Would this isolated boost topology work?

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.

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).

• Ah, the inductor is to simulate the winding inductance. The MOSFET in series will not be switching, it is just used to open the circuit when the converter is not working. – Chi Nov 18 '16 at 11:12
• In that case I have to agree with the others who have pointed out that a transformer won't pass DC :-) – frr Nov 18 '16 at 11:25
• So I should put the second mosfet in a typical fly-back topology after the boost? Would it be possible to control both mosfets with the same PI controller? – Chi Nov 18 '16 at 11:31
• Somehow you need to alternate the current through the transformer. At least switch it on/off at some radio frequency using a PWM controller. The main stage (the one with the transformer) can be a "forward" or "flyback" version - your choice, each uses slightly different design rules. In PC power supplies, the PFC is a "boost" stage (single winding flyback) and the main stage is typically forward with two windings. Google "SMPS forward vs. flyback". One specific link: onsemi.com/pub_link/Collateral/SMPSRM-D.PDF – frr Nov 18 '16 at 13:09
• BTW you need two separate PWM controllers, because in principle, there will be two regulation loops. It is a fact that in PFC-equipped PSU's, the PFC stage and the main=forward stage often run both stages in sync on a particular edge, as a clever trick to relieve the main bulk capacitor of some ripple current - but each stage needs a separate feedback loop to control its respective PWM duty cycle. There are also power supplies that use a PFC controller to drive a "single stage all the way isolated flyback converter" - typically for smaller power (below 200 W or so). – frr Nov 18 '16 at 13:17

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.

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.

• Simulation shows exactly the same results for a full bridge rectifier at the output and a single diode. In the bridge rectifier version, 2 of diodes never conduct. Notice that the current never changes direction in this converter, very much like the fly-back topology, which also doesn't need a bridge rectifier at the output. – Chi Nov 18 '16 at 1:38
• Does your simulation include saturation for the transformer core? The reason why it works for a standard fly-back design is the transformer includes the coil (transformer with air gap in the core), meaning the transformer is driven to saturation by design. In your setup, the two elements are separated. – Janka Nov 18 '16 at 1:59
• You will have a D.C. component on the output of the first stage boost. Therefore there will be constantly increasing volt-seconds on the transformer primary and you are guaranteed to saturate it. – John D Nov 18 '16 at 4:12
• Exactly. But that's not as easy to see as the saturation coming from DC on the secondary. The circuit won't work that way because the OP is using a standard transformer and not the the special flyback transformer required for this. – Janka Nov 18 '16 at 8:00
• @Janka This wouldn't work even if there was a gapped coupled inductor (flyback transformer). Flyback transformers are designed to store energy, but they are not run into saturation. – John D Nov 18 '16 at 11:31