**Basics**

The voltage impressed across the primary side winding during D is just the voltage on the input side energy transfer cap. [Q is ON, P is OFF]

But since the input inductor and primary transformer winding cannot stand off DC, the (steady-state, average, or DC) voltage on this capacitor is just Vg, the source voltage. This means that the primary winding sees DVgTs volt-seconds each time Q is ON. Likewise, the secondary sees D’VTs when P is ON (and Q is OFF.)

**Note That:**

…if the magnetizing inductance is not to saturate, i.e. the transformer’s core is not to ‘walk off’ the top of the B-H loop, DVgTs must equal D’VTs. But this is just a restatement of the relation V/Vg=D/D’, which is automatic!

**Net Excitation**

Now, one of the reasons for the variable duty cycle in a switcher is to correct for various input voltages. That is, our duty cycle changes depending on Vg, to maintain a constant output voltage.

But if D is a function of Vg, and the volt-seconds across the primary winding are just DVgTs, we should be able to find a simple relation by eliminating D!

Considering first a 1-to-1 transformer, we remember that V/Vg = D/D’ = M, where V is the output voltage. Using the relations D=M/(1+M) and D’=1/(1+M); and rewriting DVg {=D’V} as MVg/(1+M) {=V/(1+M)} we reach the surprising but altogether satisfactory conclusion that DVg = (V||Vg)!

**Adjustment When the Turns Ratio is Not One**

If the design requires a transformer to step down, with N turns at the primary side to each turn of the secondary, the expression becomes DVgTs = (NV||Vg). Naturally, any output can be used, so long as the N corresponds to the particular V chosen. Note that when stepping up, N will be <1! In general, if there are Np turns at the primary side for each Ns turns at the secondary: N = Np/Ns.

**Efficiency**

To account for inefficiencies, replace N with N/% where % is the estimated efficiency of the finished converter, not including housekeeping, control, or drive circuits.

In a multi-output converter, also exclude losses in the outputs not being used to calculate volt-second excitation. Remember that only one V and N need be used in this calculation. The one with the greatest output power is usually the best to use!

**Practically Speaking**

As explained on the old site, transient current must be provided for in designing the transformer of the Cuk Converter. [For a discussion of this circumstance, click on the icon above.]

The solution turns out to be simple, though a considerable calculation is required to reach it! Introducing a gap in the core to reduce its permeability to mu = 300 or so does the trick. Remember that the (effective) mu=lm/lg, where lm is the magnetic path length of the transformer, while lg is the gap length. Thus a transformer with a 3cm path length takes a 4 mil gap, or 2 mils in each leg!