Lecture Note. Uncontrolled and Controlled Rectifiers

Transcription

1 Lecture Note 7 Uncontrolled and Controlled Rectifiers Prepared by Dr. Oday A Ahmed Website: @uotechnology.edu.iq Scan QR

2 single-phase diode and SCR rectifiers The diode rectifiers are referred to as uncontrolled rectifiers, which make use of power semiconductor diodes to carry the load current. The diode rectifiers give a fixed dc output voltage (fixed average output voltage) and each diode rectifying element conducts for one half cycle duration (T/2 seconds), that is the diode conduction angle = or π radians. We cannot control (we cannot vary) the dc output voltage or the average dc load current in a diode rectifier circuit. Controlled SCR rectifiers are line-commutated ac to dc power converters that are used to convert a fixed voltage, fixed frequency ac power supply into variable dc output voltage. Important NOTE: The thyristor current and the load current begin to flow once the thyristors are triggered (turned ON) at angle called firing angle α. The main difference between the diode rectifier and SCR rectifier is the required trigger for firing the SCR. Hence, for ωt= α=0 0 the behaviour of SCR rectifier is look like the diode rectifier. Thus, the controlled SCR rectifier will only be explained here. The thyristor remains reverse biased during the negative half cycle of input supply. The type of commutation used in controlled rectifier circuits is referred to AC line commutation or Natural commutation or AC phase commutation. SCR Rectifier can be controlled by α from 0 0 (operate like a diode rectifier) to (provided zero DC voltage for R-load). Where the SCR conduction angle δ =π α Single quadrant ac-dc converters where the output voltage is only positive and cannot be made negative for a given polarity of output current. Single quadrant converters can also be designed to provide only negative dc output voltage. two quadrant converters so that the output voltage can be made either positive or negative for a given polarity of output load current can be achieved by using fully controlled bridge converter circuit. 1

3 Note: the operation of the rectifier has different effects on the power quality when it is work. Also its operation is effected by the loads type. These effects can be summarized below: A. The effect of RL load with low inductance on the output DC voltage and on the input current. B. The effect of RL load with heavy inductive on the input current power factor and current waveform shape. In addition, the effect on the average power that transferred from the source to load, as stated in Lec.7. C. The effect of output C filter on the input current shape and Peak value and RMS current. D. The effect of reverse recovery current on the output DC voltage of the rectifier. E. The effect of leakage inductance of the input transformer on the output voltage of the rectifier. F. The effect of RLE load on the output voltage and operation of the converter. I. Single Phase Half Wave Phase Controlled Rectifier 2

4 A. R-Load State T1 vo vt1 it1 io is io(max) For ωt = 0 α OFF 0 V For ωt = α π ON vs 0 v o R L v o R L v o R L V m R L For ωt = π 2π OFF 0 Vm o V 1 = V m sin α (hence this voltage depend on α, if α=0 V 1 = 0V) o v T1 = (0.7+V f ) for real SCR for ωt = α π o peak reverse voltage PIV = V m o i T1 = reverese leakge current when SCR OFF for real SCR o v p represents the primary supply voltage = V m sinωt. o v S represents the secondary supply voltage = V m sinωt for turns ratio n=1. The maximum value (peak value) of that flow via the R L, T 1, and secondary winding is calculated as: SCR naturally turns off when the current flowing through it falls to zero at ωt =π. 3

5 Since conduction angle δ =π α, hence maximum conduction angle when α=0 0 RMS value of input ac supply voltage across transformer secondary. Drive an Expression for output DC Voltage: For single-phase diode HWR, α=0 0 The control characteristic of SCR HWR is shown aside: DC Load current is equal to: Drive an Expression for output RMS Voltage: RMS Load current is equal to: 4

6 For diode Rectifier (α=0 0 ) Input power factor pf Since the shape input current is still sinusoidal, DF=1 and DPF=pf (Can you guess the value of fundamental components of input current, FF, RF, and CF?). B. RL-Load ON OFF T 1 ON T 1 Off For ωt = α β ωt = β 2π In practice, most of the loads are of RL type. Note: Load current increases slowly due the inductance since the inductance in the load forced the load current to lag load voltage. The load current flowing through T 1 would not fall to zero at ωt=π, when v s starts to become negative (due to magnetic field in the inductor). T 1 will continue to conduct i o until all the inductive energy stored in L is completely utilized and current via T 1 falls to zero at ωt=β. β is referred to as the Extinction angle or advance angle. β 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. conduction angle of SCR δ = β α δ depends on the firing angle and the load impedance angle ϕ 5

7 At ωt =β, i o = 0, V L =0 and V R =0 and v s appears as reverse bias across SCR. then SCR turned off peak reverse voltage PIV = V m negative in output voltage reduces V o(dc) when compared to a purely R-Load. Derive an expression for inductive load current, during α <ωt< β During ωt = α β, the load current is determined as following: i o =i s +i t where, i s steady state load current, i t : transient load current Thus, i s i t The current via L is produce from sinusoidal source but the current via it is rising slowly exponentially. Hence, the original of current is sinusoidal entered with it exponential current component. A 1 is the initial transient current which is changing with α which found as explained in Lecture09. Based on the above equation the steady-state component of load current in related to source voltage and transient components at different firing angles are plotted below: Source voltage V m sinωt 6

8 I to measure from the beginning of SCR triggered not at zerocrossing voltage. (i.e. at ωt = α φ Φ=90 0 If α<φ : δ>π, as for α 1 and α 2 If α=φ : δ=π, as for α 3 If α>φ : δ<π, as for α 4 If α= π : δ=0 Steady state load current Voltage Max I at di/dt=0, which occurs at v s =v R V m v L Voltage increasing, L stored energy v R Voltage decreasing, L realising energy π/2 λ π β The Figure shows the load impedance phase angle related to i o at α=0: current I max I to i t i o i s φ λ β i t(β) i o(β) =0 i s(β) I so 7

9 The load current is the same one for the current via T 1 and source (with no FWD) is: TO CALCULATE EXTINCTION ANGLE β As shown in Lecture09: Iterative solutions to find β The exact value for β exist only for the purely resistive load, Φ= 0, and the purely inductive load, Φ = ½π. 8

10 TO CALCULATE MAX LOAD CURRENT AT ωt=λ To find the maximum value of the load current which should be occurred at ωt=λ, di dωt = 0 = d dωt (V m [sin(ωt φ) sin (α φ)e R ωl (ωt α) ]) Z sin(a-b)=sin A cos B - cos A sin B, thus, 0 = V m Z [cos(ωt φ) + sin(α φ) L R L R e R ωl (ωt α) ], at ωt=λ cos(λ φ) = sin(α φ) e R ωl (λ α) Iteration needed for solving λ Hence, by substituting ωt=λ, to i o, the maximum current via L can be obtained. I max = V m Z [sin(λ φ) sin (α φ)e R ωl (λ α) ] TO DERIVE AN EXPRESSION FOR AVERAGE (DC) LOAD VOLTAGE Average DC Load Current The RMS load voltage is (if ωt=θ) The RMS load current: I o(rms) = 1 β 2 2π (V m [sin(ωt φ) sin (α φ)e R ωl (ωt α) ]) dωt Z α 9

11 Input power factor pf can be obtained as: C. RL-Load with Freewheeling Diode (FWD) With a RL load it was observed that the average output voltage reduces. This disadvantage can be overcome by connecting a diode across the load as shown in figure. The modes operation of the rectifier is shown below: State For ωt = 0 α For ωt = α π For ωt = π β For ωt = β 2π SCR FWD v o v T1 i T1 I FWD OFF OFF 0 V ON OFF v S 0 io(t) 0 OFF ON 0 V 2 0 io(t) OFF OFF 0 V m 0 0 H.W: What are the reasons results in SCR and FWD turned off during the periods ωt = 0 α and ωt = β 2π At ωt=π, v S falls to zero and as v S becomes negative, FWD is forward biased and SCR is turned OFF. V 2 Stored energy in L maintains the load current flow through R, L, and the FWD. During the period π to ωt=β, the load current flows through FWD (freewheeling load current) and decreases exponentially towards zero at ωt=β. The load is shorted by the conducting FWD and the load voltage is almost zero, if the FWD forward voltage drop is neglected. The mean load voltage (hence mean output current) for all conduction cases, with a passive L-R load, is 10

12 RMS output voltage for both continuous and discontinuous load current is: To find load current, the same equation used in Part B can be used. However, current via SCR is different where this current occurs between α<ωt π NOTE: The following points are to be noted. If L value is not very large, the energy stored in its able to maintain the load current only up to ωt=β, where π <β< 2π, where before the next gate pulse, the load current tends to become discontinuous. During the conduction period (δ) α to π, i o is carried by the SCR and during the freewheeling period π to β, i o is carried by the freewheeling diode. The value of β depends on R and L and the forward resistance of the FWD. If the value of L is very large, i o does not decrease to zero during the freewheeling time interval and the ripple in i o waveform decreases. 11

13 D. RLE-Load The load circuit consists of a dc source E in addition to resistance and inductance. SCR will be forward biased for anode supply voltage greater than the load dc voltage (i.e. v S >E ). If v S <E, SCR will be revered biased. The value of ωt at which the supply voltage increases and becomes equal to the load circuit dc voltage can be calculated as: o For trigger angle α < γ, the SCR conducts only from ωt= γ to β. o For trigger angle α > γ, the thyristor conducts from ωt= α to β. The load current can be found as: 12

14 The average or dc load voltage Note that: V o(dc) = E+I o R RMS Output Voltage can be calculated by using the expression For L=0H, β= π- γ, the average load current can be given as: I o = 1 2πR {V m(cosα + cos γ) E(π γ α)} The RMS load current for L=0H can be given as: I or = 1 β 2 2π (V msinωt E ) dωt R α I or = 1 2πR {(V s 2 + E 2 )(β α) V 2 s 2 (sin2β 2sin2α) 2V me(cosα cosβ)} Power deliver to the load is equal to: And the input power factor is: P dc = I 2 or R + I o E 1/2 13

15 pf = I or 2 R + I o E V s I or To find Ior, Io for half-wave diode rectifier, α = γ The PIV= Vm+E II. SINGLE PHASE FULL WAVE CONTROLLED RECTIFIERS Single-phase full wave rectifier can be divided into the following configurations: Centre tapped rectifier. Semi-Converter rectifier. Full-bridge rectifier. 1 Full-wave half-uncontrolled converter 2 Full-wave mid-point controlled converter 3 Full-wave uncontrolled converter 4 Full-wave Full -controlled converter 5 Bridge Rectifier split rail dc supplies 6 Bridge Rectifier voltage doubler 14

16 7 8 9 Full-wave half-controlled semi-converters Single phase half wave controlled rectifiers are rarely used in practice as they give low dc output and low dc output power. They are only of theoretical interest. Therefore, various full-wave rectifier configurations are studied here. NOTE: All previous configurations give a full-wave two-pulse output voltage. A) Configurations 1 and 3 Both circuits appear identical as far as the load and supply are concerned. Two fewer diodes can be employed in Config. 1, but this circuit requires a centre-tapped secondary transformer where each secondary has only a 50% copper utilisation factor. Each of the secondary windings in 1 must have the same RMS voltage rating as the single secondary winding of the transformer in 3. Rectifying diodes in 1 experience twice the reverse voltage, (2 2V s ), as that experienced by each of the four diodes in the circuit of 3, ( 2 V s ). The operation modes of two configurations above are shown below for continuous load current operation: 15

17 V s The average and RMS output voltage can be derived as shown below: π V o = 1 π V m 0 V or = 1 π π V m 2 0 V m for fullbridge, 2V m for centre-tap sinωtd(ωt) = 2V m π sinωt 2 d(ωt) = V m 2 = V s Note: V m is the maximum of secondary winding voltage With an inductive passive load, (no back EMF) continuous load current flows, which is given by: Appropriate integration of the load current squared, gives the RMS current: 16

18 This current is the same one for all full-wave uncontrolled rectifier configurations. H.W.: Derive an expression to find the average and RMS value of diode currents. Hint: the current via each diode whether centre-tap or full-bridge configuration is for the following period 0 ωt π or π ωt 2π. B) Configurations 2 and 4 The main difference between the configurations 1 and 3 and the rectifiers shown in 2 and 4 is the possibility of control the output voltage using SCR. The operation modes of 2 and 4 are similar to 1 and 3 except of when the voltage appear across the load is depending on firing angle α: There are two types of operations possible. ƒ Discontinuous load current operation, which occurs for a purely resistive load or an RL load with low inductance value. ƒ Continuous load current operation which occurs for an RL type of load with large load inductance. Discontinuous Load Current Operation (for low value of load inductance) The main waveforms with equivalent state are shown below 17

19 Same output current equation with RL load can be obtained as shown for half-wave rectifier except her the equation should be multiplied by 2. The average and RMS output voltage can be derived as shown below: The output voltage waveforms and DC output voltage with FWD are shown below: H.W: Derive an expression for RMS output voltage for full-bridge controlled rectifier with low value of load inductance. 18

20 Continuous Load Current Operation (for high value of load inductance) The main waveforms are shown below Generally the load current is continuous for large load inductance and for low trigger angles. The load current is discontinuous for low values of load inductance and for large values of trigger angles. The DC output voltage can be obtained as: The above equation can be plotted to obtain the control characteristic of a single phase full wave controlled rectifier with RL load assuming continuous load current operation. For trigger angle α in the range of 0 to 90 0 V dc is positive and the circuit operates as a controlled rectifier. For α> 90 0, V dc becomes negative but i o flows in the same positive direction. Hence the output power becomes negative. This means that the power flows from the load circuit to the input ac source. This is referred to as line commutated inverter operation. During the inverter mode load energy can be fed back from the load circuit to the input ac source. The RMS thyristor current can be calculated as The average thyristor current can be calculated as 19

21 C) Configurations 7,8, and9 These rectifier topologies are semi-converters since they have diode devices in addition to the SCRs. The key waveforms of these configurations are shown below; the circuits 7 to 9 don t require additional FWD since they inherently can remove the effect of load inductance: The DC output voltage for circuits 7,8 can be obtained as shown in circuits 2,4 with FWD which is equal to: The DC output voltage for circuit 9 is different and can be obtained as shown below: For α=0 0, this circuit operate like full-bridge uncontrolled rectifier. III. Single-phase full-wave bridge rectifier circuit with a C-filter and R-load The capacitor reduces the ripple voltage, so large voltage-polarised capacitance is used to produce an almost constant dc output voltage. 20

22 V s rises Supply provides load and simultaneously charges the capacitor charging current period capacitor voltage > V s At ωt= β, D1-D2/D3-D4 OFF capacitor supplies the load current voltage decreases with an R-C time constant until ωt = π +α The average voltage α: The start of diode conduction, β: The diode current extinction angle Θ c :diode conduction period= β-α The capacitor charging current period θ c around the ac supply extremes is short, giving a high peak to rms ratio of diode and supply current. 21

23 The average voltage Diodes conducting α ωt β v o (ωt) Diodes non-conducting β ωt π+α v o (ωt) V m sinωt See Next Page 22

24 The output voltage for (β ωt π+α) can be found as: V s = IR + 1 id(ωt) + V c (ωt = β) 0 = IR + 1 id(ωt) + V β V β = V m sinβ Taking LT: 0 = I(s)R + I(s) s + V β s By simplifying the above equation: During all diodes are turned off C supply the required current. V s = 0 V β = 1 R Taking LTI: i t = V β e (ωt β) ωr R i D (ωt)= i c (ωt)+ i R (ωt) at ωt= β, i o =0 i R(ωt) = V o(ωt) = V β R R = V msinβ R 0= V mcosβ + V m s β R The average voltage V o ωt = V β e (ωt β) ωr i (ωt) = ω dv o(ωt) dωt = V mcosβ V o ωt = V β e β ωt π+α (ωt β) ta β for (β ωt π+α) e e V β = V m sinβ t nβ = ωr V o = 1 2π α β sinωt dωt + V β e β π α (ωt β) ta β dωt V o = V m 1 cos c cosβ I o = V o R c= β- α 23

25 The maximum output voltage occurs at The Output voltage Ripple V m The minimum output voltage occurs at V m sinα output peak-to-peak ripple voltage Δv o = V m V m sinα = V m (1-sinα) At ωt = π+α Sine voltage start and exp voltage end both at the same value. The Output voltage Ripple v o (ωt)=v o (π+α)=v m sin(π+α) (ωt β) (π+α β) V o ωt = V β e ta β V o π+α = V β e ta β (π+α ) V m sin(π+α) = V β e The two equation sides gives maximum value when α and =0.5π (π) V o π+α =V m V V o π+α = V m eta β β =V m sin β= V m sin 0.5π=V m (π) Δv o =V m -V m e (π) eta β 1 π ωr Δv o = V m R 24

26 The average capacitor current is zero i D (ωt) = i c (ωt)+ i R (ωt), I o = I D, the average diode current is the same as the average load current because Since the diode is on for a short time in each cycle, the peak diode current is generally much larger than the average diode current i DPe = i cpe + i RPe, i ωt = ω dv o ωt dωt i R(ωt) = V msinωt R i Pe i cpe = V msinα R i DPe = V m cosα + s α R = V mcosα i RPe i cpe Peak current occurs when the diode turns on The source current equal to diode current Example A single-phase, full-wave, diode rectifier is supplied from a 230V ac, 50Hz voltage source and uses a capacitor output filter, 1000µF, with a resistor 100Ω load. Ignoring diode voltage drops and assume the diode start conducting at 66.5 degree, determine: Expressions for the output voltage output voltage ripple v o and the % error Expressions for the capacitor current Diode peak current The average load voltage and current Assuming the output ripple voltage is triangular, estimate: The average output voltage and rms output ripple voltage Capacitance C for v o = 2% of the maximum output voltage 25

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28 Capacitance C for v o = 2% of the maximum output voltage Δv o = V m R IV. Rectifier performance with input stray inductance It s shown that for single-phase bridge controlled rectifier when incoming SCRs T l and T 2 are fired outgoing SCRs T 3 and T 4 get turned off due to the application of reverse voltage and the current shifts to SCRs T l and T 2 instantaneously. This is possible only if the voltage source has no internal impedance. If the source impedance is R s I o R s voltage drop occur V o(dc) reduced by I o R s This drop voltage must be considered. However, the source impedance could be consist of internal resistance R s and stray inductance L s. Hence, source inductance the outgoing and incoming SCRs to conduct together results in zero voltage across the load. This period of commutation called overlap period. the effect of source inductance is :lower the mean output voltage, and distort the output voltage and current waveforms. 27

29 The effect of source inductance on the operation of single-phase full-bridge controlled rectifier will be explained below. Same analysis can be done for uncontrolled rectifier simply by applying α=0 0. When T 1, T 2 are triggered at a firing angle α, the commutation of already conducting T 3, T 4 begins. Because of the presence of source inductance L s " the current via outgoing devices T 3, T 4 decreases gradually to zero from its initial value of i o ; whereas in incoming SCRs T 1, T 2 ; the current builds up gradually from zero to full value of i o, Hence, the full-bridge equivalent circuit during this period is look like the following circuit: During overlap period μ: The current flow via L s, T 1, Load, and T 2 The current flow via T 3, load, T 4, and L s At beginning of μ, L s stored energy by i 2, while then realise energy by i 1. By applying KVL for abcda loop: 28

30 It can be re-expressed Dc output voltage to make a relation between V o and load current and value of source inductance: V o without overlap is equal to: The maximum DC voltage occur at α=0 0. Thus, Hence, output voltage with overlap period is can be found as: From load current equation it can be obtained that: Substitute the above equation in V o, then from this equation μ can be obtained 29

31 It can be seen from equation above that the voltage drop due to L s, is proportional to I o and L s. Note: for μ<π, V o is reduces due to L s, for μ=π, V o =0V since all SCRs are conducting. Note: if α=0 0, then V o can be controlled between μ α π Note: The maximum value of firing angle can be ( μ). 30

32 Exercise: A fully controlled bridge converter is connected to a 400V, 50Hz supply having a source reactance of 0.3Ω. The converter is operating at a firing angle of quarter of the one cycle. Determine the average load voltage and the overlap angle when the converter is supplying a steady-state current of 60A. 31