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Transcript
Lecture 9 Controlled Rectifiers The Controlled Half-wave Rectifier • Normal rectifiers are considered as uncontrolled rectifiers. • Once the source and load parameters are established, the dc level of the output and power transferred to the load are fixed quantities. • A way to control the output is to use SCR instead of diode. Two condition must be met before SCR can conduct: – The SCR must be forward biased (VSCR>0) – Current must be applied to the gate of SCR Controlled, Half-wave R load • A gate signal is applied at t = , where is the delay/firing angle. I o, rms Vo, rms Vs R 2R Example • Design a circuit to produce an average voltage of 40V across 100 load resistor from a 120Vrms 60 Hz ac source. Determine the power absorbed by the resistor and the power factor. Briefly describe what happen if the circuit is replaced by diode to produce the same average output. Example (Cont) • Solution In such that to achieved 40V average voltage, the delay angle must be Vs [1 cos ] 2 120 2 40 [1 cos ] 2 61.2o 1.07 rad Vo Vo, rms Vm sin( 2 ) 1 2 2 120 2 1.07 sin 2(1.07) 1 2 2 75.6V V 2 rms 75.6 2 P 57.1W R 100 pf 57.1 0.63 75.6 (120) 100 • If an uncontrolled diode is used, the average voltage would be Vo Vs 2 (120) 54V • That means, some reducing average resistor to the design must be made. A series resistor or inductor could be added to an uncontrolled rectifier, while controlled rectifier has advantage of not altering the load or introducing the losses Controlled, Half-wave R-L load • The analysis of the circuit is very much similar to that of uncontrolled rectifier. t V m [sin( wt ) sin( )e for t i ( wt ) Z 0 otherwise and L L Z R 2 (L) 2 , tan 1 , R R rms current , I rms 1 2 2 1 2 i ( t ) d ( t ) i (t ) d (t ) 2 2 and , average current , 1 Io i (t ) d (t ) 2 Controlled, Half-wave R-L load The average output voltage, Vm [cos cos ] 1 Vm sin( t )dt Vo 2 2 The average power absorbed by load , 2 P I rms R ; Controlled full-wave rectifiers Resistive load: 1 Vo Vm sin( wt )d ( wt ) Vm ( 1 cos ) Io delay angle Vo Vm (1 cos ) R R I rms 1 ( Vm sin wt ) 2 d ( wt ) R Vm 1 sin( 2 ) R 2 2 4 The power delivered to the load P I 2 rmsR The rms current in source is the same as the rms the load. current in Full wave Controlled Rectifier with RL load Discontinuous and Continuous Operations Discontinuous Mode discontinuous current : Vm io ( wt ) sin( t ) sin( )e ( t ) /( ) Z for t Z R 2 ( L )2 tan 1 ( L ) R , L R For discontinuous current Analysis of the controlled full-wave rectifier operating in the discontinuous current mode is identical to that of the controlled half-wave rectifier, except that the period for the output current is . Continuous Mode continuous current wt , i ( ) 0 sin( ) sin( )e ( ) /( ) sin( ) 1 e /( ) 0 sin( - ) 0 v0 ( wt ) Vo Vn cos( nwt n) ( - ) 0 Tan ( -1 0 n 1 L Vo ) R for continuous current 1 Vm sin wt d ( wt ) Vn an bn 2 n Tan -1 ( bn ) an 2Vm cos 2 an 2Vm cos( n 1) cos( n 1) n 1 n 1 bn 2Vm sin( n 1) sin( n 1) n 1 n 1 n 2,4,6,.... In Vn Zn Vn Irms Io 2 | R jnwL | n 2 ,4... Io Vo R ( In 2 )2 R-L Source load : Fig.4-14 The SCRS may be turned on at any forward biased, which is at an angle sin 1 (Vdc Vm) time that they are For continuous current case, the average bridge output voltage is Vo 2 Vm cos average load current is Io Vo Vdc R The ac voltage terms are unchanged from the controlled rectifier with an R-L load. The ac current terms are determined from circuit. Power absorbed by the dc voltage is Pdc Io Vdc Power absorbed by resistor in the load is P I 2 rmsR Io 2 R if L is l arg e Controlled Single-phase converter operating as inverter an Average DC output voltage Vd 2 2 Vs cos • Assuming AC side inductance is zero • Note that output voltage can go negative for alpha > 90 degrees. • This means negative power flow or inversion Copyright © 2003 by John Wiley & Sons, Inc. Chapter 6 Thyristor Converters