Thyristor Converters or Controlled Converters

EE 435- Electric Drives

Dr. Ali M. Eltamaly

Chapter 3

Thyristor Converters or Controlled Converters

3.1 Introduction The controlled rectifier circuit is divided into three main circuits:(1) Power Circuit This is the circuit contains voltage source, load and switches as diodes, thyristors or IGBTs. (2) Control Circuit This circuit is the circuit, which contains the logic of the firing of switches that may, contains amplifiers, logic gates and sensors. (3) Triggering circuit This circuit lies between the control circuit and power thyristors. Sometimes this circuit called switch drivers circuit. This circuit contains buffers, opt coupler or pulse transformers. The main purpose of this circuit is to separate between the power circuit and control circuit. The method of switching off the thyristor is known as Thyristor commutation. The thyristor can be turned off by reducing its forward current below its holding current or by applying a reverse voltage across it. The commutation of thyristor is classified into two types:1- Natural Commutation If the input voltage is AC, the thyristor current passes through a natural zero, and a reverse voltage appear across the thyristor, which in turn automatically turned off the device due to the natural behavior of AC voltage source. This is known as natural commutation or line commutation. This type of commutation is applied in AC voltage controller rectifiers and cycloconverters. 2- Forced Commutation In DC thyristor circuits, if the input voltage is DC, the forward current of the thyristor is forced to zero by an additional circuit called commutation circuit to turn off the thyristor. This technique is called forced commutation. Normally this method for turning off the thyristor is applied in choppers. There are many thyristor circuits we can not present all of them. In the following items we are going to present and analyze the most famous thyristor circuits.

3.2 Half Wave Single Phase Controlled Rectifier 3.2.1 Half Wave Single Phase Controlled Rectifier With Resistive Load The circuit with single SCR is similar to the single diode circuit, the difference being that an SCR is used in place of the diode. Most of the power electronic applications operate at a relative high voltage and in such cases; the voltage drop across the SCR tends to be small. It is quite often justifiable to assume that the conduction drop across the SCR is zero when the circuit is analyzed. It is also justifiable to assume that the current through the SCR is zero when it is not conducting. It is known that the SCR can block conduction in either direction. The explanation and the analysis presented below are based on the ideal SCR model. All simulation carried out by using PSIM computer simulation program. A circuit with a single SCR and resistive load is shown in Fig.3.1. The source vs is an alternating sinusoidal Fig.3.1 Half wave single phase controlled rectifier.

30

Chapter Three

source. If vs = Vm sin (ωt ) , vs is positive when 0 < ω t < π, and vs is negative when π < ω t 30. Fig.3.31 Three phase half wave controlled rectifier. In case of α ≤ 30, various voltages and currents of the converter shown in Fig.3.31 are shown in Fig.3.32. Fig.3.33 shows FFT components of load voltage, secondary current and primary current. As we see the load voltage contains high third harmonics and all other triplex harmonics. Also secondary current contains DC component, which saturate the transformer core. The saturation of the transformer core is the main drawback of this system. Also the primary current is highly distorted but without a DC component. The average output voltage and current are shown in equation (3.57) and (3.58) respectively. The rms output voltage and current are shown in equation (3.59) and (3.60) respectively. 3 2π

Vdc =

I dc =

Vrms

5π / 6 +α

3 3 3 Vm cos α = 0.827Vm cos α = VLL cos α = 0.675VLL cosα (3.57) 2π 2π

∫ Vm sin ω t dω t =

π / 6 +α

3 3 Vm 0.827 * Vm cos α = cos α 2 *π *R R

3 = 2π

5π / 6 +α

∫ (Vm sin ω t )

π / 6 +α

2

dω t = 3 Vm

(3.58) 1 3 + cos 2α 6 8π

(3.59)

SCR Rectifier or Controlled Rectifier

43

3 Vm 1 3 + cos 2α (3.60) R 6 8π Then the thyristor rms current is equal to secondery current and can be obtaiend as follows:

I rms =

Ir = IS =

I rms 3

=

Vm R

1 3 cos 2α + 6 8π

The PIV of the diodes is

2 VLL = 3 Vm

(3.61) (3.62)

Fig.3.32 Voltages and currents waveforms for rectifier shown in Fig.3.31 at α ≤ 30.

Fig.3.33 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.31 for α ≤ 30.

44

Chapter Three

In case of α > 30, various voltages and currents of the rectifier shown in Fig.3.31 are shown in Fig.3.34. Fig.3.35 shows FFT components of load voltage, secondary current and primary current. As we can see the load voltage and current equal zero in some regions (i.e. discontinuous load current). The average output voltage and current are shown in equation (3.63) and (3.64) respectively. The rms output voltage and current are shown in equation (3.65) and (3.66) respectively. The average output voltage is :3 Vdc = 2π I dc =

π

∫ Vm sin ω t dωt =

π / 6+α

3 Vm 2π

  π  π  1 + cos  6 + α  = 0.4775Vm 1 + cos  6 + α       

3 Vm  π  1 + cos  + α   2π R  6  π

Vrms =

3 (Vm sin ω t )2 dω t = 3 Vm 5 − α + 1 sin(π / 3 + 2α ) ∫ 24 4π 8 π 2π π / 6 +α

3 Vm 5 α 1 − + sin(π / 3 + 2α ) R 24 4π 8 π Then the diode rms current can be obtaiend as follows: I V 5 α 1 − + sin(π / 3 + 2α ) I r = I S = rms = m R 24 4π 8 π 3

I rms =

The PIV of the diodes is

2 VLL = 3 Vm

(3.63) (3.64) (3.65) (3.66)

(3.67) (3.68)


45

Fig.3.34 Various voltages and currents waveforms for converter shown in Fig.3.22 for α > 30.

Fig.3.35 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.22 for α > 30.

Example 7 Three-phase half-wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the rectfier is pure resistance of 5 Ω . The delay angle α = 25o . Calculate: The rectfication effeciency (b) PIV of thyristors Solution: From (3.57) the DC value of the output voltage can be obtained as following: V 3 3 281.5 Vdc = VLL cos α = 460 cos 25 = 281.5V Then; I dc = dc = = 56.3 A R 5 2π 2π From (3.59) we can calculate Vrms as following: Vrms = 3 Vm

1 3 1 3 + cos 2α = 2 VLL * + cos 2α 6 8π 6 8π

46

Chapter Three

Then, Vrms = 2 * 460 * Then I rms =

1 3 + cos (2 * 25) = 298.8 V 6 8π

Vrms 298.8 = = 59.76 A R 5

Then, the rectfication effeciency is, η =

Vdc I dc *100 = 88.75% Vrms I rms

PIV = 2 VLL = 2 * 460 = 650.54 V

Example 8 Solve the previous example (evample 7) if the firing angle α = 60 o Slution: From (3.63) the DC value of the output voltage can be obtained as following:  2  * 460 3  3  3 Vm    π π  π   Vdc = 1 + cos  + α  = 1 + cos  +  = 179.33 V   2π  2π  6 3  6   V 179.33 = 35.87 A Then; I dc = dc = 5 R From (3.65) we can calculate Vrms as following: Vrms = 3 Vm

5 α 1 5 π /3 1 − + − + sin(π / 3 + 2α ) = 2 * 460 * sin(π / 3 + 2π / 3 ) = 230V 24 4π 8 π 24 4π 8 π

Vrms 230 = = 46 A 5 R Then, the rectfication effeciency can be calculated as following V I η = dc dc *100 = 60.79 % and PIV = 2 VLL = 2 * 460 = 650.54 V Vrms I rms

Then I rms =

3.5 Three Phase Half Wave Controlled Rectifier With DC Load Current The Three Phase Half Wave Controlled Rectifier With DC Load Current is shown in Fig.3.36, the load voltage will reverse its direction only if α > 30. However if α < 30 the load voltage will be positive all the time. Then in case of α > 30 the load voltage will be negative till the next thyristor in the sequence gets triggering pulse. Also each thyristor will conduct for 120o if the load current is continuous as shown in Fig.3.37. Fig.3.38 shows the FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.36 for α > 30 and pure DC current load. Fig.3.36 Three Phase Half Wave Controlled Rectifier With DC Load Current

47

t=0


Fig.3.37 Various voltages and currents waveforms for the converter shown in Fig.3.36 for α > 30 and pure DC current load.

Fig.3.38 FFT components of load voltage, secondary current and supply current for the converter shown in Fig.3.36 for α > 30 and pure DC current load.

As explained before the secondary current of transformer contains DC component. Also the source current is highly distorted which make this system has less practical significance. The THD of the supply current can be obtained by the aid of Fourier analysis as shown in the following:If we move y-axis of supply current to be as shown in Fig.3.33, then the waveform can be represented as odd function. So, an=0 and bn can be obtained as the following:bn =

2

π

Then,

2π / 3

∫ I dc sin(nω t ) dω t = 0

bn =

2 I dc 3 * πn 2

2 I dc  2nπ  1 − cos  for n=1,2,3,4,… 3  πn  for n=1,2,4,5,7,8,10,…..

(3.69)

Chapter Three

48

And b n = 0 For n=3,6,9,12,….. Then the source current waveform can be expressed as the following equation

(3.70)

i p (ωt ) =

3I dc  1 1 1 1  sin ωt + sin 2ωt + sin 4ωt + sin 5ωt + sin 7ωt + ......  2 4 5 7 π  

(3.71)

The resultant waveform shown in equation (3.61) agrees with the result from simulation (Fig.3.38). The THD of source current can be obtained by two different methods. The first method is shown below:THD =

I 2p − I 2p1

Where, I p =

I 2p1

2 * I dc 3

(3.72)

The rms of the fundamental component of supply current can be obtained from equation (3.71) and it will be as shown in equation (3.74) 3I (3.74) I p1 = dc 2π Substitute equations (3.73) and (3.74) into equation (3.72), then, 2 2 9 2 I dc − I dc 3 2π 2 = 68 % (3.75) THD = 9 2 I dc 2π 2 Another method to determine the THD of supply current is shown in the following:Substitute from equation (3.71) into equation (3.72) we get the following equation:2

2

2

2

2

2

2

2

2

1 1 1 1 1  1   1   1   1  THD =   +   +   +   +   +   +   +   +   + .... ≅ 68 % (3.76)  2   4   5   7   8   10   11   13   14  The supply current THD is very high and it is not acceptable by any electric utility system. In case of full wave three-phase converter, the THD in supply current becomes much better than half wave (THD=35%) but still this value of THD is not acceptable.

Example 9 Three phase half wave controlled rectfier is connected to 380 V three phase supply via delta-way 380/460V transformer. The load of the rectfier draws 100 A pure DC current. The (b) Input power factor. delay angle, α = 30 o . Calculate: (a) THD of primary current. Solution: The voltage ratio of delta-way transformer is 380/460V. Then, the peak value of 460 2 = 121.05 A . Then, I P , rms = 121.05 * = 98.84 A . primary current is 100 * 3 380 3I 3 *121.05 I P1 can be obtained from equation (372) where I P1 = dc = = 81.74 A . 2π 2π 2

2  I P, rms  98.84    − 1 *100 =  Then, (THD )I P =   − 1 *100 = 67.98 % 81 . 74 I    P1  The input power factor can be calculated as following: I π  81.74  π π  * cos +  = 0.414 Lagging P. f = P1 * cos α +  = 6  98.84 I P, rms  6 6


49

3.7 Three Phase Full Wave Fully Controlled Rectifier Bridge 3.7.1 Three Phase Full Wave Fully Controlled Rectifier With Resistive Load Three-phase full wave controlled rectifier shown in Fig.3.42. As we can see in this figure the thyristors has labels T1, T2,……,T6. The label of each thyristor is chosen to be identical to triggering sequence where thyristors are triggered in the sequence of T1, T2,……,T6 which is clear from the thyristors currents shown in Fig.3.43.

Fig.3.42 Three-phase full wave controlled rectifier.

Fig.3.43 Thyristors currents of three-phase full wave controlled rectifier.

The operation of the circuit explained here depending on the understanding of the reader the three phase diode bridge rectifier. The Three-phase voltages vary with time as shown in the following equations: va = Vm sin (ω t ) vb = Vm sin (ω t − 120) vc = Vm sin (ω t + 120) It can be seen from Fig.3.44 that the voltage va is the highest positive voltage of the three

phase voltage when ωt is in the range 30 < ω t < 150o . So, the thyristor T1 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44 the firing angle α = 40 as an example. So, T1 takes a pulse at ω t = 30 + α = 30 + 40 = 70 o as shown in Fig.3.44. Also, it is clear from Fig3.38 that thyristor T1 or any other thyristor remains on for 120o .

Chapter Three

50

Fig.3.44 Phase voltages and thyristors currents of three-phase full wave controlled rectifier at α = 40 o .

It can be seen from Fig.3.44 that the voltage vb is the highest positive voltage of the three phase voltage when ωt is in the range of 150 < ωt < 270 o . So, the thyristor T3 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle α = 40 as an example. So, T3 takes a pulse at ω t = 150 + α = 150 + 40 = 190o . It can be seen from Fig.3.44 that the voltage vc is the highest positive voltage of the three phase voltage when ωt is in the range 270 < ω t < 390 o . So, the thyristor T5 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle α = 40 as an example. So, T3 takes a pulse at ωt = 270 + α = 310o . It can be seen from Fig.3.44 that the voltage va is the highest negative voltage of the three phase voltage when ω t is in the range 210 < ω t < 330 o . So, the thyristor T4 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.338, the firing angle α = 40 as an example. So, T4 takes a pulse at ω t = 210 + α = 210 + 40 = 250o . It can be seen from Fig.3.44 that the voltage vb is the highest negative voltage of the three phase voltage when ω t is in the range 330 < ω t < 450 o or 330 < ω t < 90 o in the next period of supply voltage waveform. So, the thyristor T6 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle α = 40 as an example. So, T6 takes a pulse at ω t = 330 + α = 370o . It can be seen from Fig.3.44 that the voltage vc is the highest negative voltage of the three phase voltage when ω t is in the range 90 < ωt < 210o . So, the thyristor T2 is forward bias during this period and it is ready to conduct at any instant in this period if it gets a pulse on its gate. In Fig.3.44, the firing angle α = 40 as an example. So, T2 takes a pulse at ωt = 90 + α = 130 o .

51


From the above explanation we can conclude that there is two thyristor in conduction at any time during the period of supply voltage. It is also clear that the two thyristors in conduction one in the upper half (T1, T3, or, T5) which become forward bias at highest positive voltage connected to its anode and another one in the lower half (T2, T4, or, T6) which become forward bias at highest negative voltage connected to its cathode. So the load is connected at any time between the highest positive phase voltage and the highest negative phase voltage. So, the load voltage equal the highest line to line voltage at any time which is clear from Fig.3.45. The following table summarizes the above explanation. Period, range of wt SCR Pair in conduction α + 30o to α + 90o

T1 and T6

α + 90o to α + 150o

T1 and T2

α + 150o to α + 210o o

o

o

o

o

o

α + 210 to α + 270 α + 270 to α + 330

T2 and T3 T3 and T4 T4 and T5 o

α + 330 to α + 360 and α + 0 to α + 30

o

T5 and T6

o

Fig.3.45 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at α = 40 .

The line current waveform is very easy to obtain it by applying kerchief's current law at the terminals of any phase. As an example I a = I T 1 − I T 4 which is clear from Fig.3.42. The input current of this rectifier for α = 40, (α ≤ 60 ) is shown in Fig.3.46. Fast Fourier transform (FFT) of output voltage and supply current are shown in Fig.3.47. Analysis of this three-phase controlled rectifier is in many ways similar to the analysis of single-phase bridge controlled rectifier circuit. The average output voltage, the rms output voltage, the ripple content in output voltage, the total rms line current, the fundamental rms current, THD in line current, the displacement power factor and the apparent power factor are to be determined. In this section, the analysis is carried out assuming that the load is pure resistance.

Vdc =

3

π / 2 +α

π π / 6∫+α

π

3 Vm sin(ω t + ) dω t = 6

3 3 Vm

π

cos α

The maximum average output voltage for delay angle α=0 is

(3.81)

52

Chapter Three

Vdm =

3 3 Vm

π The normalized average output voltage is as shown in (3.83)

Vn =

Vdc = cos α Vdm

(3.82) (3.83)

The rms value of the output voltage is found from the following equation: Vrms =

π / 2 +α

2

π   3 Vm sin(ω t + )  dω t = 3 Vm π 6   π / 6 +α 3

∫

1 3 3   + cos 2α  2 4π  

(3.84)

Fig.3.46 The input current of this rectifier of rectifier in Fig.3.42 at α = 40, (α ≤ 60) .

Fig.3.46 FFT components of output voltage and supply current of rectifier in Fig.3.42 at α = 40, (α ≤ 60) .

53


In the converter shown in Fig.3.42 the output voltage will be continuous only and only if α ≤ 60 o . If α > 60o the output voltage, and phase current will be as shown in Fig.3.47.

Fig.3.47 Output voltage along with three phase line to line voltages of rectifier in Fig.3.42 at α = 75o .

The average and rms values of output voltage is shown in the following equation: Vdc =

3

π

5π / 6

∫

3 Vm sin(ω t +

π / 6 +α

π 6

) dω t =

3 3 Vm

π

[1 + cos ( π / 3 + α )]

The maximum average output voltage for delay angle α=0 is: Vdm =

(3.85) 3 3 Vm

π

Vdc = [1 + cos ( π / 3 + α )] Vdm The rms value of the output voltage is found from the following equation:

The normalized average output voltage is

Vrms

5π / 6

Vn =

(3.87)

2

3  π  π    3 Vm sin(ω t + )  dω t = 3Vm 1 − =  2α − cos 2α +   ∫ 6  4π  6  π π / 6 +α   3

(3.86)

(3.88)

Example 10 Three-phase full-wave controlled rectifier is connected to 380 V, 50 Hz supply to feed a load of 10 Ω pure resistance. If it is required to get 400 V DC output voltage, calculate the following: (a) The firing angle, α (b) The rectfication effeciency (c) PIV of the thyristors. Solution: From (3.81) the average voltage is : 3 3 Vm 3 3 2 Vdc = cos α = * * 380 cos α = 400V . π R 3 V 400 Then α = 38.79 o , I dc = dc = = 40 A R 10 From (3.84) the rms value of the output voltage is: Vrms = 3 Vm

 1 3 3 2  + cos 2 α  = 3 * * 380 * 2 4π 3  

Vrms 412.412 = = 41.24 A R 10 400 * 40 *100 = *100 = 94.07% 412.4 * 41.24

Then, Vrms = 412.412 V Then, I rms = Then, η =

Vdc * I dc Vrms * I rms

 1 3 3  + cos (2 * 38.79 ) 2 4π  

The PIV= 3 Vm=537.4V

54

Chapter Three

Example 11 Solve the previous example if the required dc voltage is 150V.

Solution: From (3.81) the average voltage is : 2 3 3* * 380 3 3 Vm 3 Vdc = cos α = cos α = 150V . Then, α = 73o

π

π

It is not acceptable result because the above equation valid only for α ≤ 60 . Then we have to use the (3.85) to get Vdc as following:

3 3* Vdc ==

π

2 * 380 3 [1 + cos ( π / 3 + α )] = 150V . Then, α = 75.05o

Vdc 150 = = 15 A R 10 From (3.88) the rms value of the output voltage is: π  3  2  Vrms = 3Vm 1 − * 380 *  2α − cos 2α +   = 3 * 4π  6  3 

Then I dc =

 π 3   1 − − cos (2 * 75.05 + 30 )   2 * 75.05 * 180   4π 

Vrms 198.075 = = 19.8075 A R 10 150 *15 *100 = *100 = 57.35 % 198.075 *19.81

Then, Vrms = 198.075 V , Then, I rms = Then, η =

Vdc * I dc Vrms * I rms

The PIV= 3 Vm=537.4V 3.7.1 Three Phase Full Wave Fully Controlled Rectifier With pure DC Load Current Three-phase full wave-fully controlled rectifier with pure DC load current is shown in Fig.3.48. Fig.3.49 shows various currents and voltage of the converter shown in Fig.3.48 when the delay angle is less than 60o. As we see in Fig.3.49, the load voltage is only positive and there is no negative period in the output waveform. Fig.3.50 shows FFT components of output voltage of

rectifier shown in Fig.3.48 for α < 60o .

Fig.3.48 Three phase full wave fully controlled rectifier with pure dc load current

In case of the firing angle is greater than 60 o , the output voltage contains negaive portion as shown in Fig.3.51. Fig.3.52 shows FFT components of output voltage of rectifier shown in Fig.3.48 for α > 60 o . The average and rms voltage is the same as in equations (3.81) and (3.84) respectively. The line current of this rectifier is the same as line current of three-phase full-wave diode bridge rectifier typically except the phase shift between the phase voltage and phase current is zero in case of diode bridge but it is α in case of three-phase full-wave controlled rectifier


55

with pure DC load current as shown in Fig.3.53. So, the input power factor of three-phase fullwave diode bridge rectifier with pure DC load current is: I (3.89) PowerFactor = s1 cosα Is

Fig.3.49 Output voltage and supply current waveforms along with three phase line voltages for the rectifier shown in Fig.3.48 for α < 60o with pure DC current load.

Fig.3.50 FFT components of SCR, secondary, primary currents respectively of rectifier shown in Fig.3.48.

Chapter Three

56

Fig.3.51 Output voltage and supply current waveforms along with three phase line voltages for the rectifier shown in Fig.3.48 for α > 60o with pure DC current load.

Fig.3.52 FFT components of SCR, secondary, primary currents respectively of rectifier shown in Fig.3.48 for α > 60o.

Fig.3.53 Phase a voltage, current and fundamental components of phase a of three phase full bridge fully controlled rectifier with pure DC current load and α > 60 .


57

In case of three-phase full-wave controlled rectifier with pure DC load and source inductance the waveform of output voltage and line current and their FFT components are shown in Fig.3.54 and Fig.3.55 respectively. The output voltage reduction due to the source inductance is the same as obtained before in Three-phase diode bridge rectifier. But, the commutation time will differ than the commutation time obtained in case of Three-phase diode bridge rectifier. It is left to the reader to determine the commutation angle u in case of three-phase full-wave diode bridge rectifier with pure DC load and source inductance. The Fourier transform of line current and THD will be the same as obtained before in Three-phase diode bridge rectifier with pure DC load and source inductance which explained in the previous chapter.

Fig.3.54 Output voltage and supply current of rectifier shown n Fig.3.48 with pure DC load and source inductance the waveforms.

Fig.3.55 FFT components of output voltage of rectifier shown in Fig.3.48 for α > 60o and there is a source inductance.

 2ω LS I o  u = cos −1 cos(α ) −  −α V LL   Vrd = 6 fLI o

(3.99)

(3.104) The DC voltage without source inductance tacking into account can be calculated as following:

58

Chapter Three

Vdc

actual

= Vdc

without sourceinduc tan ce

− Vrd =

3 2

π

VLL cosα − 6 fLs I o

2 I o2  π u  − π  3 6  2 6 Io  u  sin   I S1 = πu 2 The power factor can be calculated from the following equation:

Then I S =

pf =

I S1 u  cos  α +  = 2 IS 

2 6 Io  u  sin   πu 2

u 2 3 * sin   u   2  cos α + u  cos  α +  =   2 2   π u  2 I o2  π u  u π −  −  3 6 π  3 6 

(3.105) (3.106) (3.110)

(3.111)

Note, if we approximate the source current to be trapezoidal as shown in Fig.3.58n, the u  displacement power factor will be as shown in (3.111) is cos α +  . Another expression for the 2  displacement power factor, by equating the AC side and DC side powers [ ] as shown in the following derivation: From (3.98) and (3.105) we can get the following equation: 3 2 3 Vdc = VLL cos α − VLL (cos α − cos(α + u )) π 2π 3 2 3 2 VLL [2 cos α − (cos α − cos(α + u ))] ∴Vdc = VLL [cos α + cos(α + u )] 2π 2π Then the DC power output from the rectifier is Pdc = Vdc I o . Then,

∴Vdc =

(3.112)

3 2 VLL * I o [cos α + cos(α + u )] (3.113) 2π (3.114) On the AC side, the AC power is: Pac = 3 VLL I S1 cos φ1 Substitute from (3.110) into (3.114) we get the following equation: 4 3 Io  u  6 2 VLL I o  u  sin   cos φ1 = sin   cos φ1 (3.115) Pac = 3 VLL πu πu 2 2 2 By equating (3.113) and (3.115) we get the following: u [cos α + cos (α + u )] cos φ1 = (3.116) u 4 * sin   2 3.7.2 Inverter Mode of Operation Once again, to understand the inverter mode of operation, we will assume that the do side of the converter can be represented by a current source of a constant amplitude I d , as shown in Fig.3.61. For a delay angle a greater than 90° but less than 180°, the voltage and current waveforms are shown in Fig.3.62a. The average value of Vd is negative according to (3.81). On the ac side, the negative power implies that the phase angle φ1 , between vs and is , is greater than 90°, as shown in Fig.3.62b.

Pdc =


59

Fig.3.61 Three phase SCR inverter with a DC current.

Fig.3.62 Waveforms in the inverter shown in Fig.3.56.

In a practical circuit shown in Fig.3.63, the operating point for a given Ed and α can be obtained from the characteristics shown in Fig.3.64. Similar to the discussion in connection with single-phase converters, the extinction angle γ = 180o − α − u must be greater than the thyristor turn-off interval ω t q in the waveforms of

(

)

Fig.3.54, where v5 is the voltage across thyristor 5.

Fig.3.63 Three phase SCR inverter with a DC voltage source.

Chapter Three

60

Fig.3.64 Vd versus I d of Three phase SCR inverter with a DC voltage source.

Inverter Startup As discussed for start up of a single-phase inverter, the delay angle α in the three-phase inverter of Fig.3.63 is initially made sufficiently large (e.g., 165°) so that id is discontinuous. Then, α is decreased by the controller such that the desired I d and Pd are obtained.