written 5.4 years ago by |
The power circuit diagram for a four-quadrant chopper is shown in Fig.1.a. It consist of four semiconductor switches $\mathrm{CH} 1$ to $\mathrm{CH} 4$ and four diodes $\mathrm{D} 1$ to $\mathrm{D} 4$ in antiparallel. Working of this chopper in the four quadrants is explained as under,
First quadrant: For first-quadrant operation of Fig.1.a, $\mathrm{CH} 4$ is kept on, $\mathrm{CH} 3$ is kept off and $\mathrm{CH} 1$ is operated. With $\mathrm{CH} 1,$ $\mathrm{CH} 4$ on, load voltage $v_{0}=V_{s}$ (source voltage) and, load current $i_{0}$ begins to flow. Here both $v_{0}$ and $i_{0}$ are positive giving first quadrant operation.
When $\mathrm{CH} 1$ is turned off, positive current freewheels through $\mathrm{CH} 4,$ $D2$. In this manner, both $V_{0}, I_{0}$ can be controlled in the first quadrant.
Second quadrant: Here $C H 2$ is operated and $\mathrm{CH} 1, \mathrm{CH} 3$ and $\mathrm{CH} 4$ are kept off. With $CH2$ on, reverse current flows through $L, \mathrm{CH} 2, \mathrm{D} 4$ and E . Inductance L stores energy during the time $\mathrm{CH} 2$ is on. When $\mathrm{CH} 2$ is turned off, current, is fed back to source through diodes $\mathrm{D} 1, D4$. Note that here $\left(E+L \frac{d i}{d t}\right)$ is more than the source voltage $V_{s}$ . As load voltage $V_{0}$ is positive and $I_{0}$ is negative, it is second quadrant operation of chopper. Also, power is fed back from load to source.
Third quadrant: For third-quadrant operation of Fig.1.a, $\mathrm{CH} 1$ is kept off, $\mathrm{CH} 2$ is kept on and $\mathrm{CH} 3$ is operated. Polarity of load emf E must be reversed for this quadrant working. With $CH3$ on, load gets connected to source $V_{s}$ so that both $v_{0}, i_{0}$ are negative leading to third quadrant operation. When $\mathrm{CH} 3$ is turned off, negative current freewheels through $\mathrm{CH} 2, \mathrm{D} 4 .$ In this manner, $v_{0}$ and $i_{0}$ can be controlled in the third quadrant.
Fourth quadrant: Here $CH4$ is operated and other devices are kept off. Load emf E must have its polarity reversed to that shown in Fig. 1.a. for operation in the fourth quadrant. With $CH4$ on, positive current flows through $\mathrm{CH} 4, \mathrm{D} 2, L$ and $E.$ Inductance L stores energy during the time $\mathrm{CH} 4$ is on. When $\mathrm{CH} 4$ is turned off, current is fed back to source through diodes $D2, D3.$ Here load voltage is negative, but load current is positive leading to the chopper operation in the fourth quadrant. Also power is fed back from load to source.