Hi,
I am having trouble with a homework problem on a DC chopper circuit. I attached the circuit as a jpeg. The diode is a freewheeling diode to prevent arcing when the switch is opened or closed. Also, it is given that E1 is greater than E2.
The question is 2-fold:
a) Show that the maximum difference between Imax and Imin (i.e. Imax - Imin) occurs when the duty cycle alpha=0.5
b) Obtain the value of the load inductance L so that Imax-Imin= 0.2*Iavg; where Iavg=200A, E1=400V, alpha=0.5 and T=10^(-3) sec
My first problem is the fact that there is no resistance in the circuit...I am having trouble writing a differential equation. If I assume the switch has been opened and closed repeatedly for a while, and then it is opened again at t=0, there should be residual current in the loop with the inductor, and I think that should be the Imin, so the equation I wrote was:
L(di/dt)-E2=0. I don't think this can be correct though, because then you would have i(t)=(E2/L)*t and that doesn't make any sense because if the switch was kept opened, the field in the inductor would dissipate and you'd be left with a short cirucited battery (E2) and ultimately no current. I don't really know how to proceed from here (I don't have much of a background in differential equations)...any help you could give would be greatly appreciated.
Thanks.
I am having trouble with a homework problem on a DC chopper circuit. I attached the circuit as a jpeg. The diode is a freewheeling diode to prevent arcing when the switch is opened or closed. Also, it is given that E1 is greater than E2.
The question is 2-fold:
a) Show that the maximum difference between Imax and Imin (i.e. Imax - Imin) occurs when the duty cycle alpha=0.5
b) Obtain the value of the load inductance L so that Imax-Imin= 0.2*Iavg; where Iavg=200A, E1=400V, alpha=0.5 and T=10^(-3) sec
My first problem is the fact that there is no resistance in the circuit...I am having trouble writing a differential equation. If I assume the switch has been opened and closed repeatedly for a while, and then it is opened again at t=0, there should be residual current in the loop with the inductor, and I think that should be the Imin, so the equation I wrote was:
L(di/dt)-E2=0. I don't think this can be correct though, because then you would have i(t)=(E2/L)*t and that doesn't make any sense because if the switch was kept opened, the field in the inductor would dissipate and you'd be left with a short cirucited battery (E2) and ultimately no current. I don't really know how to proceed from here (I don't have much of a background in differential equations)...any help you could give would be greatly appreciated.
Thanks.
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