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I am wondering whether the boost converter is possible to operate with these conditions below or not:

  1. All components or sources are ideal
  2. Initial voltage across output capacitor is zero voltage.

Another question, if the voltage source and all components are ideal, then the driving current at the output can be infinite. Is that right?

enter image description here

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    \$\begingroup\$ I'm wondering what your question really is. \$\endgroup\$ – Andy aka Mar 12 '16 at 12:21
  • \$\begingroup\$ Well, as I said above, I am wondering whether the boost converter is possible to operate with these conditions mentioned above. \$\endgroup\$ – anhnha Mar 12 '16 at 12:22
  • \$\begingroup\$ Why shouldn't it? \$\endgroup\$ – Andy aka Mar 12 '16 at 12:22
  • \$\begingroup\$ Because the load is connected while output voltage is still zero. \$\endgroup\$ – anhnha Mar 12 '16 at 12:23
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    \$\begingroup\$ Given that this is how most real-world boost converters DO work, I don't see the problem. There is no infinite current, because the coil doesn't allow it. \$\endgroup\$ – Dave Tweed Mar 12 '16 at 12:43
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When "S" is closed, the inductor is charged with energy. When S opens that energy is forced into the capacitor. This charges the capacitor up a bit and its terminal voltage rises. When S closes again, the capacitor is slightly discharged by the resistor.

The important thing to note here is that the amount by which the capacitor increases in voltage MUST be a bit more than the amount of voltage that is reduced by the discharging effect of R on C.

Eventually, the capacitor becomes charged to the "required" voltage. From this point in time the sole purpose of the energy transfer system is to keep feeding enough energy to the capacitor to increase the voltage by the same amount that the resistor discharges that voltage.

This keeps the average output voltage fixed. To do this effectively the simple circuit you show in your question MUST be controlled by another circuit that alters the mark space ratio of the switch opening and closing. This is the fundamentally important part of a boost regulator.

If you kept the mark space ratio constant the capacitor would keep charging to a really high voltage. The controller is therefore important to act as a regulator.

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  • \$\begingroup\$ Thanks, with the circuit above I meant to say that no PWM/PFM or any controller is used. The duty cycle is fixed. So, can the converter work? \$\endgroup\$ – anhnha Mar 12 '16 at 12:56
  • \$\begingroup\$ Without a controller the converter will continue to push energy each cycle into the capacitor and, after a few hundred or thousand cycles, the circuit will fail. \$\endgroup\$ – Andy aka Mar 12 '16 at 15:05
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Sure.

  • On power-up current flows through L and D charging up C until \$V_O = V_I\$.
  • When S1 closes current flows through L to ground.
  • When S1 opens current continues to flow through L, D and R. Provided R is not too low in value the output voltage will rise towards \$ V_O = I_L \cdot R\$.
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  • \$\begingroup\$ Thank you. I have three questions. 1. With IL do you mean the average current through inductor? 2. I tried to calculate voltage conversion ratio by applying Laplace transform for cases when switch is ON or OFF but couldn't get the result. When switch is ON, the current through inductor increases linearly with time and assumes that it is I0 at the end of switch time. Next switch is OFF, I model inductor as sL in series with L*I0 where s is Laplace variable. However, I couldn't get the correct result. 3. With ideal components, is driving capacitance at the output infinite? \$\endgroup\$ – anhnha Mar 12 '16 at 12:50

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