Single-Phase Half-Wave Rectifiers

Transcription

1 ectifiers Single-Phase Half-Wave ectifiers A rectifier is a circuit that converts an ac signal into a unidirectional signal. A single-phase half-way rectifier is the simplest type. Although it is not widely used in industry, it helps undertand the principle of rectifier operation. esistive Load A circuit digram with a resistive load is shown in Figure (1). Figure (1): A Single-Phase Half-Wave ectifier During the positive half cycle of V S diode D 1 conducts V L V S. During the negative half cycle of V S diode D 1 is in a blocking condition V L. Performance Parameters The performance of a rectifier is normally evaluated in terms of the following parameters The average value of the output (load) voltage, V dc. The average value of the output (load) current, I dc. The output (load) dc power 1

2 ectifiers P dc V dc I dc. The rms value of the output (load) voltage, V rms. The rms value of the output (load) current, I rms. The output (load) ac power P ac V rms I rms The efficiency (or rectification ratio) of a rectifier is defined as ξ P dc P ac The output voltage can be viewed as being composed of two components (1) the dc value; and () the ac component or ripple. The rms value of the ac component of the output voltage is V ac V rms V dc The form factor, which is a measure of the shape of the output voltage, is defined as FF V rms V dc The ripple factor, which a measure of the ripple content, is defined as F V ac V dc It therefore follows that F V rms V dc 1 FF 1 The transformer utilization factor is defined as TUF P dc V s I s where V s and I s are the rms voltage and rms current of the transformer secondary, respectively.

3 ectifiers Consider now waveforms shown in Figure (). Figure (): Waveorms of Input Voltage and Current where v s is the sinusoidal voltage i s is the instantaneous input current i s1 is the fundamental component of i s. Let Φ be the displacement angle between the fundamental components of the current and voltage. Then the displacement factor is defined as DF cos Φ The harmonic factor of the input current is defined as HF I s I s1 I s1 I s I s1 1 where I s1 is the fundamental component of the input current I s. Both of these are expressed in terms of rms. The input power factor is defined as PF V s I s1 V s I s cos Φ I s1 I s cos Φ Crest Factor CF, which is a measure of the peak input current I s(peak) as compared to its rms value I s, is defined as Notes CF I s(peak ) I s 3

4 ectifiers 1. The harmonic factor HF is a measure of the distortion of a waveform and is also known as total harmonic distortion (THD).. If the input current is purely sinusoidal, then I s I s1 and PFDF. Also the displacement ωl angle Φ becomes the impedance angle Φtan 1 for an L load. 3. Displacement factor DF is often known as displacement power factor (DPF). 4. An ideal rectifier should have 1% efficiency, V ac, F, TUF1, HFTHD, and PFDPF1. Example The ectifier in Figure (1) has a purely resistive load of. Determine: (a) the efficiency, (b) the form factor, (c) the ripple factor, (d) the transformer utilisation factor, and (e) the crest factor of the input current. Solution The average output voltage V dc is defined as V dc 1 T T v L (t)dt From Figure (1), it is clear that v L for T / t T Hence V dc 1 T T/ v m sin( ωt)dt V m ωt cos ωt 1 Since f1/t and ω πf, we obtain V dc V m π.318 V m I dc V dc.318 V m Note that the presence of this current means that the transformer will have to carry a dc current which results in dc saturation of the transformer core. The rms value of a periodic waveform is defined as 4

5 ectifiers V rms 1 T T v (t)dt L Since v L V m sin ωt for t T/, the rms value of the output voltage is V rms 1 T T/ ( V m sin ωt) dt.5v m Thus I rms V rms.5v rms ( P dc V dc I dc.318v m) ( P ac V rms I rms.5v m) (a) The efficiency ξ P dc P ac ( ) 4.5%, ( ).318V m.5v m This is considered as low efficiency. (b) The form factor FF V rms V dc (c) The ripple factor (.5V m ).318V m ( ) 157%, F FF %, This is considered high for practical application. (d) The rms voltage of the transformer secondary V s 1 T T ( V m sin ωt) dt V m.77v m, The rms value of the transformer secondary current is the same as that of the load 5

6 ectifiers I s.5v m, The volt-ampere rating of the transformer is VAV s I s V m V m V m, The transformer utilization factor is TUF P dc V s I s (.318V m ) V m.86, This is a poor utilization factor. The reciprocal of the TUF is V s I s P dc 1 1 TUF This means that the transformer must be times larger than when it is being used from a pure ac voltage. (e) I s(peak ) V m and I V m. Thus the crest factor of the input current is s CF I s(peak ) I s. Half-Wave ectifier: esistive and Inductive Load Consider the circuit shown in Figure (3a). Figure (3a): ectifier with an L load The voltage and current wave forms are shown in Figures (3b,c,d). 6

7 ectifiers Figure (3b,c,d): Voltage and current waveforms At t<, i and v s is -ve. At t, diode becomes forward biased and i begins to flow. The diode can the be replaced by a short as shown in Figure (3,e). Figure (3e): Equivalent circuit for the rectifier of fig (3a) The current is governed by v s i +L di dt The resulting currents and voltages are shown in Figures (3b,c). For <t t 1 v s >v, and v L v s v is +ve. i builds up. inductor stored energy increases. 7

8 ectifiers For t 1 <t t : v L becomes -ve. i begins to decrease. For t <t t 3 : v s becomes -ve. v d becomes -ve. i is still +ve. diode still conducts because of the inductor stored energy. To find the instant when i and the diode stops conducting we use the following analysis or v L L di dt 1 L v Ldt di Integrating both sides between t and t t 3, and using the fact that i()i( t ), gives 3 t 1 3 i(t L v dt 3 ) di L i() i(t 3 ) i() This means that or or t 3 v L dt t 1 t 3 v L dt + v L dt t1 Area A - Area B This means that i when For t >t 3 : Area A Area B. v L v. reverse polarity voltage appears across the diode. The waveforms repeat themselves strating from tt1/f. Single-Phase Full-Wave ectifier 8

9 ectifiers Consider the full-wave rectifier shown in Figure (4a). The transformer is centre-tapped. Each half of the transformer and its associated diode acts as half-wave rectifier. The output of a fullwave rectifier is shown in Figure (4b). Note that since there is no dc current flowing through the transformer, there is no dc saturation problem of the trabsformer core. Figure (4): A full-wave rectifier circuit with centre-tapped transformer The average output voltage is V dc T T/ V m sinωt dt V m π.6366v m An alternative to using a centre tapped transformer, it is possible to use four diodes as shown in Figure (5a). During the positive half-cycle of the input voltage, the power is supplied to the load through diodes D 1 and D. During the negative half-cycle, the power is supplied to the load through diodes D 3 and D 4. The waveform of the output voltage is shown in Figure (5b), which is similar to that of Figure (4b). 9

10 ectifiers Figure (5): A full-wave rectifier circuit with four diodes The peak-inverse voltage of a diode is only V m. The circuit of Figure (5a) is known as a bridge rectifier and is commonly used in industrial applications. Example If the rectifier in Figure (4a) has a purely resistive load of, determine: (a) the efficiency (b) the form factor (c) the ripple factor (d) the transformer utilization factor (e) the peak-inverse voltage of diode D 1. (f) the CF of the input current. Solution The average output voltage is V dc T T/ V m sinωt dt V m π.6366v m The average load current is I dc V dc.6366v m The rms value of the output voltage is V rms T T/ ( V m sin ωt) dt V m.77v m 1

11 ectifiers The rms value of the current is I rms V rms.77v m The output dc power is ( P dc V dc I dc.6366v m) The output ac power is ( P ac V rms I rms.77v m) (a) The efficiency is ξ P dc.6366 P ac.77 (b) The form factor is 81% FF V rms V dc (c) The ripple factor is F FF % (d) The rms voltage of the transformer secondary is V s.77v m The rms value of the transformer secondary current is I s.5v m The VA rating of the transformer is VAV s I s (.77V m )(.5V m ).77V m The transformer utilization factor is therefore 11

12 ectifiers ( ) / TUF P dc.6366v m V s I s.77v m / % (e) The peak reverse blocking voltage is PIV V m (f) The peak value of the transformer secondary current is CF I s,peak I s,rms V m /.77V m / Single-Phase Full-Wave ectifier with L Load With a resistive load, the load current is identical in shape to the output voltage. In practice, most loads are inductive to a certain extent and the load current depends on the values of the load resistance and the load inductance L. This is illustrated in Figure (6a). A battery of voltage E is added to develop generalised equations. If the input voltage is given as Figure (6): A full-wave rectifier circuit with L load v s V m sin ωt V s sin ωt then the load current i L can be obtained from L di L dt +i L +E V s sin ωt The solution to this equation is 1

13 ectifiers i L V s sin( ωt θ)+ A 1 e L t E Z where Z + ( ωl) θtan 1 ωl Case 1: Continuous load current The constant A 1 can be determined from the condition that at ωt π, i L I 1. Thus Thus A 1 I 1 + E V s sinθ e Z i L V s sin( ωt θ)+ I 1 + E Z V s sinθ e Z L π ω π L ω t Using the initial condition i L (ωt )I 1 gives I 1 V s sinθ 1+e Z 1 e π L ω π L ω E Using this in i L gives the following relationship i L V s sin( ωt θ)+ Z 1 e π L ω sinθe L t E where ω πand i L. The diode rms current can now be found as I rms,diode 1 π π i d(ωt ) L The rms output current can then be determined by combining the rms current of each diode as 13

14 ectifiers I rms I ems,diode +I ems,diode I rms,diode The average diode current can also be found as I a,diode 1 π i d(ωt ) L π Case : discontinuous load current The load current flows only during the period α ωt β. The diods start to conduct at ωt α where α is given by αsin 1 E V m At ωt α, i L (ωt ). Using this in the solution for i L gives Thus A 1 E V s sin( α θ) e Z i L V s sin( ωt θ)+ E V s sin( α θ) e Z Z L α ω α L ω t Using the condition i L (ωt β) gives V s Z sin(β θ)+ E V s sin( α θ) Z e L α β ω The value of β can be determined by an iterative method, starting from β and increasing its value by a small amount until the left-hand becomes zero. The rms diode current can be found as I rms,diode 1 π β α i L d(ωt ) The average diode current can also be found as β I a,diode 1 π i d(ωt ) L α Example A single-phase full-wave rectifier is shown in figure (6a). It has L6.5mH,.5 Ω, and E 1 V. The input current is V s 1 V at 6 Hz. Determine: 14

15 ectifiers (a) the steady-state load current I 1 at ωt. (b) the average diode current I a,diode. (c) the rms diode current I rms,diode. (d) the rms output current I output,rms. Solution Assume that the load current is continuous. ωl πfl π(6)(.65 ).45 Ω Z Ω.45 θtan o (a) The steady-state load current at ωt is I A Since this is not zero, the load current is continuous. Thus the assumption is correct. (b) Using a numerical integration method gives I a,diode A (c) Using a numerical integration method gives I rms,diode 8.5 A (d) The rs output current is I rms I rms,diode x A 15

16 Single-Phase Half-Wave ectifiers with L Load Figure (7) The Kirchhoff s voltage law when the diode is in the forward-bias mode is V m sin( ωt ) i ( t ) + L di ( t ) dt The solution to this equation is obtained as where i( t ) V m Z sin( ωt θ ) + Ae t / τ τ L time constan t The constant parameter A is evaluated from the initial condition. As the initial current is zero before the diode started conducting, we have or i( ) V m Z sin( θ ) + Ae A V m Z sin θ Using this in the current equation gives i( t ) V m Z sin( ωt θ ) + sin( θ )e t τ

17 or i( t ) V m Z sin( ωt θ ) + sin( θ )e ωt ωτ This relationship is valid only when i(t) is positive. If the current is negative then its actual value is zero. When the source voltage is positive, the diode conducts and the voltage waveform of each element is shown in Figure (7b). Figure (7b) Note that the diode remains forward biased for β > π. the source voltage is negative for the last part of the conducting period.

18 since v L L di / dt, the inductor voltage is negative when the current is decreasing. the current reaches zero when the diode turns off. The first positive value of ωt at which the current is zero is called the extinction angle and denoted by β. thus when ωt β the current is zero i( β ) V m Z sin( β θ ) + sin( θ )e ωt ωτ Example Consider the half-wave rectifier with L load shown in Figure (7a). The circuit has 1 Ω, L.1 H, ω πf 377 rad / s, and V m 1 V. Determine (a) an expression for the load current. (b) the average value of the current. (c) the rms value of the current. (d) the power absorbed by the load. (e) the power factor. Solution From the given parameters ωl 37.7 Ω z ( + ( ωl ) ) Ω ωl θ tan 1.7 o.361 rad α sin 1 ( ),±π,±π τ L 1 ms (a) Using the current relationship gives or i( ωt ) V m Z sin( ωt θ ) + V m Z sin θ e ωt / ωτ i( ωt ).936 sin( ωt.361 ) +.331e ωt /.377 A

19 This relationship is valid for ωt β where β is found from β/.377 sin( β.361 ) + sin(.361 )e Solving this equation by a numerical method gives β 3.5 rad 1 o (b) The average current is determined from the load current as or I dc 1 π I dc 1 π β i( ωt )d( ωt ) 3.5 [.936 sin( ωt.361 ) +.331e ωt.377 d( ωt ) Using a numerical integration method gives I dc.38 A (c) The rms current is or I rms 1 π 3.5 I rms.474 A [.936 sin( ωt.361 ) +.331e ωt.377 d( ωt ) (d) The power absorbed by the resistor is I rms (.474) 1.4 W. The power absorbed by the conductor is zero (e) The power factor is computed from pf p S p V s,rms I rms ( ).67 DC Converters (Choppers)

20 A dc chopper converts directly from dc to dc and is also known as a dc-to-dc converter. A chopper can be considered as a dc equivalent to an ac transformer with continuously variable turns ratio. Like a transformer it can be used to step-down or step-up a dc voltage source. Principle of Step-Down Operation Consider Figure (8a). When switch is closed for a time t 1, the input voltage source V s appears across the load. If the switch is turned and remains off for time t, the voltage across the load is zero. In practice a finite voltage drop across the chopper ranging from.5 to V is experienced, however for simplicity we neglect the voltage drop in the following analysis. The waveforms of the output voltage and load current are shown in Figure (8b). The average output voltage is Figure (8) V output,dc 1 T v o T V s f t 1 V s kv s t 1 dt t 1 The average load current is I dc V dc k V s The parameter k t 1 T is called the duty cycle of the chopper, and f is the chopping frequency. The rms value of the output voltage is V output,rms 1 T kt v o dt k V s

21 Assuming a lossless chopper, the input power to the chopper is the same as the output power and is determined from kt P input P output 1 T v i dt o k V s The effective input resistance seen by the source is emarks input V s I dc k The duty cycle k can be varied from to 1 by varying t 1, T or f. Therefore the output voltage can be varied from to V s by controlling k. As a consequence, the power flow into the circuit cab be controlled. When the chopping frequency is kept constant and the duty cycle is varied, the chopper is said to be in constant frequency operation. In this mode of operation the width of the pulse is varied and this is referred to as pulse-width-modulation (PWM). When the chopping frequency is varied and the duty cycle is kept constant, the chopper is said to be in variable frequency operation. In this mode of operation the width of the frequency is varied over a wide range to obtain the full output voltage range. This is not a practical option as it generates harmonics at unpredictable frequencies. Example Consider the dc-dc converter shown in Figure (8). The dc converter has a resistive load of 1 Ωand the input voltage is V s V. When the converter switch remains on, its voltage drop is v ch V and the frequency is f 1 khz. If the duty cycle is k5%, determine: (a) the average output voltage V o,a. (b) the rms output voltage V o,rms. (c) the converter efficiency. (d) the effective input resistance of the converter i. Solution Given: V s V ; k.5; 1Ω; v ch V (a) The average output voltage is

22 V dc 1 t 1 T v odt t 1 T ( V s v ch ) ft 1 ( V s v ch ) k ( V s v ch ).5(-) 19V. (b) The rms output voltage is V o kt 1 T v o dt k ( V s v ch ).5( ) V (c) The efficiency is ξ p o p i The output power is found from p o 1 T kt v o dt 1 T kt ( V s v ch ) dt k ( V v s ch ) (.5 x ) W The input power to the converter is found as p i 1 T kt V s idt 1 T kt.5 x x V s ( V s v ch ) dt k V ( V v s s ch ) ( ) 1 Thus the converter efficiency is 398 W ξ p 376. o 99.9% p i 398 (d) The effective input resistance of the converter is i V s I a V s kv s / k 1.5 Ω