Magnetizing inductance and Flyback converter

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likephysics
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In the flyback inductor equivalent circuit and also in general, why is magnetizing inductance represented in parallel to the actual inductance?
We can have just the Magnetizing inductance in series with the leakage inductance.

Also, in flyback converters why does the secondary voltage get reflected to the primary side?
 
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likephysics said:
In the flyback inductor equivalent circuit and also in general, why is magnetizing inductance represented in parallel to the actual inductance?
We can have just the Magnetizing inductance in series with the leakage inductance.

Also, in flyback converters why does the secondary voltage get reflected to the primary side?

What's a "flyback inductor"?

Can you post a drawing of the particular circuit you are asking about? Or give a pointer to a typical flyback transformer circuit...
 
berkeman said:
What's a "flyback inductor"?

Can you post a drawing of the particular circuit you are asking about? Or give a pointer to a typical flyback transformer circuit...

See Fig.a in this pdf.
 
likephysics said:
In the flyback inductor equivalent circuit and also in general, why is magnetizing inductance represented in parallel to the actual inductance?
We can have just the Magnetizing inductance in series with the leakage inductance.

Also, in flyback converters why does the secondary voltage get reflected to the primary side?

likephysics said:
See Fig.a in this pdf.

The Lm is in parallel with an ideal transformer, not any other "inductance". And the Lk is in series, as it is usually modeled.

The secondary voltage is "mirrored" or sensed because of the output diode. When the output winding's voltage is below the output voltage plus a diode drop, there is no current flowing in the output.
 
berkeman said:
The Lm is in parallel with an ideal transformer, not any other "inductance". And the Lk is in series, as it is usually modeled.

The secondary voltage is "mirrored" or sensed because of the output diode. When the output winding's voltage is below the output voltage plus a diode drop, there is no current flowing in the output.

The secondary voltage gets reflected to primary because it has no where else to go?
But wouldn't it take some time to get reflected to the primary side?

When the switch (mosfet) goes off, the Drain voltage is at Vdc(or Vin). After some small time when the voltage at secondary is less than output voltage plus diode drop, the voltage is reflected back to primary.
But the waveforms don't show it this way. All the waveforms I have seen show the Drain voltage is Vin+nVo at the instant the switch goes off.
Am I interpreting this right?
 
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