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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.

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    \$\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 at 17:57
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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\$.

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