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The switch, S of the DC to DC Buck converter shown in the figure is operated with a duty cycle of 0.5 at a switching frequency of 5 kHz while it is feeding a certain load.

While the input voltage of the converter is maintained at 100 V dc, the steady state average voltage at the output terminals of the converter, \$V_o\$ is found to be 70 V.

Is the converter operating under discontinuous mode of conduction ?

enter image description here

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  • \$\begingroup\$ It's clear that you don't understand what discontinuous operation is yet you ask the question. I think you need to research what discontinuous operation is before asking this question. Your question proves it isn't in discontinuous mode. \$\endgroup\$ – Andy aka Mar 25 '13 at 8:59
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    \$\begingroup\$ How the device works is that (assuming there is initially no current) when you close the switch, the current will start to ramp up. At the instant when the switch is open, the current continues to flow in the same direction (with the help of the flywheel diode D to complete the circuit), but from that point on, the current immediately starts to ramp down. Your homework is to show that the switching rate is fast enough that the ramp never "bottoms out": as the switch is open, and the current is winding down toward zero, the switch closes in time to reverse the ramp. \$\endgroup\$ – Kaz Mar 25 '13 at 9:38
  • \$\begingroup\$ -1 for just dumping a homework problem on us without any research. Asking about specific aspects of homework is OK, but just copying the whole problem here expecting us to solve it is not. \$\endgroup\$ – Olin Lathrop Mar 25 '13 at 11:55
  • \$\begingroup\$ @ Olin Lathrop:It'll be of great help if you can suggest some really GooD book for DC-DC Converters. \$\endgroup\$ – HOLYBIBLETHE Mar 25 '13 at 12:03
  • \$\begingroup\$ @RajeshKSingh Dude there is plenty of stuff on the net about switchers and discontinuous operation. I gave you a few hours to prove that you knew what the question meant (even if you didn't know the answer) so it's a -1 from me too. \$\endgroup\$ – Andy aka Mar 25 '13 at 12:54
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Yes, this converter is operating in discontinuous mode. I can tell you that because I know that if it were in continuous mode, the output voltage would be the fraction of the input voltage as defined by the duty cycle. 100V × 0.5 would equal 50V.

Rather than handing you the answer, I'll offer a few hints: During the two parts of the switching cycle, what is the voltage across the coil? What does this imply about how fast the current in the coil is ramping up or down? To simplify things, you may assume that the output capacitor is large enough so that the output voltage ripple is negligible.

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  • \$\begingroup\$ If the switcher were a synchronous type I think you'd be correct is assuming the output is 50V. It's not a synchronous type and I think it's impossible to tell where the output voltage should be because we don't know the load resistance, L and C etc.. I'm trying to figure it out now as an exercise so pls correct me if I'm wrong. On light loads the output will largely attain the input supply voltage. \$\endgroup\$ – Andy aka Mar 26 '13 at 15:00
  • \$\begingroup\$ Ignore me please I'm being stupid (again) \$\endgroup\$ – Andy aka Mar 26 '13 at 15:10
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Assuming a constant resistance, R=V/I

If V increases I must increase and vice-versa.

This is a step down converter so the output voltage \$V_o\$ is always less than the input voltage \$V_s\$.

For continuous mode of conduction,

\$V_o=D*V_s\$

D=Duty Cycle, is the ratio of the ON time, \$T_{ON}\$ to the switching time \$T\$.

In discontinuous mode of conduction,

\$V_o=\frac{V_s}{2}(\sqrt{m(m+4)} - m)\$

where, \$m = \frac{T*R}{2L}D^2\$

Here, D=0.5

As a thumb rule, to maximize efficiency, the minimum oscillator time period is \$100*T\$.

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    \$\begingroup\$ This is not a answer, and is wrong in what it does say. Clearly your equation doesn't apply in this case since it says you should get 50 V but you are actually getting 70 V. \$\endgroup\$ – Olin Lathrop Mar 25 '13 at 14:46

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