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I'm trying to sketch the currents through each transformer winding of a forward converter. I also need to sketch the current through the source. I have a sketch of the current through the magnetizing inductor, Lm, and the inductor, Lx.

To sketch the currents through the primary, secondary, and tertiary windings, I need to determine the relationship between the windings.

As far as I can tell, the current in the primary winding is equal to the current in the secondary winding minus the current in the magnetizing inductor.

Additionally, I believe the current through the tertiary winding is equal to the negative current through the source.

I do have the ratio of coils, but I need instantaneous current, not average current.

I found these sketches in the book: Power Electronics (Hart, D. W. (2011), McGraw-Hill) but I don't know how to find the amplitude of the waveforms.

unlabeled sketch of current in windings

A schematic of the forward converter:

schematic of forward converter

I'd appreciate any relevant formulas, or a break-down of the concepts behind the relationships between windings. I am new to the concepts, so I would appreciate basic terms.

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    \$\begingroup\$ Post your sketch of what you have so far. Do you understand dot notation? \$\endgroup\$
    – Andy aka
    Jan 30 at 18:08
  • \$\begingroup\$ @Andyaka, I updated the question with a graph. I am familiar with dot notation. \$\endgroup\$ Jan 30 at 18:22
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    \$\begingroup\$ A schematic is needed that relates to the waveforms. There are literally scores of different circuit configurations for a forward converter. \$\endgroup\$
    – Andy aka
    Jan 30 at 19:09
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    \$\begingroup\$ I wanted to write an answer but it's late here. Please have a look at this PDF in which I analyzed why the forward converter was having over-power issues, like a flyback. There are many details on the waveforms. If you need more, I can try to find time and explain tomorrow. \$\endgroup\$ Jan 30 at 21:56
  • \$\begingroup\$ @andyaka, I added a schematic from the textbook \$\endgroup\$ Jan 31 at 7:04

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The forward converter belongs to the buck-derived family and it can be understood when looking at the below picture which is excerpted from my book on transfer functions:

enter image description here

You can see that the transformer scales the input voltage by its turns ratio \$N\$ and drives a classic buck arrangement. Assuming a perfect transformer, when the power switch closes during the on-time or \$t_{on}\$, the input voltage is applied across the primary side and appears at the secondary side, scaled by the turns ratio: \$D_1\$ conducts and \$L_1\$ is magnetizing with a slope equal to \$\frac{NV_{in}-V_{out}}{L_1}\$. However, the transformer is not a perfect element. It can be represented in its very basic form as perfectly-coupled windings to which a primary or magnetizing inductance is added in the primary:

enter image description here

Therefore, when the power switch closes, you not only magnetize \$L_1\$ but also this magnetizing inductance \$L_m\$. \$L_m\$ offers a way to model the core magnetization process and how the flux moves in the transformer magnetic material during the applied volt-seconds. During \$t_{on}\$, the current in this inductance increases with a slope equal to \$S=\frac{V_{in}}{L_m}\$ and reaches a peak value when the switch turns off. In the transformer, the core flux density has reached also a peak value, well below the saturation level of course. Energy is stored in the magnetizing inductance but it does not participate to the power transfer.

When the switch opens, you will interrupt the current in the primary and secondary-side inductors, \$L_m\$ and \$L_1\$. In the secondary side, \$D_1\$ will block and \$D_2\$ freewheels the current and ensures \$L_1\$ amp-turns continuity as in a buck. But in the primary side, if no precaution are taken, the primary-side voltage reverses and brings the drain to an extremely high value, limited in amplitude by the transistor breakdown voltage and the parasitic capacitance at the drain: destruction is almost guaranteed if no precaution are taken. You need to offer a path for this magnetizing current to circulate at the switch opening to that the magnetizing inductance is fully de-energized during this off-cycle. And this is important because energy stored in the magnetizing inductance implies a remanent flux in the core: at the next cycle, the flux density will no longer start from almost zero but from a pedestal, the remanent flux, and will go to a higher peak value. This process will continue cycle-by-cycle until saturation occurs:

enter image description here

You thus must demagnetize the primary inductance before the next cycle takes place. This mandatory so-called core reset process has been the object of many structures to bring the most efficient and compact way to realize it (active clamp is one of them). The simplest approach is a tertiary winding, as shown in your circuit diagram. It is wired as in a flyback configuration: during the on-time, the diode \$D_3\$ is silent (reverse-biased) because of the dot arrangements between the primary and this extra winding. When the switch turns off, as the primary voltage reverses, this diode now conducts and clamps the voltage across the primary winding to \$-V_{in}\$ (there is usually a 1:1 turns ratio) and ensures demagnetization of \$L_m\$ down to 0 A.

This approach naturally limits the maximum duty ratio below 50%: if it takes say 4.5 µs to bring the magnetizing current to its peak by applying \$V_{in}\$ during \$t_{on}\$, you realize that you will need the same 4.5 µs to bring the peak down to 0 A if you apply \$-V_{in}\$ across the inductance. If you push the duty ratio beyond this 50% limit - in practice designers or (PWM controllers) clamp the excursion to 45% for safety (insertion of a dead time DT) - then saturation can potentially occur with all bad consequences. This duty ratio limits restricts the usage of forward converters to narrow input voltage ranges. I mentioned the active-clamp forward and that is a way to extend the duty ratio dynamics beyond 50% allowing operations on a wider input range as in telecom applications. You could also change the tertiary winding ratio and allow more voltage on the drain at the switch opening. The below graph illustrates this phenomenon:

enter image description here

You can visualize these operating waveforms through simulations. If you download my free 120+ ready-made templates, you can check operations through the demo version of SIMPLIS and it is quick:

enter image description here

You see that during the off-time the drain voltage jumps to twice the input voltage (the primary is clamped to \$V_{in}\$ by the tertiary winding which now comes in series with the input rail when diode \$D_3\$ conducts). The primary current in the switch peaks to the output inductor current reflected to the primary side to which you add the peak magnetizing current:

enter image description here

In the above operating point, you see how the magnetizing current returns to zero during the off time. When it is zero, the diode from the tertiary winding blocks and the voltage at the drain returns to \$V_{in}\$.

There are more things to say about the forward converter and I hope this small write-up will help you understand better how the forward operates.

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