The figure below is a diagram of a voltage-mode controlled boost converter. I am wondering how experts solve the start-up problem for this converter.


Assume the output voltage is zero at the beginning. Then from the control loop the output of the comparator will be high. This means the inductor is charged. However, there is no energy transferred to the output so the output will remain at zero forever and the inductor is charged forever.

So how do you solve this problem?

simple boost converter circuit diagram


3 Answers 3


You can see there's a "Vramp" in there. This creates a timing element.

If the Vramp is above the first op-amp's signal, the transistor will be off. So if you know, for example, that your error amplifier (the first op-amp) can never output a voltage higher than 3V, you can make the Vramp such that it goes between 0V and 4V, which guarantees that whenever it's above 3V the transistor is always off. In this case that would be a minimum off-time of 25%.

  • \$\begingroup\$ I thought about this but this would limit the duty cycle also. \$\endgroup\$
    – emnha
    Jul 2, 2018 at 6:01
  • 2
    \$\begingroup\$ @anhnha Yes, that's exactly what it will do, it will prevent the duty-cycle from saturating too close to 100% and going into lock-down. \$\endgroup\$
    – Asmyldof
    Jul 2, 2018 at 6:03
  • \$\begingroup\$ With output a voltage never higher than 3V, I can make the Vramp such that it goes between 0V and 4V, 0 and 5V, or 0 and 20V, etc. It seems that they all work. So what is the optimum Vramp? \$\endgroup\$
    – emnha
    Jul 2, 2018 at 6:09
  • \$\begingroup\$ Okay, I see how Vramp affects minimum off-time now. \$\endgroup\$
    – emnha
    Jul 2, 2018 at 6:17
  • \$\begingroup\$ @anhnha you have to see what else changes with the range of VRAMP. With constant frequency you will affect the duty cycle limit and the slew rate- which in fact is proportional to your control loop gain. You will be able to compensate by error amplifier gain. I would suggest going to 95% duty cycle just becuase it's a good number. If you go to 15% maximum duty cycle- you don't leave a chance to your system to cotrol itself, and besides normally people don't do that. \$\endgroup\$
    – user76844
    Jul 2, 2018 at 7:21

In addition to the other suggestion, use soft start. The reference voltage must slowly ramp up to its final value. This way you kind of avoid the start conditions, because you follow the reference rather than jump from somewhere to it.

  • 1
    \$\begingroup\$ A soft start is absolutely necessary for all boost (flyback) converters, or else the core of the inductor (flyback transformer) will go into saturation and the primary current will go through the roof. The same response is needed after a secondary short-circuit, so you could also solve it with an over-current protection on the primary side. \$\endgroup\$
    – StessenJ
    Jul 2, 2018 at 10:13

By the way, you have a control circuit, and they usually require tuning. So you need variable gain for the error amplifier, but also you may want to use integrator to have PI controller.

enter image description here

Disclaimer: i didn't check whether the gains are all positive or negative. You may need to invert things.


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