Trying to get 2 Amps out of 3 Amp Regulator (MIC29302WU)
Can I achieve 2A without a heat sink?


Probably yes, BUT only under quite strict conditions! We'll do the calculation later.

You need to realize that a linear regulator like this MIC29302 works by "burning off" the voltage just like a resistor would. That "burning off" means power is dissipated (as heat) by the chip. That heats up the chip. If the chip gets too hot it will protect itself by lowering the output voltage which should lower the current that is flowing and that should limit the amount of power that is dissipated by the chip. Basically the regulator will go in a "temperature regulation" mode where it tries to deliver as much power as it can without overheating.

So you want 2 A but no heatsink. First thing we have to figure out is what our power dissipation is allowed to be. For that we need values for the thermal resistance, this tells us how much the chip will heat up for a certain amount of dissipated power (usually 1 Watt).

You're using the  MIC29302WU which uses a 5 pin TO-263 package. Some thermal resistance values are given, see the datasheet, top of page 7. Unfortunately these are \$\phi_{JC}\$ values where JC means Junction-to-Case. Junction is the chip, case is the metal part of the package. There is no \$\phi_{JA}\$ (Junction-to-Ambient) value which is what you need for use without a heatsink. As there is nothing better, let's use the \$\phi_{JA}\$ for the TO-252 package instead, that will not be accurate but it is all we have.

So the thermal resistance \$\phi_{JA}\$ will be around  56 °C/W. That means the chip will get 56 °C hotter for every Watt that is dissipated. The highest junction temperature we can operate the chip at is 125°C (also top of page 7), we should stay below that 125°C as the temperature protection will kick in at around 125°C. So let's take some margin and design for a 110°C maximum.

Let's assume an ambient temperature of 40°C that gives us:

110°C - 40°C = 70°C

of allowed temperature increase. At 56 °C/W that results in a maximum power dissipation of:

70°C / 56 °C/W = 1.25 Watt

You want 2 A to flow so that means the maximum voltage drop across the regulator can be:

P = V * I => V = P / I:

\$V_{drop}\$ = 1.25 W / 2 A = 0.625 V

So the voltage drop across the regulator should not be higher than 0.625 V or else it will get too hot. Can this regulator even work with such a low drop voltage?

The answer is again on page 7, under "Dropout Voltage". For the MIC29300 the maximum dropout voltage is 600 mV at 3 A. At 2 A this voltage will be smaller which is OK.
So yes the \$V_{drop}\$ &lt; 0.625 V can be met.

But realize that the \$V_{drop}\$ is the difference between the input and output voltage of the regulator. So if you apply a high input voltage and want a low output voltage, the regulator will need to drop much more than 0.625 V!

You want 5 V at the output, that means that your input voltage cannot be more than:

\$V_{out} = V_{in} - V_{drop}\$ = 5 V = \$V_{in}\$ - 0.625 V => 

\$V_{in}\$ &lt; 5.625 V

So conclusion:

Yes you can operate the MIC29302 without a heatsink and draw up to 2 A IF you limit the input voltage to 5.625 V.

This is highly impractical as usually you do not have such a precise input voltage.

A possible solution is to use a heatsink!

But at 2 A you will be much better off using a switched mode regulator. You can design and build one yourself if you have enough experience but it is much easier and less risk (of it not working) to buy a ready made module. For hobby purposes an LM2596 based module might do the job.