How much effect does the capacitor have in a SEPIC converter?

[Diver300]

In a circuit like this:-

Where there is a 1:1 ratio on the transformer, the AC waveform on the dot ends of the two windings are very similar. My question is how much effect does the capacitor between those two points have? If the capacitor has little effect, there is a larger AC component of the winding current, so more heating. I am not quite sure how to calculate how big the capacitor has to be, or how the AC current in it can be calculated.

Any ideas, anyone?

 

[ronsimpson]

>If the transformer (all parts) were perfect (text book parts) then there will be no current flow in the capacitor. (Primary to secondary capacitor) >Then you have a part with the same voltage on each end then there probably is no current flow.
>But the transformer is not perfect!
>Leakage inductance will cause a delay between when the primary and secondary switch. There will be a voltage across the capacitor for 50nS (some amount of time) when the MOSFET first opens up or closes. The answer to your question will take a chapter in a book. To make a simple answer; for the leakage inductance time (maybe 50nS) there will be about (Vin+Vout) across the capacitor.
>The MOSFET will take time to open up. This will slow down the rise time of the voltage across the capacitor.
>If there were no other capacitors in the circuit and the MOSFET opened up infinitely fast; The current in the MOSFET (1A for example) would stop flowing, when turned off, and the current would flow in your capacitor. 1A into 100pf gives you the rise time of the voltage on the MOSFET. >If the leakage inductance time was long the voltage could get very high. BUT If the leakage inductance time was short, the secondary could switch before the primary voltage got very high. In the last case the current would be for a short time.
>One answer is that the current must be equal to or less than the current in the MOSFET.
>Many years ago I built something like this and put real capacitors across the P to S of the transformer and the circuit ran very well.

Now that you are totally confused; ask the next question.

 

[Diver300]

Thank you for that. I thought that it was fairly complicated.

 

[CapeCAD]

Maxim application note AN-1051 (SEPIC Equations and Component Ratings) covers this

 

[simonbramble]

To keep it simple... if you remove the cap you have a flyback converter. With a flyback, the current in the primary ramps up linearly with time, then collapses to zero as the energy is dumped into the secondary. That sudden negative change in current can cause problems upstream of the flyback (EMI etc). The back emf (inductive switching spike) is caused by the sudden change in current (V/L = di/dt). If dt goes down, V goes up. The coupling cap slows down the change in current and hence the negative step on the input current looks more sinusoidal and the switching spike is reduced/disappears. This is the coupling cap resonating with the primary inductance. Changing the coupling cap should change the input current waveform as the resonant frequency changes. Pick about 1uF (rated higher than Vin + Vout) and experiment from there.

The coupling cap will also change the loop response, so certain values might make the output oscillate