*Converter Operation

Choice of Converter Circuit1 Half-Wave Converter 1 Full-Wave Center-tap Converter1 Full-Wave Bridge Converter

3 Half-Wave Converter3 Full-Wave (6-pulse) Converter3 Full-Wave (12-pulse) Converter

Properties of convertor circuits:1. VA Rating of a valve = PIV Average current 2. VA Rating of Transformer = Vrms Irms In terms of DC power output Pd = Vd Id

Note: Vd, Id, Vrms, Irms and PIV are determined with the help of voltage and current waveforms.

*Assumptions:

1. Ideal AC source: no impedance and delivers sinusoidal voltage of - constant magnitude - constant frequency.2. Ideal Transformers no leakage impedance - no exciting admittance3. Ideal Valves - zero voltage-drop during conduction ( i.e zero impedance when ON)- zero current during non-conduction ( i.e infinite impedance when OFF)4. Perfect DC load:- DC current is constant i.e. ripple free- HVDC converters have large smoothing reactors (1H)

*Single phase rectifiers

A1. Half-wave Rectifier Discontinuous direct voltage DC component core saturationv1IdVd

A2. Full-wave Rectifier*e1e2T:1:1v1

*A3. Bridge RectifierVA rating of valves Everything same except PIV = Eminstead of 2Em But number of valves doubled So same VA ratingVA rating of Xmer Sec. current Id flows continuously(Further reduced)

B. Three-phase rectifiers

B1. Half-wave RectifierRipple in direct voltage - smaller in magnitude - higher in frequencyOnly one valve (having highest potential) conducts..

Once one valve conducts, other two get reverse biased.3. Each valves conducts for 1200 (i.e. 1/3 of a cycle) when associated sec. voltage is the highest one.

B2. 3-Bridge Rectifier

CircuitPulse number pValve ratingXmer rating1 HW13.142Pd2.465Pd1 FW23.142Pd1.571Pd1 Bridge23.142Pd1.111Pd3 HW32.094Pd1.481Pd3 Bridge62.094Pd1.047Pd

B3. 12 pulse Rectifier

*

*

*

*(1)(2)

*

*

*(3)(4)(5)Ideal no-load direct voltage

*

*(6)

* can vary from 00 to 1800

Cos will vary from 1 to -1

Vd will vary in the range of +Vd0 to Vd0

Id cannot reverse due to unidirectional property of the valves, therefore

+Vd and +Id: AC to DC Conversion (Rectification) -Vd and +Id: DC to AC Conversion (Inversion)

When > 600, some negative voltage periods begin to appear.

If the dc load connected is a pure resistance, the direct current through the load would be discontinuous.

A large smoothing reactor on the DC side avoids the discontinuous operation.

*

*(7)(8)

*(9)(10)(11)(from Eq. 4)(from Eq. 8)

*The converter operates as a sort of transformer a fixed current ratio (1: 0.78) variable voltage ratio which vary with the firing delay angleOperation with Grid control only: SummaryPowersV & I

pf00+Vd01Maximum, +veZero00 - 900+ve, < Vd0

*Due to source inductance , the phase currents cannot change instantly, therefore the transfer of current from one phase to other takes time called overlap or commutation time

* < 600 : Normal Operation (200 250) During commutation, three valves conduct

Between commutation, two valves conduct

Since a new commutation begins every 60 and last for angle , the angular interval when two valves conduct is 60 .

The sequence is 12, 123, 23, 234, 34, 345, 45, 456, 56, 561, 61, 612

= 600: Always a set of three valves conduct.123, 234, 345, 456, 561, 612.

600 < < 1200 (Abnormal operation)Alternately three and four valves conduct

= 1200 (Limiting case)Always four valves conduct

*(12)

(13) Normal operation ( < 600)

*During commutation,

or(14)

*(15)(16)(17)

*

*(14)Thus due to overlap, the voltage of point p rises to instead of

*

*Without overlap, the direct voltage was , therefore with overlap -(18)(19)From (17) and (18)(20)Hence from (19), (21)

*Substituting from (3) and from (16)Equivalent circuit of bridge rectifier(22)(23)Rectifier Equation:

*

*(24)Substituting in (17) and (20), we have (25)Eliminating from (24) and (25), we haveAlternately from Rectifier equation, we have(26)(27)Note: Inverter voltage, considered negative in general converter equations,Is usually taken as positive when written specially for an inverter.Inversion

*

*Based on (26) and (27), Inverter can be represented by two equivalent circuits

*Maximum DC Power possible during rectification For Pd to be maximum, (44)

*For n=1, Generally, If then . Since u > 600, it will result in abnormal operation.(45)Let then . It will result in normal operation.Thus with and , . Using (44),

*(from Eq. 4)(28)(29)(30)

*Using (20) and (22), we getUsing (30), or(31)Equations (28), (29) and (32) give approximate relations, not involving the converter angles and , for power, current and voltage, respectively.Reactive power on the AC side(32)where is given by (30) or (31).Note: There is no reactive power on the DC side.(33)

* Abnormal operation (600 1200)With the firing delay of , all above conductions are delayed by : (I) wt = to wt = ( +u) 600 : 6,1,2,3 conducts (4 valve conduction) (II) wt = (+u) 600 to wt = + 600 : 1,2,3 conducts (3 valve conduction) (III) wt = + 600 to wt = ( + u) : 1,2,3,4 conducts (4 valve conduction)Due to overloads, dc short circuit or low AC voltage600u00

**

*Average DC Voltage (Abnormal operation)Over the complete overlap period (i.e. to ) , the instantaneous voltage vd = 0: during 4-valve conduction (sub-intervals I & III) = arc of sinusoid of amplitude : during 3-valve conduction,During 1,2,3 valve conductionas(34)

*Currents during commutation (Abnormal operation)Sub-interval (I)

From wt = to wt = ( +u) 600 Four valves 6,1,2,3 conducts Applying KVL in upper loop: (35)as

AC

N

vd

eb

ec

ea

Lc

Lc

Ld

Lc

vd

ia

ic

ib

1

3

6

2

i1

i3

i6

i2

ea

ec

eb

Lc

Lc

Ld

Lc

vd

vd

ib

ic

ia

1

3

4

2

i1

i3

i4

i2

p

n

N

p

n

Id

Id

*Applying KVL in lower loop: (36)Substituting from (35)into (36)as (37)Integrating (37)

*where At the end of the subinterval i.e. at (38)

*Sub-interval (II)

From wt = ( +u) 600 to wt = + 600 Three valves 1,2,3 conducts

N

eb

ec

ea

Lc

Lc

Ld

Lc

vd

vd

ia

ic

ib

1

3

2

i1

i3

i2

p

n

Id

*Integrating, we have (39)Adding (38) & (39), The final current at the end of sub-interval II i.e. at (40)

*Sub-interval (III)

From wt = + 600 to wt = + u Four valves 1,2,3,4 conducts The phase currents are given by same equations as for sub-interval I but valve 3is now in series with phase b, so the ac component of i3 will be 900 behind eb. Also the dc component will be different from what it was in the first subinterval. The current i3, therefore, is (41)

AC

N

vd

eb

ec

ea

Lc

Lc

Ld

Lc

vd

ia

ic

ib

1

3

6

2

i1

i3

i6

i2

ea

ec

eb

Lc

Lc

Ld

Lc

vd

vd

ib

ic

ia

1

3

4

2

i1

i3

i4

i2

p

n

N

p

n

Id

Id

*As at , i3 is given by (40), thereforeSubstituting in (41), we have At the end of sub-interval, , i3 = Idas and(42)

*Hence for u>600,(34)(42)Eliminating from (34) (43)From (42), Id is max. when and Thus when andororandand

*Normal operation (u

*Minimum Ignition Angle ( min ) u>600 results in 3/4 valve conduction

Let initially valve 6,1 and 2 conducting (3 valve conduction)

Anode voltage of valve 3 =

Cathode voltage of valve 3 =

Valve 3 is the new incoming valve. (4 valve conduction)

But valve 3 can fire only when its anode potential ( ) > cathode potential( ) This will happen at point B ( )However with ueaTherefore we see that if u>600, valve 3 can fire only at point B (after 300 delay).Minimum ignition delay angle = 300 if u>600 (Note: This delay occurs even with grid control.)

a

b

c

1

3

5

2

4

6

*We have seen that for u>600 (i.e. for 3/4 valve conduction) min = 300

Therefore if u = 600 and = 300

Then It shows that at u=600, Similarly,Hence average voltage and current are same during transient phase (i.e. at u=600)

Instantaneous voltages during normal operation (u < 600)

*Waveforms for RectifierWaveforms for Inverter

*Commutation Margin InverterRectifier1: Inverse period

2. Blocking period

3. Conduction period

*Extinction advance angle angle between end of conduction & reversal of sign of sinusoidal commutation voltage

Commutation margin angle angle between end of conduction & reversal of sign of non-sinusoidal commutation voltageDue to increase in current or drop in commutation voltage, the overlap u and hence the duration of dent D increases

*