12v to 400v boost converter

[alireza7]

hi
i want to make 12v dc to 400v dc by a boost circuit in my circuit L=1900mH , C=33uF ,R=47k and i use 11n80 mosfet as switch and uf4007 diod.
there is a formula  (you can find it wikipedia) vout/vin=1/(1-D) which D is duty cycle of switch. so i need 97% duty cylcle to make 12 to 400
i tryed range of 99% to 90% pwm duty cycle and also i sweeped the pwm frequency from 20khz to 50khz and the maximum voltage that i got is 270v.

what do you think limit this voltage?
don't you think maybe the switching speed of  11n80 limited the voltage?
any suggestion?

[NiHaoMike]

I would suggest a "flyboost" topology, which is a flyback converter where the primary winding is also a boost converter choke, with the flyback output stacked on top of the boost output. In that way, the primary snubber would be "lossless".

[Zero999]

I would suggest a "flyboost" topology, which is a flyback converter where the primary winding is also a boost converter choke, with the flyback output stacked on top of the boost output. In that way, the primary snubber would be "lossless".

It sounds like you've described an autotransformer.

[NiHaoMike]

I would suggest a "flyboost" topology, which is a flyback converter where the primary winding is also a boost converter choke, with the flyback output stacked on top of the boost output. In that way, the primary snubber would be "lossless".

It sounds like you've described an autotransformer.

Just one critical difference: in a flyboost, there's a diode and capacitor stage between the primary and secondary. In theory it will not make a difference, but in practice it acts as a snubber to counteract leakage inductance. It also gives two different output voltages if your application needs it.

[Zero999]

I would suggest a "flyboost" topology, which is a flyback converter where the primary winding is also a boost converter choke, with the flyback output stacked on top of the boost output. In that way, the primary snubber would be "lossless".

It sounds like you've described an autotransformer.

Just one critical difference: in a flyboost, there's a diode and capacitor stage between the primary and secondary. In theory it will not make a difference, but in practice it acts as a snubber to counteract leakage inductance. It also gives two different output voltages if your application needs it.

Like this? I don't think I've seen that before.

[NiHaoMike]

Like this? I don't think I've seen that before.

Move the start of the secondary to the other side of the diode. Then you'll see how the energy stored in leakage inductance gets transferred to the output.

[alireza7]

Like this? I don't think I've seen that before.

Move the start of the secondary to the other side of the diode. Then you'll see how the energy stored in leakage inductance gets transferred to the output.

do you mean this?

[NiHaoMike]

Take a flyback converter, then add a diode and capacitor so the primary winding is also a boost choke. Finally, take the secondary output and connect it in series with the boost output so the voltages add.

[alireza7]

Take a flyback converter, then add a diode and capacitor so the primary winding is also a boost choke. Finally, take the secondary output and connect it in series with the boost output so the voltages add.

is this your purpose?

[Zero999]

Take a flyback converter, then add a diode and capacitor so the primary winding is also a boost choke. Finally, take the secondary output and connect it in series with the boost output so the voltages add.

is this your purpose?

That makes much more sense.

[CM800]

Dispite what a few people have said, It's certainly dooable. I've made one for a coilgun, so have several others:


http://uzzors2k.4hv.org/index.php?page=450vboostconverter

http://www.instructables.com/id/DC-DC-HV-Boost-Converter/

Just dont' expect stunning efficiency, performance or reliability I guess.

[schmitt trigger]

Something that the OP has not disclosed is the load current.

It is doable for small currents, but not for large ones.

Why? Because of the Mosfet's switching current is roughly multiplied by the voltage transformation ratio.
In this instance the transformation ratio is 33.33:1, but considering the losses, the actual switching to load current ratio would be closer to 40:1

Edit; I know that in post #1 he discussed a 47k load resistor, but is that only for testing? or is that the actual load?

[T3sl4co1l]

The advantage you get in "flyboost" over straight flyback is very small: at 50% duty cycle (ideal, minimizes peak currents), the switch only sees 24V peak, or 6% of the total output.  Nearly negligible.

There can be good reasons for using a pure flyback topology, like isolation, multiple outputs, or reducing winding capacitance.  It's not obvious if these are useful features, based on the vague OP.

Tim

[NiHaoMike]

The advantage you get in "flyboost" over straight flyback is very small: at 50% duty cycle (ideal, minimizes peak currents), the switch only sees 24V peak, or 6% of the total output.  Nearly negligible.

There can be good reasons for using a pure flyback topology, like isolation, multiple outputs, or reducing winding capacitance.  It's not obvious if these are useful features, based on the vague OP.

Tim

It gives you a lossless snubber if you don't need isolation.