I am currently using boost converter so I made all the calculation to make it work. My question is not about calculation it is about how physically is a boost converter working.

Here it is : the output voltage is depending on the duty cycle D. If we ignore the efficiency D=1-Vin/Vout then Vout = Vin/1-D. which means that when the duty cycle increase, the output votage increase.

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The functionment of the converter is as follow:

  • when the inside switch S is on there is a current charging the inductor
  • when the inside switch S is off the inductor L oppose the variation of current and then a voltage in series with Vin is created so Vout > Vin

The only thing I don't understand is : why is the voltage increase with the duty cycle?

If the duty cycle increase ILmax will be higher but for an inductor U = L*di/dt. Then the voltage across the inductor depends on the decreasing of the current and not on ILmax. Since the load (C and R) is the same the decreasing of current should be the same and then VL should be the same no?

Then, physically why is VL depending on duty cycle?

  • 1
    \$\begingroup\$ The problem with just thinking about Ldi/dt is that your simplified model has zero time anyway, which gives infinite voltage, which isn't very useful. Think about energy storage - if you need more voltage into the same load, then you need to move more energy through the converter, which means you need to store more in the inductor. \$\endgroup\$
    – user1844
    Commented Oct 21, 2015 at 8:00

2 Answers 2


"why is the voltage increase with the duty cycle?" Simple: if you increase the DutyCycle, you charge the inductor for longer, hence it contains more energy at the end of the charge. On the discharge cycle this energy is transferred via diode D into the load and the capacitor.

Yes ILmax will be higher because it will be charged more. Realize that a capacitor is charged with current and the voltage shows how much it is charged. With an inductor it's the other way round, it is charged by applying a voltage (this happens when S closes) and the current shows how much it is charged.

You focus a bit much on Vl, the voltage across the coil. But that is not so important, the inductor current is what matters. The inductor behaves as a current source when it still contains charge and S is open.

Note that for the same dutycycle if you increase the value of the load R, the output voltage V0 will increase ! A given Dutycycle does not result in a constant voltage at the output. So that's why real boost converters need a feedback circuit to control the dutycycle.

If you would remove load R, the voltage would increase to infinity ! (in theory that is)

  • \$\begingroup\$ then Vo is the resulting of the current through the load and it is not the current through the load which is the resulting of Vo, i need to think backwards what I'm used to I guess. But if the inductor is a current source why is the current going below the current needed by the load? I mean for example you want a current of 1A the ripple will oscillate between 1.3A and 0.7A. How is it possible the current goes below 1A? When the coil has no more current to give why the generator Vi does not give the current needed? \$\endgroup\$
    – damien
    Commented Oct 21, 2015 at 8:17
  • \$\begingroup\$ What matters is the average current because that is the current through the load. The current through the coil could even reach zero ! That simply means that the coil is discharged. The min and max coil current are set by the dutycycle. Charging the coil longer increases the max current ! When the coil is discharging it behaves like a current source so Vi is completely ignored. So Vi cannot deliver any current, it is the coil that determines the current. When the coil is discharged Icoil = 0, then Vi cannot give current because Vo > Vi and the diode will be in reverse. \$\endgroup\$ Commented Oct 21, 2015 at 9:20

Assume a continuous conduction boost converter (the extension to one that is not continuous is straightforward)

If you apply a voltage to an inductor, then the current through it changes. If you apply a long-term DC voltage to it (to an ideal inductor at least), it's current will increase without limit over time.

But this inductor's current doesn't increase without limit, its average stays the same. That means the time-averaged voltage across the inductor is zero.

As the time that the inductor has the output voltage across it decreases, the output voltage has to increase, to offset the large amount of time it sees the input voltage.


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