Electronic – Buck converter design problem/confusion

Increasing the value of the inductor will sooner or later turn your circuit into Continuous Conduction Mode (CCM), where there is a constant DC current flowing through the inductor, and the ripple current (the triangle waveform) being superimposed on this base DC current.

This means that the current of the inductor in this mode never goes down to zero. The energy stored in the inductor is indeed \$\dfrac{LI^2}{2}\$, however, only a part of this energy gets transferred in a cycle to the consumer: \${\dfrac{LI_{max}^2}{2}} - {\dfrac{LI_{min}^2}{2}} = \dfrac{L(I_{max}^2-I_{min}^2)}{2}\$, the rest stays in the inductor (more or less permanently). The amount of energy used by the consumer will be the sum of the energy transferred through storage in the inductor, and the energy constantly being transferred by the DC current flowing through the inductor.

The amount of the maximum possible ripple current is determined by the inductance value, the switching frequency, the ON-OFF duty ratio (which more or less equals the output to input voltage ratio), and the voltages connected to the inductor (the input voltage minus the output voltage in the ON phase, and the output voltage in the OFF phase). Increasing the inductance value decreases the maximum possible ripple current. As soon as the output current needed is more than half of the maximum possible ripple current, operation will turn into CCM.

What disadvantages are there for having a larger inductance?

• More turns, therefore higher DC resistance in the inductor, meaning larger copper losses, or having to increase wire thickness/number of strands to compensate for this, thereby increasing inductor size & cost.

• Having a larger inductance, the feedback control loop will be slower, meaning that the power supply will be less flexible adapting to quickly changing loads. This will probably show itself as larger overshoots in response to a unit-step load change.

• When using an asynchronous buck converter, in OFF-state, the free-running diode will be conducting. In DCM, the diode is conducting only as long as the energy stored in the inductor gets fully removed. In CCM, the diode is conducting during the whole OFF-phase, as the inductor current never goes to zero. This means higher losses on the free-running diode - which may be a problem especially because the losses on the FET can be decreased by using a FET with a lower R_dson, but you cannot do the same with the diode. The smaller the output to input voltage ratio is, the more grave this problem could be.

Power out = power in minus losses. Power out is 300 watts therefore power in will be about 330 watts. At 14 volts and 330 watts, current in will be 23.6 amps but this is average current. The peak current will depend on the duty cycle and looking at it simplistically, for a 50% duty cycle the peak current might be about 2 x 23.6 amps = 47 amps worst case (borderline discontinuous).

Ideally you need to pick an inductor that doesn't saturate and one with a gap is going to work better. A solid ferrite core will tend to start saturating at a flux density of about 0.4 teslas. It might have a relative permeability of (say) 1000. Using B = \$\mu H\$ allows this to be estimated: -

H = \$\dfrac{0.4}{4\pi \times 10^{-7} \times 1000}\$ = 318 A.t/m

You need to know the mean length of the magnetic field and that is found in the data sheet for the ferrite you are using. Let's say it is 100mm, this means your maximum ampere-turns is 31.8 and of course this would be unsuitable for your application because you can immediately see that it is likely your amperage and one turn is going to heavily saturate the core.

So, if you put a gap in it might reduce the effective permeability to (say) 200 and this allows more 5x H field (5x more current or 5x more turns) but inductance is related to turns squared so there is a net gain because although the \$A_L\$ value for the gapped core has gone down by a factor of 5 you only need \$\sqrt5\$ more turns to recover the inductance.