How to calculate the duty cycle range for DC-DC buck converter

Yes, crudely put, it converts the voltage into current. Ignoring the internal losses, if you halve the voltage you double the (available) current.

So if you convert a 5V 1A input into a 2V output the current capacity becomes 2.5A minus internal losses.

Say you have a switching efficiency of 85%, then you can work it out as:

$$ \frac{5V}{2V} = 2.5 \times 0.85 = 2.125 $$ $$ 1A \times 2.125 = 2.125A $$

You can think of it as the DC equivalent of a transformer for AC.

Just at first sight, your formula gives an energy (Joule), not power (Watt)...

If the "...custom equipment has an oscilloscope which is monitoring the voltage across the inductor and the current through it...", then the inductor losses can be calculated right out of the measured values (i.e. right from definition of average power) as:

\$ P_{losses} = \frac{1}{T} \int_0^Tv(t)i(t)dt \$, (average value of instant power during period)

where v(t) is the waveform of voltage across the inductor, i(t) is the waveform of current through it and T is period of these waveforms. Provided that the oscilloscope is a digitizing equipment, then, in principle, the corresponding voltage and current samples from within one period have to be multiplied, summed, multiplied by the sample interval and divided by the period (T) length.

For instance the trapezoidal integration method can be used:

If there are n equidistant samples (of \$ v_i, i_i \$, i = 1 to n) covering one period T, then the losses can be calculated as:

\$ P_{losses}= \frac{1}{(n-1)} \cdot (\frac{{v_1} \cdot {i_1} + {v_n} \cdot {i_n}}{2}+ \Sigma_{i=2}^{n-1} v_i \cdot i_i) \$

2015-04-12, \$ \textbf 1^{st} \$ appendix

As I already stated in the very beginning, your formula is not okay. At first, the T in it is superfluous (it is already incorporated in the duty cycle, D). Let's have a look at it a bit more closely. It can be rewritten (omitting the T, of course) as:

\$ P_{AC} = [D \cdot (V_{IN}-V_{OUT})-(1-D) \cdot V_{OUT}] \cdot I_{RIPPLE} = (D \cdot V_{IN}-V_{OUT}) \cdot I_{RIPPLE} \$,

but is it already okay?

You wrote "…Since the inductor has some AC losses from eddy current and hysteresis, I took the power during the charging period and subtracted the power during the discharge period and what would be left is the loss…".

In principle, this idea is right in my opinion, but:

1. The voltage across L during \$ t_{ON} \$ (term with D multiplier) is: \$ V_{L\_on} = V_{IN}-V_{PMOS\_SWITCH\_ON}-V_{OUT} \$,
   not just \$ V_{IN}-V_{OUT} \$ (the PMOS switch contribution isn't negligible).
2. The voltage across L during \$ t_{OFF} \$ (term with (1-D) multiplier) is: \$ V_{L\_off} = -(V_{OUT}+V_{DIODE\_SWITCH\_ON}) \$,
   not just \$ -V_{OUT} \$ (neither the diode switch contribution is negligible).
3. If we presume both the above voltages as constants during their time intervals and the ripple current being "pure" sawtooth waveform, then value that must be used in the calculation on the place of current is \$ I_{RIPPLE}/2 \$ (i.e. its average value – it follows from the very first formula, because if v(t) = const., then it can be factored out the integral and the rest is the ripple current average value).

The resulting formula will be then:

\$ P_{AC} = [D \cdot V_{L\_on}+(1-D) \cdot V_{L\_off}] \cdot \frac{ I_{RIPPLE}}{2} \$

( \$ V_{L\_off} \$ is negative in relation to \$ V_{L\_on} \$, we have to measure both the voltages the same way, that's why the "+" operator is used in the formula)

It is questionable, however, whether the speculated presumptions (3) are "sufficiently" valid/met and how much they affect accuracy of the result.