I am currently studying two switch isolated DC/DC converters, more specifically the Push-Pull, the Half-Bridge and the Full-Bridge DC/DC converters. Several books say that these converters cannot operate at over 50% duty cycle because, if that's the case, you would be shorting the transformer. My question is why? Why can't you operate at over 50% duty cycle? Thanks in advance.

  • 1
    \$\begingroup\$ Duty cycle larger than 50% for example in the Push-Pull means that one of the transistors is ON longer than the other one. So you create asymmetry into transform (DC) and that is a bad thing to do (flux walking and core saturation). \$\endgroup\$
    – G36
    Jan 2 '19 at 17:57

Example of a half bridge DC-DC Converter

Image credit: Batarseh, I. (2015). Power Electronic Circuits (Student ed.). New Delhi: Wiley India Pvt.

As you can see from the above image, if the half-bridge DC-DC converter is operated at \$D\ge0.5\$ then it means that at one point of time, both the switches \$S_1\$ and \$S_2\$ will remain closed at one or some point of time. Due to this, the voltage on the upper half will have a negative polarity at the dot and positive polarity at the center tapping due to which induced polarity on the secondary winding has a negative polarity at the dot and positive polarity at its bottom. Since \$S_2\$ is also closed, the dot in the lower half will have a positive polarity and its bottom will be negative inducing a positive at the dot and a negative polarity at the bottom of the secondary as opposed to the previous condition due to \$S_1\$ thus short circuiting the transformer. So, normally, these type of DC-DC converters are used only with \$D<0.5\$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.