3.6V 38khz 25 IR Led in paralel
If we assume that all 25 LEDs have burned out shorted, then that means that all of the 100 ohm resistors will be in parallel across the 3.6 volt supply.


Since for equal-valued resistors in parallel, the total resistance will be the value of any resistor divided by the number of resistors, the 25 resitors in parallel will have a value of:

$$ Rt = \frac{R}{n} = \frac{100\Omega}{25} = \text{ 4 ohms} $$

Then, from Ohm's law, the current the supply must furnish into that resistance is:

$$ I =\frac{E}{R} = \frac{3.6V}{4\Omega} = \text{0.9 ampere} $$

Of course, if all the LEDs don't burn out it'll be less, but 900 milliamperes is worst case.

Your circuit should be common source, with the LED array connected between Vcc and the transistor's drain, and its gate connected to the MCU's PWM pin.  


I'd use a logic-level MOSFET  with an Rds(on) of 40 milliohms or less with an ampere or so of drain current at a Vgs lower than the lowest Vhigh out of your MCU.