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Inverter Mode of Operation

The circuit behaves as rectifier for < /2

where Vdo>0

And as a line-commutated inverter for > /2

where Vdo< 0

The circuits will work as inverter when we

connect a dc voltage sourceESof propermagnitude and polarity across their terminals

as shown on the next slide

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Inverter Mode of Operation Contd

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Inverter Mode of Operation contdThe dc source may be

a battery,

a photovoltaic source,

a dc voltage produced by a wind-electric

system or

a dc motor operating under overhaulingload conditions (a load that turns the

armature).

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Load and SCR T1voltage waveforms forcontinuous conduction current and with the

effect of the source inductance included areshown on the next slide

vT1=vswhenT3is conducting and T1is off. vT1=0so long as T1is conducting

The figure shows that the thyristor voltage vT1is negative during the extinction angle given

by

= 180 (+ )

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.

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The extinction angle must be great enough

to allow the thyristors to turn off and regain

their forward blocking capability beforeforward voltage is reapplied, i.e.

> tq

The two equations:

= 180 (+ ) and

> tq

place a ceiling on the delay angle:

= 180 (+ ) > tq

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Its maximum value is, in practice,

max= 180 -

If the > tqcondition is not satisfied, the

commutation process will fail and thendestructive currents will occur.

Typical value of maxis 165o.

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Three-phase fully controlled bridgeconverter (6-pulse)

The figure on the next slide shows the

converter with highly inductive load.

It is the most widely used line-commutated

thyristor rectifier.

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.

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The converter also uses two groups of SCRs:

T1, T3and T5in one group have common

cathode connection and

T2, T4and T6in the other group have acommon anode connection.

The two groups operate independently of

each other.

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Here again if the gate currents to the SCRswere applied continuously, then they would

behave like diodes

And for the common cathode connection, the

SCR with its anode at the highest potentialwould conduct

And for the common anode connection, the

SCR with its cathode at the lowest potential

would conduct.

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Letvan= Vmaxsint

vbn= Vmaxsin(t120o)

vcn= Vmaxsin(t + 120o)

If they were diodes, which of them would be

conducting at t = 60o?

Note that at t = 60o

van>0, vbn

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The SCRs are fired in the following sequenceat time interval of 2/6(i.e. the period

divided by 6): T1, T2, T3, T4, T5and T6.

In the case of discontinuous current flow, it is

possible for T2say to have stoppedconducting before T3is turned on.

Then it is necessary to apply a gate pulsesimultaneously to T2and T3when T3 is being

turned on.

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This problem is solved by double pulsing,

i.e., supplying two pulses per cycle spaced60oor 2/6apart to each SCR.

That is the SCRs are fired in the followingsequence:

(T6 T1), (T1 T2), (T2 T3), (T3T4), (T4T5) ,

(T5T6), (T6 T1),

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Alternatively a long pulse, greater than 60o,

will overcome the problem, especially when

supplying an inductive load.

It is difficult to produce long pulses. Sousually a train of pulses with a frequency of

about 10 kHz is used to simulate a long pulse,

typically 120o

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Converter Feeding Highly Inductive loadThe dc output voltage is given by

vd= vPnvNn

The voltage vPn = van when T1is conducting,

= vbnwhen T3is conducting and = vcnwhen

T5is conducting.

Similarly vNn = van when T4 is conducting, =vbn when T6is conducting and = vcn when T2

is conducting.

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Line voltage vab

(T6and T1)

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Line voltage vac

(T1and T2)

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Line voltage vbc

(T2and T3)

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Line voltage vba

(T3and T4)

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Line voltage vca

(T4and T5)

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Line voltage vcb

(T5and T6)

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Converter waveforms are shown on the nextslide

The instants at which the SCRs wouldnaturally start conducting if they were diodes

are indicated in (a)

The effect of the phase angle on the

converter waveforms is shown in (b) to (d).

The current in phase ais shown in (c).

The line-to-line ac voltages and the dc output

voltage, are shown in (d).

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Output voltageThe output voltage has 6 pulses.

The mean output voltage is given by

where Vmaxis the peak of the phase voltage

and VLLis the rms value of the line voltage.

dVtdvVabdo

)6sin(33

)(3

max

2

6

2

6

cos23

cos33 max LLVV

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Output voltage contd In the time interval used for the integration

T1and T6are conducting resulting in

vd= vanvbn= vab

[we start at when T1starts conducting (30o

+) and end at when T2starts conducting

(60o +30o+= 90o+). Since T2 takes over

from T6, it is T6that conducts in theinterval].

The expression for vabis obtained using

phasor diagram

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Output voltage contdFor delay angles up to /3, the output

voltage is at all instants non-zero; hence

load current is continuous for any

passive load.

Beyond /3 the load current may be

discontinuous.

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Output voltage contdThe rms value of the output voltage is

given by

21

2

6

22max )()6(sin33

tdtVVrms

21

max 2cos4

33

2

13

V

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Input line currentsThe input currents ia,iband ichave

rectangular waveforms with amplitudeId.

The waveform of iais phase shifted from van

by the delay angle .

It can be expressed in terms of its Fourier

components with tdefined to be zero at thepositive zero crossing of van as given on the

next slide

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where the peak of the fundamental component

is given by

and the peak of the harmonics

Only nontriplen odd harmonics are present

)](19sin[)](17sin[

)](13sin[)](11sin[

)](7sin[)](5sin[)sin()(

1917

1311

751

tItI

tItI

tItItIti

mm

mm

mmma

d

m

I

I

321

h

II

m

mh

1

where

16 kh

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From the waveform of ia, the rms value of theinput current can be shown from the basic

definition of rms to be

Alternatively, we may use the fact thatia = iT1iT4from which

And hence

dS II3

2

)()()( 242

1

2RMSIRMSIRMSI

TTa

dds III )32()3(2 2