 452.4 Ω
Impedance Z = √ (R2+XL2) = √ (7002+ 452.4 2) = 833.5 Ω
Power factor = Cos θ = R/Z = 700/833.5 =0.839
Transformer Output (Real Power)
kVA x Cos θ
100kVA x 0.839
=83900W = 83.9kW Output
Now see the difference (real power i.e., in Watts)
88kW- 83.9kW = 4100 W = 4.1kW
If we do the same (As above) for the power transformer i.e, for 500kVA Transformer, the result may be huge, as below.
(Suppose everything is same, without frequency)
Power Transformer Output (When operates on 50 Hz)
500kVA x 0.88 = 44000 = 440kW
Power Transformer Output (When operates on 60 Hz)
500kVA x 0.839 = 419500 = 419.5kW
Difference in Real power i.e. in Watts
440kW – 419.5kW = 20500 = 20kVA
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