Unit-8 AC VOLTAGE CONTROLLER CIRCUITS

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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 ‘α’

V0(RMS) AC AC Vs Variable AC Input Voltage RMS O/P Voltage Voltage fs Controller

fs fS

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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).

6.1 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 ‘α’ (delay angle), the output RMS voltage across the load can be controlled. The trigger delay angle ‘α’ is defined as the phase angle (the value of ωt) at which 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 the negative half cycle the current flowing through the conducting thyristor decreases and 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 and current 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.

6.2 Type of Ac Voltage Controllers The ac voltage controllers are classified into two types based on the type of input ac supply applied to the circuit.

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• 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.

6.3 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 T1 and T2 are turned on by applying appropriate gate trigger pulses to connect the input ac supply to the

load for ‘n’ number of input cycles during the time interval tON . The thyristor switches T1

and T2 are turned off by blocking the gate trigger pulses for ‘m’ number of input cycles

during the time interval tOFF . The ac controller ON time tON usually consists of an integral number of input cycles.

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Fig 6.1: Single phase full wave AC voltage controller

Vs n m

io wt

i g1 Gate pulse of T1

ig2 Gate pulse of T2

Fig 6.2: Waveforms

RR= L = Load Resistance

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

n =Two input cycles. Thyristors are turned ON during tON for two input cycles.

m = One input cycle. Thyristors are turned OFF during tOFF for one input cycle

Fig 6.3: Power Factor

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

The thyristor T1 is turned on at the beginning of each positive half cycle by applying the gate

trigger pulses to T1 as shown, during the ON time tON . The load current flows in the positive

direction, which is the downward direction as shown in the circuit diagram when T1 conducts.

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The thyristor T2 is turned on at the beginning of each negative half cycle, by applying gating

signal to the gate of T2 , during tON . The load current flows in the reverse direction, which is

the upward direction when T2 conducts. Thus we obtain a bi-directional load current flow (alternating load current flow) in a ac voltage 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,

vs= V msinω t = 2 V S sinω t V V = RMS value of input ac supply = m = RMS phase supply voltage. S 2 If the input ac supply is connected to load for ‘n’ number of input cycles and disconnected for ‘m’ number of input cycles, then

tON= n × T, t OFF = m × T

1 Where T = = input cycle time (time period) and f f = input supply frequency.

tON = controller on time = n× T .

tOFF = controller off time = m× T .

TO = Output time period =()()tON+ t OFF = nT + mT .

We can show that,

tON t ON Output RMS voltage VVVO()() RMS= i RMS = S TTOO

Where Vi() RMS is the RMS input supply voltage = VS .

(i) To derive an expression for the rms value of output voltage, for on-off control method.

ωtON 1 2 2 Output RMS voltage VO() RMS = ∫ Vm Sinω t. d()ω t ωTO ωt=0

2 ωtON Vm 2 VO() RMS = ∫ Sinω t. d()ω t ωTO 0

1− Cos 2θ Substituting for Sin2θ = 2

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2 ωtON Vm 1− Cos 2ω t  VO() RMS = ∫   d()ω t ωTO 0 2 

2 ωtONω t ON  Vm VO() RMS =∫ d()()ω t − ∫ Cos2ω t . dω t  2ωTO 0 0 

2 ωtONω t ON  Vm Sin2ω t VO() RMS =()ω t −  2ωTO 02 0 

2 Vmsin 2ω t ON − sin 0  VO() RMS =()ω tON −0 −  2ωTO  2 

Now tON = an integral number of input cycles; Hence

tON = T,2 T ,3 T ,4 T ,5 T ,..... & ωtON = 2π ,4π ,6π ,8π ,10π ,......

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

sin 2ωtON = 0 2 Vmω t ON V m t ON VO() RMS = = 2ω TTOO2

tON t ON VVVO()() RMS= i RMS = S TTOO

V Where VV=m = = RMS value of input supply voltage; i() RMS 2 S

t t nT n ON= ON = = = k = duty cycle (d). TO t ON+ t OFF nT + mT() n + m

n V= V = V k O() RMS SS()m+ n

Performance Parameters of Ac Voltage Controllers

• RMS Output (Load) Voltage 1 2π  2 n 2 2 VO() RMS = ∫ Vm sinω t . d()ω t  2π ()n+ m 0 

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V n V=m = V k = V k O() RMS2 ()m+ n i() RMS S

VO()() RMS= V i RMS k = VS k

Where VVS = i() RMS = RMS value of input supply voltage.

• Duty Cycle t t nT k =ON = ON = TO() t ON+ t OFF () m + n T

n Where, k = = duty cycle (d). ()m+ n

• RMS Load Current

VVO() RMS O() RMS IO() RMS = = ; For a resistive load ZR= L . ZRL

• Output AC (Load) Power

2 PIROL=O() RMS ×

• Input Power Factor

PPoutput load power PF =OO = = VAinput supply volt amperes VSS I

2 IRO() RMS × L PF = ; IIS =in() RMS = RMS input supply current. VIi()() RMS× in RMS

The input supply current is same as the load current IIIin= O = L

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

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2 I× RL V V k PF=O() RMS = O() RMS = i() RMS = k VIVViRMS()()× inRMS iRMS() iRMS()

n PF= k = m+ n

• The Average Current of Thyristor IT() Avg

Waveform of Thyristor Current

iT m n

0 π 2π 3π ωt

n π IT() Avg = ∫ Im sinω t . d()ω t 2π ()m+ n 0

π nIm IT() Avg = ∫ sinω t . d()ω t 2π ()m+ n 0

π  nIm IT() Avg = −cosω t  2π ()m+ n 0 

nI I =m [] −cosπ + cos0 T() Avg 2π ()m+ n

nI m   IT Avg = −() −1 + 1  () 2π ()m+ n

n II= []2 m T() Avg 2π ()m+ n

I n k. I I =m = m T() Avg π ()m+ n π

t n k =duty cycle =ON = ()tON+ t OFF () n + m

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I n k. I I =m = m , T() Avg π ()m+ n π

Vm Where Im = = maximum or peak thyristor current. RL

• RMS Current of Thyristor IT() RMS 1 π  2 n 2 2 IT() RMS = ∫ Im sinω t . d()ω t  2π ()n+ m 0 

1 2 π  2 nIm 2 IT() RMS = ∫sinω t . d()ω t  2π ()n+ m 0 

1 2 π  2 nIm ()1− cos 2ωt IT() RMS = ∫ d()ω t  2π ()n+ m 0 2 

1 2 π π   2 nIm IT() RMS =∫ d()()ω t − ∫ cos 2ω t . dω t   4π ()n+ m 0 0  

1 2 π π   2 nIm sin 2ωt  IT() RMS =()ω t −     4π ()n+ m 02  0  

1 2  2 nIm  sin 2π − sin 0  IT() RMS =()π −0 −     4π ()n+ m  2   

1 2  2 nIm IT() RMS ={}π −0 − 0  4π ()n+ m 

1 1 2 2  2  2 nImπ nI m IT() RMS =  =   4π ()n+ m   4() n + m 

IIn I=m = m k T() RMS 2()m+ n 2

I I= m k T() RMS 2

Problem

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1. A single phase full wave ac voltage controller working on ON-OFF control technique has supply voltage of 230V, RMS 50Hz, load = 50Ω. The controller is ON for 30 cycles and off for 40 cycles. Calculate • ON & OFF time intervals. • RMS output voltage. • Input P.F. • Average and RMS thyristor currents.

VVin() RMS = 230 , VVm =2 × 230 = 325.269 V, VVm = 325.269 ,

1 1 T = = = 0.02sec , T= 20 ms . f50 Hz

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

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

tON = n × T =30 × 20 ms = 600 ms = 0.6sec

tON = n × T = 0.6sec = controller ON time.

tOFF = m × T =40 × 20 ms = 800 ms = 0.8sec

tOFF = m × T = 0.8sec = controller OFF time.

n 30 Duty cycle k = = = 0.4285 ()m+ n ()40 + 30 RMS output voltage n VV= × O()() RMS i RMS ()m+ n 30 3 VV=230 × = 230 O() RMS ()30+ 40 7

VVO() RMS =230 0.42857 = 230 × 0.65465

VVO() RMS =150.570

VVO() RMS O() RMS 150.570V IAO() RMS = = = = 3.0114 ZRL 50Ω

2 2 PIRWOL=O() RMS × =3.0114 × 50 = 453.426498

Input Power Factor P. F= k

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n 30 PF = = = 0.4285 ()m+ n 70

PF = 0.654653

Average Thyristor Current Rating In  k× I I =m ×  = m T() Avg π m+ n  π

Vm 2× 230 325.269 where Im = = = RL 50 50

IAm = 6.505382 = Peak (maximum) thyristor current.

6.505382 3  I = ×  T() Avg π 7 

IAT() Avg = 0.88745

RMS Current Rating of Thyristor IIn 6.505382 3 I=m = m k = × T() RMS 2()m+ n 2 2 7

IAT() RMS = 2.129386

6.4 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 T1 is connected to the

cathode of diode D1 and the cathode of T1 is connected to the anode of D1 . The output voltage across the load resistor ‘R’ and hence the ac power flow to the load is controlled by varying the trigger angle ‘α’. The trigger angle or the delay angle ‘α’ refers to the value of ωt or the instant at

which the thyristor T1 is triggered to turn it ON, by applying a suitable gate trigger pulse between the gate and cathode lead.

The thyristor T1 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

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positive half cycle of input supply. When T1 is triggered it conducts and the load current

flows through the thyristorT1 , the load and through the transformer secondary winding.

By assuming T1 as an ideal thyristor switch it can be considered as a closed switch when it is ON during the period ωt = α to π radians. The output voltage across the load

follows the input supply voltage when the thyristor T1 is turned-on and when it conducts from ωt = α to π radians. When the input supply voltage decreases to zero at ωt = π , for a

resistive load the load current also falls to zero at ωt = π and hence the thyristor T1 turns off at ωt = π . Between the time period ωt = π to 2π , when the supply voltage reverses and

becomes negative the diode D1 becomes forward biased and hence turns ON and conducts.

The load current flows in the opposite direction during ωt = π to 2π radians when D1 is ON and the output voltage follows the negative half cycle of input supply.

Fig 6.4: Halfwave AC phase controller (Unidirectional Controller)

Equations Input AC Supply Voltage across the Transformer Secondary Winding.

vs= V m sinω t V VV= = m = RMS value of secondary supply voltage. S in() RMS 2

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Output Load Voltage

vo= v L = 0 ; for ωt = 0 to α

vo= v L = V m sinω t ; for ωt = α to 2π .

Output Load Current

vo V m sinω t io= i L = = ; for ωt = α to 2π . RRLL

io= i L = 0 ; for ωt = 0 to α .

(i) To Derive an Expression for rms Output VoltageVO() RMS .

2π  1 2 2 VO() RMS = ∫ Vm sinω t . d()ω t  2π α 

2 2π  Vm 1− cos 2ωt  VO() RMS = ∫  . d()ω t  2π α  2  

2 2π  Vm VO() RMS =∫ ()()1 − cos 2ω t . dω t  4π α 

2π 2π  Vm VO() RMS =∫ d()ω t − ∫ cos 2ω t . dω t  2 π α α 

2π 2π  Vm sin 2ωt  VO() RMS =()ω t −    2 π α 2  α 

2π Vm sin 2ωt  VO() RMS =()2π −α −   2 π 2  α

V sin 4π sin 2α  V =m ()2π −α − −  ;sin 4π = 0 O() RMS 2 π 2 2 

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V sin 2α V =m ()2π −α + O() RMS 2 π 2

V sin 2α V =m ()2π −α + O() RMS 2 2π 2

Vm 1 sin 2α  VO RMS =()2π −α +  () 2 2π  2 

1 sin 2α  VV=()2π −α + O()() RMS i RMS 2π  2 

1 sin 2α  VV=()2π −α + O() RMS S 2π  2 

V Where, VV= = m = RMS value of input supply voltage (across the i() RMS S 2 transformer secondary winding).

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

expression for VO() RMS

PLOT OF VO() RMS VERSUS TRIGGER ANGLE α FOR A SINGLE PHASE HALF-WAVE AC VOLTAGE CONTROLLER (UNIDIRECTIONAL CONTROLLER)

Vm 1 sin 2α  VO RMS =()2π −α +  () 2 2π  2 

1 sin 2α  VV=()2π −α + O() RMS S 2π  2 

By using the expression for VO() RMS we can obtain the control characteristics, which is

the plot of RMS output voltage VO() RMS versus the trigger angle α . A typical control characteristic of single phase half-wave phase controlled ac voltage controller is as shown below

Trigger angle α Trigger angle α V in degrees in radians O() RMS V 0 0 V = m S 2

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0 π ; 1π 0.992765 V 30 6 ()6 S 0 π ; 2π 0.949868 V 60 3 ()6 S 0 π ; 3π 0.866025 V 90 2 ()6 S 0 2π ; 4π 0.77314 V 120 3 ()6 S 0 5π ; 5π 0.717228 V 150 6 ()6 S 0 π ; 6π 0.707106 V 180 ()6 S

VO(RMS) 70.7% V 100% VS S

60% VS

20% VS

0 60 120 180 Trigger angle α in degrees

Fig 6.5: Control characteristics of single phase half-wave phase controlled ac voltage controller

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 VS to 70.7% of VS when we vary the trigger angle α from zero to 180 degrees. Thus the half wave ac controller has the drawback of limited range RMS output voltage control.

(ii) To Calculate the Average Value (Dc Value) Of Output Voltage 1 2π VO() dc = ∫ Vm sinω t . d()ω t 2π α

2π Vm VO() dc = ∫ sinω t . d()ω t 2π α

2π  Vm VO() dc = −cosω t  2π α 

V V =m [] −cos 2π + cosα ; cos 2π = 1 O() dc 2π

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2V Hence V =S ()cosα − 1 dc 2π −V When ''α is varied from 0 to π . V varies from 0 to m dc π

Disadvantages of single phase half wave ac voltage controller. • The output load voltage has a DC component because the two halves of the output voltage 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 voltage control. Because the RMS value of ac output voltage can be varied from a maximum of 100%

of VS at a trigger angle α = 0 to a low of 70.7% of VS at α = π Radians.

These drawbacks of single phase half wave ac voltage controller can be over come by using a single phase full wave ac voltage controller.

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 R =50 Ω ; input ac supply voltage is 230V RMS at 50Hz. The input supply transformer has a turn’s ratio 0 of 1:1. If the thyristor T1 is triggered atα = 60 . Calculate • RMS output voltage. • Output power. • RMS load current and average load current. • Input power factor. • Average and RMS thyristor current.

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Given,

Vp = 230 V , RMS primary supply voltage. f = Input supply frequency = 50Hz.

RL =50 Ω π α =600 = radians. 3

VS = RMS secondary voltage.

VN 1 p= p = =1 VNSS 1

Therefore VVVp= S = 230

Where, N p = Number of turns in the primary winding.

N S = Number of turns in the secondary winding.

• RMS Value of Output (Load) Voltage VO() RMS

2π 1 2 2 VO() RMS = ∫ Vm sinω t . d()ω t 2π α

We have obtained the expression for VO() RMS as

1 sin 2α   VVO RMS =S ()2π −α  + () 2π 2

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1π   sin1200 VO RMS =230 2π −   + () 2π  3   2

1 V =230[] 5.669 = 230 × 0.94986 O() RMS 2π

VVVO() RMS =218.4696 ≈ 218.47

• RMS Load Current IO() RMS

VO() RMS 218.46966 IO() RMS = = = 4.36939 Amps RL 50

• Output Load Power PO

2 2 POL= IO() RMS × R =()4.36939 × 50 = 954.5799 Watts

PKWO = 0.9545799

• Input Power Factor

P PF = O VISS×

VS = RMS secondary supply voltage = 230V.

IS = RMS secondary supply current = RMS load current.

∴IS = IO() RMS = 4.36939 Amps

954.5799 W ∴PF = = 0.9498 ()230× 4.36939 W • Average Output (Load) Voltage

1 2π  VO() dc = ∫ Vm sinω t . d()ω t  2π α 

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

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V V =m []cosα − 1 O() dc 2π

2× 2300  325.2691193 VO dc =cos() 60 − 1 =[] 0.5 − 1 () 2π   2π

325.2691193 V =[] −0.5 = − 25.88409 Volts O() dc 2π

• Average DC Load Current

VO() dc −25.884094 IO() dc = = = −0.51768 Amps RL 50

• Average & RMS Thyristor Currents

iT1

π 2π 3π α (2π + α ) ωt α α

Fig 6.6: 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 IT() Avg as 1 π  IT() Avg = ∫ Im sinω t . d()ω t  2π α 

π  Im IT() Avg = ∫sinω t . d()ω t  2π α 

π  Im IT() Avg =() −cosω t  2π α 

I m   IT Avg = −cos()π + cosα  () 2π

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I I =m []1 + cosα T() Avg 2π

Vm Where, Im = = Peak thyristor current = Peak load current. RL

2× 230 I = m 50

Im = 6.505382 Amps

Vm IT() Avg =[]1 + cosα 2π RL

2× 230 0  IT Avg =1 + cos() 60 () 2π × 50   2× 230 I =[]1 + 0.5 T() Avg 100π

IT() Avg =1.5530 Amps

• RMS thyristor current IT() RMS can be calculated by using the expression

π  1 2 2 IT() RMS = ∫ Im sinω t . d()ω t  2π α 

2 π  Im ()1− cos 2ωt IT() RMS = ∫ . d()ω t  2π α 2 

2 π π  Im IT() RMS =∫ d()()ω t − ∫ cos 2ω t . dω t  4π α α 

1 π sin 2ωt  π  IT() RMS = Im ()ω t −    4π α 2  α 

1  sin 2π − sin 2α  IIT RMS =m ()π −α −    () 4π  2  

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1 sin 2α  II=()π −α + T() RMS m 4π  2 

Im 1 sin 2α  IT RMS =()π −α +  () 2 2π  2 

0 6.50538 1 π  sin 120  ()  IT() RMS =π −  + 2 2π  3  2 

1 2π  0.8660254  IT RMS =4.6   +  () 2π  3  2 

IAT() RMS =4.6 × 0.6342 = 2.91746

IT() RMS = 2.91746 Amps

6.5 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 ''α . The RMS value of load voltage can be varied by varying the trigger angle ''α . 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 ''α . Hence the full wave ac voltage controller is also referred to as to a bi-directional controller.

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Fig 6.7: Single phase full wave ac voltage controller (Bi-directional Controller) using SCRs

The thyristor T1 is forward biased during the positive half cycle of the input supply

voltage. The thyristor T1 is triggered at a delay angle of ''α ()0≤α ≤ π radians . Considering

the ON thyristor T1 as an ideal closed switch the input supply voltage appears across the load

resistor RL and the output voltage vOS= v during ωt = α to π radians. The load current

flows through the ON thyristor T1 and through the load resistor RL in the downward direction

during the conduction time of T1 from ωt = α to π radians. At ωt = π , when the input voltage falls to zero the thyristor current (which is flowing

through the load resistor RL ) falls to zero and hence T1 naturally turns off . No current flows in the circuit during ωt = π to ()π +α .

The thyristor T2 is forward biased during the negative cycle of input supply and when

thyristor T2 is triggered at a delay angle()π +α , the output voltage follows the negative

halfcycle of input from ωt =()π +α to 2π . When T2 is ON, the load current flows in the

reverse direction (upward direction) through T2 during ωt =()π +α to 2π radians. The time 0 interval (spacing) between the gate trigger pulses of T1 and T2 is kept at π radians or 180 . At ωt = 2π the input supply voltage falls to zero and hence the load current also falls to zero and

thyristor T2 turn off naturally.

Instead of using two SCR’s in parallel, a Triac can be used for full wave ac voltage control.

Fig 6.8: Single phase full wave ac voltage controller (Bi-directional Controller) using TRIAC

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Fig 6.9: Waveforms of single phase full wave ac voltage controller

Equations Input supply voltage

vS= V msinω t = 2 V S sinω t ;

Output voltage across the load resistor RL ;

vO= v L = V m sinω t ; for ωt = α to π and ωt =()π +α to 2π

Output load current

vO V m sinω t iO= = = I m sinω t ; RRLL for ωt = α to π and ωt =()π +α to 2π

(i) To Derive an Expression for the Rms Value of Output (Load) Voltage

The RMS value of output voltage (load voltage) can be found using the expression

2π 2 21 2 VO()() RMS= V L RMS = ∫ vL d()ω t ; 2π 0 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

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

π 21 2 2 VL() RMS = ∫ Vm sinω t . dω t π 0

2π 21 2 VL() RMS = ∫ vL . d()ω t ; 2π 0

vL= v O = V m sinω t ; For ωt = α to π and ωt =()π +α to 2π

Hence, π 2π  2 1 2 2 VL() RMS =∫()()()() Vmsinω t dω t + ∫ V m sinω t dω t  2π α π +α 

π 2π  1 2 2 2 2 =Vm∫sinω t . d()()ω t + V m ∫ sinω t . dω t  2π α π +α 

V 2 π 1− cos 2ωt2π 1 − cos 2ω t  =m ∫d()ω t + ∫ d()ω t  2π α 2π +α 2 

V 2 π π 2π 2π  =m ∫d()()()()ω t − ∫cos 2ω t . dω t + ∫ dω t − ∫ cos 2ω t . dω t  2π × 2 α α π +α π +α 

2 π 2π π 2π  Vm sin 2ωt   sin 2ω t  =()ωt +()ω t −  −    4π α π +α  2 α  2  π +α 

V 2 1 1  =m ()()π −α +π −α −()sin2π − sin2α −() sin4π − sin2()π +α 4π  2 2 

V 2 1 1  =m 2()π −α −() 0sin2 −α −() 0sin2 −()π +α 4π  2 2 

V 2 sin 2α sin 2()π +α  =m 2()π −α + +  4π  2 2 

V 2 sin 2α sin() 2π + 2α  =m 2()π −α + +  4π  2 2 

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V 2 sin 2α 1  =m 2()π −α + +() sin 2π .cos 2α + cos 2π .sin 2α 4π  2 2 

sin 2π = 0 & cos 2π = 1

Therefore,

V 2 sin 2α sin 2α  V 2 =m 2()π −α + + L() RMS 4π  2 2 

V 2 =m 2()π −α + sin 2α  4π

V 2 2 m   V L RMS =()2π − 2α + sin 2α  () 4π

Taking the square root, we get

V m   VL RMS =()2π − 2α + sin 2α  () 2 π

V m   VL RMS =()2π − 2α + sin 2α  () 2 2π

V 1 m   VL RMS =()2π − 2α + sin 2α  () 2 2π

Vm 1 sin 2α   VL RMS =2()π −α +   () 2 2π  2  

Vm 1 sin 2α  VL RMS =()π −α +  () 2 π 2 

1 sin 2α  VV=()π −α + L()() RMS i RMS π 2 

1 sin 2α  VV=()π −α + L() RMS S π 2 

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Maximum RMS voltage will be applied to the load whenα = 0, in that case the full sine wave appears across the load. RMS load voltage will be the same as the RMS supply V voltage = m . When α is increased the RMS load voltage decreases. 2

Vm 1 sin 2× 0  VL() RMS =()π −0 +  α =0 2 π 2 

Vm 1 0  VL() RMS =()π +  α =0 2 π 2 

Vm VVVL() RMS= = i() RMS = S α =0 2

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

with resistive load can be obtained by plotting the equation for VO() RMS

Control Characteristic of Single Phase Full-Wave Ac Voltage Controller with Resistive Load

The control characteristic is the plot of RMS output voltage VO() RMS versus the trigger angleα ; which can be obtained by using the expression for the RMS output voltage of a full- wave ac controller with resistive load.

1 sin 2α  VV=()π −α + ; O() RMS S π 2 

V Where V =m = RMS value of input supply voltage S 2

Trigger angle α Trigger angle α V % in degrees in radians O() RMS

0 0 VS 100% VS

0 π ; 1π 0.985477 VS 98.54% V 30 6 ()6 S

0 π ; 2π 0.896938 VS 89.69% V 60 3 ()6 S

0 π ; 3π 0.7071 VS 70.7% V 90 2 ()6 S

0 2π ; 4π 0.44215 VS 44.21% V 120 3 ()6 S

0 5π ; 5π 0.1698 VS 16.98% V 150 6 ()6 S

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0 π ; 6π 0 VS 0 V 180 ()6 S

VO(RMS)

0.6VS

0.2 VS

0 60 120 180

Trigger angle α in degrees

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% VS at α = 0 to a minimum of ‘0’ at α = 1800 . 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 T1 and T2 in parallel, the gating circuits (gate trigger pulse generating circuits) of

Thyristors T1 and T2 must be isolated. Figure shows a pulse transformer with two separate

windings to provide isolation between the gating signals of T1 and T2 .

G1 Gate Trigger K1 Pulse G Generator 2

K 2

Fig 6.10: Pulse Transformer

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6.6 Single Phase Full Wave Ac Voltage Controller (Bidirectional Controller) With RL Load

In this section we will discuss the operation and performance of a single phase full wave ac voltage controller with RL load. In practice most of the loads are of RL type. For example if we consider a single phase full wave ac voltage controller controlling the speed of a single phase ac induction motor, the load which is the induction motor winding is an RL type of load, where R represents the motor winding resistance and L represents the motor winding inductance. A single phase full wave ac voltage controller circuit (bidirectional controller) with an

RL load using two thyristors T1 and T2 (T1 and T2 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 6.11: Single phase full wave ac voltage controller with RL load

The thyristor T1 is forward biased during the positive half cycle of input supply. Let

us assume that T1 is triggered atωt = α , by applying a suitable gate trigger pulse to T1 during the positive half cycle of input supply. The output voltage across the load follows the input

supply voltage when T1 is ON. The load current iO flows through the thyristor T1 and through

the load in the downward direction. This load current pulse flowing through T1 can be

considered as the positive current pulse. Due to the inductance in the load, the load current iO

flowing through T1 would not fall to zero at ωt = π , when the input supply voltage starts to become negative.

The thyristor T1 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 T1 falls to zeroω at β t = , whereβ is referred to as the Extinction angle, (the value of ωt ) at which the

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load currentβ 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. δ β α φ

The thyristor T1 conducts from ωt = α to . The conduction angle of T1 is =() − , which depends on the delay angle α and the load impedance angle . The

waveforms of the input supply voltage, the gate trigger pulses of T1 and T2 , the thyristor current, the load current and the load voltage waveforms appear as shown in the figure below.

Fig 6.12: Input supply voltage & Thyristor current waveforms β is the extinction angle which depends upon the load inductance value.

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Fig 6.13: Gating Signals

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α φ β π α β α π Waveforms of single phase full wave ac voltage controller with RL load for > . Discontinuous load current operation occurs for > and <() + ; i.e., ()− < , conduction angle < π .

Fig 6.14: Waveforms of Input supply voltage, Load Current, Load Voltage and Thyristor Voltage across T1

Note • The RMS value of the output voltage and the load current may be varied by varying the trigger angle α . • This circuit, AC RMS voltage controller can be used to regulate the RMS voltage across the terminals of an ac motor (induction motor). It can be used to control the temperature of a furnace by varying the RMS output voltage. • For very large load inductance ‘L’ the SCR may fail to commutate, after it is triggered and the load voltage will be a full sine wave (similar to the applied input supply

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

the thyristors T1 and T2 . 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 φ. ω α β

(i) To Derive an Expression for the Output (Inductive Load) Current, During

t = to When Thyristor T1 Conducts

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

vS= V m sinω t = instantaneous value of the input supply voltage.

β Let us assume that the thyristor T1 is triggered by applying the gating signal to T1 at

ωt = α . The load current which flows through the thyristor T1 during ωt = α to can be found from the equation

diO  L  + RiO = V m sinω t ; dt 

The solution of the above differential equation gives the general expression for the output load current which is of the form

τ V −t i=m sin ()ω t −φ + A e ; O Z 1

Where VVm= 2 S = maximum or peak value of input supply voltage.

ω ZRL=2 + ()2 = Load impedance. ω L  φ = tan−1   = Load impedance angle (power factor angle of load). R 

τ L = = Load circuit time constant. R

Therefore the general expression for the output load current is given by the equation −R Vω φ t i=m sin () t − + A e L ; O Z 1

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The value of the constant A1 can be determined from the initial condition. i.e. initial

value of load current iO = 0 , at ωt = α . Hence from the equation for iO equating iO to zero α φ and substituting ωt = α , we get

−R t Vm L iO =0 = sin () − + A1 e α φ Z

−R t −V Therefore A e L =m sin () − 1 Z α φ

1 −Vm  A1 =−R sin () −  t Z  e L α φ +R t L −Vm  A1 = e sin () −  Z  α φ R()ω t ωL −Vm  A1 = e sin () −  Z 

By substitutingωt = α , we get the value of constant A1 as α φ R()α ωL −Vm  A1 = e sin () −  Z 

Substituting the value of constant A1 from the above equation into the expression for iO , we obtain ω φ− αR R φ()α t VVmL ω L − m  iO =sin() t − + e e  sin () −  ; ZZ 

ω φ−R()ω t R α()α φ VVmω LLω − m  iO =sin() t − + e e  sin () −  ZZ 

−R ω φ()ωt−α α φ VVmωL − m  iO =sin() t − + e  sin () −  ZZ 

Therefore we obtain the final expression for the inductive load current of a single phase full wave ac voltage controller with RL load as

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−R ()ωt−α  Vm ωL iO =sin()()ω t −φ − sin α −φ e  ; Where ≤t ≤ . Z   ω α β

The above expression also represents the thyristor current iT1 , during the conduction β time interval of thyristor T1 from t = to . β To Calculate Extinction Angle The extinction angle , which is the value of ωt at which the load current i ω β O

falls to zero and T1 is turned off can be estimated by using the condition that iO = 0 , at t = By using the above expression for the output load current, we can write

−R ()β −α  Vm ωL iO =0 = sin()()β −φ − sin α −φ e  Z   V As m ≠ 0 we can write Z −R ()β −α  sin()()β −φ − sinα −φ eωL  = 0  

β α Therefore we obtain theω expression −R ()− sin()()β −φ = sin α −φ e L β The extinction angle can be determinedβ from this transcendental equation by using the iterative methodδ of solution β α (trial and error method). After is calculated, we can β determine the thyristor conduction angle =() − . is the extinction angle which dependsβ upon the load inductance value. Conduction angle δ increases as α βis decreased α π for a known valueπ αof β. For δ < π radians, i.e., for ()− < radians, for ()+ > the load current waveform appears as a ωdiscontinuous β current waveform as shown in the figure. The output load current remains at zero during t = to ()π +α . This is referred to as discontinuous β π α load current operation which occurs for <() + . φ When the trigger angle α is decreased and made equal to the load impedance angle α φ β φ i.e., when = we obtain from the expression for sin ()− ,

β φ π sin()β −φ = 0 ; Therefore ()− = radians.

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

Conduction angle δ =()βω −α φ =π radiansπ φ = 1800 ; for the case when = π φ π φ 0 Each thyristor conducts for 180 (π radians) . T1 conducts from t = to ()+

and provides a positive load current. T2 conducts from ()+ to ()2 + and provides a α φ negative load current. Hence we obtain a continuous load current and the output voltage 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 vOS =v Vm

π 2π 3π 0 φ ωt

φ φ φ

φ ωt

Fig 6.15: Output Voltage And Outputα Current φ Waveforms For A Single Phase Full Wave Ac Voltage Controller With Rl Load For ≤ α φ Thus we observe that for trigger angle ≤ , the load current tends to flow continuously 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 αwaveform φ identical to the input supply voltage waveform. We lose the control on the output voltage for ≤ as the output voltage becomes equal to the input supply voltage and thus we obtain V α φ VV=m = ; for ≤ O() RMS 2 S Hence, α φ RMS output voltage = RMS input supply voltage for ≤

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(ii) To Derive an Expression For rms Output Voltage VO() RMS of a Single Phase Full-Wave Ac Voltage Controller with RL Load.

Whenα > O , the load current and load voltage waveforms become discontinuous as shown in the figure above. 1 1 β  2 V=  V2sin 2 ω t . d()ω t  O() RMS π ∫ m ω α β α 

Output vo= V m sinω t , for t = to , when T1 is ON.

1 2 β  2 Vm ()1− cos 2ωt VO() RMS = ∫ d()ω t  π α 2 

1 2 β β   2 Vm   VO() RMS =∫ d()()ω t − ∫ cos 2ω t . dω t   2π α α  

1 2 β β   2 Vm sin 2ωt  VO() RMS =()ω t −     2π α 2  α   1 2  2 Vm sin 2β sin 2α  VO() RMS =()β −α − +   2π  2 2  

1 1 sin 2α sin 2β   2 VVO RMS =m ()β −α + −   () 2π  2 2  

1 2 Vm 1 sin 2α sin 2β   VO RMS =()β −α + −   () 2 π 2 2  

The RMS output voltage across the load can be varied by changing the trigger angle α . For a purely resistive load L = 0 , therefore load power factor angleφ = 0 .

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ωL  φ =tan−1   = 0 ; R  Extinction angle β =π radians = 1800

Performance Parameters of A Single Phase Full Wave Ac Voltage Controller with Resistive Load

Vm 1 sin 2α  Vm • RMS Output Voltage VO RMS =()π −α +  ; = VS = RMS input () 2 π 2  2 supply voltage.

VO() RMS • IO() RMS = = RMS value of load current. RL

• IIS = O() RMS = RMS value of input supply current.

• Output load power 2 PIROL=O() RMS ×

• Input Power Factor 2 P IRIRO RMS×LL O RMS × PF =O =() = () VIVIVSSSS× × O() RMS

VO() RMS 1 sin 2α  PF = =()π −α +  VS π 2 

• Average Thyristor Current,

iT1

π 2π 3π α (2π + α ) ωt α α

Fig6.16: Thyristor Current Waveform

1π 1 π IT() Avg =∫ iT d()ω t = ∫ I m sinω t . d()ω t 2π α 2π α

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π π  IIm m IT() Avg =∫sinω t . d()ω t = − cosω t  2π α 2π α 

II I =m[] −cosπ + cosα = m [] 1 + cosα T() Avg 2π 2π • Maximum Average Thyristor Current, for α = 0, I I = m T() Avg π

• RMS Thyristor Current π  1 2 2 IT() RMS = ∫ Im sinω t . d()ω t  2π α 

Im 1 sin 2α  IT RMS =()π −α +  () 2 2π  2 

• Maximum RMS Thyristor Current, for α = 0,

I I = m T() RMS 2 In the case of a single phase full wave ac voltage controller circuit using a Triac with

resistive load, the average thyristor current IT() Avg = 0 . Because the Triac conducts in both 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 ω α β The expression for the output (load) current which flows through the thyristor, during t = to is given by

−R ()ωt−α  ω β Vm ωL iOT= i =sin()()ω t −φ − sin α −φ e  ; for ≤t ≤ 1 Z   Where,

VVm= 2 S = Maximum or peak value of input ac supply voltage.

ω ZRL=2 + ()2 = Load impedance.

ω L  φ = tan−1   = Load impedance angle (load power factor angle). R 

α = Thyristor trigger angle = Delay angle.

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β = Extinction angle of thyristor, (value of ωt ) at which the thyristor (load) current β α ω falls to zero. is calculated by solving the equation −R ()− δ β α sin()()β −φ = sin α −φ e L δ β α π α φ Thyristor Conduction Angle =() − Maximum thyristor conduction angle =() − = radians = 1800 for ≤ . α β β α RMS Output Voltage

Vm 1 sin 2 sin 2  VO RMS =() − + −  () 2 π 2 2 

The Average Thyristor Current 1 β  I=  i dω t  T() Avg ∫ T1 () 2π α 

β −R ()ωt−α   1 Vm ωL IT() Avg =∫ sin()()ω t −φ − sin α −φ e  d()ω t  2π α Z   

β β −R ()ωt−α  Vm ω L IT() Avg =∫sin()()()ω t −φ . dω t − ∫ sin α −φ e d()ω t  2π Z α α 

Maximum value of IT() Avg occur at α = 0. The thyristors should be rated for

Im  Vm maximum I =   , where Im = . T() Avg π  Z

RMS Thyristor Current IT() RMS 1 β  I=  i2 dω t  T() RMS ∫ T1 () 2π α 

Maximum value of IT() RMS occurs at α = 0. Thyristors should be rated for maximum I  I = m  T() RMS 2 

When a Triac is used in a single phase full wave ac voltage controller with RL type of I load, then I = 0 and maximum I = m T() Avg T() RMS 2

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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 = 10Ω, the delay angle of π thyristors T and T are equal, where α =α = . Determine 1 2 1 2 3

a. Conduction angle of the thyristor T1 . b. RMS output voltage. c. The input power factor. Comment on the type of operation. Given π VV= 230 , f= 50 Hz , L=10 mH , R =10 Ω , α = 600 , α =α =α = s 1 2 3 radians, .

VVVm=2 S = 2 × 230 = 325.2691193 ω ω ZRLL= Load Impedance =2 +()2 =()() 10 2 + 2

ωL=()2π fL =() 2π × 501010 × ×−3 =π = 3.14159 Ω

Z =()()102 + 3.14159 2 = 109.8696 = 10.4818 Ω

V 2× 230 IA=m = = 31.03179 m Z 10.4818 ω L  Load Impedance Angle φ = tan−1   R 

π  φ =tan−1  = tan − 1() 0.314159 = 17.44059 0 10  α φ Trigger Angle > . Hence the type of operation will be discontinuous load current operation, βwe get π α <() +

β <()180 + 60 ; β < 2400

β Therefore the range of is from 180 degrees to 240 degrees. ()1800<β < 240 0

Extinction Angleβ is calculated by using the equation

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−R β ()− sin()()β −φ = sin α −φ e L

In the exponential term the value of α and should be substituted in radians. Hence − R ()βRad−α Rad π  sin()()β −φ = sin α −φ eωL ; α =   Rad 3 

()()α −φ =60 − 17.44059 = 42.55940

−10 0 ()β −α 0 π β β α sin()β − 17.44 = sin() 42.5594 e

0 −3.183() − β π sin()− 17.44 = 0.676354e β 0 × 1800 → π radians, = Rad 1800 Assuming β = 1900 ; β 0×π 190 0 ×π β = = = 3.3161 Rad 1800 180

L.H.S: sin()() 190− 17.440 = sin 172.56 = 0.129487 π  −3.183 3.3161 −  R.H.S: 0.676354×e 3  = 4.94 × 10−4

Assuming β = 1830 ; β 0×π 183 0 ×π β = = = 3.19395 Rad 1800 180

π  ()β −α =3.19395 −  = 2.14675 3 

L.H.S: sin()()β −φ = sin 183 − 17.44 = sin165.560 = 0.24936

R.H.S: 0.676354e−3.183() 2.14675 = 7.2876 × 10−4

Assuming β ≈ 1800 ββ 0×π 180 π 0 ×π = = = Rad 1800 180

π  2π  ()β −α =π −  =   3   3  L.H.S: sin()()β −φ = sin 180 − 17.44 = 0.2997

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π  −3.183π −  R.H.S: 0.676354e 3  = 8.6092 × 10−4

Assuming β = 1960 β 0×π 196 0 ×π β = = = 3.420845 Rad 1800 180

L.H.S: sin()()β −φ = sin 196 − 17.44 = 0.02513 π  −3.183 3.420845 −  R.H.S: 0.676354e 3  = 3.5394 × 10−4

Assuming β = 1970 β 0×π 197 0 ×π β = = = 3.43829 Rad 1800 180

L.H.S: sin()()β −φ = sin 197 − 17.44 = 7.69 = 7.67937 × 10−3 π  −3.183 3.43829 −  R.H.S: 0.676354e 3  = 4.950386476 × 10−4

Assuming β = 197.420 β 0 ×π 197.42 ×π β = = = 3.4456 Rad 1800 180

L.H.S: sin()()β −φ = sin 197.42 − 17.44 = 3.4906 × 10−4 π  −3.183 3.4456 −  R.H.S: 0.676354e 3  = 3.2709 × 10−4

Conduction Angle δ =()β −α =()197.420 − 60 0 = 137.42 0

RMS Output Voltage α β β α 1 sin 2 sin 2  VV=() − + − O() RMS S π 2 2 

0 0 1 π  sin2 60 sin2 197.42  () ()  VO() RMS =230 3.4456 −  + − π 3  2 2  1   VO RMS =230() 2.39843 + 0.4330 − 0.285640  () π

VO() RMS =230 × 0.9 = 207.0445 V

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Input Power Factor P PF = O VISS×

VO RMS 207.0445 I =() = =19.7527 A O() RMS Z 10.4818

2 2 PIROL=O() RMS × =()19.7527 × 10 = 3901.716 W

VVIISS=230 , =O() RMS = 19.7527

P 3901.716 PF =O = = 0.8588 VISS×230 × 19.7527

2. A single phase full wave controller has an input voltage of 120 V (RMS) and a load resistance of 6 ohm. The firing angle of thyristor isπ 2 . Find d. RMS output voltage e. Power output f. Input power factor g. Average and RMS thyristor current.

Solution π α = =900 ,VR = 120 V, = 6 Ω 2 S

RMS Value of Output Voltage 1 1 sin 2α   2 VVOS=π −α +   π 2  

1 1π sin180   2 VO =120 π − +   π 2 2  

VO = 84.85 Volts

RMS Output Current

V 84.85 I =O = =14.14 A O R 6

Load Power 2 PIROO= ×

2 PO =()14.14 × 6 = 1200 watts

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Input Current is same as Load Current

Therefore IISO= = 14.14 Amps

Input Supply Volt-Amp =VIVASS = 120 × 14.14 = 1696.8

Therefore Load Power 1200 Input Power Factor = = = 0.707()lag Input Volt-Amp 1696.8

Each Thyristor Conducts only for half a cycle

Average thyristor current IT() Avg

1 π I= Vsinω t . d()ω t T() Avg 2π R ∫ m α α

V =m ()1 + cos ; V = 2V 2π R m S

2× 120 =[]1 + cos90 = 4.5 A 2π × 6

RMS thyristor current IT() RMS

π 2 2 1 Vm sin ω t IT() RMS = ∫ 2 d()ω t 2π α R

2 π Vm ()1− cos 2ωt = 2 ∫ d()ω t 2π R α 2

1 V 1 sin 2α   2 =m π −α +   2R π  2  

1 2V 1 sin 2α   2 =S π −α +   2R π  2  

1 2120×  1π sin180   2 =π − +   =10 Amps 2× 6π  2 2  

3. A single phase half wave ac regulator using one SCR in anti-parallel with a diode feeds 1 kW, 230 V heater. Find load power for a firing angle of 450.

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Solution π α =450 = ,V = 230 V ; PKWW= 1 = 1000 4 S O At standard rms supply voltage of 230V, the heater dissipates 1KW of output power

Therefore VVV× 2 PVI= × =OOO = OOO RR

Resistance of heater 2 V 2 ()230 R =O = =52.9 Ω PO 1000 RMS value of output voltage 1 2 1 sin 2α   0 VVOS=2π −α +   ; for firing angle α = 45 2π  2  

1 1π sin 90   2 VO =230 2π − +   = 224.7157 Volts 2π  4 2  

RMS value of output current V 224.9 I =O = = 4.2479 Amps O R 52.9 Load Power 2 2 PIROO= × =()4.25 × 52.9 = 954.56 Watts

4. Find the RMS and average current flowing through the heater shown in figure. The delay angle of both the SCRs is 450.

SCR1 io +

1-φ SCR2 1 kW, 220V 220V heater ac

Solution π α =450 = ,V = 220 V 4 S

Resistance of heater

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2 V 2 ()220 R = = =48.4 Ω R 1000

Resistance value of output voltage 1 sin 2α   VVOS=π −α +   π 2  

1π sin 90   VO =220 π − +   π 4 2  

1π 1   VO =220π − +   = 209.769 Volts π 4 2  

V 209.769 RMS current flowing through heater =O = = 4.334 Amps R 48.4

Average current flowing through the heater I Avg = 0

5. A single phase voltage controller is employed for controlling the power flow from 220 V, 50 Hz source into a load circuit consisting of R = 4 Ω and L = 6 mH. Calculate the following a. Control range of firing angle b. Maximum value of RMS load current c. Maximum power and power factor d. Maximum value of average and RMS thyristor current.

Solution

For control of output power, minimum angle of firing angle ααα is equal to the load impedance angle θθθ

α =θ , load angle

ωL  6  θ =tan−1  = tan − 1   = 56.3 0 R  4 

Maximum possible value of α is 1800 Therefore control range of firing angle is 56.30<α < 180 0

Maximum value of RMS load current occurs whenα =θ = 56.30 . At this value of α the Maximum value of RMS load current

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VS 220 IO = = = 30.5085 Amps Z 42+ 6 2

2 2 Maximum Power PIROO= =()30.5085 × 4 = 3723.077 W

Input Volt-Amp =VISO = 220 × 30.5085 = 6711.87 W

P 3723.077 Power Factor =O = = 0.5547 Input VA 6711.87

Average thyristor current will be maximum when α =θ and conduction angleγ =1800 . Therefore maximum value of average thyristor current π +α 1 Vm IT() Avg =∫ sin ()()ω t −θ dω t 2π α Z

−R ()ωt−α  Vm ωL Note: iOT= i =sin()()ω t −θ − sin α −θ e  1 Z   Atω α θ= 0, V i= i =m sin t − TO1 () π α Z

V + m   IT Avg = −cos()ω t −θ  () 2π Z α π α θ α θ V m   IT Avg = −cos()() + − + cos −  () 2π Z But α =θ , π VVV m  m m IT Avg = −cos()() + cos 0  =[] 2 = () 2π ZZZ 2π π

Vm 2× 220 ∴IT() Avg = = = 13.7336 Amps π Z π 42+ 6 2 γ π Similarly, maximum RMS value occurs when α = 0 and = .

Therefore maximum value of RMS thyristor current

π +α 2 1 Vm  ITM =∫ sin ()()ω t −θ  dω t 2π α Z 

2 π +α   Vm 1− cos() 2ωt − 2θ ITM = 2 ∫   d()ω t 2π Z α  2 

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π +α V 2 sin() 2ωt − 2θ  I=m ω t −  TM 4π Z 2 2  α

2 Vm ITM =[]π +α −α − 0 4π Z 2

V 2× 220 I =m = = 21.57277 Amps TM 2 2 2Z 2 4+ 6