Flyback Principe

Hello all,

I've been off for a while( school work). Now I need to know how the flyback transfo works.

The schematic :

**broken link removed**

1) if the N-gate Mosfet conducts , the current in the primary coil will rise linear. The voltage on the secundairy coil will be -(N2/N1)*V(coil1). The diode will not conduct.

1) the mosfet does not conduct, the coil1 wants the current to be stable and will make a reversed voltage. the polarity of the voltage on the secondary side will be reversed. The diode will conduct and the the cap will charge.

my question now is : what voltage will the secondary side have?

hopefully someone someone has an answer for me .

Thanks in advance.

Im pretty sure that the voltage output is relative to the frequency of the MOSFET pulses.

If the frequency is increased so will the voltage.

Tom

A flyback transformer uses inductive energy to generate the voltage. When the MOSFET is conducting the current increases in the transformer primary, storing energy in its inductance equal to ½LI² Joules. When the transistor turns off, this energy will generate an inductive spike across the transistor. This voltage is transfered by the transformer and appear at the secondary with value N2/N1 times the primary value. The peak voltage value is determined by the amount of inductive energy stored, and the value of the output load resistance and capacitance. For repetitive pulses of frequency f, the voltage on the output capacitor will build until the power dissipated in the load equals the power stored in the inductance (which equals ½LI² * f).

For more info Google "flyback converter" or read this.

SneaKSz said:

Hello all,

I've been off for a while( school work). Now I need to know how the flyback transfo works.

The schematic :

**broken link removed**

1) if the N-gate Mosfet conducts , the current in the primary coil will rise linear. The voltage on the secundairy coil will be -(N2/N1)*V(coil1). The diode will not conduct.

1) the mosfet does not conduct, the coil1 wants the current to be stable and will make a reversed voltage. the polarity of the voltage on the secondary side will be reversed. The diode will conduct and the the cap will charge.

my question now is : what voltage will the secondary side have?

hopefully someone someone has an answer for me .

Thanks in advance.

Click to expand...

Hi there,


One simple way to understand this kind of circuit better is to look at it almost like a boost converter, except the inductor is not coupled directly to the output but is coupled through an ideal transformer. Thus the circuit "transformer" looks instead like an inductor with a transformer connected to it where the primary is across the inductor and the output of the transformer is connected to the output of the circuit. This makes it a lot easier to understand because it acts like the boost circuit during the inductor charging phase and acts like a buck circuit while the inductor is discharging, where the current changes from a true boost or buck by the turns ratio as usual with a transformer.
If you've never dealt with a boost or buck circuit yet then i would suggest learning those two first.

A flyback transformer is a coupled inductor. Its core must store energy. A power transformer, like a push-pull design, is a power coupler, its core's job is not to store energy but provide a medium to couple the primary's changing magnetic field to secondary windings.

The MOSFET switch will charge the primary inductance then it shuts off and the energy stored in the core is dissipated in the secondary. Flyback B-H must be more linear then a power transformer. This is often accomplished by gapping the core which spreads out the B-H curve allowing a higher H excitation enabling more energy storage.

This is a good source for info on SMPS design.
**broken link removed**