Would this isolated boost topology work?

Question

I've been trying to create an isolated DC/DC converter for cell balancing but I can't seem to get the topology quite right.

In the topology shown in the picture, the cell is on the left and the battery on the right.

The MOSFET with the tag "C1" shorts the inductor, increasing its current. Then it turns off and the inductor current forces the diode to conduct, thus charging the capacitor and increasing the current in the primary of the transformer.

When the MOSFET turns off again, the capacitor would take the decreasing current from the transformer. Since the current in the transformer is always going up and down (but always with the same polarity), this would create magnetic flux in the core of the transformer and transfer energy to the secondary winding, charging the main battery.

The FET with the "BoostON" tag would only turn on when the converter is turned on, and it would be always conducting. When the converter stops the energy transfer between cell and battery, this transistor would be turned off. This is to avoid shorting the cell through the transformer and boost diode when the converter is not working.

So, are there any flaws in this topology? Are there any reasons why this would just outright fail, or would it actually work as intended?

Extra information:

This is just a typical boost converter with an extra transformer and a diode on the secondary side. The bigger circle in the transformer indicates the primary side. The transformer used would be this one, with a 1:4 turn ratio. The objective is to implement PI/PID current control to get an average inductor current of 3 A. The reason I want to avoid the typical fly-back topology is because of the control issues that may arise. The switching frequency is going to be 30kHz, and the inductor used would be somewhere around 70~90uH (calculated for the normal boost topology). The capacitance value in the image is random. Simulations show that the circuit should work as intended.

Answer 1

Your topology very much resembles a standard PC PSU with active PFC. A boost stage followed by an isolated down-converter stage. Except that:

• the inductor in series with the transformer primary... that looks a little odd... what's the purpose?

• the FET in series with the "main stage" (the isolated transformer) is N-type in your design, which matches the typical topology, but yours is connected in "common collector" fashion to the + polarity of the "midway bulk capacitor". This would force you to implement high-side N-polarity driving circuitry in the primary PWM control block... This is normally done in a different way: the N-FET is hooked up as a low-side switch (common emitter) and is driven by a PWM controller IC which also uses the bulk cap's "-" as a common ground with the FET's emitter... What voltages are we talking about on the primary side?

By the transformer's winding end marks, your converter would be a "forward" converter. That's more typical than "an isolated flyback" (which is sometimes used for low-power SMPS, and has its own pitfalls - e.g. the primary FET must cope with higher voltage spikes if memory serves).

Answer 2

No. It will generate a boosted DC voltage on one side of the transformer. Transformers only pass AC voltage, so the isolated side won't really build up any significant voltage other than maybe a transient spike at power on.

Answer 3

No, your circuit obviously won't work because the secondary of the transformer has to deliver DC. It can't do that. You need a bridge rectifier or a transformer with middle tap and two diodes for a DC output. And the diodes have to be fast-recovery types, 50/60 Hz ones won't work.