L2_3 - 0.00761 L H = For 300 d V V =(3 2 240 10 1 sin...

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EXAMPLE A three-phase bridge rectifier supplies power to DC motor with negligible armature resistance. The sours is 240V RMS, 60 Hz. The average load voltage must vary from 150 V to 300 V. The minimum average load current at any load voltage is 10 A. Continuous load current is a requirement. Find: a) required rang for α b) minimum value of inductance necessary for continuous motor current. Solution a) From Equation 0.955 cos d ml V V = for 100 d V V = : ( 29 3 100 (240) 2 cos π = 0 72.0 = For 300 d V V = : ( 29 3 300 (240) 2 cos = 0 22.24 = b) For both ends of the control range, equation ( 29 3 1 sin 2 3 ml d V I X = - ÷ applies. The minimum value of d I can be used to determine the inductance required. For 100 d V V = : 0 (3)( 2 240) 10 1 sin72.0 ( )(2 60)( ) 2 3 L × = - ÷ ÷ ×
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Unformatted text preview: 0.00761 L H = For 300 d V V = : (3)( 2 240) 10 1 sin 22.24 ( )(2 60)( ) 2 3 L × =- ÷ ÷ × 0.00303 L H = The larger of the two values is selected so that continuous current exists over the entire control range. 15 The Half-Controlled Three-Phase Bridge Rectifier ( 29 ( 29 ( 29 ( 29 ( 29 { } [ ] 6 2 3 6 6 1 1 6 6 2 6 2 6 3 cos cos 2 3 sin sin sin sin 2 3 1 cos 2 ml d ml ml V V d V V π α ϑ ϑ ϑ-+ = +-= = +--= + + +--= = + ∫ ∫ 16 Inverter mod 17...
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L2_3 - 0.00761 L H = For 300 d V V =(3 2 240 10 1 sin...

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