DESIGNING SECONDARY WINDINGS OF A FLYBACK TRANSFORMER

In a previous post we discussed the design of the primary of a switchmode transformer for a flyback converter. In the last post we described the basic topology of a flyback converter with one output. Now we want to design the secondary winding of the converter. We will use the specifications we discussed in the just before the one immediately preceding this one. In that post we described a 100 watt converter with a maximum input voltage of 1000V and a 70% efficiency. A 150 watt input power was required to provide the 100 watt output. The primary was 90 turns with an inductance of 5 mH.

Lets assume the following basic specifications for the secondary side output circuit;

12 volt output at 8.333 amps (100 watts). We will shoot for 8.5 amps for a very slight margin over 100 watts to 102 watts.

First we recognize that in a flyback converter, energy is stored in the primary side inductor and air gap during switch-on time, is then transferred to the secondary circuit, less the energy lost in lossy elements in the primary side circuit, when the switch turns off, and the core resets. Lossy elements are the transistor switch, snubber circuit, the primary winding resistance, and other lossy elements including parasitics. By conservation of energy we can write the equation for the energy balance:

Ef * Lp * Ip^2 = Ls * Is^2

where Ef if the efficiency of the primary side circuit in this simplified equation, Lp is the inductance of the primary, Ip is the primary current, Ls is the secondary winding inductance, and Is is the secondary current. From that equation we can use a little algebra to find the inductance of the secondary winding. We will use input current for a 1000v input that would be 0.15 amps for a 150 watt input capability:

Ls = Ef * Lp * ( Ip / Is )^2 = 0.7 * 0.005 * ( 0.150 / 8.5 )^2 = 1.09 * 10^-6 H

or approximately 1 microhenry for the secondary inductance.

We also have the standard transformer equations which are not totally accurate for a flyback converter but they will serve well enough for preliminary calculations:

Np / Ns = sqrt ( Lp / Ls )

In other words, the turns ratio is equal to the square root of the inductance ratio. We can use the above equation, adding the efficiency factor into the equation to solve for the number of secondary turns:

Ns = Np / sqrt ( Ef * Lp / Ls ) = 90 / sqrt ( 0.7 * 0.005 / 10^-6 ) = 1.52 turns

In the construction of a prototype sample we would probably round up the turns to 2 although there are methods of winding the equivalent of fractional turns. This winding would probably be made of copper foil instead of wire for better coupling with the primary winding.

We have now have a preliminary calculation of the secondary inductance and number of turns, but there is a lot of work left to do. Now it would be a good idea to simulate the transformer and power switching stage using SPICE or some other simulation software. You might find that the secondary inductance and number of turns needs some change in values, especially when parasitic and lossy elements are introduced.

In the above calculation we did not consider the condition of minimum input voltage. If the minimum input is say 200 volts, we would get much different answers in the above calculations and it would be necessary to use the values for the minimum input voltage to make sure we will have enough output power and voltage.

In another post we will discuss design of the remainder of the secondary circuit elements (refer to Fig. 1 in the previous post.)