CHAPTER THREE DIODE RECTIFIERS

Transcription

1 CHATE THEE DODE ECTFES 4 Three hase ectifiers Three-phase rectifiers are classified into Half-wave, and Full-wave energized loads with various impedances and back emf Applying three-phase rectifiers aims to realize smooth rectified voltage, increasing efficiency, utilization, and minimizing the parameters of the filter 4: Three-hase Half-Wave ectifier Figure 9-a illustrates three-phase half wave rectifier energized load, where three diodes ( D, D, and D) operates in series seuence, each one for a time of The operation seuence is determined by the criteria : " the diode with maximum positive voltage applied across it's terminals will conduct ", therefore each phase will pass the current for, where for the rest of time this phase will be off and doesn't participate in the rectification process Fig9-a: Three-phase half-wave rectifier- Wye connected

2 Fig9-b: Three-phase voltage, and output rectified voltage Fig9-c: hase current, and instantaneous rectified current

3 Fig9-d: Diode voltage Fig9-e: hase current, and instantaneous rectified current at L load The average, rms voltage and current: / cos ω t / MS in the case / / 8468 MS MS The efficiency and TUF: of ( cosωt) Three phase MS three 869 d ( ω t ) d( ωt) phase + sin sin ;

4 η AC AC ; η TUF ( A ) S ( A ) TUF rating MS AC ( A ) rating ( A ) ( 869 ) ( 8468 ) MS / rating S m ; S cos ω t d( ω t) rating m sin % % 66 % % 4854 m The FF, F, F and MS 846 FF 869 F FF 84% % 65% AC 846 F 6844 (A ) rating The average and rms diode current and / DA m cos ω t d ( ω t) m sin 757 m D / ( m cos ω t ) d ( ω t ) MS / 6 s Summary: Taking into account the obtained rectifier parameters we conclude:

5 The output average voltage is 8% of the phase magnitude Satisfied transformer utilization 66%, which means that the transformer must be /6655 times larger that when it is used to deliver power from a pure ac voltage good rectification efficiency 9666% 4 There is a dc component in the secondary current, therefore additional losses in the transformer core This reason explain the small value of TUF 5 Good form factor %, and Acceptable ripples factor 84% greater than that when the source is pure dc 6 The diode must % of the total average dc current, 57% of the total rms current, and must carry 7 in the reverse biasing With purpose to enhanced the rectifier parameters ( reducing the ripples, increasing TUF, increasing the rectifier capability, and elimination the dc component in the secondary current) a three-phase full-wave rectifier is applied as follows: 4: Three-hase Full-Wave ectifier Figure -a illustrates three-phase full- wave rectifier energized load, where six diodes ( D, D, D, D4, D5, & D6) operates in six groups, each group for a time of 6, while each diode operates for The operation seuence is determined by the criteria : " the diode with maximum positive voltage applied across it's terminals will conduct, and the diode with maximum negative voltage applied across it's terminals also will conduct ", therefore each phase will pass the current for 4, where for the rest of time this phase will be off and doesn't participate in the rectification process Fig-a: Three-phase Full-wave rectifier

6 Fig-b: Three-phase voltage, and output rectified voltage -c: hase current, and instantaneous rectified current

7 Fig-d: Diode voltage Fig-e: Output voltage, rectified current, and phase current at L load The effect of circuit inductance L on the output voltage Fig-f: The output voltage with illustrated effect of the circuit inductance on this voltage The rectifier mathematical euations are going to be described by discussing the following example : Example 5: Three- phase bridge rectifier with the following parameters : 75; A; 5 H; Wye connected; the voltage reduction due to circuit inductance is 5% of the output average voltage Determine: - the load resistance, and secondary current - the diode average, rms current, and - the transformer utilization 4- the form factor and ripple factor 5- the circuit inductance

8 6- the average voltage reduction, and net output voltage 7 - the average voltage reduction, and net output voltage at Hz freuency Solution: The average, rms voltage and current: / 6 / 6 6 LL ( ω t ) d ( ω t) sin( ω t + / ) d( ω t) / 6 MS MS Ω / 6 / 6 6 L L( ωt) d( ωt) ( sin( ωt + / ) ) d( ωt) / MS A The diode average, rms current and : DA m L sin 8 m L 6 m L 48 A D / 6 m sin( 45 The transformer utilization: TUF ( A ) rating ( A S s ) TUF rating 8 L s s / 6 ω t + ( m L cos( ω t ) ) / kw ka ) d ( ω t ) 6 68 d ( ω % 75 t ) kw 95 9 % The Form Factor, ipple Factor and ower Factor: MS 786 m 98 L 577 ka A A

9 FF F MS FF AC 75 6 kw F (A ) rating ka 5 The Circuit inductance: 6 fs Lc 5 % 6fs Lc 5 % mh the average voltage reduction, and net output voltage (L ) the average voltage reduction, and net output voltage at freuency of Hz ' ' 6 (L ) 5 / ' Summary: Taking into account the obtained rectifier parameters we conclude: The output average voltage is 65% of the phase magnitude High transformer utilization 954%, which means that the transformer must be /95448 times larger that when it is used to deliver power from a pure ac voltage, ie approximately there is no difference between both modes good rectification efficiency 99% 4 There is no dc component in the secondary current, therefore no additional losses in the transformer core This reason explain the high value of TUF 5 Good form factor 4%, and Acceptable ripples factor 4% greater than that when the source is pure dc 6 The diode must % of the total average dc current, 57% of the total rms current, and must carry 7 in the reverse biasing 7 The presence of circuit inductance cause additional voltage reduction, as Lc increases, the net output rectified voltage decreases 8 As the source freuency increases the voltage reduction being significant 4: Three-hase Full-Wave ectifier with -L & E Figure -a illustrates three-phase full- wave rectifier energized - L load with back emf E depending on the values of L, E and circuit initial conditions two modes of operation are defined: continuous mode Discontinuous mode, where this mode cause additional loses, reducing the circuit efficiency, and mechanical vibration Figure b illustrates the waveforms showing the effect of above mentioned parameters

10 Fig-a: Three-phase Full-wave rectifier with -L and E Fig-b: the circuit waveforms The rectifier mathematical euations are going to be described by discussing the following example : Example 6:

11 Three- phase bridge rectifier with the following parameters : ab8; E; 6 Hz; 5Ω Determine: - the steady state load current at at wt p/ - the diode average, and rms current Solution: The commutation process is repeated six time per period Fig- illustrates the euivalent circuit of one commutations where D and D6 conducts and the euilibrium voltage euation is expressed as follows: Fig-: Euivalent circuit when D,D6 conducts The steady state current di ab ( ω t) i + L + E dt The phase current obtained by solving the previous first order euation is as follows: i( ωt) abm Ae t L + abm sin( ωt θ) E + ( ω L ) nitial conditions : i(ωt/) i(ωt/) A A e ω L + abm i( ω t)? abm sin( E + abm abm sin( ω t θ) e sin( E θ) sin( sin( + θ ) θ)e E θ ) e ω L L E abm + sin( θ) e L ω E L ω ( t )

12 Now Numerical results: ωl 944 Ω; abm 8 m 588 ωl o θ tan 7 ; 57 8 A + ( ωl ) 5 9A 588 ( 77 i( ω t) 57 8 sin( ωt θ) 4 + e - The average and rms current / 6 / 6 Ω t) 4 4 : dav i( ωt) dr i ( ω t)