36663455 02 power electronics notes by punith

7/30/2019 36663455 02 Power Electronics Notes by Punith

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EDUSAT PROGRAMME

LECTURE NOTES

ON

POWER ELECTRONICS

BY

PROF. M. MADHUSUDHAN RAO

DEPARTMENT OF ELECTRONICS &

COMMUNICATION ENGG.

M.S. RAMAIAH INSTITUTE OF

TECHNOLOGY

BANGALORE 560 054

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AC VOLTAGE CONTROLLER CIRCUITS

(RMS VOLTAGE CONTROLLERS)

AC voltage controllers (ac line voltage controllers) are employed to vary the RMS

value of the alternating voltage applied to a load circuit by introducing Thyristors

between the load and a constant voltage ac source. The RMS value of alternating voltage

applied to a load circuit is controlled by controlling the triggering angle of the Thyristors

in the ac voltage controller circuits.

In brief, an ac voltage controller is a type of thyristor power converter which is

used to convert a fixed voltage, fixed frequency ac input supply to obtain a variable

voltage ac output. The RMS value of the ac output voltage and the ac power flow to the

load is controlled by varying (adjusting) the trigger angle

A CV o l t a g e

C o n t r o l l e r

V 0 ( R M S )

fS

V a r i a b lR M S O / P

A CI n p u t

V o l t a g efs

V s

fs

There are two different types of thyristor control used in practice to control the ac

power flow

On-Off control

Phase control

These are the two ac output voltage control techniques.

In On-Off control technique Thyristors are used as switches to connect the load circuit

to the ac supply (source) for a few cycles of the input ac supply and then to disconnect it

for few input cycles. The Thyristors thus act as a high speed contactor (or high speed ac

switch).

PHASE CONTROL

In phase control the Thyristors are used as switches to connect the load circuit to

the input ac supply, for a part of every input cycle. That is the ac supply voltage is

chopped using Thyristors during a part of each input cycle.The thyristor switch is turned on for a part of every half cycle, so that input supply

voltage appears across the load and then turned off during the remaining part of input half

cycle to disconnect the ac supply from the load.

By controlling the phase angle or the trigger angle (delay angle), the outputRMS voltage across the load can be controlled.

The trigger delay angle is defined as the phase angle (the value of t) atwhich the thyristor turns on and the load current begins to flow.

Thyristor ac voltage controllers use ac line commutation or ac phase commutation.

Thyristors in ac voltage controllers are line commutated (phase commutated) since the

input supply is ac. When the input ac voltage reverses and becomes negative during thenegative half cycle the current flowing through the conducting thyristor decreases and

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falls to zero. Thus the ON thyristor naturally turns off, when the device current falls to

zero.

Phase control Thyristors which are relatively inexpensive, converter grade

Thyristors which are slower than fast switching inverter grade Thyristors are normally

used.

For applications upto 400Hz, if Triacs are available to meet the voltage andcurrent ratings of a particular application, Triacs are more commonly used.

Due to ac line commutation or natural commutation, there is no need of extra

commutation circuitry or components and the circuits for ac voltage controllers are very

simple.

Due to the nature of the output waveforms, the analysis, derivations of expressions

for performance parameters are not simple, especially for the phase controlled ac voltage

controllers with RL load. But however most of the practical loads are of the RL type and

hence RL load should be considered in the analysis and design of ac voltage controller

circuits.

TYPE OF AC VOLTAGE CONTROLLERSThe ac voltage controllers are classified into two types based on the type of input

ac supply applied to the circuit.

Single Phase AC Controllers. Three Phase AC Controllers.

Single phase ac controllers operate with single phase ac supply voltage of 230V

RMS at 50Hz in our country. Three phase ac controllers operate with 3 phase ac supply of

400V RMS at 50Hz supply frequency.

Each type of controller may be sub divided into

Uni-directional or half wave ac controller.

Bi-directional or full wave ac controller.In brief different types of ac voltage controllers are Single phase half wave ac voltage controller (uni-directional controller). Single phase full wave ac voltage controller (bi-directional controller). Three phase half wave ac voltage controller (uni-directional controller). Three phase full wave ac voltage controller (bi-directional controller).

APPLICATIONS OF AC VOLTAGE CONTROLLERS

Lighting / Illumination control in ac power circuits.

Induction heating.

Industrial heating & Domestic heating. Transformer tap changing (on load transformer tap changing). Speed control of induction motors (single phase and poly phase ac induction

motor control).

AC magnet controls.

PRINCIPLE OF ON-OFF CONTROL TECHNIQUE (INTEGRAL CYCLE

CONTROL)

The basic principle of on-off control technique is explained with reference to a

single phase full wave ac voltage controller circuit shown below. The thyristor switches

1T and 2T are turned on by applying appropriate gate trigger pulses to connect the input

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ac supply to the load for n number of input cycles during the time interval ONt . The

thyristor switches 1T and 2T are turned off by blocking the gate trigger pulses for m

number of input cycles during the time interval OFFt . The ac controller ON time ONt

usually consists of an integral number of input cycles.

LR R= = Load Resistance

Fig.: Single phase full wave AC voltage controller circuit

V s

V o

i o

i g 1

i g 2

w t

w t

w t

w t

G a t e p u l s e o f T1

G a t e p u l s e o f T2

n m

Fig.: Waveforms

ExampleReferring to the waveforms of ON-OFF control technique in the above diagram,

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n = Two input cycles. Thyristors are turned ON during ONt for two input cycles.m = One input cycle. Thyristors are turned OFF during OFFt for one input cycle

Fig.: Power Factor

Thyristors are turned ON precisely at the zero voltage crossings of the input

supply. The thyristor 1T is turned on at the beginning of each positive half cycle by

applying the gate trigger pulses to 1T as shown, during the ON time ONt . The load current

flows in the positive direction, which is the downward direction as shown in the circuit

diagram when 1T conducts. The thyristor 2T is turned on at the beginning of each

negative half cycle, by applying gating signal to the gate of 2T , during ONt . The load

current flows in the reverse direction, which is the upward direction when 2T conducts.

Thus we obtain a bi-directional load current flow (alternating load current flow) in a acvoltage controller circuit, by triggering the thyristors alternately.

This type of control is used in applications which have high mechanical inertia

and high thermal time constant (Industrial heating and speed control of ac motors). Due to

zero voltage and zero current switching of Thyristors, the harmonics generated by

switching actions are reduced.

For a sine wave input supply voltage,

sin 2 sins m S

v V t V t = =

SV = RMS value of input ac supply =

2

mV

= RMS phase supply voltage.

If the input ac supply is connected to load for n number of input cycles anddisconnected for m number of input cycles, then

,ON OFF

t n T t m T = =

Where1

Tf

= = input cycle time (time period) and

f = input supply frequency.

ONt = controller on time = n T .

OFFt = controller off time = m T .O

T = Output time period = ( ) ( )ON OFF t t nT mT + = + .

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We can show that,

Output RMS voltage ( ) ( )ON ON

SO RMS i RMS

O O

t tV V V

T T= =

Where ( )i RMSV is the RMS input supply voltage = SV .

TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF OUTPUT

VOLTAGE, FOR ON-OFF CONTROL METHOD.

Output RMS voltage( ) ( )

2 2

0

1.

ONt

mO RMS

O t

V V Sin t d t T

=

=

( ) ( )2

2

0

.ONt

m

O RMS

O

VV Sin t d t T

=

Substituting for2 1 2

2

CosSin

=

( ) ( )2

0

1 2

2

ONt

m

O RMS

O

V Cos t V d t

T

=

( ) ( ) ( )2

0 0

2 .2

ON ON t t

m

O RMS

O

VV d t Cos t d t

T

=

( ) ( )2

0 0

2

22

ON ON t t

m

O RMS

O

V Sin t V t

T

=

( ) ( )2 sin 2 sin 0

02 2

m ONONO RMS

O

V tV t

T

=

Now ONt = An integral number of input cycles; Hence

,2 ,3 ,4 ,5 ,.....ON

t T T T T T = & 2 ,4 ,6 ,8 ,10 ,......ONt =

Where T is the input supply time period (T = input cycle time period). Thus we note that

sin 2 0ON

t =

( )

2

2 2m ON m ON

O RMS

O O

V t V t VT T

= =

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( ) ( )ON ON

SO RMS i RMS

O O

t tV V V

T T= =

Where ( )2

mSi RMS

VV V= = = RMS value of input supply voltage;

( )

ON ON

O ON OFF

t t nT nk

T t t nT mT n m= = = =

+ + + = duty cycle (d).

( ) ( )S SO RMS

nV V V k

m n= =

+

PERFORMANCE PARAMETERS OF AC VOLTAGE CONTROLLERS

RMS Output (Load) Voltage

( ) ( )( )

122

2 2

0

sin .2

mO RMS

nV V t d t

n m

= +

( ) ( ) ( )2m

SO RMS i RMS

V nV V k V k

m n= = =

+

( ) ( ) SO RMS i RMS V V k V k = =

Where ( )S i RMSV V= = RMS value of input supply voltage.

Duty Cycle

( ) ( )ON ON

O ON OFF

t t nT k

T t t m n T = = =

+ +

Where, ( )

nk

m n= + = duty cycle (d).

RMS Load Current

( )

( ) ( )O RMS O RMS

O RMS

L

V VI

Z R= = ; for a resistive load LZ R= .

Output AC (Load) Power

( )

2

O LO RMSP I R=

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Input Power Factor

output load power

input supply volt amperes

O O

S S

P PPF

VA V I = = =

( )

( ) ( )

2

LO RMS

i RMS in RMS

I RPF

V I

=

; ( )S in RMSI I= = RMS input supply current.

The input supply current is same as the load current in O LI I I= =

Hence, RMS supply current = RMS load current; ( ) ( )in RMS O RMS I I= .

( )

( ) ( )

( )

( )

( )

( )

2

LO RMS O RMS i RMS

i RMS in RMS i RMS i RMS

I R V V k

PF kV I V V

= = = =

nPF k

m n= =

+

The Average Current of Thyristor ( )T AvgI

0 2 3 t

I m

n m

iT

W a v e f o r m o f T h y r i s t

( ) ( )( )

0

sin .2

mT Avg

nI I t d t

m n

=+

( ) ( )( )

0

sin .2

m

T Avg

nII t d t

m n

=+

( ) ( ) 0cos

2

m

T Avg

nII t

m n

= +

( ) ( )[ ]cos cos 0

2

m

T Avg

nII

m n

= +

+

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( ) ( )( )1 1

2

m

T Avg

nII

m n= + +

( )

( )

[ ]22

mT Avg

nI I

m n

=

+

( ) ( )

.m m

T Avg

I n k II

m n = =

+

( ) ( )duty cycle ON

ON OFF

t nk

t t n m= = =

+ +

( )

( )

.m m

T Avg

I n k II

m n

= =

+

,

Wherem

m

L

VI

R= = maximum or peak thyristor current.

RMS Current of Thyristor ( )T RMSI

( ) ( )( )

12

2 2

0

sin .2

mT RMS

nI I t d t

n m

= +

( ) ( )( )

122

2

0

sin .2

m

T RMS

nII t d t

n m

= +

( ) ( )

( )( )

122

0

1 cos 2

2 2

m

T RMS

tnII d t

n m

= +

( )

( )

( ) ( )

122

0 0

cos 2 .

4

m

T RMS

nII d t t d t

n m

=

+

( ) ( )( )

122

0 0

sin2

24

m

T RMS

nI tI t

n m

= +

( ) ( )( )

122 sin 2 sin 0

04 2

m

T RMS

nII

n m

= +

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( ) ( ){ }

122

0 04

m

T RMS

nII

n m

= +

( ) ( ) ( )

1 12 22 2

4 4m m

T RMSnI nI I

n m n m

= = + +

( ) ( )2 2m m

T RMS

I InI k

m n= =

+

( ) 2

m

T RMS

II k=

PROBLEM1. A single phase full wave ac voltage controller working on ON-OFF control

technique has supply voltage of 230V, RMS 50Hz, load = 50. The controller isON for 30 cycles and off for 40 cycles. Calculate

ON & OFF time intervals. RMS output voltage. Input P.F. Average and RMS thyristor currents.

( )230

in RMSV V= , 2 230 325.269mV V= = V, 325.269mV V= ,

1 1

0.02sec50

Tf Hz

= = = , 20T ms= .

n = number of input cycles during which controller is ON; 30n = .

m = number of input cycles during which controller is OFF; 40m = .

30 20 600 0.6secON

t n T ms ms= = = =

0.6secONt n T= = = controller ON time.

40 20 800 0.8secOFF

t m T ms ms= = = =0.8sec

OFFt m T= = = controller OFF time.

Duty cycle( ) ( )

300.4285

40 30

nk

m n= = =

+ +

RMS output voltage

( ) ( ) ( )O RMS i RMS nV V

m n= +

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( ) ( )

30 3230 230

30 40 7O RMS

V V= =+

( )230 0.42857 230 0.65465

O RMSV V= =

( )150.570

O RMSV V=

( )

( ) ( ) 150.5703.0114

50

O RMS O RMS

O RMS

L

V V VI A

Z R= = = =

( )2 23.0114 50 453.426498

O LO RMSP I R W= = =

Input Power Factor .P F k

=

( )

300.4285

70

nPF

m n= = =

+

0.654653PF=

Average Thyristor Current Rating

( )m m

T Avg

I k InI

m n

= = +

where2 230 325.269

50 50

mm

L

VI

R

= = =

6.505382m

I A= = Peak (maximum) thyristor current.

( )

6.505382 3

7T Avg

I

=

( ) 0.88745T AvgI A=

RMS Current Rating of Thyristor

( ) ( )

6.505382 3

2 2 2 7

m m

T RMS

I InI k

m n= = =

+

( )2.129386

T RMSI A=

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PRINCIPLE OF AC PHASE CONTROL

The basic principle of ac phase control technique is explained with reference to a

single phase half wave ac voltage controller (unidirectional controller) circuit shown in

the below figure.

The half wave ac controller uses one thyristor and one diode connected in parallel

across each other in opposite direction that is anode of thyristor 1T is connected to the

cathode of diode 1D and the cathode of 1T is connected to the anode of 1D . The output

voltage across the load resistor R and hence the ac power flow to the load is controlled

by varying the trigger angle .The trigger angle or the delay angle refers to the value of t or the instant at

which the thyristor 1T is triggered to turn it ON, by applying a suitable gate trigger pulse

between the gate and cathode lead.

The thyristor 1T is forward biased during the positive half cycle of input ac

supply. It can be triggered and made to conduct by applying a suitable gate trigger pulse

only during the positive half cycle of input supply. When 1T is triggered it conducts andthe load current flows through the thyristor 1T, the load and through the transformer

secondary winding.

By assuming 1T as an ideal thyristor switch it can be considered as a closed switch

when it is ON during the period t = to radians. The output voltage across the loadfollows the input supply voltage when the thyristor 1T is turned-on and when it conducts

from t = to radians. When the input supply voltage decreases to zero at t = , fora resistive load the load current also falls to zero at t = and hence the thyristor 1Tturns off at t = . Between the time period t = to 2 , when the supply voltagereverses and becomes negative the diode

1

D becomes forward biased and hence turns ON

and conducts. The load current flows in the opposite direction during t = to 2radians when 1D is ON and the output voltage follows the negative half cycle of input

supply.

Fig.: Halfwave AC phase controller (Unidirectional Controller)

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Equations

Input AC Supply Voltage across the Transformer Secondary Winding.

sins m

v V t=

( )2

mS in RMS

VV V= = = RMS value of secondary supply voltage.

Output Load Voltage

0o L

v v= = ; for 0t = to

sino L m

v v V t = = ; for t = to 2 .

Output Load Current

sino m

o L

L L

v V ti i

R R

= = = ; for t = to 2 .

0o Li i= = ; for 0t = to .

TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE ( )O RMSV

( ) ( )2

2 21 sin .2

mO RMSV V t d t

=

( ) ( )22 1 cos 2

.2 2

m

O RMS

V tV d t

=

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( ) ( ) ( )22

1 cos 2 .4

m

O RMS

VV t d t

=

( ) ( )

2 2

cos 2 .2

mO RMS

VV d t t d t

=

( ) ( )2 2

sin 2

22

m

O RMS

V tV t

=

( ) ( )2

sin22

22

m

O RMS

V tV

=

( ) ( )sin 4 sin 2

2 ;sin 4 02 22

m

O RMS

VV

= =

( ) ( )sin 2

222

m

O RMS

VV

= +

( ) ( )sin2

222 2

m

O RMS

VV

= +

( ) ( )1 sin 2

22 22

m

O RMS

VV

= +

( ) ( ) ( )1 sin 2

22 2

O RMS i RMS V V

= +

( ) ( )1 sin 2

2

2 2

SO RMSV V

= +

Where, ( )2

mSi RMS

VV V= = = RMS value of input supply voltage (across the

transformer secondary winding).

Note: Output RMS voltage across the load is controlled by changing ' ' as indicated by

the expression for ( )O RMSV

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PLOT OF ( )O RMSV VERSUS TRIGGER ANGLE FOR A SINGLE PHASE HALF-

WAVE AC VOLTAGE CONTROLLER (UNIDIRECTIONAL CONTROLLER)

( ) ( )1 sin 2

2

2 22

m

O RMS

VV

= +

( ) ( )1 sin 2

22 2

SO RMSV V

= +

By using the expression for ( )O RMSV we can obtain the control characteristics,

which is the plot of RMS output voltage ( )O RMSV versus the trigger angle . A typical

control characteristic of single phase half-wave phase controlled ac voltage controller is

as shown below

Trigger angle in degrees

Trigger angle

in radians( )O RMS

V

0 02

mS

VV =

030 6 ( )1; 6 0.992765 SV

060 3 ( )2; 6 0.949868 SV

090 2 ( )3; 6 0.866025 SV

0120 2 3 ( )4; 6 0.77314 SV

0150 5 6 ( )5; 6 0.717228 SV

0180 ( )6; 6 0.707106 SV

V O ( R M S )

T r i g g e r a n g l e

0 6 0 1 2 0 1 8 0

1 0 0 % VS

2 0 % VS

6 0 % VS

7 0 . 7 % VS

Fig.: Control characteristics of single phase half-wave phase controlled ac voltage

controller

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Note: We can observe from the control characteristics and the table given above that the

range of RMS output voltage control is from 100% of SV to 70.7% of SV when we vary

the trigger angle from zero to 180 degrees. Thus the half wave ac controller has thedraw back of limited range RMS output voltage control.

TO CALCULATE THE AVERAGE VALUE (DC VALUE) OF OUTPUT

VOLTAGE

( ) ( )2

1sin .

2mO dc

V V t d t

=

( ) ( )2

sin .2

m

O dc

VV t d t

=

( )

2

cos2

m

O dc

VV t

=

( ) [ ]cos 2 cos2m

O dc

VV

= + ; cos 2 1 =

[ ]cos 12

mdc

VV

= ; 2m SV V=

Hence ( )2 cos 12S

dc VV =

When ' ' is varied from 0 to . dcV varies from 0 tom

V

DISADVANTAGES OF SINGLE PHASE HALF WAVE AC VOLTAGE

CONTROLLER.

The output load voltage has a DC component because the two halves of the outputvoltage waveform are not symmetrical with respect to 0 level. The input supply

current waveform also has a DC component (average value) which can result in

the problem of core saturation of the input supply transformer.

The half wave ac voltage controller using a single thyristor and a single diode

provides control on the thyristor only in one half cycle of the input supply. Hence

ac power flow to the load can be controlled only in one half cycle.

Half wave ac voltage controller gives limited range of RMS output voltagecontrol. Because the RMS value of ac output voltage can be varied from a

maximum of 100% of SV at a trigger angle 0 = to a low of 70.7% of SV atRadians = .

These drawbacks of single phase half wave ac voltage controller can be over come

by using a single phase full wave ac voltage controller.

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APPLICATIONS OF RMS VOLTAGE CONTROLLER

Speed control of induction motor (polyphase ac induction motor). Heater control circuits (industrial heating). Welding power control. Induction heating. On load transformer tap changing. Lighting control in ac circuits. Ac magnet controls.

Problem

1. A single phase half-wave ac voltage controller has a load resistance 50R = ,input ac supply voltage is 230V RMS at 50Hz. The input supply transformer has a

turns ratio of 1:1. If the thyristor 1T is triggered at060 = . Calculate

RMS output voltage. Output power. RMS load current and average load current. Input power factor. Average and RMS thyristor current.

Given,

0

S

230 , primary supply voltage.

Input supply frequency = 50Hz.

50

60 radians.3

V RMS secondary voltage.

p

L

V V RMS

f

R

=

==

= =

=

11

1

p p

S S

V N

V N= = =

Therefore 230p SV V V= =

Where, pN

= Number of turns in the primary winding.S

N = Number of turns in the secondary winding.

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RMS Value of Output (Load) Voltage ( )O RMSV

( ) ( )2

2 21 sin .2

mO RMSV V t d t

=

We have obtained the expression for ( )O RMSV as

( ) ( )1 sin 2

22 2

SO RMSV V

= +

( )

01 sin120230 2

2 3 2O RMS

V

= +

( ) [ ]1

230 5.669 230 0.949862

O RMSV

= =

( )218.4696 218.47

O RMSV V V=

RMS Load Current ( )O RMSI

( )

( ) 218.469664.36939

50

O RMS

O RMS

L

VI Amps

R

= = =

Output Load Power OP

( ) ( )22 4.36939 50 954.5799

O LO RMSP I R Watts= = =

0.9545799O

P KW=

Input Power Factor

O

S S

PPF

V I=

SV = RMS secondary supply voltage = 230V.

SI = RMS secondary supply current = RMS load current.

( )4.36939

S O RMSI I Amps = =

( )

954.5799 W

0.9498230 4.36939 WPF = =

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Average Output (Load) Voltage

( ) ( )2

1sin .

2mO dc

V V t d t

=

We have obtained the expression for the average / DC output voltage as,

( ) [ ]cos 12m

O dc

VV

=

( ) ( ) [ ]02 230 325.2691193cos 60 1 0.5 1

2 2O dc

V

= =

( )

[ ]325.2691193

0.5 25.88409 Volts

2O dc

V

= =

Average DC Load Current

( )

( ) 25.8840940.51768 Amps

50

O dc

O dc

L

VI

R

= = =

Average & RMS Thyristor Currents

I m

i T 1

2

( 2 + )

3

t

Fig.: Thyristor Current Waveform

Referring to the thyristor current waveform of a single phase half-wave ac voltage

controller circuit, we can calculate the average thyristor current ( )T AvgI as

( ) ( )1

sin .2

mT AvgI I t d t

=

( ) ( )sin .2m

T Avg

II t d t

=

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( ) ( )cos2m

T Avg

II t

=

( ) ( )cos cos2

m

T Avg

II

= +

( ) [ ]1 cos2m

T Avg

II

= +

Where,m

m

L

VI

R= = Peak thyristor current = Peak load current.

2 230

50m

I

=

6.505382 Ampsm

I =

( ) [ ]1 cos2m

T Avg

L

VI

R

= +

( ) ( )02 230 1 cos 60

2 50T Avg

I

= +

( ) [ ]

2 230

1 0.5100T AvgI

= +

( )1.5530 Amps

T AvgI =

RMS thyristor current ( )T RMSI can be calculated by using the expression

( ) ( )2 21 sin .

2mT RMS

I I t d t

=

( )

( )( )

2 1 cos 2.

2 2

m

T RMS

tII d t

=

( ) ( ) ( )2

cos 2 .4

m

T RMS

II d t t d t

=

( ) ( )1 sin2

24mT RMS

tI I t

=

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22/213

( ) ( )1 sin 2 sin 2

4 2mT RMS

I I

=

( ) ( )

1 sin 2

4 2mT RMSI I

= +

( ) ( )1 sin 2

2 22

m

T RMS

II

= +

( )

( )0sin 1206.50538 12 3 22

T RMSI

= +

( )

1 2 0.86602544.6

2 3 2T RMS

I

= +

( )4.6 0.6342 2.91746

T RMSI A= =

( )2.91746 Amps

T RMSI =

SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER (AC

REGULATOR) OR RMS VOLTAGE CONTROLLER WITH RESISTIVE LOAD

Single phase full wave ac voltage controller circuit using two SCRs or a single

triac is generally used in most of the ac control applications. The ac power flow to the

load can be controlled in both the half cycles by varying the trigger angle ' ' .

The RMS value of load voltage can be varied by varying the trigger angle ' ' .

The input supply current is alternating in the case of a full wave ac voltage controller and

due to the symmetrical nature of the input supply current waveform there is no dc

component of input supply current i.e., the average value of the input supply current is

zero.

A single phase full wave ac voltage controller with a resistive load is shown in the

figure below. It is possible to control the ac power flow to the load in both the half cycles

by adjusting the trigger angle ' ' . Hence the full wave ac voltage controller is also

referred to as to a bi-directional controller.

22


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Fig.: Single phase full wave ac voltage controller (Bi-directional Controller) using

SCRs

The thyristor 1T is forward biased during the positive half cycle of the input

supply voltage. The thyristor 1T is triggered at a delay angle of ' ' ( )0 radians .Considering the ON thyristor 1T as an ideal closed switch the input supply voltage

appears across the load resistor LR and the output voltage O Sv v= during t = to

radians. The load current flows through the ON thyristor 1T and through the load resistor

LR in the downward direction during the conduction time of 1T from t = to radians.

At t = , when the input voltage falls to zero the thyristor current (which isflowing through the load resistor LR ) falls to zero and hence 1T naturally turns off . No

current flows in the circuit during t = to ( ) + .

The thyristor 2T is forward biased during the negative cycle of input supply and

when thyristor 2T is triggered at a delay angle ( ) + , the output voltage follows the

negative halfcycle of input from ( )t = + to 2 . When 2T is ON, the load current

flows in the reverse direction (upward direction) through 2T during ( )t = + to 2

radians. The time interval (spacing) between the gate trigger pulses of 1T and 2T is kept at

radians or 1800. At 2t = the input supply voltage falls to zero and hence the loadcurrent also falls to zero and thyristor 2T turn off naturally.

Instead of using two SCRs in parallel, a Triac can be used for full wave ac voltagecontrol.

Fig.: Single phase full wave ac voltage controller (Bi-directional Controller) usingTRIAC

23


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25/213

For a full wave ac voltage controller, we can see that the two half cycles of output

voltage waveforms are symmetrical and the output pulse time period (or output pulse

repetition time) is radians. Hence we can also calculate the RMS output voltage byusing the expression given below.

( )2 2 2

0

1 sin .mL RMS

V V t d t

=

( ) ( )2

2 2

0

1.

2LL RMS

V v d t

= ;sin

L O mv v V t = = ; For tot = and ( ) to 2t = +

Hence,

( ) ( ) ( ) ( ) ( )

22 22 1

sin sin2m mL RMSV V t d t V t d t

+

= +

( ) ( )2

2 2 2 21 sin . sin .2

m mV t d t V t d t

+

= +

( ) ( )22 1 cos 2 1 cos 2

2 2 2

mV t t

d t d t

+

= +

( ) ( ) ( ) ( )2 22

cos 2 . cos 2 .2 2

mV d t t d t d t t d t

+ +

= +

( ) ( )2 22 sin 2 sin 2

4 2 2

mV t t

t t

+ +

= +

( ) ( ) ( ) ( )( )2 1 1

sin 2 sin 2 sin 4 sin 24 2 2

mV

= + +

( ) ( ) ( )( )2

1 12 0 sin 2 0 sin 24 2 2

mV

= +

( )( )2 sin 2sin2

24 2 2

mV

+ = + +

( )( )2 sin 2 2sin2

24 2 2

mV

+ = + +

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( ) ( )2 sin 2 1

2 sin 2 .cos 2 cos 2 .sin 24 2 2

mV

= + + +

sin 2 0 & cos 2 1 = =

Therefore,

( ) ( )2

2 sin 2 sin 224 2 2

m

L RMS

VV

= + +

( )2

2 sin 24

mV

= +

( )( )

22 2 2 sin 2

4

m

L RMS

VV

= +

Taking the square root, we get

( ) ( )2 2 sin 22

m

L RMS

VV

= +

( ) ( )2 2 sin 22 2

m

L RMS

VV

= +

( ) ( )1

2 2 sin 222

m

L RMS

VV

= +

( ) ( )1 sin 2

22 22

m

L RMS

VV

= +

( ) ( )1 sin 2

22

m

L RMS

VV

= +

( ) ( ) ( )1 sin 2

2L RMS i RMS

V V

= +

( ) ( )1 sin 2

2SL RMS

V V

= +

Maximum RMS voltage will be applied to the load when 0 = , in that case thefull sine wave appears across the load. RMS load voltage will be the same as the RMS

supply voltage 2

mV

= . When is increased the RMS load voltage decreases.

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( ) ( )0

1 sin 2 00

22

m

L RMS

VV

=

= +

( ) ( )0

1 0

22

m

L RMS

V

V =

= +

( ) ( )0 2

mSL RMS i RMS

VV V V

=

= = =

The output control characteristic for a single phase full wave ac voltage controller

with resistive load can be obtained by plotting the equation for ( )O RMSV

CONTROL CHARACTERISTIC OF SINGLE PHASE FULL-WAVE AC

VOLTAGE CONTROLLER WITH RESISTIVE LOAD

The control characteristic is the plot of RMS output voltage ( )O RMSV versus the

trigger angle ; which can be obtained by using the expression for the RMS outputvoltage of a full-wave ac controller with resistive load.

( ) ( )1 sin 2

2SO RMS

V V

= + ;

Where 2

m

S

V

V = = RMS value of input supply voltage

Trigger angle

in degrees

Trigger angle

in radians ( )O RMS

V %

0 0 SV 100% SV

030 6 ( )1; 6 0.985477 S

V 98.54% SV

060 3 ( )2; 6 0.896938 S

V 89.69% SV

0

90 2

( )3;

6

0.7071 SV

70.7% SV

0120 2 3 ( )4; 6 0.44215 S

V 44.21% SV

0150 5 6 ( )5; 6 0.1698 S

V 16.98% SV

0180 ( )6; 6 0 SV 0 SV

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V O ( R M S )


0 6 0 1 2 0 1 8 0

V S

0 . 2 VS

0 . 6 VS

We can notice from the figure, that we obtain a much better output control

characteristic by using a single phase full wave ac voltage controller. The RMS output

voltage can be varied from a maximum of 100% SV at 0 = to a minimum of 0 at0180 = . Thus we get a full range output voltage control by using a single phase full

wave ac voltage controller.

Need For Isolation

In the single phase full wave ac voltage controller circuit using two SCRs or

Thyristors 1T and 2T in parallel, the gating circuits (gate trigger pulse generating circuits)

of Thyristors 1T and 2T must be isolated. Figure shows a pulse transformer with twoseparate windings to provide isolation between the gating signals of 1T and 2T .

G 1

K 1G 2

K 2

G a t eT r i g g e r

P u l s e

G e n e r a t o r

Fig.: Pulse Transformer

SINGLE PHASE FULL-WAVE AC VOLTAGE CONTROLLER WITH

COMMON CATHODE

It is possible to design a single phase full wave ac controller with a common

cathode configuration by having a common cathode point for 1T and 2T & by adding two

diodes in a full wave ac controller circuit as shown in the figure below

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Fig.: Single phase full wave ac controller with common cathode

(Bidirectional controller in common cathode configuration)

Thyristor 1T and diode 1D are forward biased during the positive half cycle of

input supply. When thyristor 1T is triggered at a delay angle , Thyristor 1T and diode

1D conduct together from t = to during the positive half cycle.

The thyristor 2T and diode 2D are forward biased during the negative half cycle

of input supply, when trigged at a delay angle , thyristor 2T and diode 2D conduct

together during the negative half cycle from ( )t = + to 2 .In this circuit as there is one single common cathode point, routing of the gate

trigger pulses to the thyristor gates of 1T and 2T is simpler and only one isolation circuit

is required.

But due to the need of two power diodes the costs of the devices increase. Asthere are two power devices conducting at the same time the voltage drop across the ON

devices increases and the ON state conducting losses of devices increase and hence the

efficiency decreases.

SINGLE PHASE FULL WAVE AC VOLTAGE CONTROLLER USING A

SINGLE THYRISTOR

R L

T 1

A CS u p p l y

-

D1

D 4

D3

D 2

+

29


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A single phase full wave ac voltage controller circuit (bidirectional controller)

with an RL load using two thyristors 1T and 2T ( 1T and 2T are two SCRs) connected in

parallel is shown in the figure below. In place of two thyristors a single Triac can be used

to implement a full wave ac controller, if a suitable Traic is available for the desired RMS

load current and the RMS output voltage ratings.

Fig: Single phase full wave ac voltage controller with RL load

The thyristor 1T is forward biased during the positive half cycle of input supply.

Let us assume that1

T is triggered at t = , by applying a suitable gate trigger pulse to

1T during the positive half cycle of input supply. The output voltage across the load

follows the input supply voltage when 1T is ON. The load current Oi flows through the

thyristor 1T and through the load in the downward direction. This load current pulse

flowing through 1T can be considered as the positive current pulse. Due to the inductance

in the load, the load current Oi flowing through 1T would not fall to zero at t = , when

the input supply voltage starts to become negative.

The thyristor 1T will continue to conduct the load current until all the inductive

energy stored in the load inductor L is completely utilized and the load current through 1T

falls to zero at t = , where is referred to as the Extinction angle, (the value of t )

at which the load current falls to zero. The extinction angle is measured from the point

of the beginning of the positive half cycle of input supply to the point where the load

current falls to zero.

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The thyristor 1T conducts from t = to . The conduction angle of 1T is

( ) = , which depends on the delay angle and the load impedance angle . The

waveforms of the input supply voltage, the gate trigger pulses of 1T and 2T , the thyristor

current, the load current and the load voltage waveforms appear as shown in the figure

below.

Fig.: Input supply voltage & Thyristor current waveforms

is the extinction angle which depends upon the load inductance value.

Fig.: Gating Signals

32


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Waveforms of single phase full wave ac voltage controller with RL load for > .Discontinuous load current operation occurs for > and ( ) < + ;

i.e., ( ) < , conduction angle < .

Fig.: Waveforms of Input supply voltage, Load Current, Load Voltage and

Thyristor Voltage across 1T

Note

The RMS value of the output voltage and the load current may be varied byvarying the trigger angle .

This circuit, AC RMS voltage controller can be used to regulate the RMS voltageacross the terminals of an ac motor (induction motor). It can be used to control thetemperature of a furnace by varying the RMS output voltage.

33


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For very large load inductance L the SCR may fail to commutate, after it istriggered and the load voltage will be a full sine wave (similar to the applied input

supply voltage and the output control will be lost) as long as the gating signals are

applied to the thyristors 1T and 2T . The load current waveform will appear as a

full continuous sine wave and the load current waveform lags behind the output

sine wave by the load power factor angle .

TO DERIVE AN EXPRESSION FOR THE OUTPUT (INDUCTIVE LOAD)

CURRENT, DURING tot = WHEN THYRISTOR 1T CONDUCTS

Considering sinusoidal input supply voltage we can write the expression for the

supply voltage as

sinS m

v V t= = instantaneous value of the input supply voltage.

Let us assume that the thyristor 1T is triggered by applying the gating signal to 1T

at t = . The load current which flows through the thyristor 1T during t = to canbe found from the equation

sinOO m

diL Ri V t

dt

+ =

;

The solution of the above differential equation gives the general expression for the

output load current which is of the form

( ) 1sint

mO

Vi t A e

Z

= + ;

Where 2m SV V= = maximum or peak value of input supply voltage.

( )22Z R L= + = Load impedance.

1tanL

R

=

= Load impedance angle (power factor angle of load).

L

R = = Load circuit time constant.

Therefore the general expression for the output load current is given by the

equation

( )

1sin

Rt

m LO

Vi t A e

Z

= + ;

34


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The value of the constant 1A can be determined from the initial condition. i.e.

initial value of load current 0Oi = , at t = . Hence from the equation for Oi equating

Oi to zero and substituting t = , we get

( )

10 sin

R tm LO

Vi A eZ

= = +

Therefore ( )1 sinR

tmL

VA e

Z

=

( )11

sinmR

tL

VA

Ze

=

( )1 sinR

t mL VA eZ

+

=

( )

( )1 sinR t

mLV

A eZ

=

By substituting t = , we get the value of constant 1A as

( )

( )1 sinR

mLV

A e

Z

=

Substituting the value of constant 1A from the above equation into the expression for Oi ,

we obtain

( )( )

( )sin sinRR

tm mLL

O

V Vi t e e

Z Z

= +

;

( )( ) ( )

( )sin sinR t R

m mL LO

V Vi t e e

Z Z

= +

( )( )

( )sin sinR

tm mL

O

V Vi t e

Z Z

= +

Therefore we obtain the final expression for the inductive load current of a single

phase full wave ac voltage controller with RL load as

( ) ( )( )

sin sinR

tm L

O

Vi t e

Z

=

; Where t .

35


36/213

The above expression also represents the thyristor current 1Ti , during the

conduction time interval of thyristor 1T from tot = .

To Calculate Extinction Angle

The extinction angle , which is the value of t at which the load currentOi falls to zero and 1T is turned off can be estimated by using the condition that

0Oi = , at t =

By using the above expression for the output load current, we can write

( ) ( )( )

0 sin sinR

m LO

Vi e

Z

= =

As 0mV

Z we can write

( ) ( ) ( )sin sin 0R

Le

=

Therefore we obtain the expression

( ) ( )( )

sin sinR

Le

=

The extinction angle can be determined from this transcendental equation by

using the iterative method of solution (trial and error method). After is calculated, we

can determine the thyristor conduction angle ( ) = . is the extinction angle which depends upon the load inductance value.

Conduction angle increases as is decreased for a known value of .

For < radians, i.e., for ( ) < radians, for ( ) + > the load currentwaveform appears as a discontinuous current waveform as shown in the figure. The

output load current remains at zero during t = to ( ) + . This is referred to as

discontinuous load current operation which occurs for ( ) < + .When the trigger angle is decreased and made equal to the load impedance

angle i.e., when = we obtain from the expression for ( )sin ,

( )sin 0 = ; Therefore ( ) = radians.

Extinction angle ( ) ( ) = + = + ; for the case when =

Conduction angle ( ) 0radians 180 = = = ; for the case when =

Each thyristor conducts for 1800 ( radians ) . 1T conducts from t = to ( ) +

and provides a positive load current. 2T conducts from ( ) + to ( )2 + and providesa negative load current. Hence we obtain a continuous load current and the output voltage

36


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waveform appears as a continuous sine wave identical to the input supply voltage

waveform for trigger angle and the control on the output is lost.

v O

2

3

t

V m

0

I m

t

v = vO S

i O

Fig.: Output voltage and output current waveforms for a single phase full wave ac

voltage controller with RL load for

Thus we observe that for trigger angle , the load current tends to flowcontinuously and we have continuous load current operation, without any break in the

load current waveform and we obtain output voltage waveform which is a continuous

sinusoidal waveform identical to the input supply voltage waveform. We loose the controlon the output voltage for as the output voltage becomes equal to the input supplyvoltage and thus we obtain

( )2

mSO RMS

VV V= = ; for

Hence,

RMS output voltage = RMS input supply voltage for

TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE ( )O RMSV OF A

SINGLE PHASE FULL-WAVE AC VOLTAGE CONTROLLER WITH RL

LOAD.

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When O > , the load current and load voltage waveforms become discontinuousas shown in the figure above.

( ) ( )

1

22 21 sin .

mO RMSV V t d t

=

Output sino mv V t= , for tot = , when 1T is ON.

( )

( )( )

122 1 cos 2

2

m

O RMS

tVV d t

=

( ) ( ) ( )

122

cos 2 .2

m

O RMS

VV d t t d t

=

( ) ( )

122

sin2

22

m

O RMS

V tV t

=

( ) ( )

12 2sin 2 sin 2

2 2 2

m

O RMS

VV

= +

( ) ( )

121 sin 2 sin 2

2 2 2

mO RMSV V

= +

( ) ( )

121 sin 2 sin 2

2 22

m

O RMS

VV

= +

The RMS output voltage across the load can be varied by changing the trigger

angle .

For a purely resistive load 0L = , therefore load power factor angle 0 = .1tan 0

L

R

= =

;

Extinction angle0radians 180 = =

38


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PERFORMANCE PARAMETERS OF A SINGLE PHASE FULL WAVE AC

VOLTAGE CONTROLLER WITH RESISTIVE LOAD

RMS Output Voltage ( ) ( )1 sin 2

22

m

O RMS

VV

= +

;2

mS

VV= = RMS

input supply voltage.

( )( )O RMS

O RMS

L

VI

R= = RMS value of load current.

( )S O RMSI I= = RMS value of input supply current.

Output load power

( )

2

O LO RMSP I R=

Input Power Factor

( )

( )

( )2

L LO RMS O RMS O

S S S S O RMS

I R I RPPF

V I V I V

= = =

( )( )

1 sin 2

2

O RMS

S

VPF

V

= = +

Average Thyristor Current,

I m

i T 1

2

( 2 + )

3

t

Fig.: Thyristor Current Waveform

( ) ( ) ( )1 1

sin .2 2

T mT AvgI i d t I t d t

= =

( ) ( )sin . cos2 2m m

T Avg

I II t d t t

= =

( ) [ ] [ ]cos cos 1 cos2 2m m

T Avg

I II = + = +

39


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Maximum Average Thyristor Current, for 0 = ,

( )m

T Avg

II

=

RMS Thyristor Current

( ) ( )2 21 sin .

2mT RMS

I I t d t

=

( ) ( )1 sin 2

2 22

m

T RMS

II

= +

Maximum RMS Thyristor Current, for 0 = ,

( ) 2m

T RMS

II =

In the case of a single phase full wave ac voltage controller circuit using a Triac

with resistive load, the average thyristor current ( ) 0T AvgI = . Because the Triac conducts inboth the half cycles and the thyristor current is alternating and we obtain a symmetrical

thyristor current waveform which gives an average value of zero on integration.

PERFORMANCE PARAMETERS OF A SINGLE PHASE FULL WAVE AC

VOLTAGE CONTROLLER WITH R-L LOAD

The Expression for the Output (Load) Current

The expression for the output (load) current which flows through the thyristor,

during tot = is given by

( ) ( )( )

1sin sin

Rt

m LO T

Vi i t e

Z

= =

; for t

Where,

2m S

V V= = Maximum or peak value of input ac supply voltage.

( )

22Z R L= +

= Load impedance.

1tanL

R

=

= Load impedance angle (load power factor angle).

= Thyristor trigger angle = Delay angle.

= Extinction angle of thyristor, (value of t ) at which the thyristor (load)

current falls to zero.

is calculated by solving the equation

( ) ( )( )

sin sinRLe

=

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Thyristor Conduction Angle ( ) =

Maximum thyristor conduction angle ( ) = = radians = 1800 for .

RMS Output Voltage

( ) ( )1 sin 2 sin 2

2 22

m

O RMS

VV

= +

The Average Thyristor Current

( ) ( )11

2TT Avg

I i d t

=

( ) ( ) ( )

( )

( )

1

sin sin2

Rt

m L

T Avg

V

I t e d tZ

=

( ) ( ) ( ) ( )( )

( )sin . sin2

Rt

m LT Avg

VI t d t e d t

Z

=

Maximum value of ( )T AvgI occur at 0 = . The thyristors should be rated for

maximum ( )m

T Avg

II

=

, where mmV

IZ

= .

RMS Thyristor Current ( )T RMSI

( ) ( )121

2TT RMS

I i d t

=

Maximum value of ( )T RMSI occurs at 0 = . Thyristors should be rated for

maximum ( ) 2

m

T RMS

II

=

When a Triac is used in a single phase full wave ac voltage controller with RL

type of load, then ( ) 0T AvgI = and maximum ( )2

m

T RMS

II =

41


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PROBLEMS

1. A single phase full wave ac voltage controller supplies an RL load. The input

supply voltage is 230V, RMS at 50Hz. The load has L = 10mH, R = 10, the

delay angle of thyristors 1T and 2T are equal, where 1 2 3

= = . Determinea. Conduction angle of the thyristor 1T.

b. RMS output voltage.

c. The input power factor.

Comment on the type of operation.

Given

230s

V V= , 50f Hz= , 10L mH= , 10R = , 060 = ,

1 23

= = = radians, .

2 2 230 325.2691193m S

V V V= = =

( ) ( ) ( )2 2 22Load Impedance 10Z R L L = = + = +

( ) ( )32 2 50 10 10 3.14159L fL = = = =

( ) ( )2 2

10 3.14159 109.8696 10.4818Z= + = =

2 23031.03179

10.4818

mm

VI A

Z

= = =

Load Impedance Angle1tan

L

R

=

( )1 1 0tan tan 0.314159 17.4405910

= = =

Trigger Angle > . Hence the type of operation will be discontinuous loadcurrent operation, we get

( ) < +

( )180 60 < + ; 0240 <

Therefore the range of is from 180 degrees to 240 degrees.

( )0 0180 240< for RL typeof load and the thyristor 1T naturally turns off at t = .

v OV m

0

2 3

( ) + ( ) +

i O

t

t0

Fig.: Waveform for Discontinuous Load Current Operation without FWD

During the negative half cycle of the input supply the voltage at the supply line

A becomes negative whereas the voltage at line B (at the lower side of the secondary

winding) becomes positive with respect to the center point O. The thyristor 2T is

forward biased during the negative half cycle and it is triggered at a delay angle of

( ) + . The current flows through the thyristor 2T , through the load, and through the

lower part of the secondary winding when 2T conducts during the negative half cycle the

load is connected to the lower half of the secondary winding when 2T conducts.

For purely resistive loads when L = 0, the extinction angle = . The loadcurrent falls to zero at t = = , when the input supply voltage falls to zero at t = .The load current and the load voltage waveforms are in phase and there is no phase shift

between the load voltage and the load current waveform in the case of a purely resistive

load.

For low values of load inductance the load current would be discontinuous and the

extinction angle > but ( ) < + .For large values of load inductance the load current would be continuous and does

not fall to zero. The thyristor 1T conducts from ( )to + , until the next thyristor 2T

is triggered. When 2T is triggered at ( )t = + , the thyristor 1T will be reverse biased

and hence 1T turns off.

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TO DERIVE AN EXPRESSION FOR THE DC OUTPUT VOLTAGE OF A

SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER WITH RL LOAD

(WITHOUT FREE WHEELING DIODE (FWD))

The average or dc output voltage of a full-wave controlled rectifier can be

calculated by finding the average value of the output voltage waveform over one output

cycle (i.e., radians) and note that the output pulse repetition time is2T seconds where

T represents the input supply time period and1

Tf

= ; wheref= input supply frequency.

Assuming the load inductance to be small so that > , ( ) < + we obtain

discontinuous load current operation. The load current flows through 1T form

tot = , where is the trigger angle of thyristor 1T and is the extinction anglewhere the load current through 1T falls to zero at t = . Therefore the average or dcoutput voltage can be obtained by using the expression

( ) ( )2

.2

dc OO dc

t

V V v d t

=

= =

( ) ( )1

.dc OO dc

t

V V v d t

=

= =

( ) ( )1

sin .dc mO dc

V V V t d t

= =

( )cosm

dcO dc

VV V t

= =

( ) ( )cos cosm

dcO dc

VV V

= =

Therefore ( ) ( )cos cosm

O dc

VV

= , for discontinuous load current operation,

( ) < < + .When the load inductance is small and negligible that is 0L , the extinction

angle radians = . Hence the average or dc output voltage for resistive load isobtained as

( ) ( )cos cosm

O dc

VV

= ; cos 1 =

( ) ( )( )cos 1m

O dc

VV

=

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( ) ( )1 cosm

O dc

VV

= + ; for resistive load, when 0L

THE EFFECT OF LOAD INDUCTANCE

Due to the presence of load inductance the output voltage reverses and becomesnegative during the time period tot = . This reduces the dc output voltage. To

prevent this reduction of dc output voltage due to the negative region in the output load

voltage waveform, we can connect a free wheeling diode across the load. The output

voltage waveform and the dc output voltage obtained would be the same as that for a full

wave controlled rectifier with resistive load.

When the Free wheeling diode (FWD) is connected across the load

When 1T is triggered at t = , during the positive half cycle of the input supplythe FWD is reverse biased during the time period tot = . FWD remains reverse

biased and cut-off from tot = . The load current flows through the conductingthyristor 1T, through the RL load and through upper half of the transformer secondary

winding during the time period to .At t = , when the input supply voltage across the upper half of the secondary

winding reverses and becomes negative the FWD turns-on. The load current continues to

flow through the FWD from tot = .

v OV m

0

2 3

( ) + ( ) +

i O

t

t0

Fig.: Waveform for Discontinuous Load Current Operation with FWD

EXPRESSION FOR THE DC OUTPUT VOLTAGE OF A SINGLE PHASE FULL

WAVE CONTROLLED RECTIFIER WITH RL LOAD AND FWD

( ) ( )0

1.

dc OO dc

t

V V v d t

=

= =

Thyristor 1T is triggered at t = . 1T conducts from tot =

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Output voltage sin ; for O mv V t t to = =

FWD conducts from tot = and 0Ov during discontinuous load current

Therefore ( ) ( )1 sin .dc mO dcV V V t d t

= =

( )cosm

dcO dc

VV V t

= =

( ) [ ]cos cos ; cos 1m

dcO dc

VV V

= = + =

Therefore ( ) ( )1 cosmdcO dcV

V V = = +

The DC output voltage dcV is same as the DC output voltage of a single phase full

wave controlled rectifier with resistive load. Note that the dc output voltage of a single

phase full wave controlled rectifier is two times the dc output voltage of a half wave

controlled rectifier.

CONTROL CHARACTERISTICS OF A SINGLE PHASE FULL WAVE

CONTROLLED RECTIFIER WITH R LOAD OR RL LOAD WITH FWD

The control characteristic can be obtained by plotting the dc output voltage dcV

versus the trigger angle .The average or dc output voltage of a single phase full wave controlled rectifier

circuit with R load or RL load with FWD is calculated by using the equation

( ) ( )1 cosm

dcO dc

VV V

= = +

dcV can be varied by varying the trigger angle from 00 to 180 . (i.e., the range

of trigger angle is from 0 to radians).Maximum dc output voltage is obtained when 0 =

( ) ( )max2

1 cos 0m mdcdc

V VV V

= = + =

Therefore ( )max2

mdcdc

VV V

= = for a single phase full wave controlled rectifier.

Normalizing the dc output voltage with respect to its maximum value, we can

write the normalized dc output voltage as

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( )max

dc dcdcn n

dmdc

V VV V

V V= = =

( )( )

1 cos

1 1 cos2 2

m

dcn n

m

V

V VV

+= = = +

Therefore ( )1

1 cos2

dcdcn n

dm

VV V

V= = + =

( )1

1 cos2

dc dmV V= +

Trigger angle in degrees ( )

O dcV Normalized

dc output voltage Vn

02

0.636619mdm m

VV V

= = 1

030 0.593974 mV 0.9330

060 0.47746 mV 0.75

090 0.3183098 mV 0.5

0120 0.191549 mV 0.25

0150 0.04264 mV 0.066980180 0 0

V O ( d c )


0 6 0 1 2 0 1 8 0

V d m

0 . 2 Vd m

0 . 6 Vd m

Fig.: Control characteristic of a single phase full wave controlled rectifier with R

load or RL load with FWD

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CONTINUOUS LOAD CURRENT OPERATION (WITHOUT FWD)

For large values of load inductance the load current flows continuously without

decreasing and falling to zero and there is always a load current flowing at any point of

time. This type of operation is referred to as continuous current operation.

Generally the load current is continuous for large load inductance and for low

trigger angles.The load current is discontinuous for low values of load inductance and for large

values of trigger angles.

The waveforms for continuous current operation are as shown.

v OV m

0

2 3

( ) +

i O

t

t0

( )2 +

T O N1 T O N2 T O N1

Fig.: Load voltage and load current waveform of a single phase full wave controlled

rectifier with RL load & without FWD for continuous load current operation

In the case of continuous current operation the thyristor 1T which is triggered at a

delay angle of , conducts from ( )tot = + . Output voltage follows the inputsupply voltage across the upper half of the transformer secondary winding

sinO AO m

v v V t = = .

The next thyristor 2T is triggered at ( )t = + , during the negative half cycleinput supply. As soon as 2T is triggered at ( )t = + , the thyristor 1T will be reverse

biased and 1T turns off due to natural commutation (ac line commutation). The load

current flows through the thyristor 2T from ( ) ( )to 2t = + + . Output voltageacross the load follows the input supply voltage across the lower half of the transformer

secondary winding sinO BO mv v V t = = .

Each thyristor conducts for ( )0radians 180 in the case of continuous currentoperation.

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TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPUT

VOLTAGE OF SINGLE PHASE FULL WAVE CONTROLLED RECTIFIER

WITH LARGE LOAD INDUCTANCE ASSUMING CONTINUOUS LOAD

CURRENT OPERATION.

( ) ( )( )1

.dc OO dc

t

V V v d t

+

=

= =

( ) ( )( )

1sin .

dc mO dcV V V t d t

+ = =

( )

( )

cosmdcO dc

VV V t

+ = =

( ) ( )cos cosm

dcO dc

VV V

= = + ; ( )cos cos + =

( ) [ ]cos cosm

dcO dc

VV V

= = +

( )

2cosm

dcO dc

VV V

= =

The above equation can be plotted to obtain the control characteristic of a singlephase full wave controlled rectifier with RL load assuming continuous load current

operation.

Normalizing the dc output voltage with respect to its maximum value, the

normalized dc output voltage is given by

( )

( )

max

2cos

cos2

m

dcdcn n

mdc

V

VV V

VV

= = = =

Therefore cosdcn nV V = =

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Trigger angle

in degrees ( )O dc

V Remarks

0

2m

dm

VV

=

Maximum dc output voltage

( )max

2m

dmdc

VV V

= =

030 0.866 dmV060 0.5 dmV

090 0 dmV

0120 -0.5 dmV

0150 -0.866 dmV

01802

mdm

VV

=

V O ( d c )

T r i g g e r a n g l e i n

03 0 6 0 9 0

Vd m

0 . 2 Vd m

0 . 6 Vd m

- 0 . 6 Vd m

- 0 . 2 Vd m

- Vd m

1 2 0 1 5 0 1 8 0

Fig.: Control Characteristic

We notice from the control characteristic that by varying the trigger angle wecan vary the output dc voltage across the load. Thus it is possible to control the dc output

voltage by changing the trigger angle . For trigger angle in the range of 0 to 90degrees ( )0. ., 0 90i e , dcV is positive and the circuit operates as a controlledrectifier to convert ac supply voltage into dc output power which is fed to the load.

For trigger angle090 ,cos > becomes negative and as a result the average dc

output voltage dcV becomes negative, but the load current flows in the same positive

direction. Hence the output power becomes negative. This means that the power flows

from the load circuit to the input ac source. This is referred to as line commutated

inverter operation. During the inverter mode operation for 090 > the load energy canbe fed back from the load circuit to the input ac source.

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TO DERIVE AN EXPRESSION FOR RMS OUTPUT VOLTAGE

The rms value of the output voltage is calculated by using the equation

( ) ( )( )

1

2

22 .2

OO RMSV v d t

+ =

( ) ( )( )

1

2

2 21 sin .mO RMS

V V t d t

+ =

( ) ( )( )

1

222sin .m

O RMS

VV t d t

+ =

( )

( )( )

( ) 122 1 cos 2.

2

m

O RMS

tVV d t

+ =

( ) ( ) ( )( )( )

1

21

cos 2 .2

mO RMSV V d t t d t

+ + =

( ) ( )

( ) ( )1

21 sin2

22mO RMSt

V V t

+ + =

( ) ( )( )

1

2sin 2 sin 21

2 2mO RMS

V V

+ = +

( ) ( )

1

21 sin 2 cos 2 cos 2 sin 2 sin 2

2 2mO RMS

V V

+ =

( ) ( )

1

21 0 sin 2 sin 2

2 2mO RMS

V V

+ =

( ) ( )

1

21

2 2

mmO RMS

VV V

= =

Therefore

( )2mO RMS

VV = ; The rms output voltage is same as the input rms supply voltage.

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SINGLE PHASE SEMICONVERTERS

Errata: Consider diode 2D as 1D in the figure and diode 1D as 2D

Single phase semi-converter circuit is a full wave half controlled bridge converter

which uses two thyristors and two diodes connected in the form of a full wave bridge

configuration.

The two thyristors are controlled power switches which are turned on one after the

other by applying suitable gating signals (gate trigger pulses). The two diodes are

uncontrolled power switches which turn-on and conduct one after the other as and when

they are forward biased.

The circuit diagram of a single phase semi-converter (half controlled bridge

converter) is shown in the above figure with highly inductive load and a dc source in theload circuit. When the load inductance is large the load current flows continuously and

we can consider the continuous load current operation assuming constant load current,

with negligible current ripple (i.e., constant and ripple free load current operation).

The ac supply to the semiconverter is normally fed through a mains supply

transformer having suitable turns ratio. The transformer is suitably designed to supply the

required ac supply voltage (secondary output voltage) to the converter.

During the positive half cycle of input ac supply voltage, when the transformer

secondary output line A is positive with respect to the line B the thyristor 1T and the

diode 1D are both forward biased. The thyristor 1T is triggered at t = ; ( )0

by applying an appropriate gate trigger signal to the gate of 1T. The current in the circuitflows through the secondary line A, through 1T , through the load in the downward

direction, through diode 1D back to the secondary line B.

1T and 1D conduct together from tot = and the load is connected to the

input ac supply. The output load voltage follows the input supply voltage (the secondary

output voltage of the transformer) during the period tot = .At t = , the input supply voltage decreases to zero and becomes negative

during the period ( )tot = + . The free wheeling diode mD across the load

becomes forward biased and conducts during the period ( )tot = + .

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Fig:. Waveforms of single phase semi-converter for RLE load and constant load

current for > 900

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The load current is transferred from 1T and 1D to the FWD mD . 1T and 1D are

turned off. The load current continues to flow through the FWD mD . The load current

free wheels (flows continuously) through the FWD during the free wheeling time period

( )to + .

During the negative half cycle of input supply voltage the secondary line Abecomes negative with respect to line B. The thyristor 2T and the diode 2D are both

forward biased. 2T is triggered at ( )t = + , during the negative half cycle. The FWD

is reverse biased and turns-off as soon as 2T is triggered. The load current continues to

flow through 2T and 2D during the period ( ) to 2t = +

TO DERIVE AN EXPRESSION FOR THE AVERAGE OR DC OUTPUT

VOLTAGE OF A SINGLE PHASE SEMI-CONVERTER

The average output voltage can be found from

( )2

sin .2

dc mV V t d t

=

[ ]2

cos2

mdc

VV t

=

[ ]cos cos ; cos 1mdcV

V

= + =

Therefore [ ]1 cosmdcV

V

= +

dcV can be varied from

2m

V

to 0 by varying from 0 to .

The maximum average output voltage is

( )max

2m

dmdc

VV V

= =

Normalizing the average output voltage with respect to its maximum value

( )0.5 1 cosdcdcn ndm

VV V

V= = = +

The output control characteristic can be plotted by using the expression for dcV

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TO DERIVE AN EXPRESSION FOR THE RMS OUTPUT VOLTAGE OF A

SINGLE PHASE SEMI-CONVERTER

The rms output voltage is found from

( ) ( )

12

2 22 sin .2

mO RMSV V t d t

=

( ) ( ) ( )

1

2 2

1 cos 2 .2

m

O RMS

VV t d t

=

( )

1

21 sin 2

22

m

O RMS

VV

= +

SINGLE PHASE FULL CONVERTER (FULLY CONTROLLED BRIDGE

CONVERTER)

The circuit diagram of a single phase fully controlled bridge converter is shown in

the figure with a highly inductive load and a dc source in the load circuit so that the load

current is continuous and ripple free (constant load current operation).

The fully controlled bridge converter consists of four thyristors 1T, 2T , 3T and 4T

connected in the form of full wave bridge configuration as shown in the figure. Eachthyristor is controlled and turned on by its gating signal and naturally turns off when a

reverse voltage appears across it. During the positive half cycle when the upper line of the

transformer secondary winding is at a positive potential with respect to the lower end the

thyristors 1T and 2T are forward biased during the time interval 0 tot = . The

thyristors 1T and 2T are triggered simultaneously ( ); 0t = , the load is

connected to the input supply through the conducting thyristors 1T and 2T . The output

voltage across the load follows the input supply voltage and hence output voltage

sinO m

v V t= . Due to the inductive load 1T and 2T will continue to conduct beyond

t = , even though the input voltage becomes negative. 1T and 2T conduct together

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during the time period ( )to + , for a time duration of radians (conduction angleof each thyristor = 0180 )

During the negative half cycle of input supply voltage for to 2t = thethyristors 3T and 4T are forward biased. 3T and 4T are triggered at ( )t = + . As

soon as the thyristors 3T and 4T are triggered a reverse voltage appears across the

thyristors 1T and 2T and they naturally turn-off and the load current is transferred from

1T and 2T to the thyristors 3T and 4T . The output voltage across the load follows the

supply voltage and sinO mv V t= during the time period ( ) ( )to 2t = + + . In

the next positive half cycle when 1T and 2T are triggered, 3T and 4T are reverse biased

and they turn-off. The figure shows the waveforms of the input supply voltage, the output

load voltage, the constant load current with negligible ripple and the input supply current.

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During the time period tot = , the input supply voltage Sv and the inputsupply current Si are both positive and the power flows from the supply to the load. The

converter operates in the rectification mode during tot = .During the time period ( )tot = + , the input supply voltage Sv is negative

and the input supply current Si is positive and there will be reverse power flow from the

load circuit to the input supply. The converter operates in the inversion mode during the

time period ( )tot = + and the load energy is fed back to the input source.The single phase full converter is extensively used in industrial applications up to

about 15kW of output power. Depending on the value of trigger angle , the averageoutput voltage may be either positive or negative and two quadrant operation is possible.

TO DERIVE AN EXPRESSION FOR THE AVERAGE (DC) OUTPUT VOLTAGE

The average (dc) output voltage can be determined by using the expression

( ) ( )2

0

1. ;

2dc OO dc

V V v d t

= =

The output voltage waveform consists of two output pulses during the input

supply time period between 0 & 2 radians . In the continuous load current operation ofa single phase full converter (assuming constant load current) each thyristor conduct for

radians (1800) after it is triggered. When thyristors 1T and 2T are triggered at t =

1T and 2T conduct from ( )to + and the output voltage follows the input supply

voltage. Therefore output voltage sinO mv V t= ; for ( )tot = +Hence the average or dc output voltage can be calculated as

( ) ( )2

sin .2

dc mO dcV V V t d t

+ = =

( ) ( )1

sin .dc mO dc

V V V t d t

+ = =

( ) ( )sin .m

dcO dc

VV V t d t

+ = =

( ) [ ]cosm

dcO dc

VV V t

+= =

( ) ( )cos cosm

dcO dc

VV V

= = + + ; ( )cos cos + =

Therefore ( )2

cosmdcO dcV

V V = =

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The dc output voltage dcV can be varied from a maximum value of02 for 0 tom

V

= a

minimum value of02 for radians 180m

V

= =

The maximum average dc output voltage is calculated for a trigger angle 00 = and is obtained as

( ) ( )max2 2

cos 0m mdmdc

V VV V

= = =

Therefore ( )max2

mdmdc

VV V

= =

The normalized average output voltage is given by

( )

( )max

O dc dcdcn n

dmdc

VVV V

V V= = =

2cos

cos2

m

dcn nm

V

V VV

= = =

Therefore cosdcn nV V = = ; for a single phase full converter assuming continuousand constant load current operation.

CONTROL CHARACTERISTIC OF SINGLE PHASE FULL CONVERTER

The dc output control characteristic can be obtained by plotting the average or dc

output voltage dcV versus the trigger angle

For a single phase full converter the average dc output voltage is given by the

equation ( )2

cosmdcO dc

VV V

= =

Trigger angle

in degrees ( )O dc

V Remarks

02 m

dmVV

=

Maximum dc output voltage

( )max

2m

dmdc

VV V

= =

030 0.866 dmV

060 0.5 dmV

090 0 dmV

0120 -0.5 dmV

0150 -0.866 dmV

01802

mdm

VV

=

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V O ( d c )

T r i g g e r a n g l e i n

03 0 6 0 9 0

Vd m

0 . 2 Vd m

0 . 6 Vd m

- 0 . 6 Vd m

- 0 . 2 Vd m

- Vd m

1 2 0 1 5 0 1 8 0

Fig.: Control Characteristic

We notice from the control characteristic that by varying the trigger angle wecan vary the output dc voltage across the load. Thus it is possible to control the dc output

voltage by changing the trigger angle . For trigger angle in the range of 0 to 90

degrees ( )0. ., 0 90i e , dcV is positive and the average dc load current dcI is also

positive. The average or dc output power dcP is positive, hence the circuit operates as a

controlled rectifier to convert ac supply voltage into dc output power which is fed to the

load.For trigger angle

090 ,cos > becomes negative and as a result the average dcoutput voltage dcV becomes negative, but the load current flows in the same positive

direction i.e., dcI is positive . Hence the output power becomes negative. This means that

the power flows from the load circuit to the input ac source. This is referred to as line

commutated inverter operation. During the inverter mode operation for 090 > the loadenergy can be fed back from the load circuit to the input ac source

TWO QUADRANT OPERATION OF A SINGLE PHASE FULL CONVERTER

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The above figure shows the two regions of single phase full converter operation in

the dcV versus dcI plane. In the first quadrant when the trigger angle is less than 900,

anddc dc

V I are both positive and the converter operates as a controlled rectifier and

converts the ac input power into dc output power. The power flows from the input source

to the load circuit. This is the normal controlled rectifier operation where dcP is positive.When the trigger angle is increased above 900 , dcV becomes negative but dcI is

positive and the average output power (dc output power) dcP becomes negative and the

power flows from the load circuit to the input source. The operation occurs in the fourth

quadrant where dcV is negative and dcI is positive. The converter operates as a line

commutated inverter.

TO DERIVE AN EXPRESSION FOR THE RMS VALUE OF THE OUTPUT

VOLTAGE

The rms value of the output voltage is calculated as

( ) ( )2

2

0

1.

2OO RMS

V v d t

=

The single phase full converter gives two output voltage pulses during the input

supply time period and hence the single phase full converter is referred to as a two pulse

converter. The rms output voltage can be calculated as

( ) ( )22 .

2OO RMS

V v d t

+

=

( ) ( )2 21 sin .

mO RMSV V t d t

+ =

( ) ( )2

2sin .mO RMS

VV t d t

+ =

( )

( )( )

2 1 cos 2.

2

m

O RMS

tVV d t

+ =

( ) ( ) ( )2

cos 2 .2

m

O RMS

VV d t t d t

+ + =

( ) ( )

2sin2

22mO RMS

V tV t

+ + =

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( ) ( )( )2 sin 2 sin 2

2 2

m

O RMS

VV

+ = +

( ) ( ) ( ) ( )2

sin 2 2 sin 2 ; sin 2 2 sin 22 2

mO RMS VV

+ = + =

( ) ( )2 sin 2 sin 2

2 2

m

O RMS

VV

=

( ) ( )2 2

02 2 2

m m m

O RMS

V V VV

= = =

Therefore ( )2

mSO RMS

VV V= =

Hence the rms output voltage is same as the rms input supply voltage

The rms thyristor current can be calculated as

Each thyristor conducts for radians or 0180 in a single phase full converter

operating at continuous and constant load current.

Therefore rms value of the thyristor current is calculated as

( ) ( ) ( )

1

2 2T RMS O RMS O RMS

I I I

= =

( )

( )

2

O RMS

T RMS

II =

The average thyristor current can be calculated as

( ) ( ) ( )

1

2 2T Avg O dc O dc

I I I

= =

( )

( )

2

O dc

T Avg

II =

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SINGLE PHASE DUAL CONVERTER

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We have seen in the case of a single phase full converter with inductive loads the

converter can operate in two different quadrants in the versusdc dcV I operating diagram.

If two single phase full converters are connected in parallel and in opposite direction

(connected in back to back) across a common load four quadrant operation is possible.

Such a converter is called as a dual converter which is shown in the figure.

The dual converter system will provide four quadrant operation and is normallyused in high power industrial variable speed drives. The converter number 1 provides a

positive dc output voltage and a positive dc load current, when operated in the

rectification mode.

The converter number 2 provides a negative dc output voltage and a negative dc

load current when operated in the rectification mode. We can thus have bi-directional

load current and bi-directional dc output voltage. The magnitude of output dc load voltage

and the dc load current can be controlled by varying the trigger angles 1 2& of the

converters 1 and 2 respectively.

Fig.: Four quadrant operation of a dual converter

There are two modes of operations possible for a dual converter system.

Non circulating current mode of operation (circulating current free modeof operation).

Circulating current mode of operation.

NON CIRCULATING CURRENT MODE OF OPERATION (CIRCULATING

CURRENT FREE MODE OF OPERATION)

In this mode of operation only one converter is switched on at a time while the

second converter is switched off. When the converter 1 is switched on and the gate trigger

signals are released to the gates of thyristors in converter 1, we get an average output

voltage across the load, which can be varied by adjusting the trigger angle 1 of the

converter 1. If 1 is less than 900, the converter 1 operates as a controlled rectifier and

converts the input ac power into dc output power to feed the load. dcV and dcI are both

positive and the operation occurs in the first quadrant. The average output power

dc dc dcP V I= is positive. The power flows from the input ac supply to the load. When 1 is increased above 900 converter 1 operates as a line commutated inverter and dcV

becomes negative while dcI is positive and the output power dcP becomes negative. The

power is fed back from the load circuit to the input ac source through the converter 1. The

load current falls to zero when the load energy is utilized completely.

The second converter 2 is switched on after a small delay of about 10 to 20 millseconds to allow all the thyristors of converter 1 to turn off completely. The gate signals

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are released to the thyristor gates of converter 2 and the trigger angle 2 is adjusted such

that0

20 90 so that converter 2 operates as a controlled rectifier. The dc output

voltage dcV and dcI are both negative and the load current flows in the reverse direction.

The magnitude of dcV and dcI are controlled by the trigger angle 2 . The operation

occurs in the third quadrant where dcV and dcI are both negative and output power dcP is

positive and the converter 2 operates as a controlled rectifier and converts the ac supply

power into dc output power which is fed to the load.

When we want to reverse the load current flow so that dcI is positive we have to

operate converter 2 in the inverter mode by increasing the trigger angle 2 above090 .

When 2 is made greater than090 , the converter 2 operates as a line commutated

inverter and the load power (load energy) is fed back to ac mains. The current falls to zero

when all the load energy is utilized and the converter 1 can be switched on after a short

delay of 10 to 20 milli seconds to ensure that the converter 2 thyristors are completely

turned off.The advantage of non circulating current mode of operation is that there is no

circulating current flowing between the two converters as only one converter operates and

conducts at a time while the other converter is switched off. Hence there is no need of the

series current limiting inductors between the outputs of the two converters. The current

rating of thyristors is low in this mode.

But the disadvantage is that the load current tends to become discontinuous and

the transfer characteristic becomes non linear. The control circuit becomes complex and

the output response is sluggish as the load current reversal takes some time due to the

time delay between the switching off of one converter and the switching on of the other

converter. Hence the output dynamic response is poor. Whenever a fast and frequent

reversal of the load current is required, the dual converter is operated in the circulatingcurrent mode.

CIRCULATING CURRENT MODE OF OPERATION

In this mode of operation both the converte

36663455 02 power electronics notes by punith

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