Electronic – How to make any basic boost converter that actually works

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.

Although the absolute value of a digital potentiometer's resistance may vary 30%, the matching between the internal resistors is really good. This means that for a voltage divider (to wit, a potentiometer), the voltage division accuracy is quite good, since the voltage division factor relies entirely on the ratio of the resistors used, not their absolute value.

If your output voltage is less than the digital potentiometer's maximum voltage rating, you can simply use a digital potentiometer to stand in for the feedback network without any fanfare. Most digital pots can take only 5.5V, but some are rated for substantially more.

If your output voltage is higher than the digital potentiometer's maximum rating, or if you want fine adjustment, you can combine the digital potentiometer with external resistors to form a compound voltage divider. Note that this will cause the digital potentiometer's absolute value variation to come into play. Techniques exist to minimize this error, as covered here.

If you want to avoid digital potentiometers altogether, you can make the feedback divider take input from a D/A as well, as shown here by having a D/A drive \$V_{CTRL}\$:

^{simulate this circuit – Schematic created using CircuitLab}

Since the feedback voltage \$V_{FB}\$ will be regulated by the LTC3780 to 0.8V, the output voltage will be regulated to:

$$V_O=\left(1+\frac{R_2}{R_1}\right)0.8-\frac{R_2}{R_3}(V_{CTRL}-0.8)$$

Setting \$V_{CTRL}\$ to 0.8V causes no change in the output voltage; increasing \$V_{CTRL}\$ causes the output to drop, and decreasing \$V_{CTRL}\$ causes it to rise.

It should be noted that no matter what you do, you should carefully evaluate the regulator and adjust components (\$L\$, \$C_{OUT}\$, \$I_{TH}\$ network) to ensure stability over all operating conditions. When in doubt, lean to the conservative side here.