Does the output voltage depend on anything except for the Duty Cycle?
Assuming no losses across switch, inductor and caps.


The answer depends on the operating mode of the converter.

In discontinuous conduction mode, which is what you usually have with a passive rectifier and light loads, the output voltage depends on the input voltage, the duty cycle and the load. With light (or no) loads, the voltage can rise arbitrarily high.

In continuous conduction mode — and assuming "ideal" components — the output voltage depends only on the input voltage and the duty cycle.

Continuous conduction mode means that current is always flowing in the inductor. This can be achieved with a passive rectifier if the load current is always above some minimum value. It can also be achieved if you use active (synchronous) rectification (a second switch) that allows current to flow in both directions.

  • \$\begingroup\$ If a boost circuit is in continuous mode and the load changes AND the duty cycle doesn't change, the voltage will change. \$\endgroup\$ – Andy aka Nov 10 '13 at 14:38
  • \$\begingroup\$ @Andyaka: Why do you think so? \$\endgroup\$ – Dave Tweed Nov 10 '13 at 15:46
  • \$\begingroup\$ Because the energy transferred by the fixed duty cycle at so many thousands of times per second dictates that the output is constant power. Not so for a sync buck of course. \$\endgroup\$ – Andy aka Nov 10 '13 at 15:56
  • \$\begingroup\$ @Andyaka: Even in a boost converter, in continuous conduction mode, energy can flow both ways through the converter's inductor. The AC current through the inductor has a waveform that depends only on the duty cycle, while the DC current can be any value, including zero. Given a duty cycle d, the output voltage is simply $$V_{in}(1 + \frac{d}{1 - d})$$ \$\endgroup\$ – Dave Tweed Nov 10 '13 at 16:34
  • \$\begingroup\$ Dave, could you link me something about this. I'm assuming a standard boost with a diode can't do this and it seems you might be talking about a sync booster and I confess I didn't think a sync type would behave like this. \$\endgroup\$ – Andy aka Nov 10 '13 at 18:04

The output voltage will depend on the duty cycle and the load. More output current means a higher duty cycle will be necessary to generate the same output voltage.

  • \$\begingroup\$ If so, why do we still have the formula Vout = Vin/(1-D) in CCM? \$\endgroup\$ – anhnha Dec 11 '17 at 22:52
  • \$\begingroup\$ You're not always in CCM. And you don't have ideal components. \$\endgroup\$ – alex.forencich Dec 12 '17 at 8:19
  • \$\begingroup\$ Let's us consider CCM and ideal components only then does output voltage depends on load current? \$\endgroup\$ – anhnha Dec 12 '17 at 10:09
  • \$\begingroup\$ Sure. If we're going to assume all the conditions where Vout = Vin/(1-D) is true, then there is no dependence on load current. However, these assumptions are rather unrealistic so most switching power supplies will implement closed-loop control to cover all the cases where these assumptions do not hold. \$\endgroup\$ – alex.forencich Dec 12 '17 at 21:58

Output voltage depends primarily on input voltage, duty cycle and load. Each switching cycle an inductor is "charged" with energy and this energy is passed to the load. Power to the load is energy per cycle multiplied by switching frequency.

Without a load and with a fixed duty cycle, the output voltage would rise until the boost transistor or output diode failed.

For boost converter the output voltage is in series with the input voltage so it's a little bit more complex to work out compared to a buck regulator.

Energy transferred is dependent on the on time of the switch, the value of the inductor and the input voltage.


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