DC/DC Converter – Minimum Load Requirements

Look at the spec for the device. There are two types; one switches at 1.6MHz and the other switches at 600kHz. Let's say, just to make life easier on folk reading this answer that it switches at 1MHz.

How much energy can it charge the inductor with - again the spec has the answer - maximum duty cycle is (on average between the two devices) about 90%.

For the sake of mathematical convenience lets call it 1\$\mu s\$ (90% of 1MHz period is still about 1\$\mu s\$).

\$E = L\frac{di}{dt}\$ and this means \$ di = dt\frac{E}{L}\$

di is how much current the inductor is taking when the maximum on-time is reached (1us). If E = 9V and L = 33\$\mu\$H, then

di = \$ 1\mu s \times \frac{9}{33e^{-6}} = 273mA\$.

Is this current going to supply the modem when it is taking 1.2A? No

What if the inductor was lowered to (say) 4.7uH? Current would be 9/4.7 which is approximately 2A however, the internal FET is only rated at 1.8A so it looks like you need to find a part that has more muscles.

EDIT assuming better switcher and 1.7\$\mu H\$ inductor (revised due to error)

The power output requirement is about 13W and if the switcher switches at a 1MHz rate this means an energy transfer per \$\mu\$s of 13\$\mu\$J. Knowing that this energy comes from the inductor means we can calculate peak current in inductor and its duty cycle.

Energy in inductor is = \$\frac{LI^2}{2} \therefore\$ peak current is \$\sqrt{\frac{2 \times 13e^{-6}}{1.7e^{-6}}}\$ which equals just about 4A. But, the topology of this type of switcher means that the inductor is only needed to transfer enough energy to raise the output level above the input voltage level. In other words the first 6V are a given.

The power needed by the load (above the 6V level) is \$1.2A \times (10.5-6)V = 5.4W\$ and this means the inductor "charge" current is 2.52A.

How long will the inductor be "charging" for?

V = \$L\frac{di}{dt}\$ - we know V (6V minimum), L (1.7uH) and di (2.52A) therefore dt is \$\frac{1.7e^{-6}\times 2.52}{6}\$ = 0.714us or a duty of 71.4% and this seems reasonable.

Why not add a low dropout 12 volt voltage regulator instead of loading the output with resistors. It's likely to produce less heat. This will only work for very light loads like the one suggested and might keep spurious switching noise (from the boost converter) from interfering with the RF circuits and get better performance.