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s
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Rms
V
t
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(5.4)
The average thyristor voltage:
The rms thyristor voltage:
( ) [ ] ( )
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t
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t
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The average thyristor voltage:
(5.5)
R
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V
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I dc
dc
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(
)
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( =](https://image.slidesharecdn.com/chapter6-230622144207-62485464/85/Chapter-6-AC-AC-Converters-pdf-14-320.jpg)
![c) The average thyristor current
The PF is
(lagging)
707
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d) The rms value of the thyristor current
A
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![Solution
the current is expressed from the parameters
A
m
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rad
R
L
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ωτ
ωτ
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θ
θ
θ
ω
ω
ω
ω
The current is then expressed as
The extinction angle β is determined from the numerical solution of i(β) = 0 in
the preceding equation
The conduction angle γ = β – α = 2.26 rad = 1300, which is less the the limit of
1800
A
e
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β
β](https://image.slidesharecdn.com/chapter6-230622144207-62485464/85/Chapter-6-AC-AC-Converters-pdf-23-320.jpg)
![the rms load current is determined by
The rms current in each SCR is determine
[[[[ ]]]]
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t
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=
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the average SCR current is obtained
Power absorbed by the load
Power factor is determined from P/S
A
rms
SCR
I 92
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1
2
2
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P
s
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pf](https://image.slidesharecdn.com/chapter6-230622144207-62485464/85/Chapter-6-AC-AC-Converters-pdf-24-320.jpg)
More Related Content
Similar to Chapter 6 AC-AC Converters.pdf
Chapter 6 AC-AC Converters.pdf
- 1.
- 2. Introduction • It usedto transfer the ac voltage (input) to other type of ac voltage (output) with different in amplitude, frequency, phase. • Can be categorized in two types: • Can be categorized in two types: single phase controllers three phase controller • to control the output waveform, the switching mode of the power electronic must be controllable
- 3. • The controlsignals are angle, pulse width, a voltage and current signals. • Two types of conversion direct conversion – desired output indirect conversion – depend on indirect conversion – depend on application Converters that change the frequency – cycloconverters Basically the applications are for steel mill roll, light control, speed control of the motors
- 4. • If athyristor is connected between ac supply and load, the power flow can be controlled by varying the rms value of ac voltage applied to the load. • Two types of control that normally used: On-Off Control Phase-angle control • In on-off control, thyristor switches connect the load to the ac supply for a few cycles and disconnect for another few cycles. • In phase-control, thyristor switches connect the load to the ac supply for a portion of each cycle of input voltage.
- 5. Principle of HalfWave AC Phase Control The power flow to the load is controlled by delaying the firing angle of T1 D1 is used to turn on when the input in negative half cycle Gate pulses are applied at T1 to turn on the device Gate pulses are generated at external source
- 6. The output voltageand input current are asymmetric and contain dc component This circuit is suitable for low power output such as heating and lighting. The rms output voltage is found form ( ) 2 / 1 2 / 1 2 ) ( 2 sin 2 2 sin 2 2 1 ∫ ∫ + = π α π π ω ω ω ω π t td m V t td m V o V ( ) 2 / 1 ) 2 2 sin 2 ( 1 2 2 / 1 2 ) 2 2 sin ( ) 2 2 sin ( 1 2 2 / 1 2 ) ( ) 2 cos 1 ( ) 2 cos 1 ( 4 2 + − = ∫ ∫ − + − = ∫ ∫ − + − = α α π π π α π π π π α π π ω ω ω ω π m V t t t t m V t d t t d t m V (5.1)
- 7. And the averagevalue of output voltage is ( ( ( ( ) ) ) ) ) 1 (cos 2 2 2 ) ( sin 2 sin 2 2 1 − − − − = = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ + + + + = = = = α α α α π π π π π π π π α α α α π π π π π π π π ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω ω π π π π s V t td s V t td s V dc V Gating sequence work. (5.2) Gating sequence work. 1. Output will occurs when a pulse received by T1 2. T1 on during positive half cycle and turn off when the input crossing zero point 3. D1 operate due to negative half cycle due to connection of diode from negative terminal to positive terminal
- 8. b) RMS loadcurrent The rms value of the output voltage: using (5.1) Example 5.1 A single phase ac voltage controller in Figure 5.1 has a resistive load of R = 10Ω and the input voltage is Vs = 120 , 60 Hz. The delay angle of thyristor T1 is α = π /2. determine (a) the rms output voltage Vo, (b) the input PF and (c) the average input current Solution : R = 10Ω, Vs = 120, α = π /2 and Vm = √2 x 120 = 169.7V V Vo 92 . 103 4 3 120 = = = = = = = = V 92 . 103 The load power Because the input current is the same as the load current, the input VA rating is The input PF V R V Io 392 . 10 10 92 . 103 0 = = = = = = = = = = = = W x R I Po 94 . 1079 10 392 . 10 2 2 0 = = = = = = = = = = = = VA x I V I V VA o s s s 04 . 1247 392 . 10 120 = = = = = = = = = = = = = = = = (lagging) 866 . 0 04 . 1247 94 . 1079 4 3 ) 2 2 sin 2 ( 2 1 2 / 1 = = = = = = = = = = = = + + + + − − − − = = = = = = = = = = = = α α α α α α α α π π π π π π π π s o o V V VA P PF
- 9. And the averageinput current The average output voltage: using (5.2) V x Vdc 27 2 2 120 − − − − = = = = − − − − = = = = π π π π A R V I dc D 7 . 2 10 27 − − − − = = = = − − − − = = = = = = = =
- 10. Single Phase Bidirectional/Fullwave Phase Control With Resistive load Figure: single phase full wave AC-C controller with resistive load 1. During positive half cycle the power flow is control by T1 with the a gate pulse (g1). 2. For negative half cycle the power flow is control by T2 with second pulse (g2)
- 11. The operations AC-ACR load: • The SCRs cannot conduct simultaneously • The output Vo is started to generate during pulse at G1 until the input crossing the zero point. • The load voltage is the same as the source voltage when either SCR is on and the load voltage is zero when both SCRs are off. • The switch voltage Vsw is zero when either SCR is on and is equal to the source voltage when either is on. source voltage when either is on. • The output is discontinuous for every cycle. • The average current in the source and load is zero if the SCRs are on for equal time intervals. • The average current in each SCR is not zero because of unidirectional SCR current. • The rms current is each SCR is 1/√2 times the rms current if the SCRs are on for equal time.
- 12. Output voltage, outputcurrent and switching voltage waveform
- 13. If is theinput voltage with delay angle of T1 and T2 is the rms output will be ( ( ( ( ) ) ) ) 1 2 α α α α π π π π α α α α + + + + = = = = t V V s s ω ω ω ω sin 2 = = = = ( ) 2 / 1 2 2 / 1 2 2 ) ( ) 2 cos 1 ( 2 sin 2 2 − = = ∫ ∫ ω ω ω ω π π π α m m rms t d t V t td V V (5.3) 2 / 1 ) 2 2 sin ( 2 1 ) ( ) 2 cos 1 ( 4 + − = − = ∫ α α π π ω ω π α m m V t d t (5.3)
- 14. 2 / 1 2 / 1 2 2 / 1 2 2 2 2 sin 1 2 ) ( ) 2 cos 1 ( 4 2 [ ] ) ( sin 2 2 1 [ ) ( + − = − = = ∫ ∫ α α π π ω ω π ω ω π π α π α s s s Rms V t d t V t td V sw V (5.4) The average thyristorvoltage: The rms thyristor voltage: ( ) [ ] ( ) α π ω π ω ω π π α π α cos 1 2 cos 2 sin 2 1 ) ( + = − = = ∫ m m m dc V t V t td V sw V The average thyristor voltage: (5.5) R sw V sw I dc dc ) ( ) ( = R sw V sw I rms rms ) ( ) ( =
- 15. : Solution : R= 10Ω, Vs = 120 (rms), 60 Hz, α = π/2 and Vm = √2 x 120 = 169.7V Example 5.2- (Tutorial no 5) A single phase full wave ac voltage controller in Figure 5.3a has a resistive load of R = 10Ω and the input voltage Vs = 120 (rms), 60 Hz. The delay angles of thyristors T1 and T2 are equal α1 = α2 = π/2. Determine a) the rms output voltage Vo, b) the input PF, c) the average current of thyristors Ia and d ) the rms current of thyristors Ir 120 The rms value of the load current is and the load power . Because the input current is same the load current, the input VA rating is From equation, the rms output voltage : using (5.3) V Vo 85 . 84 2 120 = = = = = = = = A R V Io 485 . 8 10 85 . 84 0 = = = = = = = = = = = = W x R I Po 95 . 719 10 485 . 8 2 2 0 = = = VA x I V I V VA o s s s 2 . 1018 485 . 8 120 = = = =
- 16. c) The averagethyristor current The PF is (lagging) 707 . 0 2 . 1018 95 . 719 2 1 = = = = = = s o o V V VA P PF R V t td V R I s s A ) 1 (cos 2 2 ) ( sin 2 2 1 + + + + = = = = = = = = ∫ ∫ ∫ ∫ α α α α π π π π ω ω ω ω ω ω ω ω π π π π π π π π α α α α d) The rms value of the thyristor current A R 7 . 2 10 2 120 2 2 = = = = = = = = π π π π π π π π A x R V t d t R V t td V R I s s s R 6 10 2 120 2 2 sin 1 2 ) ( ) 2 cos 1 ( 4 2 [ ] ) ( sin 2 2 1 [ 2 / 1 2 / 1 2 2 2 / 1 2 2 2 = = = = = = = = + + + + − − − − = = = = − − − − = = = = = = = = ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ α α α α α α α α π π π π π π π π ω ω ω ω ω ω ω ω π π π π ω ω ω ω ω ω ω ω π π π π π π π π α α α α π π π π α α α α
- 17. Single Phase Fullwave Phase Control with RL loads ( ) 2 / 1 2 sin 2 2 2 ∫ = β α ω ω π t td m V rms V ( ) 2 / 1 2 2 sin 2 2 sin ( 2 1 2 / 1 ) ( ) 2 cos 1 ( 4 2 2 sin 2 − + − = ∫ − = ∫ = β α α β π β α ω ω π α ω ω π m V t d t m V t td m V rms V
- 18. T1 is triggeredduring positive half cycle at : α T2 is triggered during negative half cycle at : π+α The current will falls until it reach ωt = β β is extinction angle The conduction angle of T1 is δ=β – α The control range of the firing angle: θ < α < π θ is the load angle The minimum value of firing angle α = θ (I-output Maximum) θ= tan-1(ωL/R)
- 19. Instantaneous Thyristor current () θ ω − = + t V Ri dt di L s sin 2 1 1 ( ) V 2 (5.6) The solution of (5.6) is of the form ( ) ( )t L R s e A t Z V i / 1 0 sin 2 − + − = θ ω The constant A1 at ωt = α, i1 = 0. ( )( ) ω α θ α / / 1 ) sin( 2 L R s e Z V A − − = (5.7) (5.8)
- 20. ( ) () − − − = − − ) ( / 1 ) sin( sin 2 t L R s e t Z V i ω α θ α θ ω Instantaneous Thyristor current The RMS Thyristor current The RMS Thyristor current ( ) ( ) 2 / 1 2 ) ( / 2 / 1 2 1 ) ( ) sin( sin 1 ) ( 2 1 − − − = = ∫ ∫ − − − − β α ω α β α ω θ α θ ω π ω π t d e t Z V I t d i I t L R s Thy rms Thy rms
- 21. The RMS Outputcurrent ( ) 2 / 1 2 2 R R out I I I + = The Average Thyristor current The Average Thyristor current ( ) ( ) − − − = = ∫ ∫ − − − − β α ω α β α ω θ α θ ω π ω π ) ( ) sin( sin 2 2 ) ( 2 1 ) ( / 1 t d e t Z V I t d i I t L R s Thy dc Thy dc
- 22. Example 5.3 For thesingle phase voltage controller of Figure 5.8, the source is 120V rms at 60Hz and the load is a series RL combination with R = 20Ω and L = 50mH. The delay angle is α = 900. Determine: a) an expression for load current for the first half period, b) the rms load current, b) the rms load current, c) the rms SCR current, d) the average SCR current, e) the power deliver to the load and f) the power factor.
- 23. Solution the current isexpressed from the parameters A m V rad R L rad R L L R Z 18 . 6 2 120 943 . 0 20 05 . 0 ) 377 ( 756 . 0 20 ) 05 . 0 )( 377 ( 1 tan 1 tan 5 . 27 2 )] 05 . 0 )( 377 [( 2 ) 20 ( 2 ) ( 2 = = = = = = = = = = = = = = = = = = = = = = = = − − − − = = = = − − − − = = = = = = = = + + + + = = = = + + + + = = = = ω ω ω ω ωτ ωτ ωτ ωτ ω ω ω ω θ θ θ θ ω ω ω ω The current is then expressed as The extinction angle β is determined from the numerical solution of i(β) = 0 in the preceding equation The conduction angle γ = β – α = 2.26 rad = 1300, which is less the the limit of 1800 A e Z m V rad A Z m V 8 . 23 / ) sin( 57 . 1 0 90 18 . 6 5 . 27 2 120 = = = = − − − − = = = = = = = = = = = = = = = = ωτ ωτ ωτ ωτ α α α α θ θ θ θ α α α α α α α α β β β β ω ω ω ω α α α α ωτ ωτ ωτ ωτ ω ω ω ω ω ω ω ω ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ − − − − − − − − − − − − = = = = t for 943 . 0 / 8 . 23 ) 756 . 0 sin( 18 . 6 ) ( 0 A e t t i 0 220 83 . 3 = = = = = = = = rad β β β β
- 24. the rms loadcurrent is determined by The rms current in each SCR is determine [[[[ ]]]] A t d t e t rms I 71 . 2 83 . 3 57 . 1 ) ( 2 943 . 0 / 8 . 23 ) 756 . 0 sin( 18 . 6 1 , 0 = = = = ∫ ∫ ∫ ∫ − − − − − − − − − − − − = = = = ω ω ω ω ω ω ω ω ω ω ω ω π π π π A rms o I rms SCR I 92 . 1 2 71 . 2 2 , , = = = = = = = = = = = = the average SCR current is obtained Power absorbed by the load Power factor is determined from P/S A rms SCR I 92 . 1 2 2 , = = = = = = = = = = = = [[[[ ]]]] A t d t e t avg SCR I 04 . 1 ) ( 83 . 3 57 . 1 943 . 0 / 8 . 23 ) 756 . 0 sin( 18 . 6 2 1 , = = = = ∫ ∫ ∫ ∫ − − − − − − − − − − − − = = = = ω ω ω ω ω ω ω ω ω ω ω ω π π π π W R rms o I P 147 ) 20 ( 2 ) 71 . 2 ( , 2 = = = = = = = = = = = = 45 . 0 ) 71 . 2 )( 120 ( 147 , , = = = = = = = = = = = = = = = = rms s I rms s V P s P pf
- 25. Three Phase VoltageController It is commonly used in ac motor drives Figure below shows a three phase voltage controller with Y connection The phase voltages are The line-line voltage are ) 3 4 sin( ) 3 2 sin( sin π π π π θ θ θ θ π π π π θ θ θ θ θ θ θ θ − − − − = = = = − − − − = = = = = = = = p V CN v p V BN v p V AN v ) 6 7 sin( 3 ) 2 sin( 3 ) 6 sin( 3 π π π π θ θ θ θ π π π π θ θ θ θ π π π π θ θ θ θ − − − − = = = = − − − − = = = = + + + + = = = = P V CA v P V BC v P V AB v
- 26. If the gatesignal is continuously applied the thyristors would behave as diodes There are two gate control per pulse The delay angle in all phases is referenced to 0 conduction angle Delay angle Current Communication T5 to T1 π π π π α α α α T6 to T2 T1 to T3 T2 to T4 T3 to T5 T4 to T6 3 5 3 4 3 2 3 π π π π α α α α π π π π α α α α π π π π α α α α π π π π α α α α π π π π α α α α + + + + + + + + + + + + + + + + + + + +
- 27. Thyristors T1, T3and T5 turn on in the positive half cycles of waveforms Van, Vbn, and Vcn respectively T1 conducts from α to π, T3 conducts from α + 2 π /3 to 5 π /3 and T5 conducts from α + 4 π /3 to 7 π /3 Thyristors T4, T6 and T2 turn on in the negative half cycles of the waveforms Van, Vbn, Vcn Vbn, Vcn T4 conducts from α + π to 2 π, T6 conducts from α + 5 π /3 to 8 π /3 (= 2 π + 2 π /3) and T2 conducts from α + 7 π /3 (=α +2 π + π /3) to 10 π /3 (=2 π +4 π /3) The delay angle α can be selected in the range 0 to π
- 28. The case α> π /3 first mode, the thyristor T6 was already on while T1 is in forward blocking mode due to the input voltage Van T1 is turned at α, referenced to the zero crossing point on the Van waveform T5 is reversed biased and the output current i0 flows from phase A through T1, load and T6 to phase B and T6 to phase B At angle α + π /3 in mode 2, thyristor T2 is in mode forward blocking mode due to the input phase voltage Vcn T2 is turned on at α + π /3, α angle referenced to the zero crossing point on the positive half cycle of the Vcn waveform (figure below)
- 29. T6 is reversedbiased and the output current i0 flows from phase A through T1, load and T2 to phase C Likewise from the angle α + 2 π /3 to α +4 π /3 in mode 3 and mode 4, the current i0 flows from the input phase B to other phases The phase output voltage Van consists of two parts in the positive half cycle
- 30. π π π π θ θ θ θ α α α α π π π π α α α α π π π π θ θ θ θ α α α α /3 : 2 3 : 2 〈 〈 〈 〈 ≤ ≤ ≤ ≤ + + + + = = = = + + + + ≤ ≤ ≤ ≤ ≤ ≤ ≤ ≤ = = = = AC V AB V an V The negative halfcycle of Van is the new symmetric to the positive half cycle. Output rms phase voltage Vo is calculated as the rms of the Van waveforms over half period from α to α + π
- 31. Analysis techniques usingSCR with three phase voltage controller By using SCR the six SCRs are turned on in the sequence 1-2-3-4-5-6 at 600 intervals At any instant, three SCRs, two SCRs and no SCRs are on The instantaneous load voltages are either a line to neutral voltage (three on), one half of a line to line voltage (two on) or zero (none on) When three SCRs are on (one in each phase), all three phase voltages are connected to the course corresponding to a balanced three-phase source the course corresponding to a balanced three-phase source -S1, S2 and S6 are on, van = VAN, vbn = VBN, vcn = VCN two SCRs are on, the line to line voltage of those phases is equally divided between the two load resistors which are connected
- 33. For 0 <α < 600 At ωt = 0, S5 and S6 are conducting and three is no current in Ra making van= 0 At ωt = π/6 (300), S1 receives a gate signal and begins to conduct, S5 and S6 remain on and van = VAN The current in S5 reaches zero at 600 turning S5 off while S1 and S6 remaining on, van = VAB/2 van = VAB/2 At 900 S2 is turned on, the three SCRs S1, S2 and S6 are then on and van = VAN At 1200, S6 turns off, leaving S1 and S2 on so van = VAC/2 For internals to exist when three SCRs are on, the delay angle must be less than 600
- 35. •For 600 <α < 900 Only two SCRs conduct at any one time when the delay angle is between 600 and 900 Just prior to 750, S5 and S6 are conducting, and van = 0 S1 is turned on at 750, S6 continues to conduct but S5 must turn off because VCN is negative When S2 is turned on at 1350, S6 is forced off, and van = VAC/2 When S2 is turned on at 1350, S6 is forced off, and van = VAC/2 The next SCRs to turn on in S3 which forces S1 off and Van = 0 Load voltage is one half line to line voltages or zero •For 900 < α <150 •Only two SCRs can conduct at any one time in this mode •In the intervals just prior to 1200 no SCRs are on, and van = 0. At α = 1200, S1 is given by a gate signal, and S6 has a gate signal applied •Since VAB is positive both S1 and S6 are forward biased and begin to conduct and van = VAB/2
- 36. Both S1 andS6 become turn off when VAB becomes negative When a gate signal is applied to S2, it turns on and S1 turns on again •For α > 1500 There is no time interval when an SCR is forward biased while a gate signal is applied. Output voltage is zero at this condition
- 37. AC Voltage Controlellerswith PWM Control A ac voltage controllers rectifiers can be improved by the pulse width modulation (PWM) type of control The performance of ac voltage controllers can be improved by PWM control Switches S1 and S2 are turned on and off several times during the positive and negative half cycles of the input voltage, respectively S1’ and S2’ provide freewheeling path for the load current whereas S1 and respectively S1’ and S2’ provide freewheeling path for the load current whereas S1 and S2 respectively in the off state
- 38. For a resistiveload current resembles the output voltage With a RL load, the load current rises in the positive or negative direction when switch S1 or S2 is turned on respectively