Electronic – LT3761 LED driver explodes

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.

I agree with Austin that a voltage doubler should be more than sufficient for your mosfet - the gate uses virtually no current at all.

From your diagram, it is unclear whether the gate driver and the level shifter are controlled from different MCU outputs or a common one.

Is the "Gate Driver IC" the core of your booster? From what is shown, I don't see why you'd need the "level shifter" - drive the driver from ONE pin, and when it is high, the H-bridge FET will switch.

Unless I'm misunderstanding something in your diagram, I'd think you could simply use an optocouper driven from an MCU pin to control current into a voltage doubler circuit to drive your H-bridge. Any number of OC's would work, since you're not driving much current. You're working with 12V, which your uC doesn't run off of, so the OC is necessary.

In fact, a Dickson Charge Pump clocked from a uC pin (i.e. program a pin for PWM) would provide an isolated voltage doubler: opto + 2 caps and 2 diodes. If you're not clocking, it's not doubling and your gate won't have enough voltage to conduct.

Basically, you'd be keeping the uC (MCU), H-bridge, and V1, and the Dickson would replace the rest of the components shown there, outputting to M2, running to ground, and sourcing from the 12V Vcc, with the clock as a solitary input from the uC. Opto is < US$0.50, caps and diodes are cents apiece.