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From what I have seen, flybacks are used for step down applications, are step-up applications possible? More specifically, is converting from 4VDC to 48VDC possible using a 1:8 transformer?

I'm designing a converter from scratch (no COTS controllers); and I need to step-up from 4V to 48V. I am trying to leverage the turns ratio of the transformer (currently a 1:8).

In a typical boost converter configuration, current is continually drawn from the input; but that is not the case of flybacks. Flybacks push packets of energy kind of like buck converters; they store the energy in the magnetizing inductance during the switch ON; and push out during the off cycle. The input current is discontinuous.

This discontinuity has me thinking there are limitations of the flyback topology that I am not thinking of.

For reference: enter image description here

Edit:

I am confident in solving the controls compensation, EMI/voltage-spikes, and magnetic challenges.

Edit 2:

power would be <10W

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  • \$\begingroup\$ By "What kind of issues can I expect?", what do you mean exactly? Design complexity, transformer VS inductor tradeoffs, the effect of the topology on the control circuitry, reliablility issues, total cost, the potential for electromagnetic interference or output isolation? You need to be specific or somebody will close this as "unclear about what you are asking". Phrased like that, somebody would essentially have to write a SMPS textbook to answer your question. \$\endgroup\$ – jms Feb 26 '16 at 15:42
  • \$\begingroup\$ ok, adding more detail... \$\endgroup\$ – hassan789 Feb 26 '16 at 15:59
  • \$\begingroup\$ A flyback would be ideal for this application assuming your output power is <100W. Especially if you need the isolation. Like @jms mentioned there are all kinds of issues you can expect if you don't know how to do at least good power layout, magnetics design and component selection. \$\endgroup\$ – John D Feb 26 '16 at 16:03
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The flyback converter behaves like an inverting buck-boost converter.

inverting buck-boost

In both the voltage source is never in series with the load, with the energy being sent as "packets" proportional in magnitude to the duty cycle. Both can thus theoretically produce any output voltage (as long as the polarity is the same) from any input voltage.

The principal practical difference between the two is that both the switch and diode in an inverting buck-boost converter have to endure the full output voltage and the full peak current. This is unlike a flyback converter, where the transformer turns ratio allows the designer to select a low voltage high current switch and a high voltage low current diode for a step up converter, or the opposite for a step down converter.

Another significant advantage of the flyback converter is that the transformer inductance can be both charged and discharged in a similar amount of time, even when a high conversion ratio is desired. This reduces electromagnetic interference and the switching speed requirements for the transistor and diode.

The 1:8 ratio transformer isn't optimal, but still a very good fit for the application.

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  • \$\begingroup\$ what would you consider as the optimal ratio? \$\endgroup\$ – hassan789 Feb 28 '16 at 1:17
  • \$\begingroup\$ @hassan789 48/4 = 12. 8 will work just fine. \$\endgroup\$ – jms Feb 28 '16 at 11:30

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