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I am slightly confused about the flow of the current in the forward converter during the turn off time of the MOSFET (when diode on left is conducting) as shown below. Now I am considering that n1=n2=n3. During the turn of time there must be a current IM (im=i3) flowing in the extra winding with a diode. From the figure below it is seen that there is a voltage of Vd across the winding also considering zero drop on diode.

enter image description here

Thus how can a current flow with no potential difference as the voltage source and the coil are on same potential?? Shouldn't the potential on the coil be higher than the voltage source for the current to flow?

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  • \$\begingroup\$ Please provide a citation and link to the image. \$\endgroup\$
    – Andy aka
    Mar 1 at 19:04
  • \$\begingroup\$ @Andyaka I have already provided \$\endgroup\$
    – kam1212
    Mar 1 at 19:12
  • \$\begingroup\$ Oh, it's a video. Maybe you can find a write-up type document that explains this: plexim.com/sites/default/files/demo_models_categorized/plecs/… ||| video links tend not to be watched. \$\endgroup\$
    – Andy aka
    Mar 1 at 19:17

1 Answer 1

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Once the MOSFET deactivates, the flux in the core demagnetizes via the 3rd winding and diode. The dotted end of the 3rd winding tries to goes negative but, it's clamped at -0.7 volts by the diode and, the flux energy in the core returns as a current back to the power supply (from the undotted end of the 3rd winding).

This method saves an extra MOSFET but, these days probably more forward converters will use an extra MOSFET to drive the single primary winding in push-pull.

Thus how can a current flow with no potential difference as the voltage source and the coil are on same potential?

An inductor's voltage does not prevent it from supplying current when the only route to demagnetize the core is via the 3rd winding. For instance, consider this from here: -

enter image description here

At 10 seconds into the plot, the applied voltage on the inductor is positive but, the inductor current is negative. At 12 seconds is the point in a 3-winding forward converter when the magnetic field is discharged. Sorry, I haven't got a better example but hopefully you'll see why current can still flow back into the power supply (via your diode).

This is what inductors do in these types of circuit and in AC circuits.

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  • \$\begingroup\$ Does this mean that even if the inductor of diode winding has the same voltage/no potential difference it still can give out current due to the fact that it wants to dischage the other coil?? \$\endgroup\$
    – kam1212
    Mar 1 at 21:43
  • \$\begingroup\$ @kam1212 it's not discharging another coil; it's allowing the flux previously built up in the core to be released gracefully without generating a massive surge of voltage. If it were not connected to the voltage source it would form a spark; it's passing magnetic stored energy back to the supply. \$\endgroup\$
    – Andy aka
    Mar 2 at 20:29

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