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My push-pull converter circuit uses a flyback transformer with an air gap. If that air gap is changed (increased), the primary inductance is decreased. How should I expect the secondary voltage and the output waveform to change after increasing the flyback air gap?

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All the energy is stored in the gaps between the magnetic core particles or gap in this case. As the gap size increases, flux will start to 'spray' into the area of the windings in proximity to the gap. This will generate high levels of eddy current losses in the wire. To avoid this problem you can wind on a few turns of insulating tape around the middle of the coil former to exclude wire from the region. The main advantage is it reduces the sensitivity to thermal runaway by moving the losses from the core towards the glass gap when increased.

There is an optimal gap to minimize the sum of all tradeoffs for increased winding loss, reduced core loss , increased gap coupling leakage and Eddy current losses.

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There are other variables, you have to specify what you are keeping constant to get a valid comparison.

Is anything else other than the total output capacitance limiting the output voltage? If not, then more energy stored in the flyback means a higher voltage developed into that capacitance.

Does an increased airgap mean more or less energy stored in the flyback? It depends what else stays constant.

If the primary voltage and pulse width stays constant, then the lower inductance will mean a faster rate of rise of current. Fortunately, this increased current will not saturate the inductor, as the input volt.seconds are the same. The current is higher by the factor that the inductance is lower, but as stored energy goes as \$0.5I^2L\$, the energy is increased by this factor. The output voltage will be higher.

If the primary is switched on until the current gets to a threshold, or like in a car coil the current is limited by the coil resistance and supply voltage, the current stays the same. With reduced inductance, the stored energy is lower, and the output voltage will be lower.

The increased airgap will also lower the secondary inductance, and may well reduce the primary - secondary coupling. This means that the secondary waveform will tend to have shorter time constants.

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  • \$\begingroup\$ Everything else remains constant in my push-pull converter, input waveform, and output load... So only thing changing is the gap width in my "flyback" transformer. My simulations show that increased gap does increase the output voltage, but I didn't understand well why this was the case. Your explanation makes sense, thanks. \$\endgroup\$ – ImpliedVolatility Jun 14 at 19:20
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My push-pull converter circuit uses a flyback transformer

If your converter is a push pull type then, although the transformer you used may have been called a flyback transformer when you acquired it, the new application (as part of a push pull circuit) puts it into operating as a conventional transformer.

If your converter primary voltage(s) remain as they were after the gap is increased, then the output voltage will probably not be altered but, bear in mind that enlarging the gap can mean a larger (usually unwanted) primary side current and, depending on other primary protection circuits, the secondary voltage may fall.

If you are running it as a true flyback circuit then Neil’s answer is valid.

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  • \$\begingroup\$ Yes, my converter is a transistorized push-pull and the transformer is operating conventionally (not in flyback mode). \$\endgroup\$ – ImpliedVolatility Jun 14 at 19:22
  • \$\begingroup\$ Maybe someone can explain the down vote? \$\endgroup\$ – Andy aka Jun 19 at 8:11

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