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Four Quadrant Chopper or Class E chopper
Figure class E chopper circuit
When SW1 is ON with SW4 V0 = Vs and load current flows as shown in figure when SW1 is OFF V0 = 0 but current will continuously be flowing through SW$ and diode D2. Here both V0 and I0 will be positive hence it is first quadrant operation as shown in figure.
When SW2 is ON with diode D4 V0= 0 but the direction of current will be negative. When SW2 is OFF diode D1 and D4 will be conducting in this period power flows from load to source because V0 = E+ L di/dt exceeds Vs. Hence Vo is positive but current is negative. This give the second quadrant operation as show in figure.
In this mode the polarity off must be reverses so that third quadrant operation can be obtained. Now when SW2 is ON with SW2 , V0 = -Vs and current will be flowing in reverse direction. Both Vo and Io will be negative. When SW3 is OFF D2 will conduct with SW2 therefore both V0 and Io will be negative. Hence third quadrant operation is obtained.
Diode D2 must be conducting with SW4 On E have opposite polarity to that shown in figure now the voltage V0= 0 whenSW4 is ON and current 10 will be positive when SW4 is OFF V0 will be negative but current remains in same direction. In this condition power flows from load to source and V0 = -( E+ L di/dt) exceeds Vs.
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