Your analysis oversimplifies a few things, but works out more or less correctly in the end. Just for general edification, I'll point out a couple things.
First, Rds isn't a fixed number, it varies with Vgs, temperature, and current. Figure 1 of the datasheet shows Vds vs. Ids for various values of Vgs. The reciprocal of the slope of this curve at a given current value is Rds.
For Vgs of 5V, the graph doesn't show anything for Ids = 2A, but we can extrapolate. Picking two points on the Vgs = 5V curve, we get a slope of 30 amps per volt, or (phrasing another way) an effective Rds of 33 mOhm.
Note that this assumes a junction temperature of 25C. From figure 2, we can see that at low currents, a hotter junction actually results in an even lower voltage for a given current, translating to a lower Rds. If you were running in a sub-zero ambient (say, equipment that lived outdoors in February in Moose Jaw), your Rds would be higher. But let's assume you're indoors in standard room temperatures.
So two amps RMS through 33 mOhm dissipates 132 mW of power, P = I^2 * R. .132W x 62.5 K/W gives a junction rise above ambient of 8.25K, or 14.85F. (1K = 1.8F) You might notice it get a little warmer to the touch, but that's it.
Max operating junction temperature is listed as 175C, so as long as your ambient is below 166.75C, you'll be fine thermally. If your ambient is higher than that, you're doing something either wrong or awesome.
MOSFET is "upside down" as Mike said.
BUT
3 "fixes":
- Use P Channel FETs and all shall be wellish.
FET will be on when gate is LOW (~= 0 V).
FET will be off when gate is HIGH(~= 4.1V)
IF you can isolate the MOSFET 4.1V and gnd feeds from the rest of the PCB by cutting tracks then
- Depending on pads:
Take MOSFET
Bend leads UP
Invert MOSFET - solder MOSFET D to old S pad
Solder MOSFET S to old D pad.
Solder MOSFET G to G pad.
Or use wire tacked on leads to make it easier.
Best Answer
Static dissipation with 4.5V drive will be something like 1.3W* maximum at 5A. They don't bother specifying the thermal resistance junction-to-air, but it's typically something like 65°C/W for a TO-220, so in a 50°C enclosure (easy on a warm day) it could be running at 135°C Tj, which is well above the absolute maximum. Typically, if you are lucky, not conservative, the MOSFET will be running at 'only' 116°C, which is just within the max rating. But we've not started to switch yet. Worst case will be when the LED strip is almost at full brightness.
(By the way, it will be quite a bit better if you are using 10V drive- more like 0.8W maximum, which would be acceptable, at least in some undemanding applications (not if you need really high reliability or have to deal with hot ambients). These are designs based on worst-case specs, if you roll the dice most of the time you'll do better.)
However you have to add the switching losses- they will depend on how hard you drive the gate. If you are using a lowish PWM frequency such as 120Hz and a powerful 10V gate driver you will minimize the losses, but if you trying to drive it directly with a microcontroller port pin, I think you'll have excessive losses. It's possible to calculate the losses based on the gate charge, PWM frequency and available current from your driver. If you are designing a real product you should do so. There is a small amount of capacitance from the gate to the drain, which is magnified by the Miller effect as the MOSFET switches. Your driver has to supply that current plus charge the gate capacitor.
* Conduction power dissipation is simply \$Pd=I^2Rds(on)\$ but the maximum value of Rds(on) will be higher by as much as 50% if the junction is hot.