## Presentation on theme: "Two-quadrant converter"— Presentation transcript:

Va T1 + Vdc D1 ia Q2 Q1 Ia + Va - D2 T2 T1 conducts  va = Vdc

Va T1 + Vdc D1 ia Q2 Q1 Ia + Va - D2 T2 D2 conducts  va = 0 T1 conducts  va = Vdc Quadrant 1 The average voltage is made larger than the back emf

Va T1 + Vdc D1 ia Q2 Q1 Ia + Va - D2 T2 T2 conducts  va = 0 Quadrant 2 The average voltage is made smaller than the back emf, thus forcing the current to flow in the reverse direction

Va T1 + Vdc D1 ia Q2 Q1 Ia + Va - D2 T2 D1 conducts  va = Vdc

Chopper fed DC drive Four-quadrant Chopper Va + Vdc D1 D3 Q1 Q3 + Va  Ia D2 Q2 D4 Q4 Forward Motoring Q1 & Q4 is ON Current Flow : Vdc + _ Q1 _Motor_Q4_Vdc- Current Ia & Va are positive Operates in First Quadrant

Chopper fed DC drive Four-quadrant Chopper + Vdc D1 D3 Q1 Q3 + Va  D2 Q2 D4 Q4 Q1 is OFF & Q4 is ON. Inductor current has to flow in the same Direction. Diode D2 is FB Inductor Current freewheels through D2 & Q4 Output Voltage is Zero

Chopper fed DC drive Four-quadrant Chopper + Vdc D1 D3 Q1 Q3 + Va  D2 Q2 D4 Q4 Q4 is OFF. Q2 is ON. Load is not connected with the source. Back Emf drives the current through Q2 & D4

Chopper fed DC drive Four-quadrant Chopper Va + Vdc D1 D3 Q1 Q3 + Va  Ia D2 Q2 D4 Q4 Forward Braking Q2 is OFF. Diode D1 is FB Current flows through D4 & D1 Current Ia is negative & Va is positive. Operates in second quadrant.

Chopper fed DC drive Four-quadrant Chopper + Vdc D1 D3 Q1 Q3 - Va + D2 Q2 D4 Q4 Q3 & Q2 is ON Current Flow : Vdc + _ Q3 _Motor_Q2_Vdc- Current Ia & Va are negative Operates in third Quadrant

Chopper fed DC drive Four-quadrant Chopper + Vdc D1 D3 Q1 Q3 - Va + D2 Q2 D4 Q4 Q3 is OFF. Q2 is ON. Current has to be continuous. Diode D4 is FB Current flows through Q2,D4 & (Eb,La,Ra)