How do I choose semiconductors?

[GuilTy]

I want to built a dc-dc Zeta converter with isolation and I want to choose the semiconductors.
-the voltage on the MOSFET is 18V and the current through it is 1.125A.
-the voltage on the diode is 9V and the current through it is 2.25A.
These values have resulted from the calculation in the ideal case.

Do you have any suggestion on which parts would be best? 

Thank you,
GuilTy 

[GuilTy]

I have attached the schematic.

I haven't chosen the transfomer yet. I onlu know the value of the inductances from the primary and secondary.

[T3sl4co1l]

The diode is unconnected, floating..!?

Never draw more than three wires, meeting at an intersection, where a connection is intended.

I don't see any grounds, either, so the reference for the clock input looks undefined.

Tim

[GuilTy]

The diode is connected, it works perfect, I only have to choose the semiconductor devices.
If you look at the voltage source you will see "0[Volt]", that is the ground for the primary. The grounds should be different, in order to have isolation, right?
For the clock input there is a module, and I only have to choose the duty cycle.

I will keep in mind the advice about the connections.

Thank you,
GuilTy

[T3sl4co1l]

560u sounds like quite a lot. What frequency is this?

What control method are you looking at? (Protip: if it's not a current controlled mode, it's terrible.)

These are all necessary questions because transistor properties depend on their answers.  And more directly, the coupled inductor matters too, whether you're designing it as well, or buying an off-the-shelf part.

Tim

[GuilTy]

The frequency is 50kHz.

At this point I haven't considered using a method in order to control it. I will think about it.

The transformation ratio for the transformer is 0.5.

Guilty

[T3sl4co1l]

So I guess you've chosen L for 10% ripple in CCM (continuous conduction mode: inductor current does not return to zero each cycle)?

This is usually excessive.  Modern core materials are happy with >30% ripple, so you can save cost and size on that.

To stabilize a CCM converter, you need average current mode control, which requires sensing the inductor current at all times.  However, the current is constantly alternating between primary and secondary side, so this is complicated to do.  A current sensor on L2 may do; I'm not entirely sure.  I think it would help to have LM/L11 be a transformer (relatively high inductance) so that L2's current is the primary control variable.

A DCM converter has >100% ripple (i.e., current goes to zero each cycle), and can be operated with a peak current mode controller instead.  Which is usually cheaper, because a shunt resistor on the switching transistor is sufficient information.  Ripple is much higher, obviously, so lower-loss materials are needed, and more bypass/filtering, but this is quite practical for smaller converters, under 100W or so.

In any case, the transformer's coupling coefficient will need to be as large as possible (0.99999 is probably not possible, mind), and it will need to be damped on the primary side with an R+C or better snubber network.  This will increase Vds(pk) further, affecting transistor choice.

Tim

[GuilTy]

First of all, allow me to thank you for your piece of advice. I'm new to this stuff and every piece of advice is welcomed.

Yes, you're right, I've chosen L for 10% ripple in CCM.

I want to use this converter in CCM mode, and I've chosen the coupling coefficient 0.9999 because this is the ideal case I've simulated.

In other words, you are telling me to use an input filter in order to decrease Vds(pk)?

GuilTy

[elektroda]

 
you write
-the voltage on the MOSFET is 18V and the current through it is 1.125A. ,
but you must remember that in this circuit are jumps voltage , because transistor have load about  character inductance , and you must think how will jump voltage . I think that max voltage will be about 50V . Do good  I  see , this transistor is MOSFET-N  , I suggest to example TN0106N3-G  , TN0606N3-G ,  VN0106N3-G  etc

[GuilTy]

-the voltage on the MOSFET is 18V and the current through it is 1.125A.

I'm afraid I don't understand what you mean.

[T3sl4co1l]

k = 0.99999 is probably impossible, so you'll have to deal with nonzero leakage (non-unity coupling).

Leakage is essentially the difference from k=1.  More precisely, (1 - k^2) gives the ratio of primary referred leakage inductance to primary (total) inductance: LL/Lp.  (Or, secondary to secondary, same thing.)

Once this is known, you can transform the circuit into an ideal transformer, with a parallel inductor representing Lp, and a series inductor representing LL. Then it will be obvious that, when the transistor switches off, there is nothing preventing the drain voltage from rising very quickly, due to LL.  You can turn this into a dampened RLC circuit, by adding an R+C across the transistor.  Or you can clamp the flyback energy with a diode, delivering it into a capacitor and bleeding it away with a resistor (or TVS diode, or using a lossless snubber to reuse the energy later).

Tim

[GuilTy]

@elektroda: I've taken into the consideration the pulses due to the inductance, when I calculated the value of the inductance LM, I've set the ripple for the current at 1/10 times the current throug the inductance(LM).

@T3sl4co1l: I know that k can't be 0.99999, I assume it will be somewhere around 0.7~0.8. I haven't dimensionated the transformer yet. But, when I calculated these values I have used the ideal model for the transformer, and I've taken LM in the simulation as the primary, and L11 as the secondary.
If I will use a free-wheel diode, will this do the trick?

GuilTy

[T3sl4co1l]

If I will use a free-wheel diode, will this do the trick?

How so?

You can't simply put a diode across the primary; that will be obvious watching the scope on that node.

Tim

[GuilTy]

I thought of connecting the diode in anti-parallel with the inductance in the primary. Is this what you sugested too?

G.

[T3sl4co1l]

Yeah no.. and no.

It would look perhaps something like this:



The 2.2 + 10n (you'd need to use different values) would be the R+C case.

The 10n + (15 || diode) is a dV/dt snubber, which is an alternate RCD snubber method.  If a relatively large value capacitor is used, and the resistor is returned to a supply rather than ground, it becomes a clamp snubber instead.

The clamp arrangement would look more like what's used here,

http://seventransistorlabs.com/Images/Mag_Amp_PSU.png

The IRFZ46N is the main switch, with an SB540 snubber diode, 1uF snubber capacitor, and 10 ohm 2W "bleeder" resistor.  The voltage on the capacitor is slightly higher than the output voltage, and the resistor consumes the difference (which in this case is due to stray inductance between the transistor and SBL2040 diode).

Tim

[GuilTy]

Now it makes more sense to me. This was the first time when I've heard of a snubber.

Thank you,
G