HOW DO YOU PICK INDUCTANCE FOR A FLYBACK CONVERTER?

In the initial design phase of a flyback converter, there is the usual need to pick or design the inductor-transformer as a basic component. So what should the primary inductance be set to? Well it depends on what the global specifications are for your power supply, i.e., maximum input voltage, power required at the output, and an estimate of the switching frequency of the power supply.

The switching frequency is usually decided on the basis of the design and manufacturing technology you have available to you and is usually between 100KHZ and 1MHZ. The higher the frequency is, the smaller the filter components need to be, but the more difficult the design and manufacturing gets to be. For example, if you already know that all of the power supplies you have designed or your company designs run at approximately 100KHZ, you will probably need to operate at that frequency.

So lets say that you have the following global specifications as a minimum to start with:

maximum input voltage 1000 volts.
minimum input voltage 600 volts.
power output 100 watts.
efficiency 70% minimum

So now it is easy to compute the input power, Pin, based on the output power, Pout, required and the efficiency, e :

Pin = Pout / e = 100 / 0.7 = 143 watts

To provide a little power safety margin, lets round up to 150 watts for the required input power.

We will assume for this initial calculation that load current is reasonably constant so that we can compute the average input current that is required at the maximum input voltage, Vinmax, as

Iavg = Pin / Vinmax = 150 / 1000 = 150 ma

Lets assume further that your converter will operate at a frequency F = 100KHZ

Now my rule-of-thumb formula for the required inductance L is the following:

L = Vinmax * Dmax * T / Ipk where Dmax is the duty cycle, T is the switching period, and Ipk is the peak switch corresponding to the max duty cycle (from the equation V = L * di / dt .) Note that Ipk will never be zero in theory or in real operation.

We can re-write the equation for L in terms of frequency as

L = V * D / ( Ipk * F )

Assume that we will allow an Ipk of 1 amp before the switch is turned off. At a duty cycle of 0.25 we would have an average current input of approximately 250 ma which would allow a peak power of 250 watts input at 1000V input (0.25 * 1000) and 150 watts at 600V minimum input. This should be plenty of power capability. If the minimum input voltage is lower, more peak switch current will have to be allowed, Ipk.

Assume worst case conditions for the inductor; maximum input voltage and maximum duty cycle of 50%. Compute the required inductance at 100KHZ switching frequency:

L = Vpk * D / (Ipk * F) = 1000 * 0.5 / ( 1 * 100000 ) = 0.005 Henry = 5 mH

Now the next thing you need to consider is the fact that the inductor-transformer must not be allowed to saturate due to excessive magnetic field density in the core. If it does saturate the inductance will drop to a very low value and current will become an indefinitely large value possibly destroying the power supply.

The equation for calculating the maximum flux density ( and verifying that the value is within the inductor-transformer specifications ) is as follows:

Bmax = Vpk * 10^8 / ( 4 * N * Ac * F )

where N is the specified number of primary side turns and Ac is the core area. The above equation is a worst case situation where any air gap is very small or zero. Note that the allowed Bmax depends on the magnetic material in the core which is usually a ferrite material.