1 Ph Converters E.M.D 2013

8/11/2019 1 Ph Converters E.M.D 2013

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41 Controlled Rectifiers (H.Rashid):

Introduction:

As we know that diode rectifiers provide a fixed output voltage only. In order to obtain controlledoutput voltage, phase controlled thyristors are used instead of diodes.

The output voltage of thyristor rectifier is varied by controlling the delay or firing angle of the thyristors.

A phase controlled thyristor is turned-on by applying a short pulse to its gate and turned-off due tonatural or line commutation and in case of a highly inductive load it is turned-off by firing another thyristorof the rectifier during the negative half cycle of the input voltage.

A 3-Phase controlled rectifiers are simple and less expensive and the efficiency of these rectifiers arein general above 95 %. These are extensively used in industrial purposes especially in variable speed drivesranging from fractional horse power to mega watt power levels.

The phase controlled rectifiers can be classified into two types, depending upon the type of input voltage;

1. Single phase converters.

2. Three phase converters.

Each type can be sub-divided into;

a). Semi converter: A semi converter is one quadrant converter and it has one polarity of the outputvoltage and current.

b). Full converter: A Full converter is a two quadrant converter and the polarity of its output voltagecan be either positive or negative. However the output current of full converter has one polarity.

c). Dual converter: A Dual converter can operate in four quadrants and both the output voltage &current can be either positive or negative.

Note: in some applications converters are connected in series to operate at higher voltages and to improvethe input power factor.


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4-2 Principle of Phase controlled Converters (H.Rashid):

Operation:

Consider the figure 4-1a, with resistive load R.

During the positive half cycle of the input voltage, the thyristor anode is positive w.r.to its cathodeand the thyristor is said to be forward biased.

When thyristor T 1is fired at an angle t =, thyristor T 1conducts and the input voltage appearsacross the load.

When the input voltage starts to be negative at t = , the thyristor anode is negative w.r.to itscathode and thyristor T 1is said to be reverse biased and it is turned off.

The time after the input voltage starts to go positive until thyristor is fired at t = , is called as thedelay or firing angle .

Figure - b shows the region of converter operation, where the output voltage and current have one

polarity.

Figurec shows the wave forms for input voltage, output voltage load current and voltage across T 1.


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If V mis the peak input voltage, the average output voltage V d ccan be obtained from;

V d c=

( ) .

=> ( )

=> ( => (V m/ 2 ) {cos cos } => (V m/ 2 ) {cos + cos }

(V m/ 2 ) {(1) + cos } => V d c= (V m/ 2 ) {1 + cos }

And the voltage V d ccan be varied from (V m / ) to 0; by varying from 0 to .

The average output voltage becomes maximum, when = 0 and the maximum voltage V d mis obtained as;

V d m= (V m/ 2 ) {1 + cos }

V d m= (V m/ 2 ) {1 + cos 0}

=> (V m/ 2 ) {1 + 1} => V d m= (V m/ 2 ) {2} => V d m= (V m/ )

Now the normalized output voltage V ncan be obtained as;

V n= (V d c/ V d m)

V n= {V m(1 + cos )/ 2 }/ (V m/ )} => V n= 0.5 {(1 + cos )}

The rms output voltage V r m s can be obtained as;

V r m s = [

V m 2sin 2 d ] V r m s = V m[

sin 2 d ]

=> V m[

( ) ]

=> V m[ ( ) ]

=> V m[

*( ( ) ) +] => V m[

*( ) ( )]

=> V m[

*( ) ( )+] => V r m s= [ *( ) ( )+]


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Pa c= V r.m.s x I r.m.s

V d c= (V m/ 2 ) ( 1 + cos )

The r.m.s voltage is given by; V r.m.s= (Vm/ 2) [( sin (2)]

The r.m.s load current is given by; I r.m.s= V r.m.s/ R

Average output current I d c= V d c/ R

Pd c= V d.c x I d.c

The rectification efficiency is obtained as; = Pd c/ Pa cForm Factor = V r.m.s/ V d c

Ripple Factor = ( ) T.U.F = Pd c/ V I

V I = V Sx I S

I S = I r.m.s

PIV = V m


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Example 41: if the converter in given figure 4-1 a has a purely resistive load of R and the delay

Angle is = /2. Determine the factors of Thyristor T 1.i). rectification efficiency.

ii). Form factor.

iii). Ripple factor.iv). Transformer utilization factor.

v). peak inverse voltage.

Solution:

From the given data, the delay angle is = ;(1). V d c=

(1 + cos )

=> (1 + cos ) =>

(1 + 0)

=> => V d c= 0.1592 V m

(2). The average output current is

I d c= V d c/ R => 0.1592 V m/ R => I d c= (0.1592 V m/ R) (3). Pd c= V d.c x I d.c => (0.1592 V m) x (0.1592 V m/ R) => Pd c= (0.1592)

2V m2/ R

For Pa c= V r.m.s x I r.m.s

The r.m.s voltage can be obtained as;

(4). V r.m.s= [( sin (2)] =>

[

( () sin 2 ()]

=> [

() ] => (V m ) => V r.m.s= 0.3536 V m

(5). And the r.m.s load current is obtained as;

I r.m.s= V r.m.s/ R => 0.3536 V m/ R => I r.m.s= (0.3536 V m/ R) So that;

(6). Pa c= V r.m.s x I r.m.s => (0.3536 V m) x (0.3536 V m/ R) => Pa c= (0.3536)2V m

2/ R

(7). The rectification efficiency is obtained as; = Pd c/ Pa c - - - - - - - - - - (244)=> [(0.1592) 2V m

2/ R] / [(0.3536) 2V m2/ R] => [(0.025345] / [(0.125] => = 0.2027 or 20.27%.


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(8). Form Factor = V r.m.s/ V d c

=> [(0.3536 V m] / [(0.1592 V m] => Form Factor = 2.21

(9). Ripple Factor = {(F.F) 21}

=> {(2.21) 21} => => Ripple Factor = 1.983

(10). For T.U.F = Pd c/ V I

V I = V Sx I S

=> V S =V m/ => V S = 0.707V m&

I S = I r.m.s => I r.m.s= (0.3536 V m/ R) Now T.U.F = Pd c/ V I

=> [(0.1592) 2. (Vm) 2/ R] / [ (0.707V

m) (0.3536 V

m) / R] => T.U.F = 0.1014 or 10.14 %

(11). PIV = V m


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4-3 Single Phase Semi converters:


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The average output voltage is given by;

V d c=

( ( ) ( ) .

V d c=

(cos | => V d c= cos |0

=> {cos ( + ) cos } => {cos cos sin sin ) cos } (As sin = 0)=>

{cos cos cos } => {(1) cos cos } => {cos cos }

=> {cos + cos } => {2 cos } => V d c= ()

And the voltage V d ccan be varied from (2 V m/ ) to (2 V m/ ); by varying from 0 to .

The maximum voltage V d m= 2 V m/

Similarly the normalized output voltage V ncan be obtained as;

V n= (V d c/ V d m) => V n= {2 V m(cos )/ } / (2 V m/ )}=> V n= cos

The r.m.s value of the output voltage V r m s can be obtained as; V r m s = [

()

]

V r m s = V m[

() ]

=> V m[

( )

() ]

=> V m[

( )

() ]

=> V m[

( ( ]

=> V m[

*( ( ] => V m[

*( ) *( ( ) )]

=> V m[

{ (sin 2 cos 2 + cos 2 sin 2 ) sin 2 }] 0 1

=> V m[

{ (sin 2 cos 2 + cos 2 sin 2 ) sin 2 }] (As sin 2 = 0, cos 2 = 1)

=> V m[ { (sin 2 sin 2 }]

=> V m[

{}] => V m[ ] => V r m s= = V s With a purely resistive load T 1& T 2conducts from to and T 3& T 4will conducts from + to 2

The instantaneous output voltage will be similar to that for the semi converter. In this condition

T 1& T 2conducts from to + and T 3& T 4will conducts from + to 2 + .


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1 Ph Converters E.M.D 2013

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