help me finish simulating this circuit..

hii all
am new and i really hope find a solution to my problem here...i am designing a flyback converter (5 to 15 Vdc) i did all the design but i still didnt got the output---i jss got 6V????
so am lost dont knw wat to doo?? i attached my circuit with a waveform of the mosfit plz take a look...
thnx in advance

Ali,

how do you have the flyback transformer connected? As the drain of the MOSFET is pulled to ground the upper end of the transformer output should also swing in a negative direction. Then when the MOSFET shuts OFF and the drain flies positive the energy stored in the transformer core causes the upper end of the transformer output to swing positive.

Drawing polarity dots on the transformer one would be placed on the bottom of the primary and one would be placed on the top of the secondary.

First of all, I don't get why you have that complementary symmetry transistor driver in there. The 555 has it's own driver and should be able to drive that MOSFET not problem. Also, you need a diode across the primary winding with the cathode facing towards V1+. That's to absorb the back EMF and allow the current in the primary winding to collapse, so that energy can be transferred to the secondary. At those frequencies, use a Schottky diode that can handle whatever current is involved (you left out that detail).

It looks like the 555 is running at around 126kHz? Is your transformer designed for that frequency?

Also, you need to set the duty cycle and frequency of your oscillator to what is optimal for the transformer. And, that will change depending on the load on the secondary. Thus, unless you'll be using this with a consistent load, you'll probably need some sort of feedback to adjust the duty cycle---otherwise it won't be very efficient (unless efficiency isn't an issue, in which case, adjust the frequency/duty cycle for maximum load--and, perhaps, put a heatsink on the MOSFET).

How the transformer is connected will effect the polarity at the output. You can check this with your oscilloscope, or if the output from the rectifier is low, try reversing the secondary leads and see if that improves the output level. It doesn't matter which way you connect it at the primary, but at the secondary it makes a difference.

It just occurred to me. You have the transformer connected in a "auto-transformer" arrangement, thus the output isn't isolated. So, I'm wondering why you are using a transformer? Why not use a boost converter. Much simpler, I would think.

NO, DO NOT connect a diode across the primary winding. The primary must be allowed to "flyback" to a higher voltage than the power supply voltage.

And plenty of drive is needed to quickly switch the heavy capacitive load of a MOSFET. The 2N2222/2N2907 driver is adequate. The 555 alone will not do the job unless your flyback is running at only 20 kHz. I would add a 10 ohm gate series resistor to ensure the transistor driver current does not exceed 500 mA and to help prevent the MOSFET from going into HF/VHF oscillation.

The secondary load absorbs the energy stored in the transformer magnetizing inductance. The energy stored in the transformer leakage inductance is not absorbed by the load and will be dissipated in various ways depending on the transformer leakage inductance. Usually a snubber circuit is placed across the transformer primary to absorb this energy. This circuit often consists of a diode and capacitor in series with the diode anode to the MOSFET drain. Across the capacitor is a resistor. Note that this must be configured to protect the MOSFET from avalancing during start-up where the most energy is stored in leakage inductance. I sometimes place a zener diode across the snubber resistor to make a hard clamp for use during start-up. Some engineers allow the MOSFET to avalance during this time but I don't do that.

A flyback, unlike a buck for instance, is completely load dependent. With a flyback with each cycle you are filling a bucket with energy and pouring it into the load/capacitor. If you do not drain off the energy fast enough it will continue to fill with the output voltage rising higher and higher. So, a flyback is almost always configured with voltage feedback. An inner current feedback loop is also often used.

And there are continuous mode and discontinuous mode flybacks, each having advantages and disadvantages.

Flyback transformers are usually built such that the magnetizing energy is stored in air and not in the ferrite core. To do this an air gap is placed in the magnetic path.

Wow, talk about not knowing what you don't know...I see that I'm clearly not qualified to be advising on the topic of fly-back circuits. Thank you David..I learned a great deal

Question: is there a tradeoff between gate capacitance and drain current capacity? Would a MOSFET designed to handle lower drain current have a lower gate capacitance?

Ok this is too simple.. The 555 is just a signal generator with no feedback so it can't regulate the voltage. Therefore the duty cycle must set to:

Vo = Vin * Ns/Np * D/(1-D => Vo has to include the diode drop of approx 0.7V

So if Np = Ns the D should be 15.7/20.7 or 0.76. Clearly the duty cycle isn't 76%. So plug in your primary turns (Np), secondary turns (Ns) and then adjust the R/C values to get the correct duty cycle. You'll have to read up on the 555 to figure out how to adjust the duty cycle.

Ok so the comments mention discontinious, which is simple enough but it simply will not maintain a reasonably constant voltage with load changes.

Vo/Vi = D*SQRT(Rload/2*L) => in this case L would be the inductance of the primary or secondary if they are the same. Therefore please advise the coupled inductor parameters (what you call a transformer), like Np, Ns, cross section, gap if any, material, mag path length. Or if you got this from an app note, send the app note. Are you using ferrite or powered iron? Ferrite? How did you figure the air gap and then actually gap it and then measure it?

SQRT( Rload/s*L), sorry for the mistype.

This blog has bugs, can't even type in info between parens ( )

anyway it is the square root of the load resistance in ohms divided by the product of 2x-inductance.