1. introduction to voltage source inverters

8/13/2019 1. Introduction to Voltage Source Inverters

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How to Get AC Output From DC Input Supply?Description:

For certain applications, it is essential that it is supplied with a certain kind of input whether it is an AC or DC supply.So, the significance of the conversion of DC input to an AC output arises.

How to Get AC Output From DC Input Supply?:In order to have a single control signalfor the

transistor switches, one transistor is of n-p-n type and theother of p-n-p type and their emitters and bases are shortedas shown in the figures. Both circuits require a symmetrical

bipolar dc supply. Collector of n-p-n transistor is connected topositive dc supply (+E) and that of p-n-p transistor isconnected to negative dc supply of same magnitude (-E). Load,which has been assumed resistive, is connected between theemitter shorting point and the power supply ground.

The transistors work in active (amplifier) mode and asinusoidal control voltage of desired frequency is appliedbetween the base and emitter points. When applied basesignal is positive, the p-n-p transistor is reverse biased and then-p-n transistor conducts the load current.

Similarly for negative base voltage the p-n-p transistorconducts while n-p-n transistor remains reverse biased. A

suitable resistor in series with the base signal will limit thebase current and keep it sinusoidal provided the applied(sinusoidal) base signal magnitude is much higher than thebase to emitter conduction-voltage drop.

Under the assumption of constant gain (hfe

) of the

transistor over its working range, the load current can be seento follow the applied base signal.

Accordingly the base magnitudes of current and

power are E/R and E2/R respectively. The power loss in

switches is a considerable portion of circuits input power andhence such circuits are unacceptable for large output powerapplications.

The conducting switch remains fully on havingnegligible on-state voltage drop and the non-conductingswitch remains fully off allowing no leakage current throughit. The load voltage waveform output by switched-modecircuit of is rectangular with magnitude +E when the n-p-ntransistor is on and E when p-n-p transistor is on. In theswitched mode (inverter) mode the on and off durations ofthe two transistors are controlled so that (i) the resulting

rectangular waveform has no dc component (ii) has a fundamental (sinusoidal) component of desired frequency andmagnitude and (iii) the frequencies of unwanted harmonic voltages are much higher than that of the fundamentalcomponent.Both amplifier mode and switched mode circuits are capable of producing ac voltages of controllable magnitude andfrequency, however, the amplifier circuit is not acceptable in power-electronic applications due to high switch powerloss.On the other hand, the switched mode circuit generates significant amount of unwanted harmonic voltages along withthe desired fundamental frequency voltage.The frequency spectrum of these unwanted harmonics can be shiftedtowards high frequency by adopting proper switching pattern. These high frequency voltage harmonics can easily beblocked using small size filter and the resulting quality of load voltage can be made acceptable.

Fig: A push-pull active amplifier circuit&A push-pull switched modecircuit

Fig: Switch in amplifier mode operation

Fig: Switched mode (inverter) operation


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Title:

Description:

QUESTIONS:

1. How do you get an AC output from a DC input supply?

2. Obtain the waveforms for the Switch in amplifier mode operation& switch mode (inverter) operation and give a briefdescription.


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What If the Load Is Not Resistive? - A

Description:

Most of the circuits are assumed that the loads connected to the output are always resistive in nature. But, there arecases and mostly in all the practical applications the load connected to the various circuits are non-resistive in nature.

What if the load is not resistive?

Transistors used in the circuit of A push-pull active

amplifier circuit are meant to carry only unidirectionalcurrent (from collector to emitter) and thus if the upper (n-p-

n) transistor is on, the current must enter the star (*) markedterminal of the load and this same terminal will getconnected to the positive dc supply (+E), other load terminal

being at ground potential.

When n-p-n transistor turns off and p-n-p type turnson, the load voltage and current polarities reversesimultaneously (p-n-p transistor can only carry currentcoming out of star marked end of load). Such one to onematching between the instantaneous polarities of loadvoltage and load current can be achieved only in purelyresistive loads.

For a general load the instantaneous current polarity may be different from instantaneous load-voltage polarity. Theinverter switching-pattern fixes the output waveform irrespective of the load. Thus the magnitude, phase andfrequency of the fundamental voltage output by a VSI is independent of the nature of load. Thus it turns out that for anon-resistive load the switches in the circuit of A push-pull switched mode circuit should be able to carry bi-directionalcurrent and at the same time be controllable

If an anti-parallel diode isconnected across each transistor switchthe combination can conduct a bi-directional current. Now the transistor inanti-parallel with the diode may beconsidered as a single switch. In spite ofunidirectional voltage blocking capability,the new electronic switch suffices for theinverter application as pointed out in thefollowing paragraphs.

The push-pull circuit operation is now revisited using bi-directional current carrying switches. It may be noted thatboth IGBT and BJT type transistors, when bypassed by anti-parallel diode, qualify as bi-directional current carryingswitches. However, IGBT switch is controlled by gate voltage whereas the BJT switch is controlled using base current.In the circuit of Modified push-pull circuit, n-p-n transistor (Q

1

) together with diode (D1

) constitutes the upper switch

(SW1). Similarly lower switch (SW

2) consists of p-n-p transistor (Q

2) in anti-parallel with diode (D

2). By applying positive

base-to-emitter voltage of suitable magnitude to transistor Q 1, the upper switch is turned on. Once the upper switch

(diode D1or transistor Q

1) is conducting star end of load is at +E potential and diode D

2 of lower switch gets

reverse biased. Transistor Q 2 is also reverse biased due to application of positive base voltage to the transistors. Thus

while switch SW1is conducting current, switch SW

2 is off and is blocking voltage of magnitude 2E. Similarly when

applied base voltage to the transistors is made negative, Q 1 is reverse biased and Q

2 is forward biased. This results

in SW1 turning off and SW

2 turning on. Now SW

1 blocks a voltage of magnitude 2E. It may be interesting to see

how diodes follow the switching command given to the transistor part of the switches.READ MORE

Fig: A push-pull active amplifier circuit&A push-pull switched modecircuit

Fig: Bi-directional controlled switch&Modified push-pull circuit


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Title:

Description:


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What If the Load Is Not Resistive?- B

To illustrate this point some details of circuit operation with aninductive load, consisting of a resistor and an inductor in series, isconsidered.Current through such loads cannot change abruptly. Theelectrical inertial time constant of the load, given by its L (inductance) / R(resistance) ratio, may in general be large compared to the chosenswitching time period of the transistor switches. Thus the transistors Q

1

and Q 2 may turn-on and turn-off several times before the load current

direction changes.Let us consider the time instant when instantaneous load current

is entering the star end of the load. Now with the assumed load currentdirection when Q

1 is given turn-on signal current flows from positive dc

supply, through transistor Q 1, to load. Next, when Q

1 is turned-off and

Q 2 is turned on (but load current direction remaining unchanged) the

load current finds its path through diode of lower switch (D2).

Whether D2 or Q

2 conducts, voltage drop across SW

2 is virtually zero and it can be considered as a closed or a

fully-on switch. In the following switching cycle when Q 1 is turned on again (load current direction still unchanged)

the load current path reverts back from D2 to Q

1. It may not be difficult to see how this happens.

While current flowed through D 2 the load circuit got connected to negative emf (-E) of the supply. When Q 1conducts the positive (+E) emf supports the load current.The natural choice for load current is to move from D

2 to Q

1. In fact turning on of Q

1 will make D

2 reverse biased.

The reader may repeat a similar exercise when the instantaneous load current comes out of the star end of load. Thusit will be evident that diodes do not need a separate command to turn on and off.Irrespective of the load current direction, turning on of Q

1 makes SW

1 on and similarly turning off of Q

1 (with

simultaneous turn-on of Q 2) makes SW

2on. Q

1 and Q

2 are turned on in a complementary manner. It may not be

difficult to see that the circuit of Modified push-pull circuit will work satisfactorily for a purely resistive load and aseries connected resistor-capacitor load too.The push-pull circuit of Modified push-pull circuit has some technical demerits that have been discussed below.

First, it needs a bipolar dc supply with identical magnitudes of positive and negative supply voltages. Forpractical reasons it would have been simpler if only one (uni-polar) dc source was required. In fact somecircuit topologies realize a bi-polar dc supply by splitting the single dc voltage-source through capacitivepotential divider arrangement. [A resistive potential divider will be terribly inefficient.] Two identicalcapacitors of large magnitude are put across the dc supply and the junction point of the capacitors is usedas the neutral (ground) point of the bi-polar dc supply. Topology of a 1-phase half bridge VSI shows onesuch circuit where a single dc supply has been split in two halves. In such circuits the voltages across thetwo capacitors may not remain exactly balanced due to mismatch in the loading patterns or mismatch inleakage currents of the individual capacitors. Also, unless the capacitors are of very large magnitude, theremay be significant ripple in the capacitor voltages, especially at low switching frequencies. Therequirement of splitting a single dc source is eliminated if a full bridge circuit, as mentioned in the nextsection, is used.The second demerit of the push-pull circuit shown in Modified push-pull circuit is the requirement of twodifferent kinds of transistors, one n-p-n type and the other p-n-p type. The switching speeds of n-p-n andp-n-p transistors are widely different unless they are produced carefully as matched pairs. In powerelectronic applications, n-p-n transistors are preferred as they can operate at higher switchingfrequencies. Similarly n-channel MOSFETs and IGBTs are preferred over their p-channel counterparts. Thedifficulty in using two n-p-n transistors in the above discussed push-pull circuit is that they can no longerhave a common base and a common emitter point and thus it wont be possible to have a single basedrive signal for controlling both of them. The base signals for the individual transistors will then need to beseparate and isolated from each other. The difficulty in providing isolated base signals for the twotransistors is, often, more than compensated by the improved capability of the circuit that uses both n-p-ntransistors and n-channel IGBTs. The circuit in Topology of a 1-phase half bridge VSI shows identicaltransistors (n-channel IGBTs) for both upper and lower switches. The gate drive signals of the twotransistors (IGBTs) now need to be different and isolated as the two emitter points are at differentpotentials. The circuit in Topology of a 1-phase half bridge VSI is better known as a half bridge inverter.

Fig: Topology of a 1-phase half bridge VSI


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Title:

Description:

QUESTIONS:

1. Explain the procedure taken when the transistor or switches are connected to loads which are not resistive in nature.2. Explain the working of a switch when the load is not resistive.


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General Structure of Voltage Source InvertersDescription:

The voltage source inverter topology uses a diode rectifier thatconverts utility/line AC voltage (60 Hz) to DC. Theconverteris not controlled through electronic firing like the CSI drive. TheDC link is parallel capacitors, which regulate the DCbus voltageripple and store energy for the system.General structure of Voltage Source Inverters:

These topologies require only asingle dc source and for medium outputpower applications the preferred devicesare n-channel IGBTs. E

dc is the input dc

supply and a large dc link capacitor (Cdc

)

is put across the supply terminals.Capacitors and switches are

connected to dc bus using short leads tominimize the stray inductance betweenthe capacitor and the inverter switches.

The physical layout of positiveand negative bus lines is also importantto limit stray inductances. Q

1, Q

2, Q

3etc.

are fast and controllable switches. D1, D

2,

D3

etc. are fast recovery diodes connected in anti-parallel with the switches. A, B and C are output terminals of the

inverter that get connected to the ac load. A three-phase inverter has three load-phase terminals whereas a single-phase inverter has only one pair of load terminals.The current supplied by the dc bus to the inverter switches is referred as dc link current and has been shown as i

dc.

The magnitude of dc link current often changes in step (and sometimes its direction also changes) as the inverterswitches are turned on and off. The step change in instantaneous dc link current occurs even if the ac load at theinverter output is drawing steady power.However, average magnitude of the dc link current remains positive if net power-flow is from dc bus to ac load. Thenet power-flow direction reverses if the ac load connected to the inverter is regenerating. Under regeneration, themean magnitude of dc link current is negative.For an ideal input (dc) supply, with no series impedance, the dc link capacitor does not have any role. However apractical voltage supply may have considerable amount of output impedance. The supply line impedance, if not

bypassed by a sufficiently large dc link capacitor, may cause considerable voltage spike at the dc bus during inverteroperation.This may result in deterioration of output voltage quality, it may also cause malfunction of the inverter switches as thebus voltage appears across the non-conducting switches of the inverter. In the absence of dc link capacitor, the seriesinductance of the supply line will prevent quick build up or fall of current through it and the circuit behaves differentlyfrom the ideal VSI where the dc voltage supply is supposed to allow rise and fall in current as per the demand of theinverter circuit.The dc link capacitor should be put very close to the switches so that it provides a low impedance path to the highfrequency component of the switch currents. The capacitor itself must be of good quality with very low equivalentseries resistor (ESR) and equivalent series inductor (ESL). The length of leads that interconnect switches and diodes tothe dc bus must also be minimum to avoid insertion of significant amount of stray inductances in the circuit.The overall layout of the power circuit has a significant effect over the performance of the inverter circuit. Thus the

single phase full-bridge (often, simply called as bridge) circuit has two legs of switches, each leg consisting of anupper switch and a lower switch.Junction point of the upper and lower switches is the output point of that particular leg. Voltage between outputpoint of legs and the mid-potential of the dc bus is called as pole voltage referred to the mid potential of the dc bus.One may think of pole voltage referred to negative bus or referred to positive bus too but unless otherwise mentionedpole voltages are assumed to be referred to the mid-potential of the dc bus.The two pole voltages of the single-phase bridge inverter generally have same magnitude and frequency but their

phases are 1800

apart. Thus the load connected between these two pole outputs (between points A and B) will havea voltage equal to twice the magnitude of the individual pole voltage. The pole voltages of the 3-phase inverter bridge,

are phase apart by 1200

each.

Fig: Topology of 1- phase and 3-Phase VSI


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Title:

Description:

QUESTIONS:

1. Describe the general structure of Voltage source Inverters.2. State the significance of Voltage source Inverters.


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Need for Isolated Gate-Control Signals For the Switches

Description:

The switches in bridge configurations of inverters require isolated gate (or base) drive signals. The individual controlsignal for the switches needs to be provided across the gate (base) and source (or emitter) terminals of the particular switch.The gate control signals are low voltage signals referred to the source (emitter) terminal of the switch.

Need for isolated Gate-control signals for the switches:

The gate control signals are low voltage

signals referred to the source (emitter) terminal ofthe switch. For n-channel IGBT and MOSFETswitches, when gate to source voltage is morethan threshold voltage for turn-on, the switchturns on and when it is less than threshold voltagethe switch turns off.

The threshold voltage is generally of theorder of +5 volts but for quicker switching theturn-on gate voltage magnitude is kept around+15 volts whereas turn-off gate voltage is zero orlittle negative (around 5 volts).

The two switches of an inverter-leg are

controlled in a complementary manner. When theupper switch of any leg is on, the corresponding lower switch remains off and vice-versa. When a switch is on itsemitter and collector terminals are virtually shorted. Thus with upper switch on, the emitter of the upper switch is atpositive dc bus potential.Similarly with lower switch on, the emitter of upper switch of that leg is virtually at the negative dc bus potential.

Emitters of all the lower switches are solidly connected to the negative line of the dc bus. Since gate control signals areapplied with respect to the emitter terminals of the switches, the gate voltages of all the upper switches must befloating with respect to the dc bus line potentials.This calls for isolation between the gate control signals of upper switches and between upper and lower switches.Only the emitters of lower switches of all the legs are at the same potential (since all of them are solidly connected tothe negative dc bus) and hence the gate control signals of lower switches need not be isolated among themselves.The isolation provided between upper and lower switches must withstand a peak voltage stress equal to dc bus

voltage. Gate-signal isolation for inverter switches is generally achieved by means of optical-isolator (opto-isolator)circuits.Input stage of the IC is a light emitting diode (LED) that emits light when forward biased. The light output of the LEDfalls on reverse biased junction of an optical diode. The LED and the photo-diode are suitably positioned inside theopto-coupler chip to ensure that the light emitted by the LED falls on the photo-diode junction. The gate controlpulses for the switch are applied to the input LED through a current limiting resistor of appropriate magnitude.These gate pulses, generated by the gate logic circuit, are essentially in the digital form. A high level of the gate signalmay be taken as on command and a low level (at ground level) may be taken as off command.Under this assumption, the cathode of the LED is connected to the ground point of the gate-logic card and anode is

fed with the logic card output. The control (logic card) supply ground is isolated from the floating-supply ground of theoutput. In the figure the two grounds have been shown by two different symbols.The schematic connection shown in the figure indicates that the photo-diode is reverse biased. A resistor in series

with the diode indicates the magnitude of the reverse leakage current of the diode. When input signal to LED is high,LED conducts and the emitted light falls on the reverse biased p-n junction.Irradiation of light causes generation of significant number of electron-hole pairs in the depletion region of the

reverse biased diode. As a result magnitude of reverse leakage current of the diode increases appreciably. The resistorconnected in series with the photo-diode now has higher voltage drop due to the increased leakage current.A signal comparator circuit senses this condition and outputs a high level signal, which is amplified before being

output. Thus an isolated and amplified gate signal is obtained and may directly be connected to the gate terminal ofthe switch (often a small series resistor, as suggested by the switch manufacturer, is put between the output signaland the gate terminal of the switch).


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Title:

Description:

QUESTIONS:

1. Explain the need for an isolated gate control signals for the switches.

2. Using diagrams state the need for an isolated gate control signals for the switches.


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Classification of Voltage Source Inverters

Description:

Voltage source inverters can be classified according to different criterions. They can be classified according to numberof phases they output. Accordingly there are single-phase or three-phase inverters depending on whether they output singleor three-phase voltages.

Classification of Voltage source Inverters:

Voltage sourceinverters can be classifiedaccording to differentcriterions. They can beclassified according to numberof phases they output.

Accordingly there aresingle-phase or three-phaseinverters depending onwhether they output single or

three-phase voltages. It is alsopossible to have inverters with

two or five or any other number of output phases.

Inverters can also be classified according to their ability in controlling the magnitude of output parameters like,frequency, voltage, harmonic content etc.

Some inverters can output only fixed magnitude (though variable frequency) voltages whereas some others arecapable of both variable voltage, variable frequency (VVVF) output.

Output of some voltage source inverters iscorrupted by significant amount of many low order

harmonics like 3rd

, 5th

, 7th

, 11th

, 13th

order of the desired(fundamental) frequency voltage.

Some other inverters may be free from low orderharmonics but may still be corrupted by some high orderharmonics.

Inverters used for ac motor drive applications areexpected to have less of low order harmonics in the

output voltage waveform, even if it is at the cost ofincreased high order harmonics.

Higher order harmonic voltage distortions are, inmost ac motor loads, filtered away by the inductive

nature of the load itself.

Inverters may also be classified according to their topologies. Some inverter topologies are suitable for low andmedium voltage ratings whereas some others are more suitable for higher voltage applications.

The inverters shown are two level inverters as the pole voltages may acquire either positive dc bus or negative dc bus

potential. For higher voltage applications it may not be uncommon to have three level or five level inverters.

Fig: Topology of 1- phase and 3-Phase VSI

Fig: Bi-Directional controlled switch


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Title:

Description:

QUESTIONS:

1. Classify the different voltage source inverters.

2. Write a short note on the different classifications of voltage source inverters.

1. introduction to voltage source inverters

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