Electronic – arduino – Want to control 30V LED strip with a microcontroller, any help or ideas

For 1.5 A at 12 Volts, switched by 3.3 Volts, here is a MOSFET solution that would work well. The MOSFET suggested here is an IRLML2502 available from eBay and other sites for as little as $2.35 for 10 with free shipping.

^{simulate this circuit – Schematic created using CircuitLab}

The IRLML2502 has a maximum on-resistance of 0.08 Ohms at 2.5 Volts gate voltage, and less as the gate voltage gets closer to 3.3 Volts. It can withstand 20 Volts Drain to Source, so it will work well with a 12 Volt supply. Drain-Source current rating is greater than 3 Amperes, providing over 100% margin of safety.

At 0.08 Ohms and 1.5 Amperes, the MOSFET will dissipate 180 milliWatt when fully on. Even allowing for the switching edges of the PWM, dissipation will not exceed 250 mW or so, hence no heat sink is required for this application.

Regarding the assumptions:

• NPN transistor drop and dissipation are correct, give or take a bit due to Vce of specific transistors
• Chunky transistor (BJT), not really, but a TO-220 size would be typical, and yes, a heat sink would be required
• Yes, see suggested MOSFET above
• Not correct, there are several low-cost MOSFETs that turn on solidly well below 3.3 Volts, and can easily pass 1.5 Amperes
• No, with an NPN BJT there is always a balancing act around base current etc. MOSFETs being voltage driven devices, work with less fuss

Some of your assumptions are correct. This answer provides one better way, and I am sure there are others.

You're nearly there, a couple of things though:

The base resistor calculation ins't correct - remember you only need 1.6mA according to your calculations (the collector current is separate).
Looking at the datasheet, the minimum gain is 1000, and maximum base-emitter voltage is 2.5V, which means we need to adjust the calculations, 1.6mA will do for the base current (always good to have extra for a switch as the gain drops at saturation) but we need to use 2.5V rather than 1.3V for worst case (it's better to use worst case/maximum values to design with, though looking at the graph it seems the extra Vbe is unlikely at this current, so somewhere between the two figures below should be okay):

So:

(3.3V - 2.5V) / 1.4mA = 570Ω

or

(3.3V - 1.5V) / 1.4mA = ~1.2kΩ

This should work okay, but is not the most efficient way to do things - the transistor dissipation will be at least 0.4A * Vce(sat) which is about 0.4A * 0.75V = 0.3W, plus your R-pi needs at least a couple of mA or so to drive it.
A modern logic level MOSFET can be much smaller, be driven with (almost) no current) and have almost no dissipation. Here is an example part, the FDC637BNZ , chosen at random from thousands at Farnell: